On Weakly S-Prime Hyperideals of Multiplicative Hyperrings

1. Introduction

Hyperstructure theory was first defined by F.Marty in 1934 at the 8th Scandinavian Mathematical Congress. According to Marty’s definition of hyperstructures, the image of two elements under a defined operation is a non-empty set [1]. Hypergroups are generations of groups, and it has been stated that not every hypergroup is a group. Furthermore, hypergroup theory has been used in many areas of pure and applied mathematics. The concepts of hyperoperation, hypergroupoid, semihypergroup and quasi hypergroup and given and the conditions for being a hypergroup are specified. Novak defined the concept of a hyperring and gave examples of a subhyperring and a hyperideal of a hyperring [2]. Davvaz specified the conditions required for hyperrings to be hyperring homomorphisms [3]. Krasner defined by Krasner hyperrings [4]. Rota, in 1982, defined multiplicative hyperrings, which will be used in our study, and included their properties [5].

Corsini defined the concepts of hyperfield and hypermodule [6]. The prime hyperideals of multiplicative hyperrings were studied extensively by Dasgupta in 2012 [7]. Ay and Yesilot provided a definition of weakly prime hyperideals on the multiplicative hyperring and presented significant results in 2022, finding relationships between prime hyperideals and weakly prime hyperideals [8]. Ghiasvand and Farzalipour introduced the S− prime hyperideals of multiplicative hyperrings, which are a generalization of prime hyperideals, and gave their basic properties [9].

Our research includes four sections. The first section contains a literature review and the aim of this study; the aim is to apply the definitions and general properties of weakly prime ideals used in classical algebra to multiplicative hyperideals and to examine the relationship between prime, prime, and weakly prime hyperideals. The second section contains the basic definitions that will be used in our study. In the third section, the main part of our study, we defined the weakly S− prime hyperideal which is a proper hyperideal $\Omega$ is called to be weakly S− prime hyperideal then there exists an $s\in S$ with if whenever $\left\{0\right\}\ne \varrho \diamond \sigma$ then for all $\varrho ,\sigma \in \Gamma$ where $\Gamma$ be multiplicative hyperring $S\subseteq \Gamma$ be multiplicative closed subset of $\Gamma$ and $\Omega$ be proper hyperideal of $\Gamma$ such that $\Omega \cap S=\varnothing$ and supported it with example. We proved that weakly prime hyperideals and S− prime hyperideals are weakly S− prime hyperideals, and gave examples showing that the converse is not always true. Using the properties in Lemma $3.8$ and Lemma $3.9$, respectively, where we showed that the intersection of two weakly S− prime hyperideals and the hyperproduct of two weakly S− prime hyperideals are weakly S−prime hyperideals, we proved in Proposition $3.11$ that if $\Omega$ is a weakly S−prime hyperideal and ${\Omega }^2\ne \left\{0\right\}$, then $\Omega$ must be an S− prime hyperideal.We also gave an example showing that, unlike in classical algebra, the converse of Proposition $3.11$ is not always true and it was proven by Ghisvand and Farzalipour that the necessary condition for $(\Omega :s)$ to be a prime hyperideal is that $\Omega$ is a S− prime hyperideal of $\Gamma$[9]. Using the given proof and Proposition $3.11$, we obtained Corollary $3.13$, i.e., $(\Omega :s)$ is a prime ⇔I is a S− prime hyperideal ⇔I is a weakly S− prime for $s\in S$. Furthermore, in Proposition $3.14\left(ii\right)$, we showed that the necessary and sufficient condition for $\Omega$ to be a weakly prime hyperideal, where S consist of units of $\Gamma$ which is a multiplicative closed subset of the $\Gamma$, is that $\Gamma$ is a weakly S− prime hyperideal, and we generalized this in Proposition $3.15$. The properties of weakly S− prime hyperideals under good homomorphism, direct product hyperring, and quotient hyperring localization were examined, and different characterizations were given. Finally, in the fourth section, we created a figure showing the relationship between weakly S− prime hyperideals and existing hyperideals in the literature and presented the results.

2. Preliminaries

This section will include the basic definitions, theorems, and examples that we will use in our study. We provided examples

Suppose that ${\rm Y}$ is a nonempty subset and set ″$\diamond$″ : ${\rm Y}\times {\rm Y}\to P^{*}\left({\rm Y}\right)$ where $P^{*}\left({\rm Y}\right)$ is the set of all nonempty subsets of ${\rm Y}$. The mapping ″$\diamond$″ is called a hyperoperation on ${\rm Y}$ and $({\rm Y},\diamond )$ is said to be a hypergroupoid, where for $\mu \in {\rm Y}$ and $\Omega ,\Delta \in P^{*}\left({\rm Y}\right)$ then we have $\Omega \diamond \Lambda ={\bigcup }_{\varrho \in \Omega ,\sigma \in \Lambda }\varrho \diamond \sigma$ and $\Omega \diamond \mu =\Omega \diamond \left\{\mu \right\}$. A hypergroupoid $({\rm Y},\diamond )$ is called a semihypergroup if for all $\mu ,\varrho ,\sigma \in {\rm Y}$, then we get $(\varrho \diamond \sigma )\diamond \mu =\varrho \diamond (\sigma \diamond \mu )$. We say that a semihpergroup $({\rm Y},\diamond$) is a hypergroup if for all $\varrho \in {\rm Y}$; $\varrho \diamond {\rm Y}={\rm Y}={\rm Y}\diamond \varrho$ [3].

Definition 2.1. 

A triple $(\Gamma ,{+}^{\prime },\diamond )$ is called a multiplicative hyperring if

(i)$(\Gamma ,{+}^{\prime })$is an abelian group

(ii)$(\Gamma ,\diamond )$ is a semihypergroup

(iii) For all $\varrho ,\sigma ,\mu \in \Gamma$, then we have $\varrho \diamond \left(\sigma {+}^{\prime }\mu \right)\subseteq \varrho \diamond \sigma {+}^{\prime }\varrho \diamond \mu$

(iv) For all $\varrho ,\sigma \in \Gamma$; we have $\varrho \diamond (-\sigma )=-(\varrho \diamond \sigma )=(-\varrho )\diamond \sigma$

If in (iii) the equality holds, then we say that multiplicative hyperring is strongly distributive [5].

Definition 2.2. 

Let $(\Gamma ,{+}^{\prime },\diamond )$ be a multiplicative hyperring. If for all $\varrho ,\sigma \in \Gamma$ $\varrho \diamond \sigma =\sigma \diamond \varrho$, then we say that $(\Gamma ,{+}^{\prime },\diamond )$ is a commutative multiplicative hyperring, and if for all $\varrho \in \Gamma$ $\Gamma$ is said to be a unit, then there exists $ϵ\in \Gamma$ such that if whenever $ϵ\diamond \varrho =\varrho =\varrho \diamond ϵ$ [10].

Definition 2.3. 

Let $(\Gamma ,{+}^{\prime },\diamond )$ be a ring then corresponding to every subset $A\in P^{*}(\Gamma )=P(\Gamma )\setminus \left\{\varnothing \right\}(\left|A\right|\ge 2)$, there exists a multiplicative hyperring with absorbing zero $({\Gamma }_A,{+}^{\prime },\diamond )$ where ${\Gamma }_A=\Gamma$ and for any $\varrho ,\sigma {\Gamma }_A$, $\varrho \diamond \sigma =\left\{\varrho \diamond \sigma :a\in A\right\}$ [7].

Definition 2.4. 

Suppose that $({\Gamma }_1,{+}^{\prime },\diamond )$ and $({\Gamma }_2,{+}^{\prime },\diamond )$ are two multiplicative hyperrings. A $({\Gamma }_1\times {\Gamma }_2,{+}^{\prime },\diamond )$ is called a direct multiplicative hyperring if for all $(\varrho ,\sigma ),(\mu ,\kappa )\in {\Gamma }_1\times {\Gamma }_2$ then we get $(\varrho ,\sigma ){+}^{\prime }(\mu ,\kappa )=(\varrho {+}^{\prime }\mu ,\sigma {+}^{\prime }\kappa )$ and $(\varrho ,\sigma )\diamond (\mu ,\kappa )=\left\{(\nu ,\omega )\in {\Gamma }_1\times {\Gamma }_2:\nu \in \varrho \diamond \mu ,\omega \in \sigma \diamond \kappa \right\}$ [10].

Definition 2.5. 

