Abstract
FUZZY INFORMATION AND ENGINEERING https://doi.org/10.1080/16168658.2022.2152883 Characterization of Derived Nilpotent (Engel) Lie Ring of Fuzzy Hyperrings by Using Fuzzy Strongly Regular Relations a b a c E. Mohammadzadeh , R. A. Borzooei , F. Mohammadzadeh and S. S. Ahn a b Department of Mathematics, Faculty of Science, Payame Noor University, Tehran, Iran; Department of Mathematics, Faculty of Mathematical Sciences, Shahid Beheshti University, Tehran, Iran; Department of Mathematics Education, Dongguk University, Seoul, Korea ABSTRACT ARTICLE HISTORY Received 31 August 2021 In this paper, we determined a new characterisation of the derived Accepted 14 November 2022 nilpotent (Engel) Lie ring of fuzzy hyperrings by fuzzy strongly regu- ∗ ∗ lar relation ζ (ν ). Moreover, we proved that for a fuzzy hyperring n n,s KEYWORDS ∗ ∗ S, the quotient S/ζ (S/ν ) was a nilpotent (Engel) Lie ring. Also, we n n,s (Nilpotent) Lie ring; Engel Lie introduced the notion of an ζ -role of a fuzzy hyperring and inves- ring; fuzzy hyperring; tigated its essential properties. Basically, we stated a necessary and strongly regular relation sufficient condition for transitivity of ζ . Also, we studied the relation- MSC (2010) ship between the strongly regular relation and ζ -role of a given fuzzy 17B62; 20N20 hyperring. 1. Introduction The notion of nilpotency is the most critical concepts in the study of groups [1]. Nilpotent groups arise in Galois theory, as well as in the classification of groups. By Galois theory, specific problems in field theory replace with problems in group theory, that is in some sense better understood. The certain generalisation of nilpotent groups is Engel groups [2–4]. Hyperstructure theory introduced by Marty 1934, he defined hypergroups, investigated their properties, and applied them to groups and relational algebraic functions (see [5]). In 1971, Rosenfeld used the concept of fuzzy sets to introduce fuzzy groups(see [6]). Since then, many researchers extended the concepts of abstract algebra to the fuzzy sets (see [6–14, 19–21]). The main tool in the definition of fuzzy hyperstructures is the ‘fuzzy hyperoperation’, where the non-empty set replaces the nonzero fuzzy set. Vougiouklis stud- ied the fundamental relations on hyperrings [15]. In [16], Nozari and Fahimi introduced fundamental relation and commutative fundamental relation on fuzzy hyperrings. In this paper, we define the ζ and ν relations, which are defined in fuzzy hyperrings. (n,s) These relations connect the fuzzy hyperrins with (Engel)nilpotent-Lie rings. Especially, we introduce the notion of an ζ -role of a fuzzy hyperring and investigate its basic properties. Moreover, we study the relationship between the strongly regular relation and ζ -role of a given fuzzy hyperring. Also, we obtain Engel Lie rings from fuzzy hyperrings, by a FSR. CONTACT S. S. Ahn sunshine@dongguk.edu © 2022 The Authors. Published by Taylor & Francis Group on behalf of the Fuzzy Information and Engineering Branch of the Operations Research Society, Guangdong Province Operations Research Society of China. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons. org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 2 E. MOHAMMADZADEH ET AL. 2. Preliminaries First we recall some definitions and theorems. Definition 2.1: Let H be a non-empty set and F (H) be the set of all fuzzy subsets of H. Suppose F (H) = F (H) \{0}, where 0 is the zero fuzzy set. Then H equieped with a fuzzy hyperoperation ◦ : H × H → F (H), is called a fuzzy semihypergroup if for any x, y, z ∈ H, (x ◦ y) z = x (y ◦ z), where for any α ∈ F (H),and r ∈ H, ((x ◦ u)(r) ∧ α(u)), α = 0 (x α)(r) = u∈H 0, α = 0 (α(u) ∧ (u ◦ x)(r)), α = 0 (α x)(r) = u∈H 0, α = 0 If A, B are non-empty subsets of H and x ∈ H, then for any u ∈ H we have (x ◦ A)(u) = (x ◦ a)(u), (A ◦ x)(u) = (a ◦ x)(u)and(B ◦ A)(u) = (b ◦ a)(u). a∈A a∈A a∈A,b∈B If a fuzzy semihypergroup (H, ◦) satisfies in the condition a ◦ H = H ◦ a = χ , for any a ∈ H, where χ is the characteristic function of H, then the pair (H, ◦) is called a fuzzy hypergroup. Definition 2.2 ([8]): A non-empty set (S, +, ·), where + and · are two fuzzy hyperopera- tions, is called a fuzzy hyperring (we write by FHR) if for any a, b, c ∈ S, the following axioms are valid: (i) a ⊕ (b + c) = (a + b) ⊕ c, (ii) a + S = S + a = χ , (iii) a (b · c) = (a · b) c, (iv) a + b = b + a, (v) a (b + c) = (a · b) (a · c) and (a + b) c = (a · c) (b · c), where for fuzzy subsets α, ν of a fuzzy hypergroupoid (H, +) we have ((x + u)(r) ∧ α(u)), α = 0 (x ⊕ α)(r) = u∈H 0; α = 0 (α(u) ∧ (u + x)(r)), α = 0 (α ⊕ x)(r) = u∈H 0; α = 0 (α ν)(u) = (α(p) ∧ (p + q)(u) ∧ ν(q)); p,q∈H ((x · u)(r) ∧ α(u)), α = 0 (x α)(r) = u∈H 0; α = 0 FUZZY INFORMATION AND ENGINEERING 3 (α(u) ∧ (u · x)(r)), α = 0 (α x)(r) = u∈H 0. α = 0 The fuzzy hyperring (S, +, ·) is called commutative if for any a, b ∈ S, a · b = b · a. Also, S ⊆ S is called a subfuzzy hyperring of S if for any s , s ∈ S and x ∈ S, the following axioms are 1 2 valid: (i) if (s + s )(x)> 0, then x ∈ S , 1 2 (ii) s + S = S + s = χ , 1 1 S (iii) if (s · s )(x)> 0, then x ∈ S . 