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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 16, Issue 1

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

Morteza Norouzi
/ Irina Cristea
• Corresponding author
• Centre for Information Technologies and Applied Mathematics, University of Nova Gorica, Nova Gorica, Slovenia
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Published Online: 2018-08-24 | DOI: https://doi.org/10.1515/math-2018-0085

## 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 εm on general hyperrings, proving that its transitive closure $\begin{array}{}{\epsilon }_{m}^{\ast }\end{array}$ is a strongly regular equivalence relation smaller than the γ*-relation on some classes of hyperrings, such that the associated quotient structure modulo $\begin{array}{}{\epsilon }_{m}^{\ast }\end{array}$ is an ordinary ring. Thus, on such hyperrings, $\begin{array}{}{\epsilon }_{m}^{\ast }\end{array}$ is a fundamental relation. In this paper, we discuss the transitivity conditions of the εm-relation on hyperrings and m-idempotent hyperrings.

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 Hv-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 εm-relation was defined on a special class of (semi)hyperrings, such that εmγ and the quotient structure modulo $\begin{array}{}{\epsilon }_{m}^{\ast }\end{array}$ is an ordinary (semi)ring. Moreover, on m-idempotent hyperrings it was proved that $\begin{array}{}{\epsilon }_{m}^{\ast }\end{array}$ = γ*.

In this paper, we study the transitivity property of the εm-relation on general hyperrings. First, we introduce the notion of m-complete parts based on the εm-relation and investigate their properties, which help us to show that εm 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 a general 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 xI and rR, we have rxxrI.

We recall that a subhypergroup A of (R, ⋅) is said to be invertible on the left (on the right), if xAy (xyA), then yAx(yxA), for all x, yR. 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 xay (xya) implies that aB, for every aR and x, yB. 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, BH, AρB means that, for all xA there exists yB such that xρy, and for all vB there exists uA such that uρv. Moreover, Aρ̿B means that for all xA and for all yB, we have xρy. Accordingly, an equivalence relation ρ on a hypergroup (H, ∘) is called regular if aρb and cρd imply (ac) ρ (bd), for a, b, c, dR. Besides, ρ is called strongly regular if, under the same conditions, we have (ac) ρ̿ (bd), for a, b, c, dR.

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

#### Theorem 2.2

([17]). Consider the equivalence relation ρ on the hypergroup (H, ∘) and the hyperoperation ρ(x) ⊗ ρ(y) = {ρ(z) | zρ(x) ∘ ρ(y)} on the quotient H/ρ = {ρ(x) | xH}. Then ρ is regular (strongly regular) on H if 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, yR. 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} ⊆ $\begin{array}{}\sum _{j\in J}\left(\prod _{i\in {I}_{j}}{z}_{i}\right),\end{array}$ for some finite sets of indices J and Ij and elements ziR. Let γ* be the transitive closure of γ, that is *y if and only if there exist the elements z1, …, zn+1R, with z1 = x and zn+1 = y, such that ziγzi+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 zγ*(x) + γ*(y) and γ*(x) ⊙ γ*(y) = γ*(t), for all tγ*(x)⋅γ*(y). Hence, (R/γ*, ⊕, ⊙) is called the fundamental ring obtained by the factorization with the γ*-relation.

## 3 The εm-relation on hyperrings

In [12] the authors defined on (semi)hyperrings a new relation, denoted by εm, smaller than the γ-relation, and which is not transitive in general. Thus they found some conditions for the transitivity of the εm-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) | xR} ⊆ εm and for all a, bR define

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

where $\begin{array}{}{z}_{i}^{m}=\underset{m\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}times}{\underset{⏟}{{z}_{i}\cdot {z}_{i}\cdot \dots \cdot {z}_{i}}}.\end{array}$

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, …, AnR. Accordingly with Theorems 3.3 and 3.4 in [12], on a hyperring R satisfying condition (2), the relation $\begin{array}{}{\epsilon }_{m}^{\ast }\end{array}$ is the smallest strongly regular equivalence relation such that the quotient set R/ $\begin{array}{}{\epsilon }_{m}^{\ast }\end{array}$ 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, …, AnR, there exists xiAi(1 ≤ in) such that $\begin{array}{}\sum _{i=1}^{n}{A}_{i}^{m}\subseteq \sum _{i=1}^{n}{x}_{i}^{m}.\end{array}$

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

#### Theorem 3.1

([12]). Let (R, +, ⋅) be a hyperring satisfying the relation (2) such that there exists 0 ∈ R such that x + 0 = {x} and x ⋅ 0 = {0} for all xR. If A1, …, An are hyperideals of R, then X = $\begin{array}{}\bigcup \left\{\sum _{i=1}^{n}{A}_{i}^{m}\mid {A}_{1},\dots ,{A}_{n}\subseteq R\right\}\end{array}$ is an equivalence class of $\begin{array}{}{\epsilon }_{m}^{\ast }\end{array}$ and εm is transitive.