Suppose that $(\Gamma ,{+}^{\prime },\diamond )$ is a multiplicative hyperring and

$\varnothing \ne \Omega \subseteq R$ is a subset of $\Gamma$ . If $\Omega$ is a multiplicative hyperring with operations defined in $\Gamma$, then the hyperalgebraic structure $(\Omega ,{+}^{\prime },\diamond )$ is called a subhyperring of $(\Gamma ,{+}^{\prime },\diamond )$

Also if ;

(i)$\Omega -\Omega \subseteq \Omega$

(ii) For all $\varrho ,\sigma \in \Omega$ and for $\gamma \in \Gamma$ , $\varrho \diamond \gamma \cup \gamma \diamond \varrho \subseteq \Omega$ then $\Omega$ is called a hyperideal of $\Gamma$ [3].

Definition 2.6. 

Let $(\Gamma ,{+}^{\prime },\diamond )$ be a multiplicative hyperring ; $\Omega$ and $\Lambda$ be two hyperideals of $\Gamma$. In this case, the sets are defined as $\Omega {+}^{\prime }\Lambda =\left\{\varrho {+}^{\prime }\sigma |\varrho \in \Omega ,\sigma \in \Lambda \right\}$ and $\Omega \diamond \Lambda =\cup \left\{{\sum }_{i=1}^n{\varrho }_i\diamond {\sigma }_i|{\varrho }_i\in \Omega ,{\sigma }_i\in \Lambda \right\}$ are the hyperideals of $\Gamma$ [7].

In this article, we will denote the product of two hyperideals as $\Omega .\Lambda$

Definition 2.7. 

Suppose that $(\Gamma ,{+}^{\prime },\diamond )$ is a multiplicative hyperring of commutative with identity. If for some ${\varrho }_1\dots {\varrho }_n\in \Gamma$ and for any $\varrho \in \Omega$ there exits elements ${\varrho }_1\dots {\varrho }_n\in \Gamma$ such that $\varrho \in {\gamma }_1\diamond {\varrho }_1+\dots +{\gamma }_n\diamond {\varrho }_n$ then I is called a finitely generated hyperideal of $\Gamma$ [7].

Definition 2.8. 

Let $(\Gamma ,{+}^{\prime },\diamond )$ be a multiplicative hyperring . For any $\vartheta \in \Gamma$ if we write $〈\vartheta 〉=\left\{\phi .\vartheta |\phi \in \mathbb{Z}\right\}+$

$\left\{{\sum }_{i=1}^n{\varrho }_i+{\sum }_{j=1}^m{\sigma }_j+{\sum }_{k=1}^l{\mu }_k|\forall i,j,k;\exists {\gamma }_i,{\alpha }_j,{\varpi }_k\in \Gamma |{\varrho }_i\in {\gamma }_i\diamond \vartheta ,{\sigma }_j\in \vartheta \diamond {\alpha }_j,{\mu }_k\in {\epsilon }_k\diamond \vartheta \diamond {\varpi }_k\right\}$

then $〈\vartheta 〉$ is a principal hyperideal of $\Gamma$. If all hyperideals of $\Gamma$ are principal hyperideals, then $\Gamma$ is called a principal multiplicative hyperring. [7]

Definition 2.9. 

Assume that $(\Gamma ,{+}^{\prime },\diamond )$ is a multiplicative hypering, $\Omega$ and $\Lambda$ are hyperideals of $\Gamma$. The set $(\Omega :{}_{\Gamma }\Lambda )=\left\{\varrho \in \Gamma |\varrho \diamond \Lambda \subseteq \Omega \right\}$ is called colon hyperideal of $\Gamma$.[10]

Definition 2.10. 

A homomorphism (resp. good homomorphism) between two multiplicative hyperrings $(\Gamma ,{+}^{\prime },\diamond )$ and $(\Theta ,{+}^{\prime \prime },▿)$ is a map $\varphi$ from $\Gamma$ to $\Theta$, such that for all $\varrho ,\sigma \in \Gamma$

$\varphi \left(\varrho {+}^{\prime }\sigma \right)=\varphi \left(\varrho \right){+}^{\prime \prime }\varphi \left(\sigma \right)$ and $\varphi (\varrho \diamond \sigma )\subseteq \varphi \left(\varrho \right)▿\varphi \left(\sigma \right)$ $(resp.\varphi (\varrho \diamond \sigma )=\varphi (\varrho )▿\varphi (\sigma \left)\right)$.[3]

In this study, we will use a good homomorphism.

Definition 2.11. 

$\varphi$ is the inverse image of the kernel of $\varphi$; where $\varphi$ is good homomorphism from $\Gamma$ to $\Theta$ and $\varphi \left(\right(0\left)\right)\subseteq \left(\varphi \right)$. That is $\left(0\right)\subseteq Ker\left(\varphi \right)$.[3]

Definition 2.12. 

Suppose that $\Omega$ is a hyperideal of a multiplicative hyperring $\Gamma$ and $\Gamma /\Omega =\left\{\gamma {+}^{\prime }\Gamma |\gamma \in \Gamma \right\}$ . Define the operations ″$+$′″ and ″$*$″ on $\Gamma /\Omega$ by $(\varrho {+}^{\prime }\Omega ){+}^{\prime }(\sigma {+}^{\prime }\Omega )=\left(\varrho {+}^{\prime }\sigma \right){+}^{\prime }\Omega$ and $(\varrho {+}^{\prime }\Omega )*(\sigma {+}^{\prime }\Omega )=\cup \left\{\mu {+}^{\prime }\Omega |\mu \in \varrho \diamond \sigma \right\}$. Then $(\Gamma /\Omega ,{+}^{\prime },*)$ is called a quotient multiplicative hyperring.[3]

Definition 2.13. 

Let C be the class of all finite products of elements of a multiplicative hyperring $\Gamma$. i.e. $C=\left\{{\gamma }_1\diamond {\gamma }_{2\diamond \dots }\diamond {\gamma }_n|{\gamma }_i\in \Gamma ,n\in \mathbb{N}\right\}\subseteq P^{*}(\Gamma )$. A hyperideal $\Omega$ of $\Gamma$ is called a C− ideal of $\Gamma$ if for any $T\in C$ , $T\cap \Omega \ne \varnothing$ then $T\subseteq \Omega$.[7]

Definition 2.14. 

(i) Let $\Omega$ be a proper hyperideal of $\Gamma$ be called to be a maximal hyperideal of $\Gamma$ if for any hyperideal $\Lambda$ of $\Gamma$, $\Omega \subset \Lambda \subseteq \Gamma$ implies $\Omega =\Lambda$ or $\Lambda =\Gamma$

(ii) A proper hyperideal $\Theta$ of a multiplicative hyperring $\Gamma$ is called to be a prime hyperideal of $\Gamma$ if for any $\varrho ,\sigma \in \Gamma$ , $\varrho \diamond \sigma \subseteq T$ then $\varrho \in T$ or $\sigma \in T$.[7]

Definition 2.15. 

Let $(\Gamma ,{+}^{\prime },\diamond )$ be a multiplicative hyperring and $S\subseteq \Gamma$ be a subset of $\Gamma$. If for $\varrho ,\sigma \in S$, $\varrho \diamond \sigma \cap S\ne \varnothing$.

If $(\Gamma ,{+}^{\prime },\diamond )$ is a commutative with identitiy multiplicative hyperring, $S\subseteq R$ has the properties

(i)$1\in S$

(ii) For all $s_1,s_2\in S$ , $s_1\circ s_2\cap S\ne \varnothing$ [9]

Definition 2.16. 

Suppose that $\Omega$ is a hyperideal of multiplicative hyperring $\Gamma$. The intersection of all prime hyperideals containing the hyperideal $\Omega$ of $\Gamma$ is called the radical of $\Omega$ and is denoted by $Rad(\Omega )$. If $\Gamma$ does not have prime hyperideal of $\Gamma$ containing $\Omega$, it is denoted by $Rad(\Omega )=\Gamma$.

Furhermore for the hyperideal $D(\Gamma )=\left\{\gamma \in \Gamma |{\gamma }^n\subseteq \Omega ;n\in \mathbb{N}\right\}$; $D(\Omega )\subseteq Rad(\Omega )$ holds [7].

Definition 2.17. 

Let $\Gamma$ be a multiplicative hyperring, $S\subseteq \Gamma$ be a multiplicative closed subset of $\Gamma$ and $\Omega$ be a hyperideal of $\Gamma$ such that $\Omega \cap S=\varnothing$. If for $\varrho ,\sigma \in \Gamma$ and $\varrho \diamond \sigma \subseteq I$ then there exists an $s\in S$ such that $s\diamond \varrho \subseteq \Omega$ or $s\diamond \sigma \subseteq \Omega$. Then we say that $\Omega$ is S− prime hyperideal of $\Gamma$ [9].

