Multiplicative F ( m,n ) -Hyperrings

Abstract

In this article, after describing some examples to explain the concept of fuzzy multiplicative (m, n)-hyperrings (F^(m,n)-hyperrings), we will offer some related properties. Also, by defining F-hyperideals and regular relations, we will study the quotient of multiplicative F^(m,n)-hyperrings. Finally, we will generalize the classical isomorphism theorems of groups to multiplicative F^(m,n)-hyperrings.

1 1. Introduction and preliminaries

In this section, at first, we review some works related to the content of the paper and then we express some necessary arrangements. As we know, Marty is the first person who introduced the notion of hypergroups. One of the most natural ways that helps in understanding fundamental properties of algebraic structures is n-ary generalizations of them. In [8], Davvaz and Vougiouklis introduced the concept of n-hypergroups as a generalization of hypergroups in the sense of Marty. An interested reader can refer to [5 , 18] for more information about n-ary generalizations of hyperstructures. Hyperring theory is another important branch of hyperstructure theory. Krasner, Multiplicative and Feeble hyperrings are some famous hyperrings that have studied by a variety of authors. Multiplicative hyperring is a triple (R, + , ·), where (R, +) is an abelian group and (R, ·) is a semihypergroup, such that · is distributive with respect to + (see [6]). In [3, 16] there are some review of hyperring theory. Three isomorphism theorems that describe the relationship between quotients, homomorphisms and subobjects of algebraic structures can be stated in (m, n)-ary generalizations of hyperrings. Some versions of isomorphism theoremsfor Krasner (m, n)-hyperrings can be found in [7]. In [20], Zadeh was introduced the notion of fuzzy set which have penetrated in various branches of science including mathematics. In [1 , 24] there are some relationships between the fuzzy sets and algebraic hyperstructures. In [9], the notion of fuzzy Krasner (m, n)-hyperrings (F^(m,n)-hyperrings) by using the notion of F^m-hyperoperations and Fⁿ-operations is introduced and some related properties are investigated. Now, in this paper, we have dealt with inspiration from the concept of F-polygroups defined by Zahedi and Hasankhani (see [21, 22]) in the multiplicative F^(m,n)-hyperrings defining. Throughout this paper we shall use the notation $x_{i}^{j}$ to denote the sequence x_i, …, x_j. Let X be a non-empty set and let $P^{*} (X)$ be the set of all non-empty subsets of X and Xⁿ be the n-times cartesian product of X. In general, an n-hyperoperation on X is a mapping $f : X^{n} ⟶ P^{*} (X)$ . The couple (X, f) where f is an n-hyperoperation is called an n-hypergroupoid. An n-hypergroupoid (H, f) is called an n-semihypergroup if f is associative, i.e., the following equality holds: $f (x_{1}^{i - 1}, f (x_{i}^{n + i - 1}), x_{n + i}^{2 n - 1}) = f (x_{1}^{j - 1}, f (x_{j}^{n + j - 1}), x_{n + j}^{2 n - 1}),$ for all i, j ∈ {1, …, n} and $x_{1}^{2 n - 1} \in X$ . If for all (x₁, …, x_n) ∈ Xⁿ, the set $f (x_{1}^{n})$ is singleton, then f is called n-operation and in the case that f is associative, (X, f) is called an n-semigroup. An element e of an n-hypergroupoid (X, f) is called an identity element, if $f (\overset{(i - 1)}{e}, x, \overset{(n - i)}{e}) = {x}$ , for all x ∈ X and all i ∈ {1, …, n}. We recall the following background from [18]. Let P be a non-empty set. Let f be an n-hyperoperation on P and ^- 1 : P ⟶ P be a unitary operation. Then, (P, f) is called an n-polygroup if the following axioms hold:

f is associative,

there exists an identity element o ∈ P such that o^-1 = o,

$z \in f (x_{1}^{n})$ implies $x_{i} \in f (x_{i - 1}^{- 1}, \dots, x_{1}^{- 1}, z, x_{n}^{- 1}, \dots, x_{i + 1}^{- 1})$ , for all $z, x_{1}^{n} \in P$ .

We shall use the notation (P, f, o) to say that o is an identity element. An n-polygroup (P, f) is said to be canonical if $f (x_{1}^{n}) = f (x_{σ (1)}^{σ (n)})$ , for all $x_{1}^{n} \in P$ and for every $σ \in 𝕊_{n} .$ A canonical n-polygroup (P, f) is said to be monadic if $| f (x_{1}^{n}) | = 1$ , for all $x_{1}^{n} \in P$ . In what follows, we have a generalization of the notion of multiplicative hyperring which introduced by R. Rota [17] in 1982. Let R be a non-empty set. A multiplicative (m, n)-hyperring is an algebraic hyperstructure (R, f, g) which satisfies the following axioms:

(R, f, o) is a monadic m-polygroup,

(R, g) is an n-semihypergroup with an identity element e,

we have $\begin{matrix} g (a_{1}^{i - 1}, f (x_{1}^{m}), a_{i + 1}^{n}) \\ \subseteq f (g (a_{1}^{i - 1}, x_{1}, a_{i + 1}^{n}), \dots, g (a_{1}^{i - 1}, x_{m}, a_{i + 1}^{n})), \end{matrix}$ for all i ∈ {1, …, n} and all $a_{1}^{i - 1}, a_{i + 1}^{n}, x_{1}^{m} \in R$ ,

for all x, y ∈ R we have g (x, y^-1, e .) = g (x^-1, y, e .) = (g (x, y, e .)) ^-1,

for every $x_{2}^{n} \in R$ we have $g (o, x_{2}^{n}) = g (x_{2}, o, x_{3}^{n}) = \dots = g (x_{2}^{n}, o) = {o},$ where o is the identity element of (R, f).

If in (3) we have equality instead of inclusion, then we say that the multiplicative (m, n)-hyperring is strongly distributive. An (m, n)-hyperring (R, f, g) is called an (m, n)-ring if (R, g) is an n-semigroup.

A fuzzy subset of X is a mapping μ : X → [0, 1]. An empty fuzzy subset of X denoted by ∅ is the zero function from X to [0, 1]. We denote the set of all fuzzy subsets of X by I (X), that is I (X) = {μ | μ : X → [0, 1] isafunction}. Also, we denote the set of all non-empty fuzzy subsets of X by I^* (X). Let {μ_α} _α∈Λ be a family of fuzzy subsets of X. We define fuzzy subsets (⋃ _α∈Λμ_α) and (⋂ _α∈Λμ_α) as follows:(⋃ _α∈Λμ_α) (x) = ⋁ _α∈Λ {μ_α (x)} and ⋂_α∈Λμ_α) (x) = ⋀ _α∈Λ {μ_α (x)}, where ∨ and ∧ denote supremum and infimum, respectively. If μ ∈ I (X), then the support of μ, is defined by supp (μ) = {x ∈ H | μ (x) >0} . If A ⊆ X and t ∈ [0, 1], then by A_t we mean a fuzzy subset of X which is defined as follows: A_t (x) = t if x ∈ A and A_t (x) =0 if x ∉ A. In particular, if A is a singleton set, say {a}, then {a} _t is said to be a fuzzy point and is denoted by a_t. An Fⁿ-hyperoperation on X is a function f : Xⁿ → I^* (X). If for all (x₁, …, x_n) ∈ Xⁿ, the set $supp (f (x_{1}^{n}))$ is singleton, then f is called Fⁿ-operation. If a ∈ X and $μ_{1}^{n} \in I^{*} (X)$ , then we define $f (μ_{1}^{n}) = ⋃_{x_{i} \in supp (μ_{i})} f (x_{1}^{n}) .$ Furthermore, for $μ \in I^{*} (X), A \in P^{*} (X)$ and $x_{1}^{n} \in X$ and for i ∈ {1, …, n} we define

$f (x_{1}^{i - 1}, μ, x_{i + 1}^{n}) = f (χ_{{x_{1}}}, \dots, χ_{{x_{i - 1}}}, μ, χ_{{x_{i + 1}}}, \dots, χ_{{x_{n}}})$ ,

$f (x_{1}^{i - 1}, A, x_{i + 1}^{n}) = f (χ_{{x_{1}}}, \dots, χ_{{x_{i - 1}}}, χ_{A}, χ_{{x_{i + 1}}}, \dots, χ_{{x_{n}}})$ ,

$f (μ_{1}^{i - 1}, x, μ_{i + 1}^{n}) = f (μ_{1}^{i - 1}, χ_{{x}}, μ_{i + 1}^{n})$ ,

$f (μ_{1}^{i - 1}, A, μ_{i + 1}^{n}) = f (μ_{1}^{i - 1}, χ_{A}, μ_{i + 1}^{n})$ ,

where χ_A is the characteristic function. whenever f is an Fⁿ-hyperoperation on X, the couple (X, f) is called an Fⁿ-hypergroupoid. An Fⁿ-hypergroupoid (X, f) is called an Fⁿ-semihypergroup if f is associative, i.e., the following equality holds:

\begin{matrix} f (x_{1}^{i - 1}, f (x_{i}^{n + i - 1}), x_{n + i}^{2 n - 1}) \\ = f (x_{1}^{j - 1}, f (x_{j}^{n + j - 1}), x_{n + j}^{2 n - 1}), \end{matrix}

for all i, j ∈ {1, …, n} and

x_{1}^{2 n - 1} \in X

. An element e of an Fⁿ-hypergroupoid (X, f) is called an F-identity element, if

supp (f (\overset{(i - 1)}{e}, x, \overset{(n - i)}{e})) = {x},

for all x ∈ X and all i ∈ {1, …, n}.