1 2 Definition 2.3 ([14]): Let ρ be an equivalence relation on a fuzzy semihypergroup (H, ◦). ∗ ∗ For any α, ϑ ∈ F (H), we define two relations ρ and ρ on F (H) as follows: (i) αρν if for any a ∈ H, there exists b ∈ H such that aρb and if α(a)> 0, then ν(b)> 0and if ν(a)> 0, then α(b)> 0. (ii) αρν if every element x ∈ H such that α(x)> 0is ρ equivalent to every y ∈ H such that ν(y)> 0. Definition 2.4 ([14]): An equivalence relation ρ on a fuzzy semihypergroup (H, ◦) is called a fuzzy (strongly) regular (we write by F(S)R)if aρb and a ρb , then (a ◦ a ) ρ(b ◦ b )((a ◦ a ) ρ(b ◦ b )). Definition 2.5 ([16]): Consider (S, +, ·) be a FHR and ϕ be a F(S)R on both (S, +) and (S, ·). Then ϕ is called a F(S)R on S. Let (S, +, ·) and (S , +, ·) be two FHR and g be a map from S to S . Then for any x, y, t ∈ S we have g(x + y) ={g(t)|(x + y)(t)> 0} and g(x · y) ={g(t)|(x · y)(t)> 0}. Theorem 2.6 ([16]): Let (S, +, ·) be a FHR and ϕ be an equivalence relation on S. If for every a b , ∈ S/ϕ, ϕ ϕ a b c = : (a + b )(c)> 0, aϕa , bϕb , ϕ ϕ ϕ a b c = : (a · b )(c)> 0, aϕa , bϕb , ϕ ϕ ϕ then (i) the relation ϕ is a FR on (S, +, ·) iff (S/ϕ,,) is a hyperring, (ii) the relation ϕ is a FSR on (S, +, ·) iff (S/ϕ,,) is a ring. Definition 2.7 ([16]): Let (S, +, ·) be a FHR. A (commutative) fundamental relation on S is the smallest equivalence relations such that the quotient structure (S/ϕ,,) is a (coommutative) ring. 4 E. MOHAMMADZADEH ET AL. Assume (S, +, ·) is a FHR and relation γ on S is defined as follows: aγ b ⇔∃n ∈ N, ∃k , ... , k ∈ N, ∃(z , ... , z ) ∈ S , (i = 1, ... , n) 1 n i1 ik ⎛ ⎞ n i ⎝ ⎠ such that z (a)> 0 ij i=1 j=1 ⎛ ⎞ n i ⎝ ⎠ and z (b)> 0. ij i=1 j=1 Then γ , the transitive closure of γ , is a fundamental relation on S (See [16]). We recall that for a set X, a one to one function from X onto X is called a permutation on X. We denote the set of all permutations of X by S .If X ={x , x , ... , x }, then we write S X 1 2 n n instead of S . S forms a group. S is the symmetric group on X (see [1]). X X X Consider S be a FHR. Nozari and Fahimi in [16], introduced the relation on S as follows: b ⇔∃n ∈ N, ∃(k , ... , k ) ∈ N , ∃ ∈ S and 1 n n × [(z , z , ... , z ) ∈ S , ∃ ∈ S , (i = 1, ... , n)] i1 i2 ik i k i i ⎛ ⎞ n k n ⎝ ⎠ such that z (a)> 0and A (b)> 0, ij (i) i=1 j=1 i=1 i ∗ ∗ where A = z . The quotient S/ , where is the transitive closure of ,isa i iσ (j) j=1 commutative ring (see [14]). Definition 2.8 ([17]): Let (L, +, ·) be a vector space over a field F. Consider the operation [−, −]: L × L −→ L defined by (x, y) −→ [x, y]. Then L is called a Lie algebra if the following axioms are satisfied: (i) [x, x] = 0, (ii) [x + y, z] = [x, z] + [y, z], (iii) [[x, y], z] + [[y, z], x] + [[z, x], y] = 0. Also, φ = I ⊆ L is called a Lie ideal if [x, r] ∈ I for any x ∈ I and r ∈ L. Let (R, +, ·) be a ring. We can introduce the Lie structure on R by defining the Lie product [x, y] = x · y − y · x for x, y ∈ R. This Lie structure, denoted by R , is called the associated Lie ring of R. Also, the Lie ring R is called nilpotent of class r if we have R = l (R ) ⊃ l (R ) ⊃ ··· ⊃ l (R ) ⊃ l (R ) = 0 l 1 l 2 l r l r+1 l where for k = 1, l (R ) = [l (R ), R ] is the Lie ideal of R generated by all elements of the k l k−1 l l l form [x, y] with x ∈ l (R ) and y ∈ R . R is called nilpotent of lenth r if its Lie ring is nilpotent k−1 l l of lenth r. Clearly [l (R ), l (R )] ⊆ l (R ). k l n l n+k l The n-th Lie centre of R is defined by U (R ) ={r ∈ R | [r, x , ... , x ] = 0, ∀x ∈ R,1 i n}, n 1 n i l l l where commutator of weight n(n ∈ N) is defined by [x , x , ... , x ] = [[x , x , ... , x ], x ] 1 2 n 1 2 n−1 n (see [18]). FUZZY INFORMATION AND ENGINEERING 5 We can show that U (R ) is a (unitary) subring of R and n l l {0}= U (R ) ⊆ U (R ) ⊆ U (R ) ⊆ ··· ⊆ U (R ) ⊆ ... . 0 1 2 n l l l l If U (R ) = R for some integer m, then R is nilpotent. The smallest such integer is called m l l l the class of R (see [18]). ∗ ∗ ∗ Now, in this paper, we define two new relations ζ and ν on FHR S such that S/ζ and n n,s n S/ϑ are nilpotent, and Engel Lie rings, respectively. n,s Note. From now one, let (S, +, ·) or S be a fuzzy hyperring (i.e. FHR), unless otherwise stated. 3. Nilpotent Fundamental Relation ζ In this section, we present that combining a new FSR with the fuzzy hyperring is a nilpotent Lie ring. Note. Let x, y ∈ S. For simplify, we use xy instead of x · y and ∃I instead of: m k ∃(k , ... , k ) ∈ N , ∃ ∈ S , ∃(x , x , ... , x ) ∈ S , ∃ ∈ S , (i = 1, 2, ... , m); 1 m m i1 i2 i ik k i i (i) = i and (j) = j if x ∈ L (S). (i) ij n m n Also, we use x instead of: ij i=1 j=1 {x + x , x ⊕ (x x ), x ⊕ ((x x ) x ), (x x21) (x x ), 11 12 11 12 22 11 12 22 32 11 12 22 × (x x ) ((x x ) x ), ...