Definition 2.18. 

Suppose that $\Gamma$ is a multiplicative hyperring . For $\varrho \in \Gamma$, if $\varrho \in \varrho \diamond \vartheta \circ \varrho$ then there exists an $\vartheta \in R$, we say that $\varrho \in \Gamma$ is regular element of $\Gamma$. If all elements of a hyperring $\Gamma$ are regular elements, then $\Gamma$ is called a regular multiplicative hyperring [11].

Definition 2.19. 

A proper hyperideal $\Omega$ of a commutative with identity multiplicative hyperring is said to be semiprime if ${\varrho }^n\diamond \sigma \subseteq \Omega$ where $\varrho ,\sigma \in \Gamma$ and $n\in {\mathbb{Z}}^{+}$ then $\varrho \diamond \sigma \subseteq \Omega$ [12].

Definition 2.20. 

Suppose that $\omega$ is a multiplicative hyperring. An element $\kappa \in \Gamma$ is nilpotent if there exists an integer n such that $0\in {\kappa }^n$. Denote the set of all nilpotent elements of $\Gamma$ by $nil(\Gamma )$ [10].

Definition 2.21. 

Suppose that $\Gamma$ is a commutative multiplicative hyperring with scalar 1 and $S\subseteq \Gamma$ is a multiplicative closed subset of $\Gamma$. Then define the operations ″$\oplus$″ and ″$\otimes$″ on $S^{-1}\Gamma$ by ${\displaystyle \frac{\varrho }{\sigma }}\oplus {\displaystyle \frac{\vartheta }{\mu }}={\displaystyle \frac{\varrho \diamond \mu {+}^{\prime }\vartheta \diamond \sigma }{\sigma \diamond \mu }}$ and ${\displaystyle \frac{\varrho }{\sigma }}\otimes {\displaystyle \frac{\vartheta }{\mu }}={\displaystyle \frac{\varrho \diamond \vartheta }{\sigma \diamond \mu }}$ for all${\displaystyle \frac{\varrho }{\sigma }},{\displaystyle \frac{\vartheta }{\mu }}\in S^{-1}\Gamma$. Then $(S^{-1}\Gamma ,\oplus ,\otimes )$ is a commutative multipllicative hyperring with an identity . This multiplicative hyperring is called the fractions of multiplicative hyperings [10].

Throughout our study, $\Gamma$ will be a commutative with scalar identity 1.

3. Weakly S- Prime Hyperideals

In this section the fundamental properties of weakly S− prime hyperideals are studied.

Definition 3.1. 

Let $\Gamma$ be a multiplicative hyperring, $S\subseteq R$ be a multiplicative closed subset of $\Gamma$ and $\Omega$ be a proper hyperideal of $\Gamma$ such that $\Omega \cap S=\varnothing$.

A proper hyperideal $\Omega$ is said to be a weakly S− prime hyperideal then there exists an $s\in S$ such that if whenever $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq \Omega$ then for all $\varrho ,\sigma \in \Gamma$ then either $s\diamond \varrho \subseteq \Omega$ or $s\diamond \sigma \subseteq \Omega$.

Example 3.2. 

Let us define the hyperoperation on the ring of integers as for all $\varrho ,\sigma \in \mathbb{Z}$;$\varrho \diamond \sigma =\left\{\varrho \sigma ,2\varrho \sigma \right\}$. Let us give $S=\left\{1,2\right\}$ be a closed multiplicative subset of $(\mathbb{Z},{+}^{\prime },\diamond )$ and $3\mathbb{Z}$ be hyperideal of $(\mathbb{Z},{+}^{\prime },\diamond )$ such that $S\cap 3\mathbb{Z}=\varnothing$ We will show that $3\mathbb{Z}$ is a weakly S− prime hyperideal of $(\mathbb{Z},{+}^{\prime },\diamond )$.

Let $\left\{0\right\}\ne \varrho \diamond \sigma =\left\{\varrho \sigma ,2\varrho \sigma \right\}\subseteq 3\mathbb{Z}$ and there exists $s\in S$ such that $s\diamond \varrho ⊈3\mathbb{Z}$. Let’s find out $s\diamond \sigma \subseteq 3\mathbb{Z}$.

Therefore; $\varrho .\sigma =3.\kappa \Rightarrow \varrho =3\vartheta$,$\sigma =3.\mu \Rightarrow s\diamond \varrho ⊈3\mathbb{Z}\Rightarrow s\diamond \varrho \ne 3.\mu \Rightarrow 3\nmid \varrho \Rightarrow 3\mid \sigma \Rightarrow 3\mid s\diamond \sigma \Rightarrow s\diamond \sigma \subseteq 3\mathbb{Z}(\kappa ,\mu ,\vartheta \in \mathbb{Z})$ and hence $3\mathbb{Z}$ is a weakly S− prime hyperideal of $\Gamma$.

Now we will show weakly prime hyperideals and the relationship between S− prime hyperideals and weakly S− prime hyperideals.

Proposition 3.3 

Suppose that $\Gamma$ is a multiplicative hyperring, $S\subseteq \Gamma$ is multiplicative closed subset of $\Gamma$ which is include $1\in S$ and $\Omega$ is a proper hyperideal of $\Gamma$ such that $\Omega \cap S=\varnothing$. In this case every weakly prime hyperideal of $\Gamma$ is a weakly S− prime hyperideal.

Proof: 

Let $\Omega$ be a weakly prime hyperideal of $\Gamma$. For $\varrho ,\sigma \in \Gamma$ if $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq \Omega$ then $\varrho \subseteq \Omega$ or $\sigma \subseteq \Omega$. If we take $1=s\in S$ then we get either $s\diamond \varrho \subseteq \Omega$ or $s\diamond \sigma \subseteq \Omega$. Therefore $\Omega$ is a weakly S− prime hyperideal of $\Gamma$.

Now, let’s give an example of how the converse of a proposition is usually not true.

Example 3.4. 

Suppose that $(\Gamma ,{+}^{\prime },.)$ is a ring for a commutative with identitiy multiplicative hyperring $({\Gamma }_A,{+}^{\prime },\diamond )$ with absorbing elements such that ${\Gamma }_A=\Gamma$ for every subset $A\in P^{*}(\Gamma )$ $(\left|A\right|\ge 2)$ of the $\Gamma$, let’s define the hypermultiplication as $\varrho \diamond \sigma =\left\{\varrho .\kappa .\sigma |\kappa \in A\right\}$ for every $\varrho ,\sigma \in {\Gamma }_A$.

Now, let $T=({\Gamma }_A\left[\varrho \right],+,*)$ be a polynomial multiplicative hyperring and let us define the ″$+$″ operation and ″$*$″ hyperoperation for all $f\left(\varrho \right)={\sum }_{k=0}^n{\alpha }_k{\varrho }^k\in T$ and $g\left(\varrho \right)={\sum }_{k=0}^m{\beta }_k{\varrho }^k\in T$ as $f\left(\varrho \right)+g\left(\varrho \right)={\sum }_{k=0}({\alpha }_k+{\beta }_k){\varrho }^k$ and $f\left(\varrho \right)*g\left(\varrho \right)=\left\{{\sum }_{k=0}^{n+m}{\vartheta }_k{x}^k|{\vartheta }_k\in {\sum }_{i+j=k}{\alpha }_i\diamond {\beta }_j\right\}$ respectively. Suppose that $A=\left\{1,2,3\right\}$ is a set and ${\Gamma }_A=(\mathbb{Z},{+}^{\prime },\diamond )$ is a multiplicative hyperring. Let $T=({\mathbb{Z}}_A\left[x\right],+,*)$ be a multiplicative polynomial hyperring and $S=\left\{3^n|n\in \mathbb{N}\right\}$ be closed multiplicative subset of T and when we take $\Theta =〈6\varrho 〉$ with $\Theta \cap S=\varnothing$

Although $\Theta$ is a weakly S− prime hyperideal of T, it is not a weakly prime hyperideal for $s=3$. Because $3\diamond 2\varrho =\left\{6\varrho ,12\varrho ,18\varrho \right\}\subseteq \Theta$ but $3\notin \Theta$ and $2\varrho \notin \Theta$.

Proposition 3.5. 

Suppose that $\Gamma$ is a multiplicative hyperring, $S\subseteq \Gamma$ is a multiplicative closed subset of $\Gamma$ and $\Omega$ is a proper hyperideal of $\Gamma$ such that $\Omega \cap S=\varnothing$ . In this case every S− prime hyperideal is a weakly S− prime hyperideal.