#### Example 3.2

Define on R = {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 all A1, …, AnR, there exist xiAi (1 ≤ in), such that $\begin{array}{}\sum _{i=1}^{n}{A}_{i}^{m}\subseteq \sum _{i=1}^{n}{x}_{i}^{m},\end{array}$ relation equivalently with (2).

We end this section emphasizing the fact that if the hyperring (R, +, ⋅) does not satisfy condition (2), then the relation εm is not transitive, while its transitive closure $\begin{array}{}{\epsilon }_{m}^{\ast }\end{array}$ is not strongly regular on (R, ⋅) [12].

## 4 Transitivity of the relation εm on m-idempotent hyperfields

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

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 ki and arbitrary elements zi1, …, zikiR, 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 subset A of a (semi)hyperring (R, +, ⋅) is an m-complete part of R if $\begin{array}{}A\cap \sum _{i=1}^{n}{z}_{i}^{m}\ne \mathrm{\varnothing }\end{array}$ implies that $\begin{array}{}\sum _{i=1}^{n}{z}_{i}^{m}\subseteq A\end{array}$, for all n ∈ ℕ and z1, …, znR. The intersection of all m-complete parts of R containing a nonempty subset A of R is called the m-complete closure of A and it is denoted by 𝓒m(A).

#### Example 4.2

Consider the following hyperoperations on the set R = {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 all z1, …, znR, we have $\begin{array}{}\sum _{i=1}^{n}{z}_{i}^{m}=\left\{b,d\right\}.\end{array}$ It follows that the subsets A1 = {b, d}, A2 = {a, b, d} and A3 = {c, b, d} are all proper m-complete parts of R, i.e. m-complete parts of R, different by R.

#### Theorem 4.3

Let ρ be a strongly regular equivalence relation on R. Then ρ(a) is an m-complete part of R, for all aR.

#### 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 aR and $\begin{array}{}\rho \left(a\right)\cap \sum _{i=1}^{n}{z}_{i}^{m}\ne \mathrm{\varnothing },\end{array}$ for arbitrary elements z1, …, znR. Hence, there exists $\begin{array}{}y\in \sum _{i=1}^{n}{z}_{i}^{m}\end{array}$ such that ρ(y) = ρ(a). Consider the strong homomorphism π : RR/ρ defined by π(x) = ρ(x), for all xR, 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 $\begin{array}{}\sum _{i=1}^{n}{z}_{i}^{m}\subseteq \rho \left(a\right).\end{array}$ 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 xR and any natural number n, for simplicity we denote $\begin{array}{}{K}_{n}^{m}\left(\left\{x\right\}\right)={K}_{n}^{m}\left(x\right).\end{array}$

#### Lemma 4.4

For any nonempty subset A of a hyperring R, the set Km(A) is an m-complete part of R.

#### Proof

Let Km(A) ∩ $\begin{array}{}\sum _{i=1}^{n}{z}_{i}^{m}\end{array}$ ≠ ∅, for arbitrary elements z1, … znR. Then, there exists t ∈ ℕ such that $\begin{array}{}{K}_{t}^{m}\end{array}$(A) ∩ $\begin{array}{}\sum _{i=1}^{n}{z}_{i}^{m}\end{array}$ ≠ ∅, which implies that $\begin{array}{}\sum _{i=1}^{n}{z}_{i}^{m}\subseteq {K}_{t+1}^{m}\left(A\right)\subseteq {K}^{m}\left(A\right).\end{array}$ Thus, Km(A) is an m-complete part of R, containing A.□

#### Theorem 4.5

Km(A) = 𝓒m(A), for any nonempty subset A of R.