}. 11 21 12 22 32 Definition 3.1: We consider L (S) = S and for any k ≥ 0, L (S) ={t ∈ S|∃r ∈ S s.t. (xy)(r)> 0, and (t ⊕ (yx))(r)> 0, for some x ∈ L (S) and y ∈ S} k+1 k For any n ∈ N,welet ζ ={(x, x)|x ∈ S} and for any a, b ∈ S and n, m ∈ N, we define 1,n ζ as follows: m,n m k aζ b ⇔∃(k , ... , k ) ∈ N , ∃ ∈ S , ∃(x , x , ... , x ) ∈ S , m,n 1 m m i1 i2 ik ∃ ∈ S , (i = 1, 2, ... , m) (i) = i and (j) = j if x ∈ L (S),suchthat a ∈ A and b ∈ A . (i) ij n 1 2 where ⎧ ⎫ ⎧ ⎫ ⎛ ⎞ ⎛ ⎞ m k m (i) ⎨ i ⎬ ⎨ ⎬ ⎝ ⎠ ⎝ ⎠ A = z ∈ S| x (z)> 0 , A = z ∈ S| x (z)> 0 . 1 ij 2 (i) (j) (i) ⎩ ⎭ ⎩ ⎭ i=1 j=1 i=1 j=1 i.e. aζ b ⇔∃I such that a ∈ A and b ∈ A . m,n n 1 2 Now, suppose that ζ = ζ .If aζ b, then aζ b for some m ∈ N.Thus, ∃(k , ... , k ) n m,n n m,n 1 m m≥1 m k ∈ N , ∃ ∈ S , ∃(x , x , ... , x ) ∈ S , ∃ ∈ S , (i = 1, 2, ... , m); m i1 i2 ik i k i i (i) = i and (j) = j if x ∈ L (S),(I) (i) ij n 6 E. MOHAMMADZADEH ET AL. such that ⎛ ⎞ ⎛ ⎞ m k m (i) ⎝ ⎠ ⎝ ⎠ x (a)> 0, and x (b)> 0. ij (i) (j) (i) i=1 j=1 i=1 j=1 −1 Put I = (i) and J = (j).So (I) = i and (i) −1 −1 j = ( ) (J) = ( ) (J). (II) −1 (i) ( (I)) m −1 k −1 Thus, by (I) and (II), for (k , ... , k ) ∈ N , ∈ S , (x , x , ... , x ) ∈ S , ∈ S , (I = 1 m m I1 I2 Ik I k I I −1 −1 1, 2, ... , m).If x ∈ L (S), then (I) = i = (i) = I and J = (j) = j = ( −1 ) (J) IJ n (i) ( (I)) and so ⎛ ⎞ ⎛ ⎞ m (I) m k ⎝ ⎠ ⎝ ⎠ x −1 −1 (J) (a)> 0, x (b)> 0. (I) IJ −1 ( )(I) I=1 J=1 I=1 J=1 Hence, bζ a. Therefore, ζ is symmetric. n n Define for any a ∈ S, a(a) = (χ )(a) = 1. Then ζ is reflexive. Hence ζ , the transitive a n closure of ζ , is an equivalence relation. Theorem 3.2 ([8]): Let (R,, ◦) be a ring. Then (R, +, ·), where + and · are defined as follows is a fuzzy hyperring. For any a, b ∈ R, a + b = χ and a · b = χ . {a,b} a◦b Example 3.3: Let S = (Z ,, ◦). By Theorem 3.2, (S, +, ·) is a FHR. Then, L (S) = Z ,and 2 0 2 L (S) ={t ∈ Z |∃r ∈ Z ; (xy)(r)> 0 and (t ⊕ (yx))(r)> 0, for some x, y ∈ Z }. 1 2 2 2 Let r = 0, x = 0and y = 1. Then (xy)(r) = χ (0) = 1 > 0and (t ⊕ (yx))(r) = (t ⊕ χ )(0) = ((t + s)(0)) ∧ (χ (s)) s∈Z = ((χ )(0)) ∧ (χ (s)) {t,s} 0 s∈Z = (((χ )(0)) ∧ (χ (0))) ∨ (((χ )(0)) ∧ (χ (1))) {t,0} 0 {t,1} 0 = 1 > 0. Thus, L (S) = Z . 1 2 Lemma 3.4: Let be a permutation of S ,k = k + 1 and (i), ∀i ∈{1, 2, ... , k} (j) = k + 1, i = k + 1. Thus is a permutation of S k FUZZY INFORMATION AND ENGINEERING 7 Proof: We show that is one to one. For this let r, s ∈{1, 2, ... , (k + 1)} and (r) = (s). By the definition of for the case r ∈{1, 2, ... , k} and s = k + 1, we have (r) = k + 1. Since ∈ S ,wehave (r) ≤ k and so k + 1 = (r) ≤ k which is a contradiction. For s ∈ {1, 2, ... , k} and r = k + 1, we have k + 1 = (s) ≤ k, which is a contradiction. Finally, for r, s ∈{1, 2, ... , k},wehave (r) = (s),and so r = s. Therefore, is one to one. It is clear that, is onto. Therefore, is a permutation of S . Theorem 3.5: For each x, y, z ∈ Sifxζ y, then: ∗ ∗ (i) (xz) ζ (yz) and (zx) ζ (zy); n n ∗ ∗ (ii) (x + z)ζ (y + z) and (z + x)ζ (z + y). n n Proof: (i) If xζ y, then there exists m ∈ N such that xζ y. Then there exist (k , ... , k ) ∈ n m,n 1 m m k N , ∈ S and there exists (x , x , ... , x ) ∈ S , ∈ S , (i = 1, 2, ... , m);( (i) = m i1 i2 ik i k i i i and (j) = j if x ∈ L (S)), such that (i) ij n ⎛ ⎞ ⎛ ⎞ m k m (i) ⎝ ⎠ ⎝ ⎠ x (x)> 0and x (y)> 0. ij (i) (j) (i) i=1 j=1 i=1 j=1 Let z ∈ S and r, s ∈ S such that (xz)(r)> 0and (yz)(s)> 0. Then ⎛ ⎛ ⎞ ⎞ ⎧ ⎛ ⎞ ⎫ k k m m i ⎨ i ⎬ ⎝ ⎝ ⎠ ⎠ ⎝ ⎠ x z (r) = x (p) ∧ (pz)(r) . ij ij ⎩ ⎭ i=1 j=1 p∈S i=1 j=1 m k Let p = x, then ((( )( x ) z))(r)> 0. Also ij i=1 j=1 ⎛ ⎛ ⎞ ⎞ ⎧ ⎛ ⎞ ⎫ k k (i) (i) m m ⎨ ⎬ ⎝ ⎝ ⎠ ⎠ ⎝ ⎠ x z (s) = x (q) ∧ (qz)(s) . (i) (j) (i) (j) (i) (i) ⎩ ⎭ i=1 j=1 i=1 j=1 q∈S m (i) Let q = y, then ((( )( x ) z))(s)> 0. (i) (j) i=1 j=1 (i) We set k = k + 1, and we define (j), ∀j ∈{1, 2, ... , k } i i (j) = k + 1, j = k + 1. i i By Lemma 3.4, is a permutation of S .Thusfor z = x and any r, s ∈ S; k ik i i ⎛ ⎞ ⎛ ⎞ m m (i) ⎜ ⎟ ⎝ ⎠ x (r)> 0, and x (s) > 0 ij ⎝ ⎠ (i) (i) i=1 j=1 i=1 j=1 such that (i) = i and (j) = j if x ∈ L (S). Therefore, for any r, s ∈ S such that (xz)(r)> ij n (i) 0and (yz)(s)> 0, we have rζ s.Thus (xz)ζ (yz). In the simillar way, we can show that m,n (zx)ζ (zy). n 8 E. MOHAMMADZADEH ET AL. Now, if xζ y, then there exists k ∈ N and u = x, u , ... , u = y ∈ S such that 0 1 k u = xζ u ζ u ζ ...ζ u = y. 0 n 1 n 2 n n k By the above result we have ∗ ∗ ∗ ∗ (u z) = (xz)ζ (u z)ζ (u z)ζ ... ζ (u z) = (yz), 0 1 2 k n n n n ∗ ∗ and so (xz)ζ (yz). By the simillar way, (zx)ζ (zy). n n ∗ ∗ (ii) By the same manipulation we can prove (x + z)ζ (y + z) and (z + x)ζ (z + y). n n Corollary 3.6: For any n ∈ N, the relation ζ is a FSR. ∗ ∗ Proof: Let a, b, a , b ∈ S and aζ a , bζ b . Then by Theorem 3.5, we have (ab)ζ (a b) and n n n ∗ ∗ ∗ (a b)ζ (a b ). Then (ab)ζ (a b ). Similarly, we have (a + b)ζ (a + b ). n n n ∗ ∗ Proposition 3.7: For any n ∈ N, we have ζ ⊆ ζ . n+1 Proof: Let xζ y, then there exist k ∈ N, u = x, u , ... , u = y ∈ S such that 0 1 n+1 (u = x)ζ u ζ u ζ ...