Lemma 1.1.Let (X, f) be anFⁿ-semihypergroup containing anF-identity elemente. Then, we have $f (\overset{(i - 1)}{e}, x, \overset{(n - j - 1)}{e}, y, \overset{(j - i)}{e}) = f (x, y, \overset{(n - 2)}{e}),$ for allx, y ∈ Xandi ∈ {1, …, n - 1}, j ∈ {i, …, n - 1}.

Proof. Let x, y be arbitrary elements of X and i ∈ {1, …, n - 1}, j ∈ {i, …, n - 1}. Then, we have $f (\overset{(i - 1)}{e}, x, \overset{(n - j - 1)}{e}, y, \overset{(j - i)}{e})$ $= f (\overset{(i - 1)}{e}, f (x, \overset{(n - 1)}{e}), \overset{(n - j - 1)}{e}, y, \overset{(j - i)}{e})$ $= f (f (\overset{(i - 1)}{e}, x, \overset{(n - i)}{e}), \overset{(n - j + i - 2)}{e}, y, \overset{(j - i)}{e})$ $= f (x, \overset{(n - j + i - 2)}{e}, y, \overset{(j - i)}{e})$ $= f (x, \overset{(n - j + i - 2)}{e}, f (y, \overset{(n - 1)}{e}), \overset{(j - i)}{e})$ $= f (x, f (\overset{(n - j + i - 2)}{e}, y, \overset{(j - i + 1)}{e}), \overset{(n - 2)}{e})$ $= f (x, y, \overset{(n - 2)}{e})$ . □

2 2. Multiplicative F^(m,n)-hyperrings

We recall the following background from [10]. Let P be a non-empty set. Let f be an F^m-hyperoperation on P and ^- 1 : P → P be a unitary operation. Then, (P, f) is called an F^m-polygroup if the following axioms hold:

f is associative,

there exists an F-identity element o ∈ P such that o^-1 = o,

$z \in supp (f (x_{1}^{m}))$ $x_{i} \in supp (f (x_{i - 1}^{- 1}, \dots, x_{1}^{- 1}, z, x_{m}^{- 1}, \dots, x_{i + 1}^{- 1}))$ , for all $z, x_{1}^{m} \in P$ .

We shall use the notation (P, f, o) to say that o is an F-identity element. An F^m-polygroup (P, f) is said to be canonical if $f (x_{1}^{m}) = f (x_{σ (1)}^{σ (m)})$ , for all $x_{1}^{m} \in P$ and for every $σ \in 𝕊_{m}$ . A canonical F^m-polygroup (P, f) is said to be monadic if $| supp (f (x_{1}^{m})) | = 1$ , for all $x_{1}^{m} \in P$ . Sometimes, in a monadic F^m-polygroup, we will denote the singleton set $supp (f (x_{1}^{m}))$ by its element. It is easy to see that in a monadic F^m-polygroup (P, f) we have supp (f (x, x^-1, o .)) = {o}, for all x ∈ P. In general case, since f is associative, we have $supp (f (f (x_{m}^{- 1}, \dots, x_{2}^{- 1}, o), x_{2}^{m})) = {o}$ , for all $x_{2}^{m} \in P$ .

Example 1. Let (G, *) be an abelian group and t ∈ (0, 1]. Let ∘ be an F²-operation on G which is defined as follows: x ∘ y = (x * y) _t, for all x, y ∈ G. If the inverse of each element of G is attributed to that element by the unitary operation ^- 1 : G → G, then (G, ∘) is a monadic F²-polygroup.

Lemma 2.1.If (P, f) is anF^m-polygroup, $μ_{1}^{m}, μ \in I_{*}^{P}$ and $x_{1}^{m}, x$ are arbitrary elements ofP, Then

(x^-1) ^-1 = x,

$(supp (f (x_{1}^{m})))^{- 1} = supp (f (x_{m}^{- 1}, \dots, x_{1}^{- 1}))$ , whereA^-1 = {a^-1 | a ∈ A},

$supp (f (x_{1}^{i - 1}, μ, x_{i + 1}^{m})) = ⋃_{t \in supp (μ)} supp (f (x_{1}^{i - 1}, t, x_{i + 1}^{m}))$ , for alli ∈ {1, …, m} ,

supp (f (μ₁, …, μ_m)) = ⋃ _{x_i∈supp(μ_i)}supp (f (x₁, …, x_m)).

Proof. It is straightforward. □

Definition 2.2.Let R be a non-empty set. A multiplicative F^(m,n)-hyperring is a fuzzy algebraic hyperstructure (R, f, g) which satisfies the following axioms:

(R, f, o) is a monadic F^m-polygroup,

(R, g) is an Fⁿ-semihypergroup with an F-identity element e,

for all x, y ∈ R we have $\begin{matrix} supp (g (x, y^{- 1}, e .)) = supp (g (x^{- 1}, y, e .)) \\ = (supp (g (x, y, e .)))^{- 1}, \end{matrix}$

for every $x_{2}^{n} \in R$ we have $\begin{matrix} supp (g (o, x_{2}^{n})) = supp (g (x_{2}, o, x_{3}^{n})) \\ = \dots = supp (g (x_{2}^{n}, o)) = {o}, \end{matrix}$ where o is the F-identity element of (R, f).

An F^(m,n)-hyperring (R, f, g) is called an F^(m,n)-ring if (R, g) is an Fⁿ-semigroup. If in (3) we have equality instead of inclusion, then we say that the multiplicative F^(m,n)-hyperring is strongly distributive.

Example 2.Consider the F²-operations ∘ and ∗ on R = {0, 1} as follows:

\begin{matrix} \circ & 0 & 1 \\ 0 & 0_{t_{2}} & 1_{t_{2}} \\ 1 & 1_{t_{2}} & 0_{t_{2}} \end{matrix} \begin{matrix} * & 0 & 1 \\ 0 & 0_{t_{1}} & 1_{t_{1}} \\ 1 & 1_{t_{1}} & 0_{t_{1}} \end{matrix}

where 0 < t₁ ≤ t₂ ≤ 1. If we define the unitary operation ^- 1 on R by x^-1 = x, then (R, ∘ , 0) is a monadic F²-polygroup. Also, (R, ∗) is an F²-semigroup and 1 is its F-identity element. It is easy to check that (R, ∘ , ∗) is a multiplicative F^(2,2)-ring. Obviously, (R, ∘ , ∗) is not strongly distributive whenever t₁ < t₂.

Example 3. Let G = {e, a, b, c} be the Klein 4-group and t ∈ (0, 1]. We define the F^m-operation f on G as follows: $f (x_{1}^{m}) (z) = {\begin{matrix} t & if z = x_{1} \dots x_{m} \\ 0 & else \end{matrix}, for all x_{1}^{m}, z \in G .$

Let ∗ be an F²-hyperoperation defined on G with the following table: $\begin{matrix} * & e & a & b & c \\ e & e_{t} & e_{t} & e_{t} & e_{t} \\ a & e_{t} & a_{t} & b_{t} & c_{t} \\ b & e_{t} & b_{t} & G_{t} & G_{t} \\ c & e_{t} & c_{t} & G_{t} & G_{t} \end{matrix}$

Then, (G, f, ∗) is a multiplicative F^(m,2)-hyperring.

Proposition 2.3.If (R, f, g) is a multiplicativeF^(m,n) -hyperring, then for allx, y, z ∈ R, $\begin{matrix} g (x, f (y, z^{- 1}, o .), e .) \\ \subseteq f (g (x, y, e .), (supp (g (x, z, e .)))^{- 1}, o .) and \\ g (f (y, z^{- 1}, o .), x, e .) \\ \subseteq f (g (y, x, e .), (supp (g (z, x, e .)))^{- 1}, o .) . \end{matrix}$

If (R, f, g) is strongly distributive, then for allx, y, z ∈ R, $\begin{matrix} g (x, f (y, z^{- 1}, o .), e .) \\ = f (g (x, y, e .), (supp (g (x, z, e .)))^{- 1}, o .) and \\ g (f (y, z^{- 1}, o .), x, e .) \\ = f (g (y, x, e .), (supp (g (z, x, e .)))^{- 1}, o .) . \end{matrix}$

Proof. The statement follows from conditions (3), (4) and (5) of Definition 2.2. □

Theorem 2.4.Let (R, f, g) be a strongly distributive multiplicativeF^(m,n)-hyperring. Then, the following assertions hold:

|supp (g (x, y, e .)) |=1, for allx, y ∈ R,

(R, f, g) is anF^(m,n)-ring.