Proof: 

Let $\Omega$ be a S− prime hyperideal of $\Gamma$ for $\varrho ,\sigma \in \Gamma$ and let we take $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq \Omega$. Since $\Omega$ is a S− prime hyperideal of $\Gamma$ then there exists $s\in S$ such that $s\diamond \varrho \subseteq \Omega$ or $s\diamond \sigma \subseteq \Omega$.

Therefore $\Omega$ is a weakly S− prime hyperideal of $\Gamma$.

Let us give an example to show that the converse of the above statement is generally not true.

Example 3.6. 

Let $\overline{\varrho }\diamond \overline{\sigma }=\left\{\overline{\varrho }\overline{\sigma },\overline{2}\overline{\varrho }\overline{\sigma },\dots ,\overline{5}\overline{\varrho }\overline{\sigma }\right\}$ be the hyperoperation defined on the multiplicative hyperring $\Gamma =(\mathbb{Z}/6\mathbb{Z},{+}^{\prime },\diamond )$ for every $\overline{\varrho },\overline{\sigma }\in \Gamma$. Let $S=\left\{\overline{1},\overline{5}\right\}$ be a multiplicative closed subset of $\Gamma$ and

$\Omega =\left(\overline{0}\right)=\left\{{\sum }_{i=1}^n{\overline{\varrho }}_i|{\overline{\varrho }}_i\in {\overline{\gamma }}_i\diamond \overline{0};{\overline{\gamma }}_i\in \mathbb{Z}/6\mathbb{Z}\right\}$ be a hyperideal of $\Gamma$ such that $\Omega \cap S=\varnothing$. Eventhough $\Omega$ is a weakly S− prime hyperideal of $\Gamma$, it is not a S− prime hyperideal of $\Gamma$. Because for $\overline{2},\overline{3}\in \Gamma$, $\overline{2}\diamond \overline{3}\subseteq \Omega$ but $\overline{1}\diamond \overline{2}=\overline{2}\diamond \overline{5}=\left(\overline{2}\right)⊈\Omega$ and $\overline{3}\diamond \overline{1}=\overline{3}\diamond \overline{5}=\left(\overline{3}\right)⊈\Omega$.

Lemma 3.7. 

Suppose that $\Gamma$ is a multiplicative hyperring, $S\subseteq \Gamma$ is a multiplicative closed subset of $\Gamma$, $\Omega$ and $\Lambda$ are two hyperideals of $\Gamma$ such that $\Omega \cap S=\varnothing$ and $\Lambda \cap S=\varnothing$. If $\Omega$ and $\Lambda$ are two S− prime hyperideals of $\Gamma$ then $\Omega \cap \Lambda$ is a S− prime hyperideal of $\Gamma$.

Proof: 

Clear.

Lemma 3.8. 

Let $\Gamma$ be a multiplicative hyperring, $S\subseteq \Gamma$ be a multiplicative subset of $\Gamma$, $\Omega$ and $\Lambda$ be two hyperideals of $\Gamma$ such that $\Omega \cap S=\varnothing$ and $\Lambda \cap S=\varnothing$. If I and J be two weakly S− prime hyperideals of $\Gamma$ then $\Omega \cap \Lambda$ be a weakly S− prime hyperideal of $\Gamma$.

Proof: 

Since every S− prime hyperideal is a weakly S− prime hyperideal so $\Omega \cap \Lambda$ is a weakly S− prime hyperideal of $\Gamma$ .

In the following proposition, we will prove that the hyperproduct of two weakly S− prime hyperideals is also a weakly S− prime hyperideal.

Proposition 3.9. 

Suppose that $\Gamma$ is a multiplicative hyperring, $S\subseteq \Gamma$ is a multiplicative closed subset of $\Gamma$, $\Omega$ and $\Lambda$ are two hyperideals of $\Gamma$ such that $\Omega \cap S=\varnothing$ and $\Lambda \cap S=\varnothing$. If $\Omega$ and $\Lambda$ are weakly S− prime hyperideals of $\Gamma$ then $\Omega \Lambda =\bigcup \left\{{\sum }_{i=1}^n{\varrho }_i\diamond {\sigma }_i|{\varrho }_i\in \Omega ,{\sigma }_i\in \Lambda \right\}$ is a weakly S− prime hyperideal of $\Gamma$.

Proof: 

Let $\left\{0\right\}\ne ({\sum }_{i=1}^n{\varrho }_i\diamond {\sigma }_i)({\sum }_{i=1}^n{\vartheta }_i\diamond {\mu }_i)=({\varrho }_1\diamond {\sigma }_1{+}^{\prime }\dots {+}^{\prime }{\varrho }_n\diamond {\sigma }_n)\diamond ({\vartheta }_1\diamond {\mu }_1{+}^{\prime }\dots {+}^{\prime }{\vartheta }_n\diamond {\mu }_n)\subseteq ({\varrho }_1\diamond {\sigma }_1\diamond {\vartheta }_1\diamond {\mu }_1{+}^{\prime }\dots {+}^{\prime }{\varrho }_1\diamond {\sigma }_1\diamond {\vartheta }_n\diamond {\mu }_n{+}^{\prime }\dots {+}^{\prime }{\varrho }_n\diamond {\sigma }_n\diamond {\varrho }_1\diamond {\mu }_1{+}^{\prime }\dots {+}^{\prime }{\varrho }_n\diamond {\sigma }_n\diamond {\vartheta }_n\diamond {\mu }_n)=({\varrho }_1\diamond {\vartheta }_1)\diamond ({\sigma }_1\diamond {\mu }_1){+}^{\prime }\dots {+}^{\prime }({\varrho }_1\diamond {\vartheta }_n)\diamond ({\sigma }_1\diamond {\mu }_n){+}^{\prime }\dots +({\varrho }_n\diamond {\vartheta }_1)\diamond ({\sigma }_n\diamond {\mu }_1){+}^{\prime }\dots {+}^{\prime }({\varrho }_n\diamond {\vartheta }_1)\diamond ({\sigma }_n\diamond {\mu }_1){+}^{\prime }\dots {+}^{\prime }({\varrho }_n\diamond {\vartheta }_n)\diamond ({\sigma }_n\diamond {\mu }_n)\subseteq \Omega \Lambda$ for all $i\in \left\{1,2,\dots ,n\right\}$ and ${\varrho }_i,{\sigma }_i,{\vartheta }_i,{\mu }_i\in \Gamma$. Because $\Omega$ and $\Lambda$ are weakly S− prime hyperideals of $\Gamma$ there exists $s,t\in S$ such that $s\diamond {\varrho }_i\subseteq \Omega$ or $s\diamond {\vartheta }_i\subseteq \Omega$ and $t\diamond {\sigma }_i\subseteq \Lambda$ or $t\diamond {\mu }_i\subseteq \Lambda$ and $\Omega \Lambda$ is a hyperideal of $\Gamma$ by [7]. Therefore we get $k\diamond [({\varrho }_1\diamond {\vartheta }_1)\diamond ({\sigma }_1\diamond {\mu }_1){+}^{\prime }\dots {+}^{\prime }({\varrho }_1\diamond {\vartheta }_n)\diamond ({\sigma }_1\diamond {\mu }_n){+}^{\prime }\dots {+}^{\prime }({\varrho }_n\diamond {\vartheta }_1)\diamond ({\sigma }_n\diamond {\mu }_1){+}^{\prime }\dots {+}^{\prime }({\varrho }_n\diamond {\vartheta }_n)\diamond ({\sigma }_n\diamond {\mu }_n)\subseteq k\diamond {\vartheta }_1\diamond {\vartheta }_1\diamond {\sigma }_1\diamond {\mu }_1{+}^{\prime }k\diamond {\varrho }_1\diamond {\vartheta }_n\diamond {\sigma }_1\diamond {\mu }_n{+}^{\prime }\dots {+}^{\prime }k\diamond {\varrho }_n\diamond {\vartheta }_1\diamond {\sigma }_n\diamond {\mu }_1{+}^{\prime }\dots {+}^{\prime }k\diamond {\varrho }_n\diamond {\vartheta }_n\diamond {\sigma }_n\diamond {\mu }_n=k\diamond \left[({\sum }_{i=1}^n{\varrho }_i\diamond {\sigma }_i)({\sum }_{i=1}^n{\vartheta }_i\diamond {\mu }_i)\right]\subseteq \Omega \Lambda$ for $k\in (s\diamond t)\cap S$ . According to Lemma $3.8$ the intersection two weakly S− prime hyperideals is also weakly S− prime hyperideal and since $\Omega \Lambda \subseteq \Omega \cap \Lambda$ we get either $k\diamond {\sum }_{i=1}^n{\varrho }_i\diamond {\sigma }_i\subseteq \Omega \Lambda$ or $k\diamond {\sum }_{i=1}^n{\vartheta }_i\diamond {\mu }_i\subseteq \Omega \Lambda$ . Therefore $\Omega \Lambda$ is a weakly S− prime hyperideal of $\Gamma$.