ζ (u = y). 0 n+1 1 n+1 2 n+1 n+1 We show that for any z, t ∈ S, if zζ t, then zζ t. (I) n+1 n Since zζ t, there exists m ∈ N such that zζ t and so ∃(k , ... , k ) ∈ N , ∃ ∈ S , n+1 m,n+1 1 m m ∃(x , x , ... , x ) ∈ S , ∃ ∈ S , (i = 1, 2, ... , m); ((i) = i and (j) = j if x ∈ i1 i2 ik i k (i) ij i i L (S)).Suchthat n+1 ⎛ ⎞ ⎛ ⎞ m k m (i) ⎝ ⎠ ⎝ ⎠ x (z)> 0, and x (t)> 0. ij (i) (j) (i) i=1 j=1 i=1 j=1 Now, let = and x = x , = and x ∈ / L (S).Since L (S) ⊆ L (S),wehave ij ij i ij n n+1 n m k x ∈ / L (S). Hence, for (k , ... , k ) ∈ N , ∈ S , (x , x , ... , x ) ∈ S , ∈ S , (i = ij n+1 1 m m i1 i2 ik i k i i 1, 2, ... , m); ( (i) = i and (j) = j if x ∈ L (S)),suchthat ij n (i) ⎛ ⎞ ⎛ ⎞ m k m (i) ⎝ ⎠ ⎝ ⎠ x (z)> 0, and x (t)> 0. ij (i) (j) (i) i=1 j=1 i=1 j=1 Therefore, zζ t. Consequently, by (I) we have (u = x)ζ u ζ u ζ ...ζ (u = y). Then xζ y. 0 n 1 n 2 n n ∗ ∗ Corollary 3.8: If S is a commutative FHR, then = ζ . ∗ ∗ Proof: It is clear that, ζ ⊆ . It is enough to show that if S is a commutative, then ζ . For this, let a b. Then for some m ∈ N, ∃(k , ... , k ) ∈ N , ∃ ∈ S , ∃(x , x , ... , x ) ∈ n 1 m m i1 i2 ik S , ∃ ∈ S , (i = 1, 2, ... , m),suchthat a ∈ A and b ∈ A . i k 1 2 i FUZZY INFORMATION AND ENGINEERING 9 Since S is commutative, for any i, j ∈ N, each element x can commute with others, (i) (j) (i) and so we can consider , such that (i) = i and (j) = j if x ∈ L (S), then b ∈ A i (i) ij n 2 which implies that aζ b and so ⊆ ζ . n n ∗ ∗ Example 3.9: Let S be FHR as Example 3.3. Then, by Corollary 3.8, we have = ζ and so ∗ ∗ S/ζ = S/ S. Definition 3.10: Let R be a Lie ring. We define M (R ) = R and l 0 l l M (R ) ={[x, y] | x ∈ M (R ), y ∈ R }. k+1 l k l l Remark 3.11: Let ϕ be a FSR on S. Then, by Theorem 2.6, without loss of generality, one can assume that (S/ϕ,,) is a Lie ring. Theorem 3.12: S/ζ is a nilpotent Lie ring of class at most n + 1. Proof: We show that for a FSR ϕ on S and any k ∈ N, t s M (S/ϕ) = , |t ∈ L (S), s ∈ S . (I) k+1 k ϕ ϕ The proof is based on induction on k.Put R = S/ϕ and x = , for any x ∈ S.If k = 0, then M (R) ={[t, s] | t ∈ L (S), s ∈ S}. Now, put a = [t, s] where t ∈ L (S) and s ∈ S,so 1 0 k+1 there exists x ∈ L (S), y ∈ S and r ∈ S such that (xy)(r)> 0and (t ⊕ (yx))(r)> 0. Then by Theorm 2.6, x y = r and t (y x) = r. Therefore t = [x, y]. By induction hypotheses we have t ∈ M (R). Hence, a = [t, s] ∈ M (R). k+1 k+2 Conversely, let a ∈ M (R). Then by Definition 3.10, a = [x, y], where x ∈ M (R) and k+2 k+1 y ∈ R. So induction hypotheses implies that x = [u, v], where u ∈ L (S) and v ∈ S.Since uv, vu ∈ F (S),wehave (uv)(c)> 0and (vu)(b)> 0, for some c, b ∈ S. By Definition 2.2, for b, c,wehave χ (c) = (S + b)(c), then there exists t ∈ S such that (t + b)(c)> 0. Thus (t ⊕ (vu))(c) = (t + b)(c) ∧ (vu)(b)> 0, and so t ∈ L (S). Hence, by (uv)(c)> 0, k+1 b∈S (t ⊕ (vu))(c)> 0 and Theorem 2.6, we have c = t (v u) = u v.Thus, t = [u, v] = x, where t ∈ L (S) and y ∈ S. Hence, k+1 a = [x, y] = [t, y] ∈{[t, s]|∈ L (S), s ∈ S}. k+1 Therefore M (S/ϕ)) ={[t, s]|t ∈ L (S), s ∈ S}. k+1 k ∗ x Now, let R = S/ζ and x = , for any x ∈ S. By induction on i, we show that M (R) ⊆ n+1−i U (R).Let i = 0, and t, s ∈ S/ζ , t ∈ L (S) and s ∈ S. For any r, s ∈ S such that (ts)(r)> i n n, 0, (st)(p)> 0. Assume that m = 1, i = 1, k =2, ∈S , x =t, x =s, ∈ S , (1) = 2, (2) = 1. 1 1 11 12 1 2 1 1 Then 0 <(ts)(r) = (x x )(r),and0 <(st)(p) = (x x )(p) = x x . Then 11 12 12 11 (1) (1) (1) (2) (1) (1) rζ p. Therefore, r = p, and so by Theorem 2.6, t s = r = p = s t. Then (t s) − n, (s t) = 0. Thus, [t, s] = (t s) − (s t) = 0. Hence, M (R) ={[t, s]|t ∈ L (S), s ∈ S}= n+1 k {0}. Therefore, {0}= M (R) ⊆ U (R) ={0}. Now let M (R) ⊆ U (R) and a ∈ M (R). n+1 0 n+1−i i n−i 10 E. MOHAMMADZADEH ET AL. Then for any y ∈ R we have [a, y] ∈ M (R) ⊆ U (R). Hence for any y ∈ R,[[a, y], y] = 0, n+1−i i i and so by definition of U (R),weget a ∈ U (R). Therefore, M (R) ⊆ U (R). i+1 i+1 n+1−i i Now, let i = n + 1. Then M (R) ⊆ U (R),thatis R = M (R) = U (R). Hence n+1−(n+1) n+1 0 n+1 R is nilpotent of class at most n + 1. The relation ζ which is defined in Definition 3.1, can also be introduced for rings. Note that every ring is a FHR. Here, we introduce the ζ relation, which is defined on a finite FHR. This relation connects the FHRs with nilpotent Lie rings. Definition 3.13: Let S be a finite FHR. Then we define the relations ζ and ζ on S by ∗ ∗ ζ = ζ and ζ = ζ . n≥1 n≥1 Theorem 3.14: Consider S be a finite FHR. Then the relation ζ is a nilpotent fundamental relation. ∗ ∗ Proof: We show that ζ is a FSR on S such that S/ζ is a nilpotent Lie ring, and if ϕ is a FSR on S such that S/ϕ is a nilpotent Lie ring (of class n + 1), then ζ ⊆ ϕ. ∗ ∗ ∗ Since ζ = ζ , by Corollary 3.6, it is easy to see that ζ is a FSR on S.Since S is finite, n≥1 