Transitivity of the εm-relation on (m-idempotent) hyperrings

Morteza Norouzi; Irina Cristea

doi:10.1515/math-2018-0085

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Transitivity of the ε_m-relation on (m-idempotent) hyperrings

Morteza Norouzi and Irina Cristea

Published/Copyright: August 24, 2018

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$Open Mathematics$

From the journal Open Mathematics Volume 16 Issue 1

Abstract

On a general hyperring, there is a fundamental relation, denoted γ^*, such that the quotient set is a classical ring. In a previous paper, the authors defined the relation ε_m on general hyperrings, proving that its transitive closure εm∗ is a strongly regular equivalence relation smaller than the γ^*-relation on some classes of hyperrings, such that the associated quotient structure modulo εm∗ is an ordinary ring. Thus, on such hyperrings, εm∗ is a fundamental relation. In this paper, we discuss the transitivity conditions of the ε_m-relation on hyperrings and m-idempotent hyperrings.

Keywords: Hyperring; m-idempotent hyperring; m-complete part

MSC 2010: 20N20; 16Y99

1 Introduction

The quotient set has played an important role in the algebraic hyperstructures theory since its beginning for at least two reasons. The first one concerns the motivation of the definition of hypergroup, very well pointed out by F. Marty in his pioneering paper on hypergroups from 1934. It is well known that the quotient of a group G by an arbitrary subgroup H of G is a group if and only if H is a normal subgroup, while Marty showed that the quotient structure G/H is always a hypergroup. More generally, as Vougiouklis proved in [1], if one factorizes the group G by any partition S of G, then the quotient G/S is an H_v-group (i.e. a reproductive hypergroupoid satisfying the weak associativity). Secondly, the quotient set represents the bridging element between the classical algebraic structures and the corresponding hyperstructures, as mentioned in [2]. The first step in this direction was made by Koskas [3], when he used the β-relation and its transitive closure β^* to obtain a group (as a quotient structure of a hypergroup modulo β^*). Later on, the study of this correspondence between classical structures and hyperstructures with similar behaviour has been extended and new equivalence relations have been defined and called fundamental relations. They are the smallest strongly regular relations defined on a hyperstructure such that the quotient set is a classical structure, having similar properties. If on a (semi)hypergroup one considers the β^*-relation, then the quotient set is a (semi)group. Besides, the quotient set modulo the γ^*-relation, introduced by Freni [4], is a commutative (semi)group. Similarly, other fundamental relations have been defined on hypergroups in order to obtain nilpotent groups [5], engel groups [6], or solvable groups [7]. The same approach was used also for ring-like hyperstructures. It started in 1991, when Vougiouklis [1] defined the γ-relation on a general hyperring R (addition and multiplications are both hyperoperations) such that the quotient R/γ^* is a ring. Even if they are denoted in the same way (this could create confusion for the new readers of the algebraic hyperstructure theory, while it is already accepted for the researchers of this field), the fundamental relation defined by Freni [4] on semihypergroups is different by the fundamental relation γ defined by Vougiouklis [1] on hyperrings. Later on, the β^*-relation [8] has been introduced to obtain a commutative ring. More recently, other fundamental relations have been defined obtaining Boolean rings [9] or commutative rings with identity [10] as associated quotient structures. We end this brief recall of the fundamental relations with those in hypermodule theory, where, for example, the θ-fundamental relation [11] leads to commutative modules by the same method of factorization.

The authors of this note proposed in [12] a new perspective of the study of fundamental relations on hyperstructures. The γ^*-relation defined on a general hyperring R is the smallest strongly regular relation such that the quotient R/γ^* is a ring. The paper [12] deals with the question: Under which conditions can a fundamental relation smaller than γ be defined on a general hyperring, such that its transitive closer behaves similar to γ^*? To answer to this question, the ε_m-relation was defined on a special class of (semi)hyperrings, such that ε_m ⊊ γ and the quotient structure modulo εm∗ is an ordinary (semi)ring. Moreover, on m-idempotent hyperrings it was proved that εm∗ = γ^*.

In this paper, we study the transitivity property of the ε_m-relation on general hyperrings. First, we introduce the notion of m-complete parts based on the ε_m-relation and investigate their properties, which help us to show that ε_m is transitive on m-idempotent hyperfields.

2 Regular and fundamental relations on hyperstructures

In this section we review some basic definitions and properties regarding fundamental relations on general hyperrings. For further details, the readers are referred to [2], [10, 13, 14, 15, 17].

Definition 2.1

([16]). An algebraic system (R, +, ⋅) is said to be ageneral hyperring(by short a hyperring), if (R, +) is a hypergroup, (R, ⋅) is a semihypergroup, and ”⋅” is distributive with respect to ”+”.