Proposition 3.10. 

Let $\Gamma$ be a multiplicative hyperring $S\subseteq \Gamma$ be a multiplicative closed subset of $\Gamma$ and $\Omega$ be a hyperideal of $\Gamma$ such that $S\cap \Omega =\varnothing$. If $\Omega$ is a weakly S− prime hyperideal of $\Gamma$ then ${\Omega }^2$ is a weakly S− prime hyperideal of $\Gamma$.

Proof: 

According to Proposition $3.9$ , ${\Omega }^2$ is a weakly S− prime hyperideal of $\Gamma$.

We proved in Proposition $3.5$ that every S− prime hyperideal is a weakly S− prime hyperideal, and we showed in Example $3.6$ that the converse is generally not true. In the following proposition, we will prove under what conditions the converse is true.

Proposition 3.11. 

Suppose that $\Gamma$ is a multiplicative hyperring, $S\subseteq \Gamma$ is a multiplicative closed subset of $\Gamma$ and $\Omega$ is a weakly S− prime hyperideal of $\Gamma$ such that $\Omega \cap S=\varnothing$ and ${\Omega }^2\ne \left\{0\right\}$ if and only if $\Omega$ is a S− prime hyperideal of $\Gamma$.

Proof: 

$(\Rightarrow ):$ If $\Omega$ is a weakly S− prime hyperideal of $\Gamma$ then by assumption for $\varrho ,\sigma \in \Gamma$, if $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq \Omega$ then there exists $s\in S$ such that $s\diamond \varrho \subseteq \Omega$ or $s\diamond \sigma \subseteq \Omega$. Let ${\Omega }^2\ne \left\{0\right\}$. We will show that a weakly S− prime hyperideal is an S− prime hyperideal. Suppose that $\varrho \diamond \sigma \subseteq \Omega$ for $\varrho ,\sigma \in \Gamma$. Let us consider the case where $\varrho \diamond \sigma =\left\{0\right\}$. If $\varrho \diamond \Omega \ne \left\{0\right\}$ then there exists $p\in \Omega$ such that $\varrho \diamond p\ne \left\{0\right\}$.

In this case $\left\{0\right\}\ne \varrho \diamond \left(\sigma {+}^{\prime }p\right)\subseteq \varrho \diamond p\subseteq \Omega$ gives $s\diamond \varrho \subseteq \omega$ or $s\diamond \left(\sigma {+}^{\prime }p\right)\subseteq \Omega$. If $s\diamond \left(\sigma {+}^{\prime }p\right)\subseteq \Omega$ then $s\diamond \left(\sigma {+}^{\prime }p\right)\subseteq s\diamond \sigma {+}^{\prime }s\diamond p\subseteq \Omega$. Because $p\in \Omega$ and $\Omega$ is a hyperideal of $\Gamma$, that is, if $\gamma \in \Gamma$ and $\vartheta \in \Omega$, then we find $\gamma \diamond \vartheta \subseteq \Omega$. Hence, we get $s\diamond p\subseteq \Omega$. For an arbitrary $\mu \in s\diamond \sigma$ and $\nu \in s\diamond p$, we find $\mu {+}^{\prime }\nu \in s\diamond \left(\sigma {+}^{\prime }p\right)\subseteq s\diamond \sigma {+}^{\prime }s\diamond p\subseteq \Omega$ and $\nu \in s\diamond p\subseteq \Omega$ . Since $\Omega$ is hyperideal of $\Gamma$ $(\Omega {-}^{\prime }\Omega \subseteq \Omega )$ , $\mu =\left(\mu {+}^{\prime }\nu \right){-}^{\prime }\nu \subseteq \Omega$. Therefore $s\diamond \sigma \subseteq \Omega$.

Similarly; if $\left\{0\right\}\ne \sigma \diamond \Omega$ then there exists $\kappa \in \Omega$ such that $\left\{0\right\}\ne \sigma \diamond \kappa$. Therefore $\left\{0\right\}\ne \sigma \diamond \left(\varrho {+}^{\prime }\kappa \right)\subseteq \sigma \diamond \kappa \subseteq \Omega$ we get either $s\diamond \sigma \subseteq \Omega$ or $s\diamond \left(\kappa {+}^{\prime }\varrho \right)\subseteq \Omega$. If $s\diamond \left(\kappa {+}^{\prime }\varrho \right)\subseteq \Omega$ then $s\diamond \left(\kappa {+}^{\prime }\varrho \right)\subseteq s\diamond \kappa {+}^{\prime }s\diamond \varrho \subseteq \Omega$. Because $\kappa \in \Omega$ and $\Omega$ is a hyperideal of $\Gamma$ so we have $s\diamond \kappa \subseteq \Omega$ and $s\diamond \varrho \subseteq \Omega$ is obtained using similar steps to those described above.

Finally; let we consider $\left\{0\right\}=\varrho \diamond \Omega$ and $\left\{0\right\}=\sigma \diamond \Omega$. Since ${\Omega }^2=\Omega \diamond \Omega =\bigcup \left\{{\sum }_{i=1}^n{p}_i\diamond {\kappa }_i|p_i,{\kappa }_i\in \Omega \right\}\ne \left\{0\right\}$ there exists $p_i,{\kappa }_i\in \Omega$ such that $p_i\diamond {\kappa }_i\ne \left\{0\right\}$ In this situation; there exists an $s\in S$ such that if $\left\{0\right\}\ne \left(\varrho {+}^{\prime }p\right)\diamond \left(\sigma {+}^{\prime }\kappa \right)\subseteq \varrho \diamond \sigma {+}^{\prime }\varrho \diamond \kappa {+}^{\prime }p\diamond \sigma {+}^{\prime }p\diamond \kappa \subseteq \Omega$ then $s\diamond \left(\varrho {+}^{\prime }p\right)\subseteq s\diamond \varrho {+}^{\prime }s\diamond p=s\diamond \varrho \subseteq \Omega$ or $s\diamond \left(\sigma {+}^{\prime }\kappa \right)\subseteq s\diamond \sigma {+}^{\prime }s\diamond \kappa =s\diamond \sigma \subseteq \Omega$.

Therefore, either $s\diamond \varrho \subseteq \Omega$ or $s\diamond \sigma \subseteq \Omega$ is used to obtain the desired result.

$(\Leftarrow ):$ Clear.

The converse of the proposition is generally not true, meaning that a hyperideal with ${\Omega }^2=\left\{0\right\}$ does not necessarily have to be a weakly S− prime.

Example 3.12. 

Let we define the $\overline{\varrho }\diamond \overline{\sigma }=\left\{\overline{\varrho }\overline{\sigma },\overline{2}\overline{\varrho }\overline{\sigma },\dots ,\overline{7}\overline{\varrho }\overline{\sigma }\right\}$ hyperoperation on the $\Gamma =\mathbb{Z}/8\mathbb{Z}$ multiplicative hyperring for each $\overline{\varrho },\overline{\sigma }\in \Gamma$, $S=\left\{\overline{1},\overline{5}\right\}\subseteq \Gamma$ is a multiplicative closed subset of $\Gamma$ and $\Omega =4\mathbb{Z}/8\mathbb{Z}$ is a hyperideal of $\Gamma$ such that $\Omega \cap S=\varnothing$. Although ${\Omega }^2=\Omega \diamond \Omega =\bigcup \left\{{\sum }_{i=1}^7{\omega }_i\diamond {\mu }_i|{\omega }_i,{\mu }_i\subseteq \Omega \right\}=\left\{\overline{0}\right\}$, $\Omega$ is not S− prime hyperideal of $\Gamma$. Because $\overline{2}\diamond \overline{2}\subseteq \Gamma$ while $\overline{1}\diamond \overline{2}⊈\Omega ,$ and $\overline{5}\diamond \overline{2}⊈\Omega ,$ are not for $\overline{2}\in \Gamma$ . Therefore it is not weakly S− prime.

Corollary 3.13. 

Suppose that $\Gamma$ is a multiplicative hyperring, $S\subseteq \Gamma$ is a multiplicative closed subset and $\Omega$ is a $\Omega$ is a S− prime hyperideal of $\Gamma$ such that $\Omega \cap S=\varnothing$. Let ${\Omega }^2\ne \left\{0\right\}$ and $(\Omega :s)$ is a prime hyperideal of $\Gamma$.