n ∗ ∗ ∗ by Proposition 3.7, there exists k ∈ N such that ζ = ζ .Thus ζ∗= ζ , for some m ∈ N k+1 k and so by Theorem 3.12, S/ζ is a nilpotent Lie ring. Suppose ϕ is a FSR on S such that K = S/ϕ is a nilpotent Lie ring of class n + 1. We show that for any n ∈ N, ζ ⊆ ϕ.Let xζ y. Then there exists m ∈ N such that xζ y. Then ∃I such n n m,n n that ⎛ ⎞ ⎛ ⎞ k (i) m i m ⎝ ⎠ ⎝ ⎠ x (x)> 0and x (y)> 0. ij (i) (j) (i) i=1 j=1 i=1 j=1 Hence, by Theorem 2.6, for x, y ∈ S, m k m (i) x x y (i) (j) ij (i) = and = . ϕ ϕ ϕ ϕ i=1 j=1 i=1 j=1 Since S/ϕ is a nilpotent Lie ring of class n + 1, by the proof of Theorem 3.12, we have 0 t s 0 t s t s s t = M (S/ϕ) ={[ , ]|t ∈ L (S), s ∈ S},and so = [ , ] = ( ) ( ). Then n+1 n ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ x x t s s t ij v v ij = . Thus, for x ∈ L (S) and v ∈ S,weget = . Moreover, if x ∈ ij n ij ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ij L (S), then commutes with others. Also, if x ∈ L (S) then (i) = i and (j) = j,and so n ij n (i) (i) (j) ij (i) x y = . Consequently, = , which implies that xϕy. Therefore, ζ ⊆ ϕ. ϕ ϕ ϕ ϕ ∗ ∗ ∗ Now, we can prove that ζ ⊆ ϕ. For this, let z, t ∈ S and zζ t. Then for any n ∈ N, zζ t andsothereexist z , z , ... , z ∈ S (k ∈ N)suchthat (z = z )ζ z ζ ··· ζ (z = t). Then we 0 1 k 0 n 1 n n k have (z = z )ϕz ϕ ··· ϕ(z = t). Hence, ζ ⊆ ϕ. 0 1 k FUZZY INFORMATION AND ENGINEERING 11 4. ζ -Role of a Fuzzy Hyperring In this section, we introduce the concept of ζ -role and we determine necessary and sufficient conditions such that the relation ζ be transitive. Note. For simplify we use ∀I instead of the following assumption: ∀m ∈ N, k ∈ r≥1 i N(i = 1, 2, ... , m), (x , ... , x ) ∈ S and ∀ ∈ S , ∈ S ; (i) = i and (j) = j,if x ∈ i1 ik m i k (i) ij i i L (S). r≥1 Definition 4.1: Let φ = X ⊆ S. Then we say that X is an ζ -role of S if the following implica- tion holds: ∀I if there exists x ∈ X such that x ∈ A , then for any y ∈ S \ X we have y ∈ A , r≥1 1 2 where ⎧ ⎫ ⎧ ⎫ ⎛ ⎞ ⎛ ⎞ m k m (i) ⎨ i ⎬ ⎨ ⎬ ⎝ ⎠ ⎝ ⎠ A = z ∈ S| x (z)> 0 and A = z ∈ S| x (z)> 0 . 1 ij 2 (i) (j) (i) ⎩ ⎭ ⎩ ⎭ i=1 j=1 i=1 j=1 m k Note that by Definition 4.1, there exists x ∈ X such that ( x )(x)> 0, and for ij i=1 j=1 m (i) any y ∈ X,wehave ( x )(y) = 0. (i) (j) i=1 (i) j=1 Remark 4.2: Let ϕ be a FSR on S, z ∈ ϕ(z) be the class of z module ϕ. Then ϕ(z) is an ζ -role of S. Proof: Let ∀I , there exists x ∈ ϕ(z) such that x ∈ A .Iffor y ∈ S\ϕ(z) we have y ∈ A , r≥1 1 2 then ϕ is a FSR implies that xϕy.Thus, y ∈ ϕ(z), is a contradiction. Theorem 4.3: Let φ = X ⊆ Sandx ∈ X. Then the following conditions are equivalent: (i) Xisan ζ -role of S, (ii) if xζ y, then y ∈ X, (iii) if xζ y, then y ∈ X. Proof: (i ⇒ ii) If (x, y) is a pair of S ,suchthat x ∈ X and xζ y, then for any n ∈ N, ∃I such that x ∈ A and y ∈ A .If y ∈ X, then y ∈ S \ X and so by Definition 4.1, y ∈ A , which is a 1 2 2 contradiction. Thus y ∈ X. 2 ∗ (ii ⇒ iii) Suppose that (x, y) is a pair of S such that x ∈ X and xζ y. Then there exists (z , ... , z ) ∈ S (k ∈ N)suchthat (x = z )ζ z ζ ··· ζ(z = y). By (ii) (k-times) we obtain y ∈ 1 0 1 k k X. (iii ⇒ i) Let x ∈ X and k ∈ N(i = 1, 2, ... , m), (x , x , ... , x ) ∈ S such that x ∈ A . i i1 i2 1 ik Then for any ∈ S and any ∈ S such that ((i) = i,and (j) = j,if x ∈ m i k (i) ij L (S)) and any y ∈ S \ X such that y ∈ A ,wehave xζ y.Thus xζ y and by (iii), we obtain r 2 r≥1 y ∈ X, which is a contradiction. Therefore, X is an ζ -role of S. Theorem 4.4: For any a ∈ S, ζ(a) is an ζ -role of S if and only if ζ is transitive. 12 E. MOHAMMADZADEH ET AL. ∗ k Proof: (⇒) Suppose xζ y. Then there exists (z , ... , z ) ∈ S (k ∈ N)suchthat (x = 1 k z )ζ z ζ ··· ζ(z = y).Since ζ(z ),for any0 ≤ i ≤ k,isan ζ -role, by Theorem 4.3, we have 0 1 k i z ∈ ζ(z ).Thus y ∈ ζ(x) and so xζ y. Hence ζ = ζ . Therefore, ζ is transitive. i i−1 (⇐) Suppose x, y ∈ S, z ∈ ζ(x) and zζ y.Since ζ is transitive, we conclude that y ∈ ζ(x), and so by Theorem 4.3, ζ(x) is a ζ -role of S. Remark 4.5: Let θ be a FSR on S and z ∈ S. Then θ(z) is an ζ -role of S. Definition 4.6: Consider A be a non-empty subset of S. The intersection of any ζ -role of S, which contains A is called ζ -closure of A in S. We denote the ζ -closure of A by K(A). Note.Weuse ∃I instead of the following assumption: r≥1 m k ∃(k , ... , k ) ∈ N , ∃(x , ... , x ) ∈ S (i = 1, ... , m),and ∃ ∈ S , ∈ S ; (i) = 1 m i ik m i k 1 i i i, (j) = j if x ∈ L (S). (i) ij r r≥1 Theorem 4.7: Let m ∈ N, φ = A ⊆ Sand (i) E (A) = A; (ii) E (A) ={x ∈ S|∃I such that x ∈ A and for some a ∈ E (A) we have a ∈ A }; n+1 r≥1 1 n 2 (iii) E(A) = E (A). Then E(A) = K(A), n≥1 Proof: By Definition 4.6, it is enough to prove E(A) is an ζ -role of S.If A ⊆ B and B is an ζ -role, then E(A) ⊆ B. Suppose there exists a ∈ E(A) such that a ∈ A ,and ∃ ∈ S , ∈ S 1 m i k such that (i) = i, (j) = j,if x ∈ L (S).Since E = E , there exists n ∈ N such (i) ij r n r≥1 n≥1 that a ∈ E (A) and a ∈ A . Now, if there exists t ∈ S \ E(A) such that t ∈ A , then t ∈ E (A). n 1 2 n+1 Thus, t ∈ E (A) = E(A) which is a contradiction. Thus for any t ∈ S \ E(A), t ∈ A and n 2 n≥1 so E(A) is an ζ -role of S. We prove the second role by induction on n.Wehave E (A) = A ⊆ B. Suppose E (A) ⊆ B. We show that E (A) ⊆ B.If z ∈ E (A), then ∃I such that z ∈ A n n+1 n+1 r≥1 1 and t ∈ A , for some t ∈ E (A).Since E (A) ⊆ B we have t ∈ B and t ∈ A .Moreover, B is an 2 n n 2 ζ -role of S,and z ∈ A , then z ∈ B. Theorem 4.8: If φ = A ⊆ S, then K(A) = K(a). a∈A Proof: It is easy to see that for any a ∈ A, K(a) ⊆ K(A). By Theorem 4.7, we get K(A) = E (A) and E (A) = A = {a}. Suppose E (A) = E (a) holds for n and z ∈ n 1 n n n≥1 a∈A a∈A E (A), then ∃I such that z ∈ A and a ∈ A , for some a ∈ E (A). By the hypotheses of n+1 r≥1 1 2 n induction, E (A) = E (b) and so for a ∈ E (A) we have a ∈ E (b), then a ∈ E (b) n n n n n b∈A b∈A for some b ∈ A.Thus, a ∈ A for some a ∈ E (b). Hence, z ∈ E (b),and so E (A) ⊆ 2 n n+1 n+1 E (b). Therefore, K(A) = K(a). n+1 b∈A a∈A Lemma 4.9: For any n ≥ 2 and x, y, z ∈ S we have (i) E (E (z)) = E (z), n 2 n+1 (ii) x ∈ E (y) if and only if y ∈ E (x). n n FUZZY INFORMATION AND ENGINEERING 13 Proof: (i) Let z ∈ S. Then by Theorem 4.7, we have E (E (z)) ={x ∈ S|∃I such that x ∈ A and for some a ∈ E (z), a ∈ A }= E (z). 2 2 r≥1 1 2 2 3 Now, we prove by induction on n. Suppose E (E (z)) = E (z). Then n 2 n+1 E (E (z)) ={x ∈ S|∃I such that x ∈ A and for some a ∈ E (E (z)) a ∈ A } n+1 2 r≥1 1 n 2 2 = E (z). n+2 (ii) We prove by induction on n.Itiseasytoseethat x ∈ E (y) if and only if y ∈ E (x). Suppose 2 2 x ∈ E (y). Then ∃I such that x ∈ A and for some a ∈ E (y),wehave a ∈ A . Clearly, n+1 r≥1 1 n 2 x ∈ E (x). Also, x ∈ A and a ∈ A , then a ∈ E (x). By hypotheses of induction, since a ∈ 1 1 2 2 E (y) we have y ∈ E (a) and so by (i), y ∈ E (E (x)) = E (x). Therefore, x ∈ E (y) if and n n n 2 n+1 n only if y ∈ E (x). Theorem 4.10: Let x, y ∈ S.Then xEy if and only if x ∈ E({y}) is an equivalence relation on S. Proof: By Theorem 4.7, W{x}= K{x} and by Definition 4.6, we have x ∈ E{x},thus xEx. Then E is reflexive. For transitivity, let x, y, z ∈ S such that xEy and yEz. Then by Theorem 4.7, x ∈ K(y) and y ∈ K(z). Thus, for any ζ -role P which contains z i.e. z ∈ P,wehave K(z) ⊆ P and since y ∈ K(z),wehave y ∈ P. Then K(y) ⊆ P and so x ∈ P. Thus for any ζ -role of S,wehave x ∈ P, i.e. x ∈ K(z). Hence, by Theorem 4.7, xEz. Then E is transitive. The symmetrically of E follows directly from Lemma 4.9. Theorem 4.11: For all a, b ∈ S, aEb if and only if aζ b. Proof: (⇐) Let aζ b. Then for any n ∈ N, there exists m ∈ N,suchthat aζ b. Then ∃I such m,n n that a ∈ A , b ∈ A ,so a ∈ E (b). Thus, by Lemma 4.9, b ∈ E (a). Hence aEb and so ζ ⊂ E. 1 2 2 2 (⇒) If xEy, then x ∈ E (y) for some n ∈ N. Then ∃I such that x ∈ A and for some x ∈ n r≥1 1 1 E (y) we have x ∈ A .Thus, xζ x . Hence continuing this method ∃x , ... , x ∈ S such n−1 1 2 n 1 2 n−1 that x ∈ E (y) and x ζ x . Then (x = x )ζ x ζ ...ζ (x = y). Therefore, E ⊆ ζ . i n−i i−1 n i 0 n 1 n n n−1 Remark 4.12: By Theorem 2.6, if (S, +, ·) is a FHR, then (S/ζ ,,) is a ring. We define 0 −1 ∗ ω by ω ={x ∈ S | ψ(x) = }= ψ (0 ), where ψ : S → S/ζ is the canonical projec- S S S/ζ tion. −1 Lemma 4.13: If M is an ζ -role of S, then ψ (ψ (M)) = M. −1 −1 Proof: We know M ⊆ ψ (ψ (M)).If x ∈ ψ (ψ (M)), then there exists b ∈ M such that x b ∗ ψ(x) = ψ(b). For any y ∈ S, ψ(y) = and so = .Thus, xζ b.Since M is an ζ role of S, ∗ ∗ ∗ ζ ζ ζ xζ b and b ∈ M, then by Theorem 4.3, we have x ∈ M. Theorem 4.14: ω is a fuzzy subhyperring of S, which is also an ζ -role of S. Proof: It is clear that, ω ⊆ S.Let x, y ∈ ω and for any r ∈ S, r = and (x + y)(r)> 0((x · S S ∗ y)(r)> 0). Then by Theorem 2.6, x y = r (x y = r). Then r = 0, and so r ∈ ω.Weshow that for any y ∈ ω , ω + y = χ .Let x, y ∈ ω . Then, by S + y = χ , there exists u ∈ S such S S ω S S S 14 E. MOHAMMADZADEH ET AL. u x u 0 that (u + y)(x)> 0. Therefore, by Theorem 2.6, = . Then = .Thus u ∈ ω . ∗ ∗ ∗ ∗ ∗ ζ ζ ζ ζ ζ Consequently, ω + y = χ . (I) S ω Hence, ω is a fuzzy subhyperring of S. Now we prove that −1 x ∈ ψ (ψ ({y})) ⇔ (ω + y)(x)> 0. (II) For this let x ∈ S, t ∈ ω and (t + y)(x)> 0. Then by Theorem 2.6, ψ(x) = ψ(t) ψ(y) = 0 ψ(y) = ψ(y) S/ζ −1 −1 and so x ∈ ψ (ψ ({y})). Conversely, let x ∈ ψ (ψ ({y})). Then ψ(x) = ψ(y).Since S + y = χ , there is a ∈ S such that (a + y)(x)> 0andso ψ(y) = ψ(x) = ψ(a) ψ(y) (since by ∗ ∗ −1 ∗ ∗ Remark 2.6, S/ζ is a ring and ψ(y) ∈ S/ζ ). Then ψ(a) = 0 and so a ∈ ψ (0 ) = ω . S/ζ S/ζ S Thus by (a + y)x > 0, we have (ω + y)(x)> 0. Also, since −1 z ∈ ψ (ψ ({y})) ⇐⇒ ψ(z) = ψ(y) z y ⇐⇒ = ∗ ∗ ζ ζ ⇐⇒ zζ y ⇐⇒ z ∈ = E({y}) = K(y). (by Theorems 4.10, 4.11, 4.7, ) −1 we have K(y) = ψ (ψ ({y})) which by (I), (II) implies that K(y) = ω ,and so ω is a ζ -role S S of S. By Lemma 4.13 and Theorem 4.14, we have the following corollary. −1 