Proof. (1) Let x, y ∈ R be arbitrary elements. Since (R, f) is a monadic F^m-polygroup, it follows that supp (f (y, y^-1, o .)) = {o} and therefore by using Proposition 2.3 we have $\begin{matrix} {o} = supp (g (x, o, e .)) \\ = supp (g (x, f (y, y^{- 1}, o .), e .)) \\ = supp (f (g (x, y, e .), g (x, y^{- 1}, e .), o .)) \\ = supp (f (g (x, y, e .), (g (x, y, e .))^{- 1}, o .)) . \end{matrix}$

So, for each u, v ∈ supp (g (x, y, e .)) we have o ∈ supp (f (u, v^-1, o .)). This implies u ∈ supp (f (v, o .)) = {v}. Whence, u = v and supp (g (x, y, e .)) contains only one element.

(2) By (1), (R, g) is an Fⁿ-semigroup and so (R, f, g) is an F^(m,n)-ring. □

Notice that in Example 2, in the case that t₁ < t₂, we have a multiplicative F^(2,2)-ring which is not strongly distributive and therefore the converse of Theorem 2.4 is not true in general.

Definition 2.5.Let (R, f, g) be a multiplicative F^(m,n)-hyperring and I be a non-empty subset of R. We say that I is an F-subhyperring of (R, f, g) if

supp (f (I, I^-1, o .)) ⊆ I,

supp (g (x, y, e .)) ⊆ I, for all x, y ∈ I.

An F-subhyperring I of (R, f, g) is called an F-hyperideal if supp (g (x, r, e .)) ∪ supp (g (r, x, e .)) ⊆ I, for all x ∈ I and all r ∈ R.

Lemma 2.6.LetIbe anF-hyperideal of a multiplicativeF^(m,n)-hyperring (R, f, g). Then, the following assertions hold:

supp (f (I, x, o .)) = I, for allx ∈ I.

$supp (f (I, x, a_{3}^{m})) = supp (f (I, o, a_{3}^{m}))$ , for allx ∈ Iand $a_{3}^{m} \in R$ .

$supp (f (I, x_{2}^{m})) = I$ , for all $x_{2}^{m} \in I$ .

$x \in supp (f (I, a_{2}^{m}))$ implies that $supp (f (I, a_{2}^{m})) = supp (f (I, x, o .))$ , where $a_{2}^{m} \in R$ .

Proof. (1) Let x ∈ I be an arbitrary element. Since I is an F-hyperideal, we have supp (f (I, x, o .)) ⊆ I. Also, we have $\begin{matrix} I & = supp (f (I, o .)) = supp (f (I, f (x^{- 1}, x, o .), o .)) \\ = supp (f (f (I, x^{- 1}, o .), x, o .)) \\ \subseteq supp (f (I, x, o .)) . \end{matrix}$

(2) Let x ∈ I and $a_{3}^{m} \in R$ be arbitrary elements. Then, we have $\begin{matrix} supp (f (I, x, a_{3}^{m})) & = & supp (f (I, f (x, o .), a_{3}^{m})) \\ = & supp (f (f (I, x, o .), o, a_{3}^{m})) \\ = & supp (f (I, o, a_{3}^{m})) . \end{matrix}$

(3) With frequent use of (2) it can be achieved.

(4) Let $a_{2}^{m}$ be arbitrary elements of R and $x \in supp (f (I, a_{2}^{m}))$ . Then, we have $\begin{matrix} supp (f (I, x, o .)) \subseteq supp (f (I, f (I, a_{2}^{m}), o .)) \\ = supp (f (f (I, I, o .), a_{2}^{m})) = supp (f (I, a_{2}^{m})) . \end{matrix}$

Now, we prove the reverse inclusion. By assumption, (R, f) is an F^m-polygroup and therefore from $x \in supp (f (I, a_{2}^{m}))$ it follows that $a_{2} \in supp (f (I, x, a_{m}^{- 1}, \dots, a_{3}^{- 1})) .$

Thus, by using (2) and that f is associative we have $\begin{matrix} supp (f (I, a_{2}^{m})) \\ \subseteq supp (f (I, f (I, x, a_{m}^{- 1}, \dots, a_{3}^{- 1}), a_{3}^{m})) \\ = supp (f (I, x, f (a_{m}^{- 1}, \dots, a_{3}^{- 1}, a_{3}, o), a_{4}^{m})) \\ = supp (f (I, x, f (a_{m}^{- 1}, \dots, a_{4}^{- 1}, o .), a_{4}^{m})) \\ ⋮ \\ = supp (f (I, x, o .)) . \end{matrix}$

This completes the proof. □

Lemma 2.7.Let $I_{1}^{m}$ beF-hyperideals of a multiplicativeF^(m,n)-hyperring (R, f, g). Then, the following assertions hold:

$supp (f (I_{1}^{m}))$ is anF-hyperideal of (R, f, g).

$supp (f (I_{1}^{j - 1}, o, I_{j + 1}^{m}))$ is anF-hyperideal of $supp (f (I_{1}^{m}))$ , for allj ∈ {1, …, m}.

I_jis anF-hyperideal of $supp (f (I_{1}^{m}))$ , for allj ∈ {1, …, m}.

Proof. (1) Let $x, y \in supp (f (I_{1}^{m}))$ and r ∈ R be arbitrary elements. Then, there exist w_j, z_j ∈ I_j, 1 ≤ j ≤ m, such that $supp (f (w_{1}^{m})) = {x}$ and $supp (f (z_{1}^{m})) = {y}$ . Now, we have $\begin{matrix} supp (f (x, y^{- 1}, o .)) \\ = supp (f (f (w_{1}, \dots, w_{m}), f (z_{m}^{- 1}, \dots, z_{1}^{- 1}), o .)) \\ = supp (f (f (w_{1}, z_{1}^{- 1}, o .), \dots, f (w_{m}, z_{m}^{- 1}, o .))) \\ \subseteq supp (f (I_{1}^{m})) . \end{matrix}$

Also, we have $\begin{matrix} supp (g (x, r, e .)) \subseteq supp (g (f (w_{1}^{m}), r, e .)) \\ \subseteq supp (f (g (w_{1}, r, e .), \dots, g (w_{m}, r, e .))) \\ \subseteq supp (f (I_{1}^{m})) . \end{matrix}$

Similarly, we can show $supp (g (r, x, e .)) \subseteq supp (f (I_{1}^{m})) .$

Therefore, $supp (f (I_{1}^{m}))$ is an F-hyperideal of (R, f, g).

The proofs of (2) and (3) are easy. □

Any intersection of F-subhyperrings of a multiplicative F^(m,n)-hyperring (R, f, g) is an F-subhyperring of R. Also, any intersection of F-hyperideals of a multiplicative F^(m,n)-hyperring (R, f, g) is an F-hyperideal of R. In this manner, we can consider the F-hyperideal generated by any subset S of R which is the intersection of all F-hyperideals of R containing S as a subset. We denote by I ⊕ J the F-hyperideal generated by I ∪ J, where I and J are F-hyperideals of (R, f, g).

Theorem 2.8.LetIandJbe twoF-hyperideals of a multiplicativeF^(m,n)-hyperring (R, f, g). Then,I ⊕ J = supp (f (I, J, o .)).

Proof. First, we show that o ∈ I ∩ J. Since I≠ ∅ we can assume that a ∈ I, for some a ∈ R. By condition (1) of Definition 2.5, we have supp (f (a, a^-1, o .)) ⊆ I.On the other hand, since (R, f) is an F^m-polygroup, from a ∈ supp (f (a, o .)) it implies that o ∈ supp (f (a, a^-1, o .)) and therefore we have o ∈ I.In a similar manner we can show that o ∈ J.Set S = supp (f (I, J, o .)). Since we have x ∈ supp (f (x, o .)) ⊆ S, for all x ∈ I, we have I ⊆ S. Similarly, we have J ⊆ S. Thus, I ∪ J ⊆ S. Now, we show that S is an F-hyperideal of R. Let u and v be arbitrary elements of S. Then, there exist x, y ∈ I and z, t ∈ J such that supp (f (x, z, o .)) = {u} and supp (f (y, t, o .)) = {v}. Since I and J are F-hyperideals, we have supp (f (x, y^-1, o .)) ⊆ I and supp (f (z, t^-1, o .)) ⊆ J. So, by using Lemma 2.1, we have $\begin{matrix} supp (f (u, v^{- 1}, o .)) \\ = supp (f (f (x, z, o .), (supp (f (y, t, o .)))^{- 1}, o .)) \\ = supp (f (f (x, z, o .), f (y^{- 1}, t^{- 1}, o .), o .)) \\ = supp (f (f (x, y^{- 1}, o .), f (z, t^{- 1}, o .))) \\ \subseteq supp (f (I, J, o .)) = S . \end{matrix}$