Therefore; $(\Omega :s)$ is a prime $\iff \Omega$ is a S− prime by [9]$\iff \Omega$ is a weakly S− prime for $s\in S$.

Proposition 3.14. 

Assume that $\Gamma$ is a multiplicative hyperring and $S\subseteq \Gamma$ is a multiplicative closed subset. Then the following statements are held.

(i) Let $S_1\subseteq S_2$ be two multiplicative closed subsets of $\Gamma$. If $\Omega$ be a weakly $S_1$− prime hyperideal and $\Omega \cap S_2=\varnothing$ then $\Omega$ be a weakly $S_2$− prime hyperideal of $\Gamma$.

(ii) If S comprise of units of $\Gamma$ then a proper hyperideal $\Omega$ of $\Gamma$ is weakly prime if and only if $\Omega$ is a weakly S− prime.

Proof: 

(i) Since $S_1\subseteq S_2$ and $\Omega \cap S_2=\varnothing$ for $\varrho ,\sigma \in \Gamma$ if $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq \Omega$ then there exists $s\in S_1\subseteq S_2$ such that $s\diamond \varrho \subseteq \Omega$ or $s\diamond \sigma \subseteq \Omega$ and hence $\Omega$ is a weakly $S_2$− prime hyperideal of $\Gamma$.

(ii)

$(\Rightarrow )$: Let $\Omega$ be a weakly S− prime hyperideal of $\Gamma$ so we have if $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq \Omega$ then there exists $s\in S$ such that $s\diamond \varrho \subseteq \Omega$ or $s\diamond \sigma \subseteq \Omega$ for $\varrho ,\sigma \in \Gamma$. Since S consist of units of $\Gamma$ for $s^{-1}\in S$ we get either $s^{-1}\diamond (s\diamond \varrho )=(s^{-1}\diamond s)\diamond \varrho \subseteq s^{-1}\diamond \Omega \subseteq \Omega \Rightarrow \varrho \subseteq \Omega$ or $s^{-1}\diamond (s\diamond \sigma )=(s^{-1}\diamond s)\diamond \sigma \subseteq s^{-1}\sigma \Omega \subseteq \Omega \Rightarrow \sigma \subseteq \Omega$. Hence $\Omega$ is a weakly prime.

$(\Leftarrow ):$ Every weakly prime hyperideal is a weakly S− prime hyperideal.

In Proposition $3.14$ $\left(ii\right)$, we proved under what conditions a weakly S− prime hyperideal is a weakly prime hyperideal. Now, let us prove the general condition for this in the following proposition.

Proposition 3.15. 

Let $\Gamma$ be a multiplicative hyperring, $S\subseteq \Gamma$ be a multiplicative closed subset of $\Gamma$ composed of the units of $\Gamma$ and ${\Omega }_1,{\Omega }_2,\dots ,{\Omega }_n$ be proper hyperideals of $\Gamma$ with ${\Omega }_i\cap S=\varnothing$ for all $i\left\{1,2,\dots ,n\right\}$. ${\Omega }_i$ is a weakly prime hyperideal.

$(\Rightarrow ):$ Let ${\Omega }_i$ be weakly S− prime hyperideal of $\Gamma$ for all $i\in \left\{1,2\dots ,n\right\}$. If $\left\{0\right\}\ne {\varrho }_i\diamond {\sigma }_i\subseteq {\Omega }_i$ then there exists $s\in S$ with $s\diamond {\varrho }_i\subseteq {\Omega }_i$ or $s\diamond {\sigma }_i\subseteq {\Omega }_i$ for $i\in \left\{1,2,\dots ,n\right\}$ and for ${\varrho }_i,{\sigma }_i\in \Gamma$ Because S be consits of the units of $\Gamma$ we get $s^{-1}\diamond (s\diamond {\varrho }_i)=(s^{-1}\diamond s)\diamond {\varrho }_i\subseteq s^{-1}\diamond {\Omega }_i\subseteq {\Omega }_i\Rightarrow {\varrho }_i\subseteq {\Omega }_i$ or $s^{-1}\diamond (s\diamond {\sigma }_i)=(s^{-1}\diamond s)\diamond {\sigma }_i\subseteq s^{-1}\diamond {\Omega }_i\subseteq {\Omega }_i\Rightarrow {\sigma }_i\subseteq {\Omega }_i$.

$(\Leftarrow ):$ Every weakly prime hyperideal is a weakly S− prime hyperideal.

Ghiasvand and Farzalipour proved that a necessary and sufficient condition for $(\Omega :s)$ to be a prime hyperideal is that $\Omega$ must be an S− prime hyperideal. [9] Now we will generalize this to weakly S− prime hyperideals.

Proposition 3.16. 

Suppose that $\Gamma$ is a multiplicative hyperring, $S\subseteq \Gamma$ is a multiplicative closed subset of $\Gamma$ and $\Omega$ is a proper hyperideal of $\Gamma$ such that $\Omega \cap S=\varnothing$. A proper hyperideal $\Omega$ is a weakly S− prime hyperideal if and only if $(\Omega :s)$ is a weakly prime hyperideal of $\Gamma$ for some $s\in S$.

Proof: 

$(\Rightarrow ):$ Let $\Omega$ be a weakly S− prime hyperideal of $\Gamma$. We have if $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq \Omega$ then there exists $s\in S$ such that $s\diamond \varrho \subseteq \Omega$ or $s\diamond \sigma \subseteq \Omega$. Consider $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq (\Omega :s)$ so we get $s\diamond (\varrho \diamond \sigma )=(s\diamond \varrho )\diamond \sigma \subseteq \Omega$. Let for any $t\in s\diamond \varrho$ so we have $t\diamond \sigma \subseteq \Omega$. Because $\Omega$ is a weakly S− prime hyperideal of $\Gamma$ there exists $s_1\in S$ we get either $s_1\diamond t\subseteq \Omega$ or $s_1\diamond \sigma \subseteq \Omega$. If $s_1\diamond t\subseteq \Omega$ then $s_1\diamond (s\diamond \varrho )=(s_1\diamond s)\diamond \varrho \subseteq \Omega$. As $(s_1\diamond s)\cap S\ne \varnothing$ for $u\in (s_1\diamond s)\cap S$ we get $u\diamond \varrho \subseteq \Omega$ and hence we get either $s\diamond u\subseteq \Omega$ or $s\diamond \varrho \subseteq \Omega$. Since $\Omega$ is a weakly S− prime hyperideal then $s\diamond u⊈\Omega$.

Therefore; we get either $s\diamond \varrho \subseteq \Omega$ or $s\diamond \sigma \subseteq \Omega$ leads to $\varrho \subseteq (\Omega :s)$ or $\sigma \subseteq (\Omega :s)$.

$(\Leftarrow ):$ Suppose that $(\Omega :s)$ is a weakly prime hyperideal of $\Gamma$. Let $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq \Gamma$ for $\varrho ,\sigma \in \Gamma$. Because $\Omega$ is a hyperideal of $\Gamma$ so we have $\Omega \subseteq (\Omega :s)$ and $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq (\Omega :s)$ Since $(\Omega :s)$ is a weakly prime hyperideal thus $\varrho \subseteq (\Omega :s)\Rightarrow s\diamond \varrho \subseteq \Omega$ or $\sigma \subseteq (\Omega :s)\Rightarrow s\diamond \sigma \subseteq \Omega$. Hence $\Omega$ is a weakly S− prime hyperideal of $\Gamma$.

Example 3.17. 

In Example $3.2$, we showed that $\Omega =3\mathbb{Z}$ is a weakly S− prime hyperideal of the multiplicative hyperring $\Gamma$. Now, using this Example $3.2$, let us show that $(\Omega :s)$ is a weakly prime hyperideal of $\Gamma$. Let $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq (\Omega :s)$ and $\varrho ⊈(\Omega :s)$. We will show that $\sigma \subseteq (\Omega :s)$. If $\varrho ⊈(\Omega :s)$ then $s\diamond \varrho ⊈\Omega$. We know $\Omega$ is a weakly S− prime hyperideal of $\Gamma$ so we have $s\diamond \sigma \subseteq \Omega$. Therefore, we get $\sigma \subseteq (\Omega :s)$ and hence $(\Omega :s)$ is a weakly prime hyperideal of $\Gamma$. Conversely, if $(\Omega :s)$ is weakly prime, then it is clear that $\Omega$ is weakly S− prime.

Proposition 3.18. 