In the above definition, if (R, +) is a semihypergroup, then (R, +, ⋅) is called a semihyperring. A nonempty subset I of a hyperring (R, +, ⋅) is a hyperideal, if (I, +) is a subhypergroup of (R, +) and, for all x ∈ I and r ∈ R, we have r ⋅ x ∪ x ⋅ r ⊆ I.

We recall that a subhypergroup A of (R, ⋅) is said to be invertible on the left (on the right), if x ∈ A ⋅ y (x ∈ y ⋅ A), then y ∈ A ⋅ x(y ∈ x ⋅ A), for all x, y ∈ R. A subhypergroup is invertible, if it is invertible on the left and on the right. Moreover, a subhypergroup B of (R, ⋅) is called closed on the left (on the right), if x ∈ a ⋅ y (x ∈ y ⋅ a) implies that a ∈ B, for every a ∈ R and x, y ∈ B. We say B is closed, if it is closed on the left and on the right. It is easy to see that every invertible subhypergroup of (R, ⋅) is closed.

Let ρ be an equivalence relation on a hypergroup (H, ∘). For A, B ⊆ H, AρB means that, for all x ∈ A there exists y ∈ B such that xρy, and for all v ∈ B there exists u ∈ A such that uρv. Moreover, Aρ̿B means that for all x ∈ A and for all y ∈ B, we have xρy. Accordingly, an equivalence relation ρ on a hypergroup (H, ∘) is called regular if aρb and cρd imply (a ∘ c) ρ (b ∘ d), for a, b, c, d ∈ R. Besides, ρ is called strongly regular if, under the same conditions, we have (a ∘ c) ρ̿ (b ∘ d), for a, b, c, d ∈ R.

The main role of the (strongly) regular relations on hypergroups is reflected by the following result.

Theorem 2.2

([17]). Consider the equivalence relationρon the hypergroup (H, ∘) and the hyperoperationρ(x) ⊗ ρ(y) = {ρ(z) | z ∈ ρ(x) ∘ ρ(y)} on the quotientH/ρ = {ρ(x) | x ∈ H}. Thenρis regular (strongly regular) onHif and only if (H/ρ, ⊗) is a hypergroup (group).

An equivalence relation ρ is (strongly) regular on a hyperring (R, +, ⋅), if it is (strongly) regular with respect to both hyperoperations ”+” and ”⋅”. One example of strongly regular relation on (semi)hyperrings is the γ-relation defined by Vougiouklis in [1] as follows. Let (R, +, ⋅) be a (semi)hyperring and x, y ∈ R. Then xγy if and only if {x, y} ⊆ u, where u is a finite sum of finite products of elements of R. In other words, xγy if and only if {x, y} ⊆ ∑j∈J(∏i∈Ijzi), for some finite sets of indices J and I_j and elements z_i ∈ R. Let γ^* be the transitive closure of γ, that is xγ^*y if and only if there exist the elements z₁, …, z_n+1 ∈ R, with z₁ = x and z_n+1 = y, such that z_iγz_i+1, for i ∈ {1, …, n}. In [1] it was shown that γ^* is the smallest strongly regular relation on a hyperring R such that the quotient (R/γ^*, ⊕, ⊙) is a classical ring with the operations defined as: γ^*(x) ⊕ γ^*(y) = γ^*(z), for all z ∈ γ^*(x) + γ^*(y) and γ^*(x) ⊙ γ^*(y) = γ^*(t), for all t ∈ γ^*(x)⋅γ^*(y). Hence, (R/γ^*, ⊕, ⊙) is called the fundamental ring obtained by the factorization with the γ^*-relation.

3 The ε_m-relation on hyperrings

In [12] the authors defined on (semi)hyperrings a new relation, denoted by ε_m, smaller than the γ-relation, and which is not transitive in general. Thus they found some conditions for the transitivity of the ε_m-relation on hyperrings. In this section we recall its definition and main properties.

Let (R, +, ⋅) be a semihyperring and select a constant m, such that 2 ≤ m ∈ ℕ. Put {(x, x) | x ∈ R} ⊆ ε_m and for all a, b ∈ R define

aεmb⟺∃n∈N,∃(z1,…,zn)∈Rn:{a,b}⊆∑i=1nzim,(1)

where zim=zi⋅zi⋅…⋅zi⏟mtimes.

Now, let (R, +, ⋅) be a hyperring such that (R, ⋅) is commutative and the following implication holds:

B⊆∑i=1nAim⟹∃xi∈Ai(1≤i≤n):B⊆∑i=1nxim,(2)

for all B, A₁, …, A_n ⊆ R. Accordingly with Theorems 3.3 and 3.4 in [12], on a hyperring R satisfying condition (2), the relation εm∗ is the smallest strongly regular equivalence relation such that the quotient set R/ εm∗ is a ring, thus it is a fundamental relation on R. Besides, we note that relation (2) is valid if and only if, for all A₁, …, A_n ⊆ R, there exists x_i ∈ A_i(1 ≤ i ≤ n) such that ∑i=1nAim⊆∑i=1nxim.