#### Proof

By Lemma 4.4, we have 𝓒m(A) ⊆ Km(A). Now, let M be an m-complete part of R containing A. Clearly, $\begin{array}{}{K}_{1}^{m}\end{array}$(A) = AM. Suppose that $\begin{array}{}{K}_{n}^{m}\end{array}$(A) ⊆ M. Let x$\begin{array}{}{K}_{n+1}^{m}\end{array}$(A). Then there exist the elements z1, …, znR such that $\begin{array}{}x\in \sum _{i=1}^{n}{z}_{i}^{m}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\mathrm{\varnothing }\ne {K}_{n}^{m}\left(A\right)\cap \sum _{i=1}^{n}{z}_{i}^{m}\subseteq M\cap \sum _{i=1}^{n}{z}_{i}^{m}.\end{array}$ Since M is an m-complete part, it follows that $\begin{array}{}\sum _{i=1}^{n}{z}_{i}^{m}\end{array}$M and so $\begin{array}{}{K}_{n+1}^{m}\end{array}$(A) ⊆ M. Hence, Km(A) ⊆ M, and thus Km(A) ⊆ 𝓒m(A).□

#### Example 4.6

Consider the semihyperring R in Example 4.2. One obtains that the m-complete closure of A2 is A2 itself, for any natural number m ≥ 2. Moreover, if we consider the γ-relation on R, then we have ∑∏zi = {b, c} := P or ∑∏zi = {b, d} := Q, for any finite hypersums of finite hyperproducts of elements ziR. Since PQ ≠ ∅ and PQ and QP, then P and Q are not complete parts of R. Besides, A2P ≠ ∅, but PA2. This means that A2 is not a complete part, but only an m-complete part.

#### Theorem 4.7

For all nonempty subsets A of R, it holds 𝓒m(A) = $\begin{array}{}\bigcup _{a\in A}\end{array}$𝓒m(a).

#### Proof

Clearly we have the inclusion 𝓒m(a) ⊆ 𝓒m(A), for all aA. Hence, $\begin{array}{}\bigcup _{a\in A}\end{array}$𝓒m(a) ⊆ 𝓒m(A).

Conversely, we show that $\begin{array}{}{K}_{n}^{m}\left(A\right)\subseteq \bigcup _{a\in A}{K}_{n}^{m}\left(a\right),\end{array}$ by induction on ”n”. For n = 1, we have $\begin{array}{}{K}_{1}^{m}\end{array}$(A) = A = $\begin{array}{}\bigcup _{a\in A}\left\{a\right\}=\bigcup _{a\in A}{K}_{1}^{m}\left(a\right).\end{array}$. Now, suppose that $\begin{array}{}{K}_{n}^{m}\left(A\right)\subseteq \bigcup _{a\in A}{K}_{n}^{m}\left(a\right)\end{array}$ and take an arbitrary x$\begin{array}{}{K}_{n+1}^{m}\end{array}$(A). Then, x$\begin{array}{}\sum _{i=1}^{n}{z}_{i}^{m}\end{array}$ and $\begin{array}{}\mathrm{\varnothing }\ne {K}_{n}^{m}\left(A\right)\cap \sum _{i=1}^{n}{z}_{i}^{m},\end{array}$ for some elements z1, …, znR. Thus, there exists yR such that x, y$\begin{array}{}\sum _{i=1}^{n}{z}_{i}^{m}\end{array}$ and $\begin{array}{}y\in {K}_{n}^{m}\left(A\right)\subseteq \bigcup _{a\in A}{K}_{n}^{m}\left(a\right).\end{array}$ This implies that there exists a′ ∈ A such that $\begin{array}{}y\in {K}_{n}^{m}\left({a}^{\prime }\right)\cap \sum _{i=1}^{n}{z}_{i}^{m}\ne \mathrm{\varnothing },\end{array}$ meaning that x$\begin{array}{}{K}_{n+1}^{m}\end{array}$(a′), and thus $\begin{array}{}{K}_{n+1}^{m}\left(A\right)\subseteq \bigcup _{a\in A}{K}_{n+1}^{m}\left(a\right).\end{array}$

If x ∈ 𝓒m(A), it follows that xKm(A) = $\begin{array}{}\bigcup _{n\ge 1}{K}_{n}^{m}\left(A\right)\subseteq \bigcup _{n\ge 1}\left(\bigcup _{a\in A}{K}_{n}^{m}\left(a\right)\right).\end{array}$ Then, for some a′ ∈ A and n ≥ 1, we have x$\begin{array}{}{K}_{n}^{m}\end{array}$(a′) ⊆ Km(a′) = 𝓒m(a′) ⊆ ⋃aA 𝓒m(a). This completes the proof. □

In the following we will give an equivalent description of the relation $\begin{array}{}{\epsilon }_{m}^{\ast }\end{array}$ on hyperrings, using the notion of m-complete part. First we will prove some properties of the m-complete parts.