Let $\Gamma$ be a multiplicative hyperring, $S\subseteq \Gamma$ be a multiplicatice closed subset of R and $\Omega$ be a hyperideal of $\Gamma$ such that $\Omega \cap S=\varnothing$. Then

(i) If $\Lambda$ is a hyperideal of such that $\Lambda \cap S\ne \varnothing$ and $\Omega$ is a weakly S− prime hyperideal of R then $\Lambda \Omega$ is a weakly S− prime hyperideal of $\Gamma$.

(ii) Let $\Gamma \subseteq {\Gamma }^{\prime }$ be a extension of $\Gamma$. If $\Theta$ is a weakly S− prime hyperideal of ${\Gamma }^{\prime }$ then $\Theta \cap \Gamma$ is a weakly S− prime.

Proof: 

(i) Let $\vartheta \in \Lambda \cap S$. Let $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq \Lambda \diamond \Omega \subseteq \Omega$ where $\varrho ,\sigma \in \Gamma$. Therefore $\vartheta \diamond s\diamond \varrho \subseteq \Lambda \Omega$ Because $(\vartheta \diamond s)\cap S\ne \varnothing$ so there is $\mu \in (\vartheta \diamond s)\cap S$. Thus $\mu \diamond \varrho \subseteq \Lambda \Omega$ or $\mu \diamond \sigma \subseteq \Lambda \Omega$ where $\Lambda \Omega$ is a weakly S− prime hyperideal of $\Gamma$.

(ii) Let $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq \Theta \cap \Gamma$ for $\varrho ,\sigma \in \Gamma$. Since $\Theta$ is a weakly S− prime of ${\Gamma }^{\prime }$ there exists $s\in S$ such that $s\diamond \varrho \subseteq \Theta$ or $s\diamond \sigma \subseteq \Theta$ so we get either $s\diamond \varrho \subseteq \Theta \cap \Gamma$ or $s\diamond \sigma \subseteq \Theta \cap \Gamma$. Thus, $\Theta \cap \Gamma$ is a weakly S− prime hyperideal of $\Gamma$.

Theorem 3.19. 

Assume that $\Gamma$ is a multiplicative hyperring, $S\subseteq \Gamma$ is a multiplicative closed subset of $\Gamma$ and let $\lambda :\Gamma \to {\Gamma }^{\prime }$ be a good homomorphism such that $\lambda \left(S\right)$ does not contain zero. If $\Theta$ be a weakly- $\lambda \left(S\right)$ prime hyperideal of ${\Gamma }^{\prime }$ then ${\lambda }^{-1}(\Theta )$ be a weakly S− prime hyperideal of $\Gamma$.

Proof: 

Let $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq {\lambda }^{-1}(\Theta )$ where $\varrho ,\sigma \in \Gamma$.

We get;

$\left\{0\right\}\ne \varrho \diamond \sigma \subseteq {\lambda }^{-1}(\Theta )\Rightarrow \lambda (\varrho \diamond \sigma )\subseteq \lambda \left({\lambda }^{-1}(\Theta )\right)\Rightarrow \lambda (\varrho \diamond \sigma )=\lambda \left(\varrho \right)▿\lambda \left(\sigma \right)\subseteq \Theta$.

Because $\Theta$ is a weakly $\lambda \left(S\right)$− prime hyperideal there exists $\lambda \left(s\right)\in \lambda \left(S\right)$ such that if $\left\{0\right\}\ne \lambda (\varrho \diamond \sigma )=\lambda \left(\varrho \right)▿\lambda \left(\sigma \right)\subseteq \Theta$ then $\lambda (s\diamond \varrho )=\lambda \left(s\right)▿\lambda \left(\varrho \right)\subseteq \Theta$ or $\lambda (s\diamond \sigma )=\lambda \left(s\right)▿\lambda \left(\sigma \right)\subseteq \Theta$.

We find, there exists $s\in S$ such that $s\diamond \varrho \subseteq {\lambda }^{-1}(\Theta )$ or $s\diamond \sigma \subseteq {\lambda }^{-1}(\Theta )$. Thus ${\lambda }^{-1}(\Theta )$ is a weakly S−prime hyperideal.

Theorem 3.20. 

Suppose that $\lambda$ is a good epimorphism between two multiplicative hyperrings $\Gamma$ and ${\Gamma }^{\prime }$ is a map $\lambda$ from $\Gamma$ to ${\Gamma }^{\prime }$. If $\Theta$ is a weakly S− prime hyperideal of $\Gamma$ and $Ker\lambda \subseteq \Theta$ then $\lambda (\Theta )$ is a weakly $\lambda \left(S\right)$− prime hyperideal of ${\Gamma }^{\prime }$.

Proof: 

Let $\left\{0\right\}\ne \mu ▿\nu \subseteq \lambda (\Gamma )$ for $\mu ,\nu \in {\Gamma }^{\prime }$. Since $\lambda$ is a good epimorphism there exists $\varrho ,\sigma \in \Gamma$ such that $\mu \subseteq \lambda \left(\varrho \right)$ and $\nu \subseteq \lambda \left(\sigma \right)$. Therefore we get $\mu ▿\nu =\lambda (\varrho \diamond \sigma )=\lambda \left(\varrho \right)▿\lambda \left(\sigma \right)\subseteq \lambda (\Theta )$. For $\vartheta \in \varrho \diamond \sigma \Rightarrow \lambda (\varrho \diamond \sigma )\subseteq \lambda (\Theta )$ and $\omega \in \Theta$ we find $\lambda \left(\vartheta \right)=\lambda \left(\omega \right)\Rightarrow \lambda \left(\vartheta {-}^{\prime }\omega \right)=0\Rightarrow \vartheta {-}^{\prime }\omega \in Ker\lambda \subseteq \Theta$.

According to our assumption; $\Theta$ is a weakly S− prime hyperideal of $\Gamma$ so if $\left\{0\right\}\ne \mu \diamond \nu \subseteq \Theta$ then there exists $s\in S$ with $s\diamond \mu \subseteq \Theta$ or $s\diamond \nu \subseteq \Theta$.

Finally; we find if $\lambda \left(\left\{0\right\}\right)\ne \lambda (\mu \diamond \nu )=\lambda \left(\mu \right)▿\lambda \left(\nu \right)\subseteq \lambda (\Theta )$ then there exists $\lambda \left(s\right)\in \lambda \left(S\right)$ with $\lambda (s\diamond \mu )=\lambda \left(s\right)\diamond \lambda \left(\mu \right)\subseteq \lambda (\Theta )$ or $\lambda (s\diamond \nu )=\lambda \left(s\right)▿\lambda \left(\nu \right)\subseteq \lambda (\Theta )$. Thus, the desired is achieved.

Proposition 3.21. 

Assume that $\Gamma$ is a multiplicative hyperring, $S\subseteq \Gamma$ is a multiplicative closed subset of $\Gamma$; $\Omega$ is a hyperideal with $\Omega \cap S=\varnothing$. Let $\Psi$ be proper hyperideal of $\Gamma$ that includes I such that $\Psi \cap \overline{S}=\varnothing$. $\Psi$ is a weakly S− prime hyperideal of $\Gamma$ if and only if $\Psi /I$ is a weakly $\overline{S}$− hyperideal of quotient $\Gamma /\Omega$ multiplicative hyperring.

Proof: 

$(\Rightarrow ):$ Suppose that $\Psi$ is a weakly S− prime hyperring of $\Gamma$. For $\varrho ,\sigma \in \Gamma$ if $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq P$ then there exists $s\in S$ with $s\diamond \varrho \subseteq \Psi$ or $s\diamond \sigma \subseteq \Psi$ Now let $\left\{0_{\Gamma /\Omega }\right\}\ne (\varrho {+}^{\prime }\Omega )*(\sigma {+}^{\prime }\Omega )=(\varrho \diamond \sigma ){+}^{\prime }\Omega \subseteq \Psi /\Omega$ as if $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq \Psi$ then there exists $s\in S$ either $s\diamond \varrho \subseteq \Psi$ or $s\diamond \sigma \subseteq \Psi$. We find there exists $s{+}^{\prime }\Omega \in \overline{S}$ either $(s\diamond \varrho ){+}^{\prime }\Omega =(s{+}^{\prime }\Omega )*(\varrho {+}^{\prime }\Omega )\subseteq \Psi /\Omega$ or $(s\diamond \sigma ){+}^{\prime }\Omega =(s{+}^{\prime }\Omega )*(\sigma {+}^{\prime }\Omega )\subseteq \Psi /\Omega$ . Thus, $\Psi /\Omega$ is a weakly $\overline{S}$− prime hyperideal of $\Gamma /\Omega$.