The next result provides sufficient conditions for the transitivity of the relation ε_m.

Theorem 3.1

([12]). Let (R, +, ⋅) be a hyperring satisfying the relation(2)such that there exists 0 ∈ Rsuch that x + 0 = {x} andx ⋅ 0 = {0} for allx ∈ R. IfA₁, …, A_nare hyperideals ofR, thenX = ⋃{∑i=1nAim∣A1,…,An⊆R}is an equivalence class ofεm∗andε_mis transitive.

Example 3.2

Define onR = {0, a, b} two hyperoperations as follows:

+0ab0{0}{a}{b}a{a}{a,b}Rb{b}R{a,b}⋅0ab0{0}{0}{0}a{0}RRb{0}RR

Then, (R, +, ⋅) is a hyperring [18].

It is easy to check that, for allA₁, …, A_n ⊆ R, there existx_i ∈ A_i (1 ≤ i ≤ n), such that∑i=1nAim⊆∑i=1nxim,relation equivalently with(2).

We end this section emphasizing the fact that if the hyperring (R, +, ⋅) does not satisfy condition (2), then the relation ε_m is not transitive, while its transitive closure εm∗ is not strongly regular on (R, ⋅) [12].

4 Transitivity of the relation ε_m on m-idempotent hyperfields

Since the conditions in Theorem 3.1 are not immediate, we aim to find some particular hyperrings, where the relation ε_m is transitive. For doing this, we will first define the concept of m-complete part and then we will prove that ε_m is transitive on m-idempotent hyperfields.

The main role of the complete parts of a semihypergroup, introduced by Koskas [3] and very well recalled by Antampoufis et al. in the survey [2], is played in finding the β^* class of each element. In particular, a nonempty subset A of a semihypergroup (H, ⋅) is called a complete part of H if, for any nonzero natural number n and any elements a₁, …, a_n of H, the following implication holds:

A∩∏i=1nai≠∅⇒∏i=1nai⊆A.

In other words, the complete part A absorbs every hyperproduct containing at least one element of A. In particular, for any element x ∈ A, the class β^*(x) is a complete part of H. Moreover, the intersection of all complete parts of H containing A is called the complete closure of A in H, denoted by C(A). Besides, β^*(x) = C(x), for any x ∈ H.

As already mentioned before, Vougiouklis [16] defined the relation γ on a hyperring R, proving that its transitive closure γ^* is the smallest strongly regular relation defined on R such that the quotient R/γ^* is a ring. Later on Mirvakili et al. [19] studied the transitivity property of this relation, introducing the notion of complete part on hyperrings as follows: a nonempty subset M of a hyperring R is a complete part if, for any natural number n, any i = 1, 2, …, n, any natural number k_i and arbitrary elements z_i1, …, z_{ik_i} ∈ R, we have

M∩∑i=1n(∏j=1kizij)≠∅⇒∑i=1n(∏j=1kizij)⊆M.

Now we will extend these definitions to the case of hyperrings, aiming to prove that the class ε^*(x) of an element x in the hyperring R is an m-complete part of R.

Definition 4.1

We say that a nonempty subsetAof a (semi)hyperring (R, +, ⋅) is anm-complete part ofRifA∩∑i=1nzim≠∅implies that∑i=1nzim⊆A, for alln ∈ ℕ andz₁, …, z_n ∈ R. The intersection of allm-complete parts ofRcontaining a nonempty subsetAofRis called them-complete closure ofAand it is denoted by 𝓒_m(A).

Example 4.2

Consider the following hyperoperations on the setR = {a, b, c, d}:

+abcda{b,c}{b,d}{b,d}{b,d}b{b,d}{b,d}{b,d}{b,d}c{b,d}{b,d}{b,d}{b,d}d{b,d}{b,d}{b,d}{b,d}⋅abcda{b,d}{b,d}{b,d}{b,d}b{b,d}{b,d}{b,d}{b,d}c{b,d}{b,d}{b,d}{b,d}d{b,d}{b,d}{b,d}{b,d}

Then (R, +, ⋅) is a semihyperring. For every m ≥ 2 and for allz₁, …, z_n ∈ R, we have∑i=1nzim={b,d}.It follows that the subsetsA₁ = {b, d}, A₂ = {a, b, d} andA₃ = {c, b, d} are all properm-complete parts ofR, i.e. m-complete parts ofR, different byR.

Theorem 4.3

Letρbe a strongly regular equivalence relation onR. Thenρ(a) is anm-complete part ofR, for alla ∈ R.