#### Lemma 4.8

$\begin{array}{}{K}_{n}^{m}\left({K}_{2}^{m}\left(x\right)\right)={K}_{n+1}^{m}\left(x\right),\end{array}$ for all xR and n ≥ 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 $\begin{array}{}{K}_{n-1}^{m}\left({K}_{2}^{m}\left(x\right)\right)={K}_{n}^{m}\left(x\right)\end{array}$ and take an arbitrary element $\begin{array}{}y\in {K}_{n}^{m}\left({K}_{2}^{m}\left(x\right)\right).\end{array}$ Then there exist z1, …, znR such that $\begin{array}{}y\in \sum _{i=1}^{n}{z}_{i}^{m}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\mathrm{\varnothing }\ne {K}_{n-1}^{m}\left({K}_{2}^{m}\left(x\right)\right)\cap \sum _{i=1}^{n}{z}_{i}^{m}.\text{\hspace{0.17em}}\text{Thus},\text{\hspace{0.17em}}\mathrm{\varnothing }\ne {K}_{n}^{m}\left(x\right)\cap \sum _{i=1}^{n}{z}_{i}^{m},\text{\hspace{0.17em}}\text{and so}\text{\hspace{0.17em}}y\in {K}_{n+1}^{m}\left(x\right).\end{array}$ Hence, $\begin{array}{}{K}_{n}^{m}\left({K}_{2}^{m}\left(x\right)\right)\subseteq {K}_{n+1}^{m}\left(x\right).\end{array}$ Similarly, we have $\begin{array}{}{K}_{n+1}^{m}\left(x\right)\subseteq {K}_{n}^{m}\left({K}_{2}^{m}\left(x\right)\right),\end{array}$ that completes the proof. □

#### Lemma 4.9

For all n ≥ 2 and x, yR, x$\begin{array}{}{K}_{n}^{m}\end{array}$(y) if and only if y$\begin{array}{}{K}_{n}^{m}\end{array}$(x).

#### Proof

We prove the result by induction on ”n”. Let n = 2. Then x$\begin{array}{}{K}_{2}^{m}\end{array}$(y) if and only if, for z1, …, znR, we have x$\begin{array}{}\sum _{i=1}^{n}{z}_{i}^{m}\end{array}$ and ∅ ≠ {y} ∩ $\begin{array}{}\sum _{i=1}^{n}{z}_{i}^{m}\end{array}$, meaning that, for z1, …, znR, y$\begin{array}{}\sum _{i=1}^{n}{z}_{i}^{m}\end{array}$ and ∅ ≠ {x} ∩ $\begin{array}{}\sum _{i=1}^{n}{z}_{i}^{m}\end{array}$. This is equivalent with y$\begin{array}{}{K}_{2}^{m}\end{array}$(x). Suppose now that x$\begin{array}{}{K}_{n-1}^{m}\end{array}$(y) if and only if y$\begin{array}{}{K}_{n-1}^{m}\end{array}$(x) and take x$\begin{array}{}{K}_{n}^{m}\end{array}$(y); then there exist z, z1, …, znR such that $\begin{array}{}x\in \sum _{i=1}^{n}{z}_{i}^{m},z\in {K}_{n-1}^{m}\left(y\right)\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}z\in \sum _{i=1}^{n}{z}_{i}^{m}.\end{array}$ Hence, by induction procedure, $\begin{array}{}y\in {K}_{n-1}^{m}\left(z\right)\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}z\in {K}_{2}^{m}\left(x\right),\end{array}$ which implies that $\begin{array}{}y\in {K}_{n-1}^{m}\left({K}_{2}^{m}\left(x\right)\right)={K}_{n}^{m}\left(x\right)\end{array}$ by Lemma 4.8. Similarly, y$\begin{array}{}{K}_{n}^{m}\end{array}$(x) implies that x$\begin{array}{}{K}_{n}^{m}\end{array}$(y).□

Define on a hyperring R the relation θ as follows: xθy if and only if xKm(y), for all x, yR.

#### Corollary 4.10

The relation θ is an equivalence on R.