Therefore, the condition (1) of Definition 2.5 holds. Now, let r ∈ R be an arbitrary element. Since I and J are F-hyperideals, we have supp (g (x, r, e .)) ⊆ I and supp (g (z, r, e .)) ⊆ J. Hence, $\begin{matrix} supp (g (u, r, e .)) = supp (g (f (x, z, o .), r, e .)) \\ \subseteq supp (f (g (x, r, e .), g (z, r, e .), o .)) \subseteq S . \end{matrix}$

In a similar manner we can show that supp (g (r, u, e .)) = S. Finally, suppose that K is an F-hyperideal of R such that I ∪ J ⊆ K. We have to show that S ⊆ K. Let s ∈ S be an arbitrary element. Then, there exist x ∈ I and y ∈ J such that supp (f (x, y, o .)) = {s}. Since I ∪ J ⊆ K we have x, y ∈ K and since K is an F-hyperideal we conclude that s ∈ K. This completes the proof. □

3 3. Quotient multiplicative F^(m,n)-hyperrings

The goal of this section is to introduce an equivalence relation on a multiplicative F^(m,n)-hyperring and to construct a quotient multiplicative F^(m,n)-hyperring.

Let ρ be a binary relation on a non-empty set H. If A and B are non-empty subsets of H, then

we write $A \bar{ρ} B$ if for every a ∈ A, there exists b ∈ B such that a ρ b and for every b ∈ B there exists a ∈ A such that b ρ a,

$A \bar{\bar{ρ}} B$ means that for every a ∈ A and for every b ∈ B, one has a ρ b.

An equivalence relation ρ defined on a multiplicative F^(m,n)-hyperring (R, f, g) is called regular if the following implications hold: $\begin{matrix} x_{1} ρ y_{1}, \dots, x_{m} ρ y_{m} \Rightarrow supp (f (x_{1}^{m})) ρ supp (f (y_{1}^{m})), \\ x_{1} ρ y_{1}, \dots, x_{n} ρ y_{n} \Rightarrow supp (g (x_{1}^{n})) \bar{ρ} supp (g (y_{1}^{n})) \end{matrix}$ and ρ is called strongly regular if the following implications hold: $\begin{matrix} x_{1} ρ y_{1}, \dots, x_{m} ρ y_{m} \Rightarrow supp (f (x_{1}^{m})) ρ supp (f (y_{1}^{m})), \\ x_{1} ρ y_{1}, \dots, x_{n} ρ y_{n} \Rightarrow supp (g (x_{1}^{n})) \bar{\bar{ρ}} supp (g (y_{1}^{n})) . \end{matrix}$

In the sequel, ρ [x] is the equivalence class of x. Also, ρ [A] means {ρ [x] | x ∈ A}, where A is a set.

Theorem 3.1.Let (R, f, g) be a multiplicativeF^(m,n)-hyperring and letρbe an equivalence relation onR. Then, the following assertions are equivalent:

ρ is regular.

Them-operationf|_ρandn-hyperoperation g|_ρdefined onR/ρ = {ρ [x] | x ∈ R} as follows are well-defined. $\begin{matrix} f |_{ρ} (ρ [x_{1}], \dots, ρ [x_{m}]) = ρ [supp (f (x_{1}^{m}))], \\ g |_{ρ} (ρ [x_{1}], \dots, ρ [x_{n}]) = {ρ [z] | z \in supp (g (x_{1}^{n}))}, \end{matrix}$ for all $x_{1}^{n} \in R$ .

Proof. (1⇒2) Let ρ be a regular relation and ρ [x_i] = ρ [y_i], for all i ∈ {1, …, m}. Then, $\begin{matrix} f |_{ρ} (ρ [x_{1}], \dots, ρ [x_{m}]) = ρ [supp (f (x_{1}^{m}))] \\ = ρ [supp (f (y_{1}^{m}))] = f |_{ρ} (ρ [y_{1}], \dots, ρ [y_{m}]) . \end{matrix}$

Therefore, f|_ρ is well-defined. Now, suppose that ρ [x_i] = ρ [y_i], for all i ∈ {1, …, n}. Set $A = {ρ [z] | z \in supp (g (x_{1}^{n}))}$ and $B = {ρ [z] | z \in supp (g (y_{1}^{n}))}$ . We show that A = B. Let ρ [z] ∈ A be an arbitrary element. Then ρ [z] = ρ [z′] for some $z^{'} \in supp (g (x_{1}^{n}))$ . Since ρ is regular, there exists $w \in supp (g (y_{1}^{n}))$ such that ρ [z′] = ρ [w]. Thus, ρ [z] ∈ B and so A ⊆ B. Similarly, we can show that B ⊆ A. Hence, A = B and g|_ρ is well-defined.

(2⇒1) Let ρ [x_i] = ρ [y_i], for all i ∈ {1, …, m}. Since f|_ρ is well-defined, we have $f |_{ρ} (ρ_{[x_{1}]}^{[x_{m}]}) = f |_{ρ} (ρ_{[y_{1}]}^{[y_{m}]})$ and therefore $ρ [supp (f (x_{1}^{m}))] = ρ [supp (f (y_{1}^{m}))] .$ Now, suppose that ρ [x_i] = ρ [y_i], for all i ∈ {1, …, n}. Since g|_ρ is well-defined, we have A = B, where A and B are the sets defined in previous part. We show that $supp (g (x_{1}^{n})) \bar{ρ} supp (g (y_{1}^{n}))$ . Let $z \in supp (g (x_{1}^{n}))$ be an arbitrary element. Since ρ [z] ∈ A and A = B, there exists $w \in supp (g (y_{1}^{n}))$ such that ρ [z] = ρ [w]. Similarly, for each $z \in supp (g (y_{1}^{n}))$ there exists $w \in supp (g (x_{1}^{n}))$ such that ρ [z] = ρ [w]. Thus $supp (g (x_{1}^{n})) \bar{ρ} supp (g (y_{1}^{n}))$ . □

Remark 1. Whenever ρ is a strongly regular relation, reflexivity of ρ implies that for every $z, z^{'} \in supp (g (x_{1}^{m}))$ we have ρ [z] = ρ [z′] and so ${ρ [z] | z \in supp (g (x_{1}^{m}))}$ is a singleton set. Thus, g|_ρ defined in Theorem 3.1 is an n-operation. Conversely, if g|_ρ is an n-operation, then ${ρ [z] | z \in supp (g (x_{1}^{m}))}$ is a singleton set and ρ is a strongly regular relation.

Corollary 3.2.Let (R, f, g) be a multiplicativeF^(m,n)-hyperring and letρbe an equivalence relation onR. Then,

ρis regular if and only if (R/ρ, f|_ρ, g|_ρ) is a multiplicative (m, n)-hyperring,

ρis strongly regular if and only if (R/ρ, f|_ρ, g|_ρ) is a multiplicative (m, n)-ring.

Now, let I be an F-hyperideal of a multiplicative F^(m,n)-hyperring (R, f, g). We define the relation ρ_I on R as follows: x ρ_I y ⇔ supp (f (x, y^-1, o .)) ⊆ I .It is not difficult to see that the relation ρ_I is an equivalence relation.

Lemma 3.3.Let (R, f, g) be a multiplicativeF^(m,n)-hyperring andx, y ∈ R. Then, the following assertions hold:

x ρ_I yif and only ifx ∈ supp (f (I, y, o .)).

ρ_I [x] = supp (f (I, x, o .)).

For every $a_{2}^{m} \in R$ we have $supp (f (I, a_{2}^{m})) = ρ_{I} [x]$ , where $x \in supp (f (I, a_{2}^{m}))$ .

Proof. (1) Let x ρ_I y, for some x, y ∈ R. By definition of ρ_I, there exists z ∈ I such that supp (f (x, y^-1, o .)) = {z}. Since (R, f) is an F^m-polygroup, we havex ∈ supp (f (z, y, o .)) ⊆ supp (f (I, y, o .)).Conversely, assume that x ∈ supp (f (I, y, o .)), for some x, y ∈ R. Then, there exists z ∈ I such that x ∈ supp (f (z, y, o .)). Again, since (R, f) is an F^m-polygroup, we have supp (f (x, y^-1, o .)) = {z}. Therefore, supp (f (x, y^-1, o .)) ⊆ I.

(2) By using (1) we have $\begin{matrix} ρ_{I} [x] = {z \in R | z ρ_{I} x} \\ = {z \in R | z \in supp (f (I, x, o .))} = supp (f (I, x, o .)) . \end{matrix}$

(3) It is a direct consequence of part (2) and Lemma 2.6 (4). □

Corollary 3.4.IfIis anF-hyperideal of a multiplicativeF^(m,n)-hyperring (R, f, g), then

$supp (f (I, a_{2}^{m})) \cap supp (f (I, b_{2}^{m})) \neq \emptyset$ implies $supp (f (I, a_{2}^{m})) = supp (f (I, b_{2}^{m}))$ ,

x ρ_I yif and only if there exist $r_{2}^{m} \in R$ such that $x, y \in supp (f (I, r_{2}^{m}))$ .