Proof

Since ρ is a strongly regular relation on R, it follows that the quotient R/ρ is a ring (with the addition “⊕” and the multiplication “⊙”). Let a ∈ R and ρ(a)∩∑i=1nzim≠∅, for arbitrary elements z₁, …, z_n ∈ R. Hence, there exists y∈∑i=1nzim such that ρ(y) = ρ(a). Consider the strong homomorphism π : R ⟶ R/ρ defined by π(x) = ρ(x), for all x ∈ R, where R/ρ is a ring (a trivial hyperring). Thus,

π(∑i=1nzim)=⨁i=1n(ρ(zi)⊙…⊙ρ(zi)⏟mtimes)=ρ(y)=ρ(a),

which implies that ∑i=1nzim⊆ρ(a). This completes the proof. □

For a nonempty subset A of a (semi)hyperring R, denote

K1m(A)=AKn+1m(A)={x∈R∣∃(z1,…,zn)∈Rn;x∈∑i=1nzimandKnm(A)∩∑i=1nzim≠∅} Km(A)=⋃n≥1Knm(A).

Moreover, for any x ∈ R and any natural number n, for simplicity we denote Knm({x})=Knm(x).

Lemma 4.4

For any nonempty subsetAof a hyperringR, the setK^m(A) is anm-complete part ofR.

Proof

Let K^m(A) ∩ ∑i=1nzim ≠ ∅, for arbitrary elements z₁, … z_n ∈ R. Then, there exists t ∈ ℕ such that Ktm(A) ∩ ∑i=1nzim ≠ ∅, which implies that ∑i=1nzim⊆Kt+1m(A)⊆Km(A). Thus, K^m(A) is an m-complete part of R, containing A.□

Theorem 4.5

K^m(A) = 𝓒_m(A), for any nonempty subsetAofR.

Proof

By Lemma 4.4, we have 𝓒_m(A) ⊆ K^m(A). Now, let M be an m-complete part of R containing A. Clearly, K1m(A) = A ⊆ M. Suppose that Knm(A) ⊆ M. Let x ∈ Kn+1m(A). Then there exist the elements z₁, …, z_n ∈ R such that x∈∑i=1nzim and ∅≠Knm(A)∩∑i=1nzim⊆M∩∑i=1nzim. Since M is an m-complete part, it follows that ∑i=1nzim ⊆ M and so Kn+1m(A) ⊆ M. Hence, K^m(A) ⊆ M, and thus K^m(A) ⊆ 𝓒_m(A).□

Example 4.6

Consider the semihyperringRin Example 4.2. One obtains that them-complete closure ofA₂isA₂itself, for any natural number m ≥ 2. Moreover, if we consider theγ-relation onR, then we have ∑∏z_i = {b, c} := Por ∑∏z_i = {b, d} := Q, for any finite hypersums of finite hyperproducts of elementsz_i ∈ R. Since P ∩ Q ≠ ∅ and P ⊈ Qand Q ⊈ P, thenPandQare not complete parts ofR. Besides, A₂ ∩ P ≠ ∅, but P ⊈ A₂. This means thatA₂is not a complete part, but only anm-complete part.

Theorem 4.7

For all nonempty subsetsAofR, it holds 𝓒_m(A) = ⋃a∈A𝓒_m(a).

Proof

Clearly we have the inclusion 𝓒_m(a) ⊆ 𝓒_m(A), for all a ∈ A. Hence, ⋃a∈A𝓒_m(a) ⊆ 𝓒_m(A).

Conversely, we show that Knm(A)⊆⋃a∈AKnm(a), by induction on ”n”. For n = 1, we have K1m(A) = A = ⋃a∈A{a}=⋃a∈AK1m(a).. Now, suppose that Knm(A)⊆⋃a∈AKnm(a) and take an arbitrary x ∈ Kn+1m(A). Then, x ∈ ∑i=1nzim and ∅≠Knm(A)∩∑i=1nzim, for some elements z₁, …, z_n ∈ R. Thus, there exists y ∈ R such that x, y ∈ ∑i=1nzim and y∈Knm(A)⊆⋃a∈AKnm(a). This implies that there exists a′ ∈ A such that y∈Knm(a′)∩∑i=1nzim≠∅, meaning that x ∈ Kn+1m(a′), and thus Kn+1m(A)⊆⋃a∈AKn+1m(a).

If x ∈ 𝓒_m(A), it follows that x ∈ K^m(A) = ⋃n≥1Knm(A)⊆⋃n≥1(⋃a∈AKnm(a)). Then, for some a′ ∈ A and n ≥ 1, we have x ∈ Knm(a′) ⊆ K^m(a′) = 𝓒_m(a′) ⊆ ⋃_a∈A 𝓒_m(a). This completes the proof. □

In the following we will give an equivalent description of the relation εm∗ on hyperrings, using the notion of m-complete part. First we will prove some properties of the m-complete parts.