#### Proof

For all xR, we have x$\begin{array}{}{K}_{1}^{m}\end{array}$(x) ⊆ Km(x). Hence, θ is reflexive. Now, let xKm(y), for x, yR. By Theorem 4.5, there exists n ≥ 1 such that x$\begin{array}{}{K}_{n}^{m}\end{array}$(y), which implies that y$\begin{array}{}{K}_{n}^{m}\end{array}$(x), by Lemma 4.9. Then, y$\begin{array}{}{K}_{n}^{m}\end{array}$(x) ⊆ Km(x). Similarly, the converse is valid. Thus, θ is symmetric. Moreover, let xθy and yθz for x, y, zR. Hence, x ∈ 𝓒m(y) and y ∈ 𝓒m(z). Let A be an m-complete part of R containing z. Since y ∈ 𝓒m(z) and 𝓒m(z) ⊆ A, it follows that yA. Hence, 𝓒m(y) ⊆ A and thus xA. Therefore, x$\begin{array}{}\bigcap _{z\in A}\end{array}$A = 𝓒m(z) = Km(z) and so xθz. □

#### Theorem 4.11

For all xR, $\begin{array}{}{\epsilon }_{m}^{\ast }\end{array}$(x) = θ(x).

#### Proof

If my, then x$\begin{array}{}\sum _{i=1}^{n}{z}_{i}^{m}\end{array}$ and ∅ ≠ {y} ∩ $\begin{array}{}\sum _{i=1}^{n}{z}_{i}^{m}\end{array}$, for some elements z1, …, znR. Hence, x$\begin{array}{}{K}_{2}^{m}\end{array}$(y) ⊆ Km(y). Thus, εmθ and $\begin{array}{}{\epsilon }_{m}^{\ast }\end{array}$θ* = θ.

Conversely, let xθy, that is, xKm(y), which implies that x$\begin{array}{}{K}_{n+1}^{m}\end{array}$(y), for n ∈ ℕ. Then, there exist x1, z1, …, znR such that $\begin{array}{}x\in \sum _{i=1}^{n}{z}_{i}^{m},{x}_{1}\in {K}_{n}^{m}\left(y\right)\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}{x}_{1}\in \sum _{i=1}^{n}{z}_{i}^{m}.\text{\hspace{0.17em}}\text{Hence},\text{\hspace{0.17em}}\left\{x,{x}_{1}\right\}\subseteq \sum _{i=1}^{n}{z}_{i}^{m}\end{array}$ and so m x1. Similarly, since x1$\begin{array}{}{K}_{n}^{m}\end{array}$(y), there exists x2$\begin{array}{}{K}_{n-1}^{m}\end{array}$(y) such that x1εm x2. By continuing this process, we can obtain $\begin{array}{}{x}_{3}\in {K}_{n-2}^{m}\left(y\right),\dots ,{x}_{n-1}\in {K}_{2}^{m}\left(y\right)\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}{x}_{n}\in {K}_{1}^{m}\left(y\right)=\left\{y\right\}\end{array}$ such that m x2εm x3εmxn–1εm xn = y. Hence, $\begin{array}{}x{\epsilon }_{m}^{\ast }y\text{\hspace{0.17em}}\text{and so}\text{\hspace{0.17em}}\theta \subseteq {\epsilon }_{m}^{\ast }.\end{array}$ 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 : RS such that f(x + y) = f(x) ⊕ f(y) and f(xy) = f(x) ⊙ f(y), for all x, yR. Considering the εm relation on R, it can be seen that the map φm : RR/ $\begin{array}{}{\epsilon }_{m}^{\ast }\end{array}$ 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/ $\begin{array}{}{\epsilon }_{m}^{\ast }\end{array}$ has a unit element denoted by $\begin{array}{}{1}_{R/{\epsilon }_{m}^{\ast }}.\text{\hspace{0.17em}}\text{Set}\text{\hspace{0.17em}}{\omega }_{R}^{m}={\phi }_{m}^{-1}\left({1}_{R/{\epsilon }_{m}^{\ast }}\right)=\left\{x\in R\mid {\phi }_{m}\left(x\right)={1}_{R/{\epsilon }_{m}^{\ast }}\right\}.\end{array}$ We will state some properties of the m-complete parts of hyperfields satisfying relation (2).

#### Theorem 4.12

If (R, +, ⋅) is a hyperfield and AR, then $\begin{array}{}{\phi }_{m}^{-1}\end{array}$(φm(A)) = A$\begin{array}{}{\omega }_{R}^{m}\end{array}$.