Corollary 4.15: ψ (ψ (ω )) = ω . S S 5. Engel Lie Rings Derived from Fuzzy Hyperrings In this section, continuing our previous work, we define a FSR,ona FHR S such that the quotient is an Engel Lie ring. Let s be a fixed element of S, unless we notify. Definition 5.1: We define L (S) = S,and for k ≥ 0, (0,s) L (S) ={t ∈ S |∃r ∈ Ss.t (x · s)(r)> 0 and (t ⊕ (s · x))(r)> 0, forsome x ∈ L (S)}. (k+1,s) (k,s) Now, let n ∈ N and ν be the diagonal relation on S. For every integer m > 1, define (1,n,s) ν as follows: (m,n,s) aν b ⇔∃k , ... , k ∈ N(m ∈ N) and ∃ ∈ S , (x , x , ... , x ) ∈ S , ∃ ∈ S ; (m,n,s) 1 m m i1 i2 ik i k i i ((i) = i, (j) = j if x ∈ L (S)) and a ∈ A , b ∈ A (i) ij (n,s) 1 2 i.e. FUZZY INFORMATION AND ENGINEERING 15 ⎛ ⎞ m k m (i) ⎝ ⎠ (x )(a)>0and (x (b)> 0 ij (i) (j) (i) i=1 j=1 i=1 j=1 Consider ν = ν . Then ν is symmetric (Similar to Definition ζ ). Define for (n,s) (m,n,s) (n,s) n m≥1 any a ∈ S, a(a) = (χ )(a) = 1, thus ν is reflexive. Then ν , the transitive closure of a (n,s) (n,s) ν , is an equivalence relation on S. (n,s) In the same way of the Corollary 3.6, we have the following theorem. Theorem 5.2: For any n ∈ N the relation ν is a FSR. (n,s) ∗ ∗ ∗ Corollary 5.3: For a fixed element s ∈ S we have ν ⊆ ζ ⊆ (n,s) Proof: Let zν t. Then there exists m ∈ N such that zν t and so ∃(k , ... , k ) ∈ (n,s) (m,n,s) 1 m m k N , ∃ ∈ S , ∃(x , x , ... , x ) ∈ S , ∃ ∈ S , (i = 1, 2, ... , m); ((i) = i and (j) = m i1 i2 ik i k (i) i i j,if x ∈ L (S)). Then ij (n,s) ⎛ ⎞ ⎛ ⎞ m k m (i) ⎝ ⎠ ⎝ ⎠ x (z)> 0, and x (t)> 0. ij (i) (j) (i) i=1 j=1 i=1 j=1 Now, let = and x = x , = and x ∈ / L (S).By L (S) ⊆ L (S),wehave ij ij i ij n (n,s) n m k x ∈ / L (S). Thus, for (k , ... , k ) ∈ N , ∈ S , (x , x , ... , x ) ∈ S , ∈ S , (i = ij (n,s) 1 m m i1 i2 ik k i i i 1, 2, ... , m),such that ( (i) = i and (j) = j if x ∈ L (S)),suchthat ij n (i) ⎛ ⎞ ⎛ ⎞ k (i) ⎝ ⎠ ⎝ ⎠ x (z)> 0, and x (t)> 0. ij (i) (j) i=1 (i) j=1 i=1 j=1 ∗ ∗ ∗ ∗ Hence, zζ t and so ν ⊆ ζ . By the similar way, ζ ⊆ n n n (n,s) Similar to the Corollary 3.8, we have: ∗ ∗ ∗ Corollary 5.4: Let S be a commutative FHR. Then ν = ζ = (n,s) ∗ ∗ Proposition 5.5: For any n ∈ N we have ν ⊆ ν . (n+1,s) (n,s) Proof: The proof is similar to Proposition 3.7. Definition 5.6: Let R be a Lie ring and n ∈ N. Then, R is an n-Engel if for any x, y ∈ l l R ,[x, y] = 0. Also, for a fixed element s ∈ R we define the n-th Lie centre of R as follows: l l l U (R ) ={0} and U (R ) ={r ∈ R ;[r, s, ... , s ] = 0}. (0,s) l (n,s) l l ! " n−times We can conclude that, R is an n-Engel if and only if for any s ∈ S, U (R ) = R . l (n,s) l l Also, let M (R ) = R and M (R ) ={[x, s] | x ∈ M (R )}. Now by Definition 5.6, (0,s) l l (k+1,s) l (k,s) l we have the following theorem. 16 E. MOHAMMADZADEH ET AL. Theorem 5.7: Let n ∈ N and l (R ) be the Lie ideal generated by M (R ). Then Lie ring R (n,y) l (n,y) l l is an n-Engel if and only if for any y ∈ R ,l (R ) ={0}. l (n,y) l Proof: (⇒)Let y be a fixed element of n-Engel Lie ring R and z ∈ l (R ). Then z ∈ [x , y] l (n,y) l 1 for some x ∈ l (R ),and so x ∈ [x , y] for some x ∈ l (R ). Then, z ∈ [x , y] ⊆ 1 (n−1,y) l 1 2 2 (n−2,y) l 1 [x , y, y]. Continuing this we have z ∈ [x , y], for some x ∈ R . By hypotheses for any x, y ∈ 2 n n n l R ,wehave[x , y] = 0. Hence, z = 0andso l (R ) ={0}. l n n (n,y) l (⇐) Let for any y ∈ R , l (R ) ={0} and x be arbitary elements of R .By[x, y] ∈ l (R ), l (n,y) l l (1,y) l we have [[x, y], y] ∈ l (R ). Continuing this we have [x, y] ∈ l (R ) = 0, and so [x, y] = (2,y) l n (n,y) l n 0, which implies that R is an n-Engel Lie ring. Remark 5.8: Let ϕ be a FSR on S. Then by Theorem 2.6, without loss of generality, we can assume that (S/ϕ,,) is a Lie ring. Theorem 5.9: If ϕ is a FSR on S, then for any k ∈ N, # $ S t s M s = , |t ∈ L (S) . k,s k+1, ϕ ϕ ϕ Proof: Put R = S/ϕ and x = , for all x ∈ S.If k = 0, then M (R) ={[t, s] | t ∈ L (S)}. 1,s (0,s) Now, we show that t s M (R) ⊇ , |t ∈ L (S) . k+2,s k+1,s ϕ ϕ Put a = [t, s], where t ∈ L (S),sothereexist x ∈ L (S) and r ∈ S such that (xs)(r)> 0 (k+1,s) (k,s) and (t ⊕ (sx))(r)> 0. Then by Theorem 2.6, x s = r and t (s x) = r.Thus t = [x, s]. By induction hypotheses we have t ∈ M (R). Hence a = [t, s] ∈ M (R). (k+1,s) (k+2,s) Conversely, let a ∈ M (R). Then a = [x, s], where x ∈ M (R). So hypotheses of k+2,s (k+1,s) induction implies that x = [u, s], where u ∈ L (S).Let b, c ∈ S, (us)(c)> 0and (su)(b)> 0. (k,s) For b, c we have, 1 = χ (c) = (S + b)(c). Then there exists t ∈ S such that (t + b)(c)> 0. Thus (t ⊕ su)(c) = (t + b)(c) ∧ (su)(b)> 0, b∈S and so t ∈ L (S). Hence, by Theorem 2.6, c = t (s u) = u s.Thus, t = [u, s] = x. (k+1,s) Then, a = [x, s] = [t, s] ∈{[t, s]|t ∈ L (S), }. (k+1,s) Therefore M (S/ϕ) ={[t, s]|t ∈ L (S)}. (k+2,s) (k+1,s) Theorem 5.10: S/ν is an (n + 1)-Engel Lie ring. (n,s) Proof: For any x ∈ S let R = S/ν and x = . By induction on i,weshow (n,s) ν (n,s) that M (R) ⊆ U (R).Let i = 0, then {0}= M (R) ⊆ U (R) ={0}.Let a ∈ (n+1−i,s) (i,s) (n+1,s) (0,s) FUZZY INFORMATION AND ENGINEERING 17 M (R). Then [a, s] ∈ M (R). By hypotheses, we have M (R) ⊆ U (R). (n−i,s) (n+1−i,s) (n+1−i,s) i,s Hence [[a, s], s] = 0, and so a ∈ U (R).Thus, M (R) ⊆ U (R).If i = n + 1, then i (i+1,s) (n−i,s) (i+1,s) M (R) ⊆ U (R) and so R = M (R) = U (R). Therefore, R is an (n + (n+1−(n+1),s) (n+1,s) (0,s) (n+1,s) 1)-Engel Lie ring. 6. Conclusion The fundamental relations on hyperrings were studied by Vougiouklis. Then, commutative fundamental relations on fuzzy hyperrings were introduced by Nozari and Fahimi. Now in this paper, first the smallest equivalence relation ζ on a FHR S, such that the set of equiva- lence classes was a nilpotent Lie ring, was introduced. Then, the relation ν was defined, (n,s) as the FSR, so that the quotient would be an Engel Lie ring. Finally, theses two different relations were compared. List of Abbreviations F (H), the set of all fuzzy subsets of H, FHR, a fuzzy hyperoperations, FR, a fuzzy regular, FSR, a fuzzy strongly regular, F (S)R, a fuzzy (strongly) regular. Disclosure Statement No potential conflict of interest was reported by the author(s). Notes on contributors E. Mohammadzadeh is assistant professor at the Payame Noor University, Iran. She has published more than 20 papers in the international journals. Her main scientific interests are algebraic logic, algebraic hyperstructures and fuzzy algebras. R. A. Borzooei is full professor at the Shahid Beheshti University, Tehran, Iran. He is currently, Editor In- Chief and founder of “Iranian Journal of Fuzzy Systems” and “Journal of Algebraic Hyperstructures and Logical”, editorial board of six international journals. He has published more than 330 papers in the international journals. His main scientific interests are algebraic logics, ordered algebraic structures, algebraic hyperstructures, fuzzy algebras and fuzzy graphs. F. Mohammadzadeh is assistant professor at the Payame Noor University, Tehran, Iran. Her main scientific interests are algebraic hyperstructures and fuzzy algebras. S. S. Ahn is full professor at the Dongguk University, Korea. She works in the Department of Mathe- matics Education from 1993 to present. She has published more than 150 papers in the international journals. Her main scientific interests are logical algebras, ordered algebras, algebraic hyperstructures, fuzzy algebras and fuzzy graphs. ORCID S. S. Ahn http://orcid.org/0000-0003-4129-9814 18 E. MOHAMMADZADEH ET AL. References [1] Robinson DJS. A course in the theory of groups. New York: Springer-Verlag; 1980. [2] Golod ES. Some problems of burnside type. In Proceeding of the Internal Congress of Mathe- maticians, Moscow: 1966. p. 284–289. [3] Heineken H. Engelsche elemente der lange drei. Illionis J Math. 1961;5:681–707. [4] Kappe LC, Kappe WP. On three-Engle groups. Bulletin Australian Math Soc. 1972;7:391–405. [5] Marty F. Sur une generalization de la notion de groupe. In: 8th Congress of Mathematics, Stockholm, Sweden: Scandenaves; 1934. p. 45–49. [6] Rosenfeld A. Fuzzy groups. J Math Anal Appl. 1971;35:512–517. [7] Ameri R, Borzooei RA, Mohammadzadeh E. Engel fuzzy subgroups. Ital J Pure Appl Math. 2015;34:251–262. [8] Leoreanu-Foteaa V, Davvaz B. Fuzzy hyperrings. Fuzzy Sets Syst. 2009;160:2366–2378. [9] Mohammadzadeh E, Borzooei RA. On Engel fuzzy subpolygroups. New Math Natural Comput. 2017;13(2):195–206. [10] Mohammadzadeh E, Borzooei RA. Nilpotent fuzzy subgroups. Mathematics. 2018;6(2):27. [11] Mohammadzadeh E, Borzooei RA. Results on (Engel, solvable, nilpotent) fuzzy subpolygroups. Kragujev J Math. 2021;45(5):667–684. [12] Mohammadzadeh E, Jun YB, Borzooei RA. Results on Engel fuzzy subgroups. Algebra Struct Appl. 2017;4(2):1–14. [13] Mordeson JN, Malik MS. Fuzzy commutative algebra. J Fuzzy Math. 1995;3:1–15. [14] Sen MK, Ameri R, Chowdhury G. Fuzzy semihypergroups. Soft Comput. 2008;12:891–900. [15] Vougiouklis T. Hyperstructures and their representations. Palm Harbor (FL): Hadronic Press; 1994. [16] Nozari T, Fahimi N. Fuzzy hyperings and fundamental relation. J Intell Fuzzy Syst. 2016;30: 1311–1317. [17] Davvaza B, Santilli RM, Vougiouklis T. Algebra hyperalgebra and Lie–Santilli theory. J Gen Lie Theory Appl. 2015;9(2):1–5. [18] Sharma RK, Srivastava JB. Lie solvable rings. Proc Am Math Soc. 1985;94(1):1–8. [19] Davvaz B. Polygroup theory and related systems. Singapore: World Scientific; 2013. [20] Mohammadzadeh E, Mohammadzadeh F. On nilpotent and solvable pair of polygroups. J Algebra Hyperstructures Log Algebra. 2021;2(2):1–12. [21] Mohammadzadeh E, Muhiuddin G, Zhan J, et al. Nilpotent fuzzy Lie ideals. J Intell Fuzzy Syst. 2020;39(3):4071–4079.
Journal
Fuzzy Information and Engineering
– Taylor & Francis
Published: Oct 2, 2022
Keywords: (Nilpotent) Lie ring; Engel Lie ring; fuzzy hyperring; strongly regular relation; 17B62; 20N20