Lemma 4.8

Knm(K2m(x))=Kn+1m(x),for allx ∈ Randn ≥ 2.

Proof

We prove it by induction on ”n”. For n = 2, we have

K2m(K2m(x))={x∈R∣∃(z1,…,zn)∈Rn;x∈∑i=1nzimandK1m(K2m(x))∩∑i=1nzim≠∅} ={x∈R∣∃(z1,…,zn)∈Rn;x∈∑i=1nzimandK2m(x)∩∑i=1nzim≠∅} =K3m(x).

Now, suppose that Kn−1m(K2m(x))=Knm(x) and take an arbitrary element y∈Knm(K2m(x)). Then there exist z₁, …, z_n ∈ R such that y∈∑i=1nzim and ∅≠Kn−1m(K2m(x))∩∑i=1nzim. Thus, ∅≠Knm(x)∩∑i=1nzim, and so y∈Kn+1m(x). Hence, Knm(K2m(x))⊆Kn+1m(x). Similarly, we have Kn+1m(x)⊆Knm(K2m(x)), that completes the proof. □

Lemma 4.9

For alln ≥ 2 andx, y ∈ R, x ∈ Knm(y) if and only ify ∈ Knm(x).

Proof

We prove the result by induction on ”n”. Let n = 2. Then x ∈ K2m(y) if and only if, for z₁, …, z_n ∈ R, we have x ∈ ∑i=1nzim and ∅ ≠ {y} ∩ ∑i=1nzim, meaning that, for z₁, …, z_n ∈ R, y ∈ ∑i=1nzim and ∅ ≠ {x} ∩ ∑i=1nzim. This is equivalent with y ∈ K2m(x). Suppose now that x ∈ Kn−1m(y) if and only if y ∈ Kn−1m(x) and take x ∈ Knm(y); then there exist z, z₁, …, z_n ∈ R such that x∈∑i=1nzim,z∈Kn−1m(y) and z∈∑i=1nzim. Hence, by induction procedure, y∈Kn−1m(z) and z∈K2m(x), which implies that y∈Kn−1m(K2m(x))=Knm(x) by Lemma 4.8. Similarly, y ∈ Knm(x) implies that x ∈ Knm(y).□

Define on a hyperring R the relation θ as follows: xθy if and only if x ∈ K^m(y), for all x, y ∈ R.

Corollary 4.10

The relationθis an equivalence onR.

Proof

For all x ∈ R, we have x ∈ K1m(x) ⊆ K^m(x). Hence, θ is reflexive. Now, let x ∈ K^m(y), for x, y ∈ R. By Theorem 4.5, there exists n ≥ 1 such that x ∈ Knm(y), which implies that y ∈ Knm(x), by Lemma 4.9. Then, y ∈ Knm(x) ⊆ K^m(x). Similarly, the converse is valid. Thus, θ is symmetric. Moreover, let xθy and yθz for x, y, z ∈ R. Hence, x ∈ 𝓒_m(y) and y ∈ 𝓒_m(z). Let A be an m-complete part of R containing z. Since y ∈ 𝓒_m(z) and 𝓒_m(z) ⊆ A, it follows that y ∈ A. Hence, 𝓒_m(y) ⊆ A and thus x ∈ A. Therefore, x ∈ ⋂z∈AA = 𝓒_m(z) = K^m(z) and so xθz. □

Theorem 4.11

For allx ∈ R, εm∗(x) = θ(x).

Proof

If xε_my, then x ∈ ∑i=1nzim and ∅ ≠ {y} ∩ ∑i=1nzim, for some elements z₁, …, z_n ∈ R. Hence, x ∈ K2m(y) ⊆ K^m(y). Thus, ε_m ⊆ θ and εm∗ ⊆ θ^* = θ.

Conversely, let xθy, that is, x ∈ K^m(y), which implies that x ∈ Kn+1m(y), for n ∈ ℕ. Then, there exist x₁, z₁, …, z_n ∈ R such that x∈∑i=1nzim,x1∈Knm(y) and x1∈∑i=1nzim. Hence, {x,x1}⊆∑i=1nzim and so xε_mx₁. Similarly, since x₁ ∈ Knm(y), there exists x₂ ∈ Kn−1m(y) such that x₁ε_mx₂. By continuing this process, we can obtain x3∈Kn−2m(y),…,xn−1∈K2m(y) and xn∈K1m(y)={y} such that xε_mx₂ε_m x₃…ε_mx_n–1ε_m x_n = y. Hence, xεm∗y and so θ⊆εm∗. Therefore the proof is completed.□

Now, we recall that a hyperring (R, +, ⋅) is said to be a hyperfield, if (R, ⋅) is a hypergroup. Moreover, a strong homomorphism from a hyperring (R, +, ⋅) to a hyperring (S, ⊕, ⊙) is a map f : R ⟶ S such that f(x + y) = f(x) ⊕ f(y) and f(x ⋅ y) = f(x) ⊙ f(y), for all x, y ∈ R. Considering the ε_m relation on R, it can be seen that the map φ_m : R ⟶ R/ εm∗ is a strong homomorphism.