#### Proof

Let x$\begin{array}{}{\phi }_{m}^{-1}\end{array}$(φm(A)). Then there exists yA such that φm(x) = φm(y). Since (R, ⋅) is a hypergroup, it follows that there exists tR such that xyt, which implies that φm(x) = φm(y) ⊙ φm(t). Since (R/ $\begin{array}{}{\epsilon }_{m}^{\ast }\end{array}$, ⊙) is a group and φm(x) = φm(y), it results φm(t) = $\begin{array}{}{1}_{R/{\epsilon }_{m}^{\ast }}\end{array}$. Thus, $\begin{array}{}t\in {\phi }_{m}^{-1}\left({1}_{R/{\epsilon }_{m}^{\ast }}\right)={\omega }_{R}^{m},\end{array}$ and so xytA$\begin{array}{}{\omega }_{R}^{m}\end{array}$. Then $\begin{array}{}{\phi }_{m}^{-1}\end{array}$(φm(A)) ⊆ A$\begin{array}{}{\omega }_{R}^{m}\end{array}$.

Conversely, let xA$\begin{array}{}{\omega }_{R}^{m}\end{array}$. Then xay, for some aA and y$\begin{array}{}{\omega }_{R}^{m}\end{array}$, that is φm(y) = $\begin{array}{}{1}_{R/{\epsilon }_{m}^{\ast }}\end{array}$. Hence φm(x) = φm(ay) = φm(a) ⊙ φm(y) = φm(a) ⊙ $\begin{array}{}{1}_{R/{\epsilon }_{m}^{\ast }}\end{array}$ = φm(a) ∈ φm(A), and so x$\begin{array}{}{\phi }_{m}^{-1}\end{array}$(φm(A)). This completes the proof. □

#### Theorem 4.13

If (R, +, ⋅) is a hyperfield and AR, then 𝓒m(A) = A$\begin{array}{}{\omega }_{R}^{m}\end{array}$.

#### Proof

It is easy to see that $\begin{array}{}{\phi }_{m}^{-1}\end{array}$(φm(A)) = {xR | ∃aA : φm(x) = φm(a)}. Also, we have φm(x) = φm(a) if and only if $\begin{array}{}{\epsilon }_{m}^{\ast }\end{array}$(x) = $\begin{array}{}{\epsilon }_{m}^{\ast }\end{array}$(a), equivalently with θ(x) = θ(a), meaning that xKm(a) = 𝓒m(a), by Theorem 4.11. Hence, x ∈ 𝓒m(A) if and only if x$\begin{array}{}{\phi }_{m}^{-1}\end{array}$(φm(A)), thus xA$\begin{array}{}{\omega }_{R}^{m}\end{array}$, by Theorem 4.7 and Theorem 4.12. Therefore, 𝓒m(A) = A$\begin{array}{}{\omega }_{R}^{m}\end{array}$.□

#### Corollary 4.14

Let R be a hyperfield. A is an m-complete part of R if and only if A = A$\begin{array}{}{\omega }_{R}^{m}\end{array}$.

#### Proof

If A is an m-complete part, then 𝓒m(A) = A. Hence, A = 𝓒m(A) = A$\begin{array}{}{\omega }_{R}^{m}\end{array}$, by Theorem 4.13. Conversely, if A = A$\begin{array}{}{\omega }_{R}^{m}\end{array}$, then A = 𝓒m(A) by Theorem 4.13, and so A is an m-complete part of R.□

By Theorem 4.12, we have $\begin{array}{}{\omega }_{R}^{m}\cdot {\omega }_{R}^{m}={\phi }_{m}^{-1}\left({\phi }_{m}\left({\omega }_{R}^{m}\right)\right)={\omega }_{R}^{m}.\text{\hspace{0.17em}}\text{Hence},\text{\hspace{0.17em}}{\omega }_{R}^{m}\end{array}$ 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 (AB) ⋅ $\begin{array}{}{\omega }_{R}^{m}\end{array}$ = (A$\begin{array}{}{\omega }_{R}^{m}\end{array}$) ⋅ B = AB, by Corollary 4.14. Hence, AB is an m-complete part of R, by Corollary 4.14.

#### Theorem 4.15

Let (R, +, ⋅) be a hyperfield. Then every m-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 xAy for x, yR. Thus, xay, for aA, which implies that φm(x) = φm(a) ⊙ φm(y). Since φm(A) is a subgroup of R/ $\begin{array}{}{\epsilon }_{m}^{\ast }\end{array}$, we have φm(y) = φm(a)–1φm(x) ∈ φm(A)⊙ φm(x) = φm(Ax). Besides, Ax is an m-complete part of R, hence y$\begin{array}{}{\phi }_{m}^{-1}\end{array}$(φm(Ax)) = 𝓒m(Ax) = Ax. 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 𝓢Cm(R) be the set of all m-complete parts of R which are subhypergroups of (R, ⋅). Then, $\begin{array}{}{\omega }_{R}^{m}=\bigcap _{A\in {\mathcal{S}}_{{C}_{m}}\left(R\right)}A.\end{array}$