Proposition 3.5.LetI be anF -hyperideal of a multiplicativeF^(m,n) -hyperring (R, f, g). Then, for all $x_{1}^{m} \in R$ we have

$ρ_{I} [supp (f (x_{1}^{m}))]$ = supp (f (ρ_I [x₁] , …, ρ_I [x_m])),

ρ_I [supp (f (ρ_I [x₁] , …, ρ_I [x_m]))] = supp (f (ρ_I [x₁] , …, ρ_I [x_m])).

Proof. (1) By using Lemma 3.3 (2), we have $\begin{matrix} supp (f (ρ_{I} [x_{1}], \dots, ρ_{I} [x_{m}])) \\ = supp (f (f (I, x_{1}, o .), \dots, f (I, x_{m}, o .))) \\ = supp (f (I, f (x_{1}^{m}), o .)) \\ = ρ_{I} [supp (f (x_{1}^{m}))] . \end{matrix}$

(2) By using (1), we have $\begin{matrix} ρ_{I} [supp (f (ρ_{I} [x_{1}], \dots, ρ_{I} [x_{m}]))] \\ = ρ_{I} [ρ_{I} [supp (f (x_{1}^{m}))]] = ρ_{I} [supp (f (x_{1}^{m}))] . \end{matrix}$ □

Lemma 3.6.LetIbe anF-hyperideal of a multiplicativeF^(m,n)-hyperring (R, f, g) andx, y ∈ R. Then, the following assertions are equivalent:

x ρ_I y,

supp (f (x, y^-1, o .)) ⊆ I,

supp (f (x, y^-1, o .))∩ I ≠ ∅.

Proof. (1⇒2) By Lemma 3.3 (1) we have x ∈ supp (f (I, y, o .)). So, $\begin{matrix} supp (f (x, y^{- 1}, o .)) \subseteq supp (f (f (I, y, o .), y^{- 1}, o .)) \\ = supp (f (I, f (y, y^{- 1}, o .), o .)) = supp (f (I, o .)) = I . \end{matrix}$

(2⇒3) It is obvious.

(3⇒1) Let z ∈ supp (f (x, y^-1, o .)) ∩ I. Since(R, f) is an F^m-polygroup we have x ∈ supp (f (z, y, o .)) ⊆ supp (f (I, y, o .)). That is, x ρ_I y. □

Corollary 3.7.For eachF-hyperideals IandJof a multiplicativeF^(m,n)-hyperring we haveI ⊆ Jif and only ifρ_I ⊆ ρ_J.

Theorem 3.8.The relationρ_Iis a regular relation on (R, f, g).

Proof. Let x₁ ρ_I y₁, …, x_m ρ_I y_m. Let $supp (f (x_{1}^{m})) = {x}$ and $supp (f (y_{1}^{m})) = {y}$ . Then, $\begin{matrix} x \in supp (f (x_{1}^{m})) \\ \subseteq supp (f (f (I, y_{1}, o .), \dots, f (I, y_{m}, o .))) \\ = supp (f (I, f (y_{1}^{m}), o .)) = supp (f (I, y, o .)) . \end{matrix}$

Therefore, by using Lemma 3.3 (1) we have x ρ_I y. Now, consider x₁ ρ_I y₁, …, x_n ρ_I y_n. We show that $supp (g (x_{1}^{n})) \bar{ρ} supp (g (y_{1}^{n}))$ . By Lemma 3.6, for each i ∈ {1, …, n} there exists z_i ∈ I such that $supp (f (x_{i}, y_{i}^{- 1}, o .)) = {z_{i}}$ . Since (R, f) is an F^m-polygroup we have supp (f (z_i, y_i, o .)) = {x_i}, for all i ∈ {1, …, n}. Hence, we have $\begin{matrix} supp (g (x_{1}^{n})) = supp (g (f (z_{1}, y_{1}, o .), \dots, f (z_{n}, y_{n}, o .))) \\ \subseteq supp (f (I, g (y_{1}^{n}), o .)) . \end{matrix}$

Consequently, for any $w \in supp (g (x_{1}^{n}))$ there exists $w^{'} \in supp (g (y_{1}^{n}))$ such that w ∈ supp (f (I, w′, o .)) which means that w ρ_I w′. Similarly, by writing $supp (f (x_{i}, z_{i}^{- 1}, o .)) = {y_{i}}$ , where i ∈ {1, …, n}, it follows that for any $w^{'} \in supp (g (y_{1}^{n}))$ there exists $w \in supp (g (x_{1}^{n}))$ such that w′ ρ_I w. This completes the proof. □

Corollary 3.9.There exists a one to one correspondence between the set of regular relations, for which the equivalence class ofois anF-hyperideal and the set ofF-hyperideals of a multiplicativeF^(m,n)-hyperring.

Proof. Suppose that A is the set of F-hyperideals of a multiplicative F^(m,n)-hyperring (R, f, g) and B is the set of regular relations for which the equivalence class of o is an F-hyperideal. We define ψ : A → B by ψ (I) = ρ_I. Clearly, ψ is well-defined. Also, by using Lemma 3.3 (1), we have ρ_I [o] = I, for all I ∈ A. Let I, J ∈ A and ψ (I) = ψ (J). Then, ρ_I = ρ_J and so we have I = ρ_I [o] = ρ_J [o] = J. Therefore, ψ is one to one. It suffices to show that ψ is onto. Assume that ρ is a regular relation for which ρ [o] is an F-hyperideal. Set I = ρ [o]. We show that ρ_I = ρ. Let x ρ_I y. By Lemma 3.3 (1), there exists z ∈ I such that supp (f (x, y^-1, o .)) = {z}. From zρ o it follows that supp (f (x, y^-1, o .)) ρ o. Since y ρ y and ρ is regular we have supp (f (f (x, y^-1, o .) , y, o .)) ρy. But we have $\begin{matrix} supp (f (f (x, y^{- 1}, o .), y, o .)) \\ = supp (f (x, f (y^{- 1}, y, o .), o .)) \\ = supp (f (x, o .)) = {x} \end{matrix}$ which implies that x ρ y. Conversely, assume that x ρ y. From regularity of ρ and that x^-1 ρ x^-1 we conclude that supp (f (x, x^-1, o .)) ρ supp (f (y, x^-1, o .)) which means that supp (f (y, x^-1, o .)) ρ o. Hence, supp (f (y, x^-1, o .)) ⊆ I and by Lemma 3.6 we have y ρ_I x. □

Theorem 3.10. (1) Let (R, f′, g′) be a multiplicative (m, n)-hyperring. Then, (R, f, g) is a multiplicativeF^(m,n)-hyperring, where $\begin{matrix} f (x_{1}^{m}) = χ_{f^{'} (x_{1}^{m})}, for all x_{1}^{m} \in R, \\ g (x_{1}^{n}) = χ_{g^{'} (x_{1}^{n})}, for all x_{1}^{n} \in R . \end{matrix}$

(R, f, g) is called the multiplicativeF^(m,n)-hyperring induced from (R, f′, g′).

(2) Let (R, f, g) be a multiplicativeF^(m,n)-hyperring. Then, (R, f′, g′) is a multiplicative (m, n)-hyperring, where $\begin{matrix} f^{'} (x_{1}^{m}) = supp (f (x_{1}^{m})), for all x_{1}^{m} \in R, \\ g^{'} (x_{1}^{n}) = supp (g (x_{1}^{n})), for all x_{1}^{n} \in R . \end{matrix}$

(R, f′, g′) is called the multiplicative (m, n)-hyperring extracted from (R, f, g).