In the following we will consider R a hyperfield satisfying relation (2) (this is a crucial assumption in the proofs of the next results) such that R/ εm∗ has a unit element denoted by 1R/εm∗. Set ωRm=φm−1(1R/εm∗)={x∈R∣φm(x)=1R/εm∗}. We will state some properties of the m-complete parts of hyperfields satisfying relation (2).

Theorem 4.12

If (R, +, ⋅) is a hyperfield andA ⊆ R, thenφm−1(φ_m(A)) = A ⋅ ωRm.

Proof

Let x ∈ φm−1(φ_m(A)). Then there exists y ∈ A such that φ_m(x) = φ_m(y). Since (R, ⋅) is a hypergroup, it follows that there exists t ∈ R such that x ∈ y ⋅ t, which implies that φ_m(x) = φ_m(y) ⊙ φ_m(t). Since (R/ εm∗, ⊙) is a group and φ_m(x) = φ_m(y), it results φ_m(t) = 1R/εm∗. Thus, t∈φm−1(1R/εm∗)=ωRm, and so x ∈ y ⋅ t ⊆ A ⋅ ωRm. Then φm−1(φ_m(A)) ⊆ A ⋅ ωRm.

Conversely, let x ∈ A ⋅ ωRm. Then x ∈ a ⋅ y, for some a ∈ A and y ∈ ωRm, that is φ_m(y) = 1R/εm∗. Hence φ_m(x) = φ_m(a ⋅ y) = φ_m(a) ⊙ φ_m(y) = φ_m(a) ⊙ 1R/εm∗ = φ_m(a) ∈ φ_m(A), and so x ∈ φm−1(φ_m(A)). This completes the proof. □

Theorem 4.13

If (R, +, ⋅) is a hyperfield andA ⊆ R, then 𝓒_m(A) = A ⋅ ωRm.

Proof

It is easy to see that φm−1(φ_m(A)) = {x ∈ R | ∃a ∈ A : φ_m(x) = φ_m(a)}. Also, we have φ_m(x) = φ_m(a) if and only if εm∗(x) = εm∗(a), equivalently with θ(x) = θ(a), meaning that x ∈ K^m(a) = 𝓒_m(a), by Theorem 4.11. Hence, x ∈ 𝓒_m(A) if and only if x ∈ φm−1(φ_m(A)), thus x ∈ A ⋅ ωRm, by Theorem 4.7 and Theorem 4.12. Therefore, 𝓒_m(A) = A ⋅ ωRm.□

Corollary 4.14

LetRbe a hyperfield. Ais anm-complete part ofRif and only ifA = A ⋅ ωRm.

Proof

If A is an m-complete part, then 𝓒_m(A) = A. Hence, A = 𝓒_m(A) = A ⋅ ωRm, by Theorem 4.13. Conversely, if A = A ⋅ ωRm, then A = 𝓒_m(A) by Theorem 4.13, and so A is an m-complete part of R.□

By Theorem 4.12, we have ωRm⋅ωRm=φm−1(φm(ωRm))=ωRm. Hence, ωRm is an m-complete part of the hyperfield (R, +, ⋅), by Corollary 4.14.

Moreover, notice that, for two subsets A and B of the hyperfield R such that one of them is an m-complete part of R (assume that A is so), we have (A ⋅ B) ⋅ ωRm = (A ⋅ ωRm) ⋅ B = A ⋅ B, by Corollary 4.14. Hence, A ⋅ B is an m-complete part of R, by Corollary 4.14.

Theorem 4.15

Let (R, +, ⋅) be a hyperfield. Then everym-complete part subhypergroup of (R, ⋅) is invertible. Moreover, it is closed.

Proof

Let A be an m-complete part of R such that (A, ⋅) is a subhypergroup of (R, ⋅). Take x ∈ A ⋅ y for x, y ∈ R. Thus, x ∈ a ⋅ y, for a ∈ A, which implies that φ_m(x) = φ_m(a) ⊙ φ_m(y). Since φ_m(A) is a subgroup of R/ εm∗, we have φ_m(y) = φ_m(a)^–1 ⊙ φ_m(x) ∈ φ_m(A)⊙ φ_m(x) = φ_m(A ⋅ x). Besides, A ⋅ x is an m-complete part of R, hence y ∈ φm−1(φ_m(A ⋅ x)) = 𝓒_m(A ⋅ x) = A ⋅ x. Then, A is invertible on the left. Similarly, we can show that A is invertible on the right. Therefore, A is invertible, and by consequence it is also closed.□

Theorem 4.16

Let (R, +, ⋅) be a hyperfield and 𝓢_{C_m}(R) be the set of allm-complete parts ofRwhich are subhypergroups of (R, ⋅). Then, ωRm=⋂A∈SCm(R)A.