#### Proof

We know that $\begin{array}{}{\omega }_{R}^{m}\end{array}$ is an m-complete part of R. Let x$\begin{array}{}{\omega }_{R}^{m}\end{array}$. For all t, y$\begin{array}{}{\omega }_{R}^{m}\end{array}$, we have φm(x) = $\begin{array}{}{1}_{R/{\epsilon }_{m}^{\ast }}\end{array}$ = φm(t) ⊙ φm(y) = φm(ty). Hence, x$\begin{array}{}{\phi }_{m}^{-1}\end{array}$(φm(tωm)) = 𝓒m(t$\begin{array}{}{\omega }_{R}^{m}\end{array}$) = t$\begin{array}{}{\omega }_{R}^{m}\end{array}$, and so $\begin{array}{}{\omega }_{R}^{m}\subseteq t\cdot {\omega }_{R}^{m}\text{\hspace{0.17em}}\text{for all}\text{\hspace{0.17em}}t\in {\omega }_{R}^{m}.\end{array}$ Clearly, $\begin{array}{}t\cdot {\omega }_{R}^{m}\subseteq {\omega }_{R}^{m}.\text{\hspace{0.17em}}\text{Then}\text{\hspace{0.17em}}{\omega }_{R}^{m}=t\cdot {\omega }_{R}^{m},\text{\hspace{0.17em}}\text{for all}\text{\hspace{0.17em}}t\in {\omega }_{R}^{m},\end{array}$ which implies that $\begin{array}{}{\omega }_{R}^{m}\end{array}$ is a subhypergroup of (R, ⋅). Thus, $\begin{array}{}\bigcap _{A\in {\mathcal{S}}_{{C}_{m}}\left(R\right)}A\subseteq {\omega }_{R}^{m}.\end{array}$ Now, let A ∈ 𝓢Cm(R). By Corollary 4.14, A = A$\begin{array}{}{\omega }_{R}^{m}\end{array}$. Hence, for every x$\begin{array}{}{\omega }_{R}^{m}\end{array}$, there exist a, bA such that abx, and so aAx. By Theorem 4.15, A is invertible, and therefore xAaA. Then $\begin{array}{}{\omega }_{R}^{m}\end{array}$A, and thus $\begin{array}{}{\omega }_{R}^{m}\subseteq \bigcap _{A\in {\mathcal{S}}_{{C}_{m}}\left(R\right)}A.\end{array}$ 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 xxm, for all xR.

#### Example 4.17

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

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

1. For every odd number m ∈ ℕ, we have 0m = 0, am = a and bm = b. Hence, R is m-idempotent, for all odd natural numbers m.

2. Besides, since a2 = aa = b, it follows that R is not an 2-idempotent hyperring. Similarly, one proves that, for all even numbers m ∈ ℕ, the hyperring R is not m-idempotent.

#### Example 4.18

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

#### Example 4.19

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

$⊞010{0}R1R{1}⊡010{0}{0}1{0}R$

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

#### Example 4.20

Similarly, take the same support set R = {0, 1} and define on R the two hyperoperations as follows:

$⊕010{0}R1RR⊙010{0}R1{0}R$

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

Now we give an example of m-idempotent hyperfield.

#### Example 4.21

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

$+010{0}R1R{1}⋅010{0}R1R{1}$

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

Now, for all aR, put $\begin{array}{}\mathcal{X}\left(a\right)=\bigcup \left\{\sum _{i=1}^{n}{A}_{i}^{m}\mid a\in \sum _{i=1}^{n}{A}_{i}^{m}\right\},\end{array}$ where A1, …, AnR.

#### Theorem 4.22

Let R be an m-idempotent hyperfield. Then 𝓧(a) is an m-complete part of R, for all aR.