Proof. (1) Since $supp (f (x_{1}^{m})) = f^{'} (x_{1}^{m})$ and (R, f′) is a monadic m-polygroup, it follows that (R, f) is a monadic F^m-polygroup. Similarly, from $supp (g (y_{1}^{n})) = g^{'} (y_{1}^{n})$ and that (R, g′) is an n-semi-group with identity element e we conclude that (R, g) is an Fⁿ-semigroup with F-identity element e. Let i ∈ {1, …, n} and $a_{1}^{i - 1}, a_{i + 1}^{n}, x_{1}^{m} \in R$ be arbitrary elements. Then, from $\begin{matrix} g (a_{1}^{i - 1}, f (x_{1}^{m}), a_{i + 1}^{n}) & = & ⋃_{t \in supp (f (x_{1}^{m}))} g (a_{1}^{i - 1}, t, a_{i + 1}^{n}) \\ = & χ_{g^{'} (a_{1}^{i - 1}, f^{'} (x_{1}^{m}), a_{i + 1}^{n})} . \end{matrix}$ and that (R, f′, g′) is distributive, we obtain that (R, f, g) is distributive. For each x, y ∈ R we have $\begin{matrix} supp (g (x, y^{- 1}, e .)) = g^{'} (x, y^{- 1}, e .) \\ = g^{'} (x^{- 1}, y, e .) = supp (g (x^{- 1}, y, e .)) . \end{matrix}$

In a similar manner we can show that $supp (g (x, y^{- 1}, e .)) = (supp (g (x, y, e .)))^{- 1} .$ Finally, for each $x_{2}^{n} \in R$ we have $supp (g (o, x_{2}^{n})) = g^{'} (o, x_{2}^{n}) = {o}$ . Similarly, we have $supp (g (x_{2}, o, x_{3}^{n})) = \dots = supp (g (x_{2}^{n}, o)) = {o} .$

(2) An argument similar to that in (1) establishes (2).□

Let I be an F-hyperideal of a multiplicative F^(m,n)-hyperring (R, f, g). Since ρ_I is a regular relation, by Corollary 3.2, (R/ρ_I, f|_{ρ
_I}, g|_{ρ
_I}) is a multiplicative (m, n)-hyperring. The multiplicative F^(m,n)-hyperring induced from (R/ρ_I, f|_{ρ
_I}, g|_{ρ
_I}) denoted by R/I is called the quotient multiplicative F^(m,n)-hyperring.

Definition 3.11. Let (R₁, f₁, g₁) and (R₂, f₂, g₂) be two multiplicative F^(m,n) -hyperrings. A homomorphism from R₁ to R₂ is a mapping φ : R₁ → R₂ such that

φ (o) = o,

$φ (supp (f_{1} (x_{1}^{m})))$ = supp (f₂ (φ (x₁) , …, φ (x_m))),

$φ (supp (g_{1} (y_{1}^{n})))$ ⊆supp (g₂ (φ (y₁) , …, φ (y_n))),

for all

x_{1}^{m}, y_{1}^{n} \in R_{1}

, where

φ_{a_{i}}^{a_{j}}

denotes the sequence φ (a_i) , …, φ (a_j). If in (3) we have equality instead of inclusion, then we say that φ is a good homomorphism.

An injective homomorphism is called a monomorphism and an onto homomorphism is called an epimorphism. An injective and onto homomorphism is called an isomorphism. We say that R₁ is isomorphic to R₂, denoted by R₁ ≅ R₂, if there exists an isomorphism from R₁ to R₂.

Lemma 3.12.Suppose $R_{1} = (R_{1}, f_{1}, g_{1})$ and $R_{2} = (R_{2}, f_{2}, g_{2})$ are two multiplicativeF^(m,n)-hyperrings andφ : R₁ → R₂is a homomorphism. Then, the following assertions hold:

φ (x^-1) = (φ (x)) ^-1, for allx ∈ R₁.

φis injective if and only if ker φ = {o}, where ker φ = {x ∈ R₁ | φ (x) = o}.

ker φis anF-hyperideal of $R_{1}$ .

Ifφis a good homomorphism andφ (e) = e, then Im φis anF^(m,n)-subhyperring of $R_{2}$ .

ρ_I = {(x, y) ∈ R₁ × R₁ | φ (x) = φ (y)}, whereI = ker φ.

Proof. (1) Let x ∈ R₁ be an arbitrary element. Since φ is a homomorphism, we have $\begin{matrix} {o} = φ ({o}) = φ (supp (f_{1} (x^{- 1}, x, o .)) \\ = supp (f_{2} (φ (x^{- 1}), φ (x), o .)) . \end{matrix}$

Now, since (R₁, f₁) is an F^m-polygroup and o ∈ supp (f₂ (φ (x^-1) , φ (x) , o .)) we have $φ (x^{- 1}) \in supp (f_{2} ((φ (x))^{- 1}, o .)) = {(φ (x))^{- 1}} .$ .

(2) Let φ is injective and x ∈ ker φ. Then, we have φ (x) = o = φ (o) which implies that x = o. Conversely, assume that ker φ = {o} and x, y are arbitrary elements of R₁ such that φ (x) = φ (y). Then, we have $\begin{matrix} φ (supp (f_{1} (x, y^{- 1}, o .)) = supp (f_{2} (φ (x), (φ (y))^{- 1}, o .)) \\ = supp (f_{2} (φ (x), (φ (x))^{- 1}, o .)) = {o} . \end{matrix}$

Therefore, we have supp (f₁ (x, y^-1, o .)) ⊆ ker φ = {o}. Since (R₁, f) is a monadic F^m-polygroup we have x ∈ supp (f₁ (y, o .)) = {y}. Thus, x = y. Consequently, φ is injective.

(3) Let x, y ∈ ker φ and r ∈ R₁ be arbitrary elements. Since $\begin{matrix} φ (supp (f_{1} (x, y^{- 1}, o .)) \\ = supp (f_{2} (φ (x), (φ (y))^{- 1}, o .)) \\ = supp (f_{2} (o .)) = {o}, \end{matrix}$ we have supp (f₁ (x, y^-1, o .)) ⊆ ker φ. Also, we have $\begin{matrix} φ (supp (g_{1} (x, y, e .)) \\ \subseteq supp (g_{2} (φ (x), φ (y), φ (e), \dots, φ (e))) \\ = supp (g_{2} (o, o, φ (e), \dots, φ (e))) = {o} . \end{matrix}$

Thus, supp (g₁ (x, y, e .)) ⊆ ker φ. Finally, we have $\begin{matrix} φ (supp (g_{1} (x, r, e .))) \\ \subseteq supp (g_{2} (φ (x), φ (r), φ (e), \dots, φ (e))) = {o}, \end{matrix}$ which implies that supp (g (x, r, e .)) ⊆ ker φ. In a similar way we can show that supp (g (r, x, e .)) ⊆ ker φ.

(4) Let φ (x) and φ (y) be arbitrary elements of Im φ. Then, we have $\begin{matrix} supp (f_{2} (φ (x), φ (y^{- 1}), o .)) \\ = φ (supp (f_{1} (x, y^{- 1}, o .))) \subseteq Im φ, \\ supp (g_{2} (φ (x), φ (y), e .)) \\ = φ (supp (g_{1} (x, y, e .))) \subseteq Im φ . \end{matrix}$

(5) Assume that (x, y) ∈ ρ_I is an arbitrary element. By Lemma 3.6 we have supp (f₁ (x, y^-1, o .)) ⊆ I.This implies that φ (supp (f₁ (x, y^-1, o .)) ⊆ φ (I) = {o}. Since φ is a homomorphism, we have $supp (f_{2} (φ (x), (φ (y))^{- 1}, o .)) = {o}$ and since (R₂, f₂) is an F^m-polygroup we haveφ (x) ∈ supp (f₂ (φ (y) , o .)) = {φ (y)}. Thus, φ (x) = φ (y). Conversely, suppose that φ (x) = φ (y), for somex, y ∈ R₁. So, we have supp (f₂ (φ (x) , (φ (y)) ^-1, o .)) = supp (f₂ (φ (x) , (φ (x)) ^-1, o .)) = {o} which implies that φ (supp (f₁ (x, y^-1, o .)) = {o}. Thus, $supp (f_{1} (x, y^{- 1}, o .)) \subseteq \ker φ = I$ and by Lemma 3.6 we have (x, y) ∈ ρ_I. This completes the proof. □

Let I be an F-hyperideal of a multiplicative F^(m,n)-hyperring (R, f, g). As is well known, the natural map π : R → R/I by π (x) = ρ_I [x] is an epimorphism. π is called canonical homomorphism.

Lemma 3.13.Let (R₁, f₁, g₁) and (R₂, f₂, g₂) be two multiplicativeF^(m,n)-hyperrings and letφ : R₁ → R₂be a homomorphism. Then, there exists a monomorphismψ : R₁/I → R₂such thatψ ∘ π = φ, where I = ker φ.