Proof

We know that ωRm is an m-complete part of R. Let x ∈ ωRm. For all t, y ∈ ωRm, we have φ_m(x) = 1R/εm∗ = φ_m(t) ⊙ φ_m(y) = φ_m(t ⋅ y). Hence, x ∈ φm−1(φ_m(t ⋅ ω_m)) = 𝓒_m(t ⋅ ωRm) = t ⋅ ωRm, and so ωRm⊆t⋅ωRm for all t∈ωRm. Clearly, t⋅ωRm⊆ωRm. Then ωRm=t⋅ωRm, for all t∈ωRm, which implies that ωRm is a subhypergroup of (R, ⋅). Thus, ⋂A∈SCm(R)A⊆ωRm. Now, let A ∈ 𝓢_{C_m}(R). By Corollary 4.14, A = A ⋅ ωRm. Hence, for every x ∈ ωRm, there exist a, b ∈ A such that a ∈ b ⋅ x, and so a ∈ A ⋅ x. By Theorem 4.15, A is invertible, and therefore x ∈ A ⋅ a ⊆ A. Then ωRm ⊆ A, and thus ωRm⊆⋂A∈SCm(R)A. Hence, the proof is complete.□

We recall that a hyperring (R, +, ⋅) is said to be m-idempotent ([12]) if there exists a constant m, 2 ≤ m ∈ ℕ, such that x ∈ x^m, for all x ∈ R.

Example 4.17

[12] Consider the Krasner hyperringR = {0, a, b} with the hyperaddition and the multiplication defined as follows [20]:

+0ab0{0}{a}{b}a{a}{a,b}Rb{b}R{a,b}⋅0ab0000a0bab0ab

For every odd numberm ∈ ℕ, we have 0^m = 0, a^m = aandb^m = b. Hence, Rism-idempotent, for all odd natural numbersm.
Besides, sincea² = a ⋅ a = b, it follows thatRis not an 2-idempotent hyperring. Similarly, one proves that, for all even numbersm ∈ ℕ, the hyperringRis notm-idempotent.

Example 4.18

The hyperring defined in Example 3.2 is anm-idempotent hyperring (satisfying relation(2)), for allm, 2 ≤ m ∈ ℕ [12].

Example 4.19

Define on the setR = {0, 1} two hyperoperations as follows:

⊞010{0}R1R{1}⊡010{0}{0}1{0}R

Then, (R, ⊞, ⊡) is anm-idempotent hyperring satisfying relation(2), for allm, 2 ≤ m ∈ ℕ.

Example 4.20

Similarly, take the same support setR = {0, 1} and define onRthe two hyperoperations as follows:

⊕010{0}R1RR⊙010{0}R1{0}R

The hyperring (R, ⊕, ⊙) ism-idempotent, for allm, 2 ≤ m ∈ ℕ, and satisfies relation(2).

Now we give an example of m-idempotent hyperfield.

Example 4.21

Define onR = {0, 1} two hyperoperations as follows:

+010{0}R1R{1}⋅010{0}R1R{1}

Then, (R, +, ⋅) is anm-idempotent hyperfield satisfying relation(2).

Now, for all a ∈ R, put X(a)=⋃{∑i=1nAim∣a∈∑i=1nAim}, where A₁, …, A_n ⊆ R.

Theorem 4.22

LetRbe anm-idempotent hyperfield. Then 𝓧(a) is anm-complete part ofR, for alla ∈ R.

Proof

Suppose that a ∈ R and ∑i=1nzim ∩ 𝓧(a) ≠ ∅, for some z₁, …, z_n ∈ R. Then there exists z ∈ R such that z ∈ ∑i=1nzim and z ∈ A, where A = ∑i=1nAim, for A₁, …, A_n ⊆ R. Since (R, ⋅) is a hypergroup, there exist w, b ∈ R such that z_n ∈ w ⋅ a and a ∈ z ⋅ b. Also, for all 1 ≤ i ≤ n – 1 we have z_i ∈ z_i ⋅ R. Hence,

∑i=1nzim⊆∑i=1n−1zim+(w⋅a)m⊆(z1⋅R)m+…+(zn−1⋅R)m+(w⋅A⋅b)m,

since z ∈ A. Moreover, R is m-idempotent and we have b ∈ b^m, thus

a∈z⋅b⊆(∑i=1nzim)⋅b⊆z1m⋅bm+…+zn−1m⋅bm+(w⋅a⋅b)m ⊆(z1⋅R)m+…+(zn−1⋅R)m+(w⋅A⋅b)m.