#### Proof

Suppose that aR and $\begin{array}{}\sum _{i=1}^{n}{z}_{i}^{m}\end{array}$ ∩ 𝓧(a) ≠ ∅, for some z1, …, znR. Then there exists zR such that z$\begin{array}{}\sum _{i=1}^{n}{z}_{i}^{m}\end{array}$ and zA, where A = $\begin{array}{}\sum _{i=1}^{n}{A}_{i}^{m}\end{array}$, for A1, …, AnR. Since (R, ⋅) is a hypergroup, there exist w, bR such that znwa and azb. Also, for all 1 ≤ in – 1 we have ziziR. Hence,

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

since zA. Moreover, R is m-idempotent and we have bbm, 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, (z1R)m + … + (zn–1R)m + (wAb)m ⊆ 𝓧(a) and so $\begin{array}{}\sum _{i=1}^{n}{z}_{i}^{m}\end{array}$ ⊆ 𝓧(a). Then, 𝓧(a) is an m-complete part of R.□

#### Theorem 4.23

Let R be an m-idempotent hyperfield. Then 𝓧(a) = $\begin{array}{}{\omega }_{R}^{m}\end{array}$, for every a$\begin{array}{}{\omega }_{R}^{m}\end{array}$.

#### Proof

It is not difficult to see that $\begin{array}{}a\cdot {\omega }_{R}^{m}={\omega }_{R}^{m},\end{array}$ for all a$\begin{array}{}{\omega }_{R}^{m}\end{array}$. Hence, $\begin{array}{}{\omega }_{R}^{m}\end{array}$ = a$\begin{array}{}{\omega }_{R}^{m}\end{array}$ = 𝓒m(a) ⊆ 𝓧(a), by Theorem 4.13 and Theorem 4.22. Now, let a$\begin{array}{}{\omega }_{R}^{m}\end{array}$ and x ∈ 𝓧(a). Then there exists A = $\begin{array}{}\sum _{i=1}^{n}{A}_{i}^{m}\end{array}$ ⊆ 𝓧(a) such that xA. Since aA, then {x, a} ⊆ A. Since R satisfies relation (2), there exist xiAi, for 1 ≤ in, such that {x, a} ⊆ $\begin{array}{}\sum _{i=1}^{n}{x}_{i}^{m}\end{array}$ and so $\begin{array}{}{\epsilon }_{m}^{\ast }\end{array}$(x) = $\begin{array}{}{\epsilon }_{m}^{\ast }\end{array}$(a). Then φm(x) = φm(a). Hence, x$\begin{array}{}{\phi }_{m}^{-1}\end{array}$(φm(a)) = a$\begin{array}{}{\omega }_{R}^{m}\end{array}$ = $\begin{array}{}{\omega }_{R}^{m}\end{array}$. Therefore, the proof is complete.□

#### Theorem 4.24

The relation εm is transitive on m-idempotent hyperfields.

#### Proof

Let R be an m-idempotent hyperfield and $\begin{array}{}x{\epsilon }_{m}^{\ast }y,\end{array}$ for x, yR. Hence, x$\begin{array}{}{\phi }_{m}^{-1}\end{array}$(φm(y)) = y$\begin{array}{}{\omega }_{R}^{m}\end{array}$. Similarly, we have yx$\begin{array}{}{\omega }_{R}^{m}\end{array}$ which implies that {x, y} ⊆ x$\begin{array}{}{\omega }_{R}^{m}\end{array}$. Then, there exist t, z$\begin{array}{}{\omega }_{R}^{m}\end{array}$ such that xxt and yxz. By Theorem 4.23, z$\begin{array}{}{\omega }_{R}^{m}\end{array}$ = 𝓧(t) and thus {t, z} ⊆ A = $\begin{array}{}\sum _{i=1}^{n}{A}_{i}^{m}\end{array}$ for A ⊆ 𝓧(t). Since R is m-idempotent, {x, y} ⊆ xA$\begin{array}{}\sum _{i=1}^{n}\end{array}$(xAi)m. So, there exist zixAi for every 1 ≤ in such that {x, y} ⊆ $\begin{array}{}\sum _{i=1}^{n}{x}_{i}^{m}\end{array}$. Therefore, my and so $\begin{array}{}{\epsilon }_{m}^{\ast }\end{array}$ = εm.□

## 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, $\begin{array}{}{\epsilon }_{m}^{\ast }\end{array}$-relation [12], smaller than γ*, while on m-idempotent hyperrings satisfying relation (2), we have $\begin{array}{}{\epsilon }_{m}^{\ast }\end{array}$ = γ* [12]. In general, εm is not transitive. This paper provides a detailed study on the transitivity property of εm. Using m-complete parts of a hyperring, we have proved that εm is transitive on m-idempotent hyperfields satisfying relation (2).

## 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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irina.cristea@ung.si

m.norouzi@ub.ac.ir

Accepted: 2018-07-04

Published Online: 2018-08-24

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 1012–1021, ISSN (Online) 2391-5455,

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