Proof. We define ψ (ρ_I [x]) = φ (x). By Lemma 3.12 (5), ψ is well-defined and one to one. We show that ψ is a homomorphism. We have ψ (ρ_I [o]) = φ (o) = {o}, where ρ_I [o] is an F-identity element of $(R / I, f_{1}^{'} |_{ρ_{I}})$ . Let ρ_I [x₁] , …, ρ_I [x_m] be arbitrary elements of R/I. Then, $\begin{matrix} ψ (supp (f_{1}^{'} |_{ρ_{I}} (ρ_{I} [x_{1}], \dots, ρ_{I} [x_{m}]))) \\ = ψ (f_{1} |_{ρ_{I}} (ρ_{I} [x_{1}], \dots, ρ_{I} [x_{m}])) \\ = ψ (ρ_{I} [supp (f_{1} (x_{1}^{m}))]) \\ = φ (supp (f_{1} (x_{1}^{m})) \\ = supp (f_{2} (φ (x_{1}), \dots, φ (x_{m}))) \\ = supp (f_{2} (ψ (ρ_{I} [x_{1}]), \dots, ψ (ρ_{I} [x_{m}]))) . \end{matrix}$

For each ρ_I [y₁] , …, ρ_I [y_n] ∈ R/I we have $\begin{matrix} ψ (supp (g_{1}^{'} |_{ρ_{I}} (ρ_{I} [y_{1}], \dots, ρ_{I} [y_{n}]))) \\ = ψ (g_{1} |_{ρ_{I}} (ρ_{I} [y_{1}], \dots, ρ_{I} [y_{n}])) \\ = {ψ (ρ_{I} [z]) | z \in supp (g_{1} (y_{1}^{n}))} \\ = {φ (z) | z \in supp (g_{1} (y_{1}^{n}))} \\ = φ (supp (g_{1} (y_{1}^{n}))) \\ \subseteq supp (g_{2} (φ (y_{1}), \dots, φ (y_{n}))) \\ = supp (g_{2} (ψ (ρ_{I} [y_{1}]), \dots, ψ (ρ_{I} [y_{n}]))) . \end{matrix}$

Finally, for each x ∈ R₁ we have ψ ∘ π (x) = ψ (ρ_I [x]) = φ (x). □

Notice that in the above lemma, ψ is a good monomorphism if φ is a good homomorphism.

Theorem 3.14.LetIandJbeF-hyperideals of a multiplicativeF^(m,n)-hyperring (R, f, g) such thatI ⊆ J. Then, there exists anF-hyperidealLofR/Isuch that (R/I)/Lis good isomorphic toR/J.

Proof. We define the map φ : R/I → R/J by φ (ρ_I [x]) = ρ_J [x]. By Corollary 3.7, φ is well-defined. For ρ_I [x₁] , …, ρ_I [x_m] ∈ R/I we have $\begin{matrix} φ (supp (f^{'} |_{ρ_{I}} (ρ_{I} [x_{1}], \dots, ρ_{I} [x_{m}]))) \\ = φ (f |_{ρ_{I}} (ρ_{I} [x_{1}], \dots, ρ_{I} [x_{m}])) \\ = φ (ρ_{I} [supp (f (x_{1}^{m}))]) \\ = ρ_{J} [supp (f (x_{1}^{m}))] \\ = f |_{ρ_{J}} (ρ_{J} [x_{1}], \dots, ρ_{J} [x_{m}]) \\ = supp (f^{'} |_{ρ_{J}} (φ (ρ_{I} [x_{1}]), \dots, φ (ρ_{I} [x_{m}]))) . \end{matrix}$

Also, for ρ_I [y₁] , …, ρ_I [y_n] ∈ R/I we have $\begin{matrix} φ (supp (g^{'} |_{ρ_{I}} (ρ_{I} [y_{1}], \dots, ρ_{I} [y_{n}]))) \\ = φ (g |_{ρ_{I}} (ρ_{I} [y_{1}], \dots, ρ_{I} [y_{n}])) \\ = {φ (ρ_{I} [z]) | z \in supp (g (y_{1}^{n}))} \\ = {ρ_{J} [z] | z \in supp (g (y_{1}^{n}))} \\ = g |_{ρ_{J}} (ρ_{J} [y_{1}], \dots, ρ_{J} [y_{n}]) \\ = supp (g^{'} |_{ρ_{J}} (φ (ρ_{I} [y_{1}]), \dots, φ (ρ_{I} [y_{n}]))) . \end{matrix}$

Therefore, φ is a good homomorphism. Now, if L = ker φ, then by Lemma 3.13, there exists a good monomorphism ψ : (R/I)/J → R/J such that ψ ∘ π = φ, and so ψ is a good isomorphism. □

Lemma 3.15Let (R, f, g), (R₁, f₁, g₁) and (R₂, f₂, g₂) be three multiplicativeF^(m,n)-hyperrings and letφ₁ : R → R₁ and φ₂ : R → R₂be epimorphisms such that $φ_{1}^{- 1} \circ φ_{1} \subseteq φ_{2}^{- 1} \circ φ_{2}$ . Then, there exists a unique epimorphismψ : R₁ → R₂such thatψ ∘ φ₁ = φ₂.

Proof. We define ψ : R₁ → R₂ by ψ (z) = φ₂ (x), where $x \in φ_{1}^{- 1} (z)$ . For each y ∈ φ^-1 (z) we have $(x, y) \in φ_{1}^{- 1} \circ φ_{1} \subseteq φ_{2}^{- 1} \circ φ_{2}$ , and so φ₂ (x) = φ₂ (y). This proves that ψ is well-defined. We prove that ψ is an epimorphism. Clearly, we have ψ (o) = o. Let $x_{1}^{m} \in R_{1}$ be arbitrary elements. Since φ₁ is onto, there exist $y_{1}^{m} \in R$ such that φ₁ (y_i) = x_i, (1 ≤ i ≤ m). Set $supp (f (y_{1}^{m})) = {t}$ . Since φ₁ is a homomorphism, $\begin{matrix} φ_{1} (t) \in φ_{1} (supp (f (y_{1}^{m}))) \\ = supp (f (φ_{1} (y_{1}), \dots, φ_{1} (y_{m}))) = supp (f (x_{1}^{m})), \end{matrix}$ which implies that $supp (f (x_{1}^{m})) = {φ_{1} (t)}$ . Thus, $\begin{matrix} supp (f_{2} (ψ (x_{1}), \dots, ψ (x_{m}))) \\ = supp (f_{2} (φ_{2} (y_{1}), \dots, φ_{2} (y_{m}))) \\ = φ_{2} (supp (f (y_{1}^{m}))) = {φ_{2} (t)} \\ = {ψ (φ_{1} (t))} = ψ (supp (f_{1} (x_{1}^{m}))) . \end{matrix}$

Also, for every $x_{1}^{n} \in R_{1}$ , there exist $y_{1}^{n} \in R$ such that φ₁ (y_i) = x_i, (1 ≤ i ≤ n) and we have $\begin{matrix} supp (g_{2} (ψ (x_{1}), \dots, ψ (x_{n}))) \\ = supp (g_{2} (φ_{2} (y_{1}), \dots, φ_{2} (y_{n}))) \\ \subseteq φ_{2} (supp (g (y_{1}^{n}))) = ψ (φ_{1} (supp (g (y_{1}^{n})))) \\ = ψ (supp (g_{1} (x_{1}^{n}))) . \end{matrix}$

Clearly, ψ is onto and ψ ∘ φ₁ = φ₂. The uniqueness is evident. □

Theorem 3.16.IfIandJareF-hyperideals of a multiplicativeF^(m,n)-hyperring (R, f, g) such thatI ⊆ J, then there exists an epimorphismR/I → R/J.

Proof. Let π₁ : R → R/I and π₂ : R → R/J be canonical homomorphisms. We show that $π_{1}^{- 1} \circ π_{1} \subseteq π_{2}^{- 1} \circ π_{2}$ . Let $(x, y) \in π_{1}^{- 1} \circ π_{1}$ be an arbitrary element. Then, we have π₁ (x) = π₁ (y) which implies that ρ_I [x] = ρ_I [y]. Since I ⊆ J, by Corollary 3.7 we have ρ_J [x] = ρ_J [y] and therefore $(x, y) \in π_{2}^{- 1} \circ π_{2}$ . Now, by Lemma 3.15 the proof is completed.□

We present now the first isomorphism theorem for multiplicative F^(m,n)-hyperrings.

Theorem 3.17.(First Isomorphism Theorem). Let (R₁, f₁, g₁) and (R₂, f₂, g₂) be two multiplicativeF^(m,n)-hyperrings andφ : R₁ → R₂be a good homomorphism such thatφ (e) = e. Then,R₁/I ≅ Im φ, whereI = ker φ.