Therefore, (z₁ ⋅ R)^m + … + (z_n–1 ⋅ R)^m + (w ⋅ A ⋅ b)^m ⊆ 𝓧(a) and so ∑i=1nzim ⊆ 𝓧(a). Then, 𝓧(a) is an m-complete part of R.□

Theorem 4.23

LetRbe anm-idempotent hyperfield. Then 𝓧(a) = ωRm, for everya ∈ ωRm.

Proof

It is not difficult to see that a⋅ωRm=ωRm, for all a ∈ ωRm. Hence, ωRm = a ⋅ ωRm = 𝓒_m(a) ⊆ 𝓧(a), by Theorem 4.13 and Theorem 4.22. Now, let a ∈ ωRm and x ∈ 𝓧(a). Then there exists A = ∑i=1nAim ⊆ 𝓧(a) such that x ∈ A. Since a ∈ A, then {x, a} ⊆ A. Since R satisfies relation (2), there exist x_i ∈ A_i, for 1 ≤ i ≤ n, such that {x, a} ⊆ ∑i=1nxim and so εm∗(x) = εm∗(a). Then φ_m(x) = φ_m(a). Hence, x ∈ φm−1(φ_m(a)) = a ⋅ ωRm = ωRm. Therefore, the proof is complete.□

Theorem 4.24

The relationε_mis transitive onm-idempotent hyperfields.

Proof

Let R be an m-idempotent hyperfield and xεm∗y, for x, y ∈ R. Hence, x ∈ φm−1(φ_m(y)) = y ⋅ ωRm. Similarly, we have y ∈ x ⋅ ωRm which implies that {x, y} ⊆ x ⋅ ωRm. Then, there exist t, z ∈ ωRm such that x ∈ x ⋅ t and y ∈ x ⋅ z. By Theorem 4.23, z ∈ ωRm = 𝓧(t) and thus {t, z} ⊆ A = ∑i=1nAim for A ⊆ 𝓧(t). Since R is m-idempotent, {x, y} ⊆ x ⋅ A ⊆ ∑i=1n(x ⋅ A_i)^m. So, there exist z_i ∈ x ⋅ A_i for every 1 ≤ i ≤ n such that {x, y} ⊆ ∑i=1nxim. Therefore, xε_my and so εm∗ = ε_m.□

5 Conclusions

The fundamental relation γ^* defined by Vougiouklis [16] on a general hyperring R is the smallest equivalence relation on R such that the quotient structure R/γ^* is a ring. If we consider a special type of hyperrings, i.e. those satisfying relation (2), we can define another fundamental relation on R, εm∗-relation [12], smaller than γ^*, while on m-idempotent hyperrings satisfying relation (2), we have εm∗ = γ^* [12]. In general, ε_m is not transitive. This paper provides a detailed study on the transitivity property of ε_m. Using m-complete parts of a hyperring, we have proved that ε_m is transitive on m-idempotent hyperfields satisfying relation (2).

m.norouzi@ub.ac.ir

irina.cristea@ung.si

Acknowledgement

The first author was partially supported by a grant from University of Bojnord. The second author acknowledges the financial support from the Slovenian Research Agency (research core funding No. P1 - 0285).

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Received: 2018-04-17

Accepted: 2018-07-04

Published Online: 2018-08-24

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License.

Articles in the same Issue

https://doi.org/10.1515/math-2018-0085

Keywords for this article

Hyperring; m-idempotent hyperring; m-complete part

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Transitivity of the εm-relation on (m-idempotent) hyperrings

Article

Abstract

1 Introduction

2 Regular and fundamental relations on hyperstructures

Definition 2.1

Theorem 2.2

3 The εm-relation on hyperrings

Theorem 3.1

Example 3.2

4 Transitivity of the relation εm on m-idempotent hyperfields

Definition 4.1

Example 4.2

Theorem 4.3

Proof

Lemma 4.4

Proof

Theorem 4.5

Proof

Example 4.6

Theorem 4.7

Proof

Lemma 4.8

Proof

Lemma 4.9

Proof

Corollary 4.10

Proof

Theorem 4.11

Proof

Theorem 4.12

Proof

Theorem 4.13

Proof

Corollary 4.14

Proof

Theorem 4.15

Proof

Theorem 4.16

Proof

Example 4.17

Example 4.18

Example 4.19

Example 4.20

Example 4.21

Theorem 4.22

Proof

Theorem 4.23

Proof

Theorem 4.24

Proof

5 Conclusions

Acknowledgement

References

Articles in the same Issue

Articles in the same Issue

Articles in the same Issue

Transitivity of the ε_m-relation on (m-idempotent) hyperrings

3 The ε_m-relation on hyperrings

4 Transitivity of the relation ε_m on m-idempotent hyperfields