Proof. By Lemma 3.12 (4), Im φ is an F^(m,n)-subhyper ring of (R₂, f₂, g₂). We define the map ψ : R₁/I → Im φ by ψ (ρ_I [x]) = φ (x). By the following argument ψ is well-defined. $\begin{matrix} ρ_{I} [x] = ρ_{I} [y] \\ \Leftrightarrow supp (f_{1} (I, x, o .)) = supp (f_{1} (I, y, o .)) \\ \Rightarrow φ (supp (f_{1} (I, x, o .))) = φ (supp (f_{1} (I, y, o .))) \\ \Leftrightarrow supp (f_{2} (φ (I), φ (x), o .)) = supp (f_{2} (φ (I), φ (y), o .)) \\ \Leftrightarrow supp (f_{2} (φ (x), o .)) = supp (f_{2} (φ (y), o .)) \\ \Leftrightarrow φ (x) = φ (y) . \end{matrix}$ Obviously, ψ (ρ_I [o]) = φ (o) = o and for every ρ_I [x₁] , …, ρ_I [x_m] ∈ R₁/I, we have $\begin{matrix} ψ (supp (f_{1}^{'} |_{ρ_{I}} (ρ_{I} [x_{1}], \dots, ρ_{I} [x_{m}]))) \\ = ψ (f_{1} |_{ρ_{I}} (ρ_{I} [x_{1}], \dots, ρ_{I} [x_{m}])) = ψ (ρ_{I} [supp (f_{1} (x_{1}^{m}))]) \\ = φ (supp (f_{1} (x_{1}^{m}))) = supp (f_{2} (φ (x_{1}), \dots, φ (x_{m}))) \\ = supp (f_{2} (ψ (ρ_{I} [x_{1}]), \dots, ψ (ρ_{I} [x_{m}]))), \end{matrix}$

Also, for every ρ_I [y₁] , …, ρ_I [y_n] ∈ R₁/I, we have $\begin{matrix} ψ (supp (g_{1}^{'} |_{ρ_{I}} (ρ_{I} [y_{1}], \dots, ρ_{I} [y_{n}]))) \\ = ψ (g_{1} |_{ρ_{I}} (ρ_{I} [y_{1}], \dots, ρ_{I} [y_{n}])) \\ = {ψ (ρ_{I} [z]) | z \in supp (g_{1} (y_{1}^{n}))} \\ = {φ (z) | z \in supp (g_{1} (y_{1}^{n}))} \\ = φ (supp (g_{1} (y_{1}^{n}))) \\ = supp (g_{2} (φ (y_{1}), \dots, φ (y_{n}))) \\ = supp (g_{2} (ψ (ρ_{I} [y_{1}]), \dots, ψ (ρ_{I} [y_{n}]))) . \end{matrix}$

Therefore, ψ is a good homomorphism. Evidently, ψ is onto. Finally, we have $\begin{matrix} \ker φ & = {ρ_{I} [x] \in R_{1} / I | ψ (ρ_{I} [x]) = o} \\ = {ρ_{I} [x] \in R_{1} / I | φ (x) = o} \\ = {ρ_{I} [x] \in R_{1} / I | x \in I} = {ρ_{I} [o]} . \end{matrix}$

Therefore, ψ is an isomorphism and the desired result follows. □

We are now in a position to state and prove the second and third isomorphism theorems in multiplicative F^(m,n)-hyperrings.

Theorem 3.18.(Second Isomorphism Theorem). If $I_{1}^{m}$ areF-hyperideals of a multiplicativeF^(m,n)-hyperring (R, f, g), then for eachj ∈ {1, …, m} we have $\begin{matrix} supp (f (I_{1}^{j - 1}, o, I_{j + 1}^{m})) / (supp (f (I_{1}^{j - 1}, o, I_{j + 1}^{m})) \cap I_{j}) \\ ≅ supp (f (I_{1}^{m})) / I_{j} . \end{matrix}$

Proof. Let j ∈ {1, …, m} be an arbitrary element. By Lemma 2.7, $supp (f (I_{1}^{m})) / I_{j}$ and $supp (f (I_{1}^{j - 1}, o, I_{j + 1}^{m})) / (supp (f (I_{1}^{j - 1}, o, I_{j + 1}^{m})) \cap I_{j})$ are defined. We consider the map $ψ : supp (f (I_{1}^{j - 1}, o, I_{j + 1}^{m})) \to supp (f (I_{1}^{m})) / I_{j}$ by ψ (x) = ρ_{I
_j} [x]. For each $x_{1}^{m} \in supp (f (I_{1}^{j - 1}, o, I_{j + 1}^{m}))$ we have $\begin{matrix} ψ (supp (f (x_{1}^{m}))) & = & ρ_{I_{j}} [supp (f (x_{1}^{m}))] \\ = & f |_{ρ_{I_{j}}} (ρ_{I_{j}} [x_{1}], \dots, ρ_{I_{j}} [x_{m}]) \\ = & supp (f^{'} |_{ρ_{I_{j}}} (ψ (x_{1}), \dots, ψ (x_{m}))) . \end{matrix}$

Also, for each $y_{1}^{n} \in supp (f (I_{1}^{j - 1}, o, I_{j + 1}^{m}))$ we have $\begin{matrix} ψ (supp (g (y_{1}^{n}))) = ρ_{I_{j}} [supp (g (y_{1}^{n}))] \\ = g |_{ρ_{I_{j}}} (ρ_{I_{j}} [y_{1}], \dots, ρ_{I_{j}} [y_{n}]) \\ = supp (g^{'} |_{ρ_{I_{j}}} (ψ (y_{1}), \dots, ψ (y_{n}))) . \end{matrix}$

Therefore, ψ is a good homomorphism. Let $ρ_{I_{j}} [a] \in supp (f (I_{1}^{m})) / I_{j}$ be an arbitrary element. Then, there exist a_i ∈ I_i, 1 ≤ i ≤ m, such that $a \in supp (f (a_{1}^{m}))$ . By using Lemmas 2.6 (1) and 3.3 (2), for $x \in supp (f (a_{1}^{j - 1}, o, a_{j + 1}^{m})) \subseteq supp (f (I_{1}^{j - 1}, o, I_{j + 1}^{m}))$ we have $\begin{matrix} ψ (x) = ρ_{I_{j}} [x] = supp (f (I_{j}, x, o .)) \\ = supp (f (I_{j}, f (a_{1}^{j - 1}, o, a_{j + 1}^{m}), o)) \\ = supp (f (a_{1}^{j - 1}, f (I_{j}, o .), a_{j + 1}^{m})) \\ = supp (f (a_{1}^{j - 1}, f (I_{j}, a_{j}, o .), a_{j + 1}^{m})) \\ = supp (f (I_{j}, f (a_{1}^{m}), o .)) \\ = ρ_{I_{j}} [supp (f (a_{1}^{m}))] = ρ_{I_{j}} [a] . \end{matrix}$

Therefore, ψ is onto. We have $\begin{matrix} \ker ψ = {x \in supp (f (I_{1}^{j - 1}, o, I_{j + 1}^{m})) | ρ_{I_{j}} [x] = ρ_{I_{j}} [o]} \\ = {x \in supp (f (I_{1}^{j - 1}, o, I_{j + 1}^{m})) | x \in I_{j}} \\ = supp (f (I_{1}^{j - 1}, o, I_{j + 1}^{m})) \cap I_{j} . \end{matrix}$ Evidently, ρ_{I
_j} [e] is an F-identity element of $(supp (f (I_{1}^{m})) / I_{j}, g^{'} |_{ρ_{I_{j}}})$ and ψ (e) = ρ_{I
_j} [e]. So, by the first isomorphism theorem the desired result holds. □

Theorem 3.19.(Third Isomorphism Theorem). IfIandJareF-hyperideals of a multiplicativeF^(m,n)-hyperring (R, f, g) such thatI ⊆ J, thenJ/Iis anF-hyperideal ofR/Iand (R/I)/(J/I) ≅ R/J.

Proof. First, we show that J/I is an F-hyperideal of R/I. Let x, y ∈ J and r ∈ R be arbitrary elements. Since J is an F-hyperideal of (R, f, g), we have supp (f (x, y^-1, o .)) ⊆ J and supp (g (x, r, e .)) ∪ supp (g (r, x, e .)) ⊆ J and so we have $\begin{matrix} supp (f^{'} |_{ρ_{I}} (ρ_{I} [x], ρ_{I} [y^{- 1}], ρ_{I} [o], \dots, ρ_{I} [o])) \\ = f |_{ρ_{I}} (ρ_{I} [x], ρ_{I} [y^{- 1}], ρ_{I} [o], \dots, ρ_{I} [o]) \\ = ρ_{I} [supp (f (x, y^{- 1}, o .))] \subseteq J / I . \end{matrix}$

Also, we have $\begin{matrix} supp (g^{'} |_{ρ_{I}} (ρ_{I} [x], ρ_{I} [r], ρ_{I} [e], \dots, ρ_{I} [e])) \\ = g |_{ρ_{I}} (ρ_{I} [x], ρ_{I} [r], ρ_{I} [e], \dots, ρ_{I} [e]) \\ = ρ_{I} [supp (g (x, r, e .))] \subseteq J / I . \end{matrix}$

In a similar manner we can show that $supp (g^{'} |_{ρ_{I}} (ρ_{I} [x], ρ_{I} [r], ρ_{I} [e], \dots, ρ_{I} [e])) \subseteq J / I .$ So far we have proved that J/I is an F-hyperideal of R/I. It is not difficult to see that ψ : R/I → R/J defined by ψ (ρ_I [x]) = ρ_J [x] is an onto good homomorphism and ker φ = J/I. Therefore, by the First Isomorphism Theorem the desired result follows easily. □

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Multiplicative F ( m,n ) -Hyperrings

Abstract

1 1. Introduction and preliminaries

2 2. Multiplicative F(m,n)-hyperrings

3 3. Quotient multiplicative F(m,n)-hyperrings

References

2 2. Multiplicative F^(m,n)-hyperrings

3 3. Quotient multiplicative F^(m,n)-hyperrings