$(\Leftarrow ):$ According to our acceptance; $\Psi /\Omega$ is a weakly S− prime of $\Gamma /\Omega$ and $(\Psi /\Omega )\cap \overline{S}=\varnothing$ hence $\Psi \cap S=\varnothing$. Let $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq \Psi$ for $\varrho ,\sigma \in \Gamma$. If $\left\{0_{\Gamma /\Omega }\right\}\ne (\varrho {+}^{\prime }\Omega )*(\sigma {+}^{\prime }\Omega )=(\varrho \diamond \sigma ){+}^{\prime }\Omega \subseteq Psi/\Omega$ then there exists $\overline{s}=s{+}^{\prime }\Omega \in \overline{S}$ either $(s{+}^{\prime }\Omega )*(\varrho {+}^{\prime }\Omega )=(s\diamond \varrho ){+}^{\prime }\Omega \subseteq \Psi /\Omega$ or $(s{+}^{\prime }\Omega )*(\sigma {+}^{\prime }\Omega )=(s\diamond \sigma ){+}^{\prime }\Omega \subseteq \Psi /\Omega$ Thus we find there exists $s\in S$ with $s\diamond \varrho \subseteq \Psi$ or $s\diamond \sigma \subseteq \Psi$. That is, $\Psi$ is a weakly S− prime hyperideal of $\Gamma$.

Example 3.22. 

From Example $3.4$, we know that $\Theta =〈6x〉$ is a weakly S− prime hyperideal of $T=({\Gamma }_A\left[\varrho \right],+,*)$ . Let we use this Example $3.4$ and let $\left\{0\right\}\ne \Omega \Lambda \subseteq \Theta$ for hyperideals $\Lambda =〈2x〉$ and $\Omega =〈3x〉$ hyperideals of T. then $s\diamond \Omega ⊈\Theta$ but $s\diamond \Lambda \subseteq \Theta$.

Now, let us prove that the example we gave is generally true.

Theorem 3.23. 

Let $\Gamma$ be a multiplicative hyperring $S\subseteq \Gamma$ be multiplicative closed subset of $\Gamma$ and $\Phi$ be a hyperideal of $\Gamma$ with $\Phi \cap S=\varnothing$. $\Phi$ is a weakly S− prime hyperideal of $\Gamma$ if and only if for each $\Omega ,\Lambda$ hyperideals of $\Gamma$ if $\left\{0\right\}\ne \Omega \Lambda \subseteq \Phi$ then there exists $s\in S$ either $s\diamond \Omega \subseteq \Phi$ or $s\diamond \Lambda \subseteq \Phi$.

Proof: 

$(\Rightarrow ):$ Let $\Phi$ be a weakly S− prime hyperideal of $\Gamma$ so we have there exists an $s\in S$ such that for $\varrho ,\sigma \in \Gamma$ with $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq \Phi$, then $s\diamond \varrho \subseteq \Phi$ or $s\diamond \sigma \subseteq \Phi$. Now, for $\Omega$ and $\Lambda$ be two hyperideals of $\Gamma$. Let for all $s\in S$ , $\left\{0\right\}\ne \Omega \Lambda \subseteq \Phi$ both $s\diamond \Omega ⊈\Phi$ and $s\diamond \Lambda ⊈\Phi$. Therefore for $\varrho ,\sigma \in \Gamma$; while $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq \Phi$ ; $s\diamond \varrho ⊈\Phi$ and $s\diamond \sigma ⊈\Phi$ where $\varrho \in \Omega ,\sigma \in \Lambda$. We deduce that $\Phi$ is a weakly S− prime hyperideal $\Gamma$, which is a contradiction.

$(\Leftarrow ):$ For $\varrho ,\sigma \in \Gamma$ let $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq \Phi$ We have $〈\left\{0\right\}〉\ne 〈\varrho \diamond \sigma 〉\subseteq \Phi$ for $〈\varrho 〉\in \Omega$, $〈\sigma 〉\in \Lambda$ by [7] .

According to hypothesis since $s\in S$ such that $s\diamond 〈\varrho 〉\subseteq \Phi$ or $s\diamond 〈\sigma 〉\subseteq \Phi$ so we get either $s\diamond \varrho \subseteq \Phi$ or $s\diamond \sigma \subseteq \Phi$ is obtained and desired result is achieved.

Corollary 3.24. 

Suppose that $\Gamma$ is a multiplicative hyperring , $S\subseteq \Gamma$ is a multiplicative closed subset of $\Gamma$ and $\Phi$ is a hyperideal of $\Gamma$ with $\Phi \cap S=\varnothing$ $\Phi$ is a weakly S− prime hyperideal of $\Gamma$ if only if for some $i\in \left\{1,2,\dots ,n\right\}$ and for all ${\Omega }_1,{usmega}_2,\dots ,{\Omega }_n$ are hyperideals of $\Gamma$ if $\left\{0\right\}\ne {\Omega }_1{\Omega }_2\dots {\Omega }_n\subseteq \Phi$ then there exists $s\in S$ such that $s\diamond {\Omega }_i\subseteq \Phi$.

Proof: 

$(\Rightarrow ):$ Let $\Phi$ be a weakly S− prime hyperideal of $\Gamma$ Now let we apply the inductive theorem for $n\ge 3$. Let us assume our claim is true for $n-1$ and

let $\left\{0\right\}\ne {\Omega }_1{\Omega }_2\dots {\Omega }_n$ for ${\Omega }_1,{\Omega }_2,\dots ,{\Omega }_n$ hyperideals for $\Gamma$. We have $\left\{0\right\}\ne ({\Omega }_1{\Omega }_2\dots {\Omega }_{n-1}){\Omega }_n\subseteq \Phi$ then there exists $s\in S$ with $s\diamond {\Omega }_n\subseteq \Phi$. If $s\diamond ({\Omega }_1{\Omega }_2\dots {\Omega }_{n-1})\subseteq \Phi$ then $(s\diamond s)\diamond {\Omega }_1\subseteq \Phi$ or $s\diamond {\Omega }_i\subseteq \Phi (i\in \left\{2,\dots ,n-1\right\})$ and if $(s\diamond s)\diamond {\Omega }_1\subseteq \Phi$; then there exists $t\in s\circ s\cap S$ such that $t\diamond {\Omega }_1\subseteq \Phi$. In this situation, $s\diamond t\subseteq \Phi$ or $s\diamond {\Omega }_1\subseteq \Phi$ but for some $i\in \left\{1,2,\dots ,n\right\},(s\diamond t)\cap S\subseteq \Phi \cap S=\varnothing$ must be $s\diamond {\Omega }_i\subseteq \Phi$.

$(\Leftarrow ):$ Let $\left\{0\right\}\ne {\varrho }_1\diamond {\varrho }_2\diamond \dots \diamond {\varrho }_{n-1}\diamond {\varrho }_n\subseteq \Phi$ for $i\in \left\{1,2,\dots ,n\right\}$ and ${\varrho }_1,{\varrho }_2,\dots ,{\varrho }_{n-1},{\varrho }_n\in \Gamma$. We have if $〈\left\{0\right\}〉\ne 〈{\varrho }_1〉\diamond 〈{\varrho }_2〉\diamond \dots \diamond 〈{\varrho }_{n-1}〉\diamond 〈{\varrho }_n〉\subseteq P$. Thus we find there exists an $s\in S$ with $s\diamond {\varrho }_1\subseteq \Phi ,\dots ,$ or $s\diamond {\varrho }_n\subseteq \Phi$ $(s\diamond {\varrho }_i\subseteq \Phi )$ for $i\in \left\{1,2,\dots ,n\right\}$. Therefore $\Phi$ is a weakly S− prime hyperideal of $\Gamma$.

Proposition 3.25. 

Let $\Gamma$ be a multiplicative hyperring, $S\subseteq \Gamma$ be a multiplicative closed subset of $\Gamma$ and $\Omega$ be a hyperideal of $\Gamma$ with $\Omega \cap S=\varnothing$. $\Omega$ be a weakly S− prime hyperideal of $\Gamma$ if and only if $\left\{0\right\}\ne {\varrho }_1\diamond {\varrho }_2\diamond \dots \diamond {\varrho }_n\subseteq \Omega$ then there exists $s\in S$ such that $s\diamond {\varrho }_i\subseteq \Omega$ for some $i\in \left\{1,2,\dots ,n\right\}$ and ${\varrho }_1,{\varrho }_2,\dots ,{\varrho }_n\in \Gamma$.

Proof: 

$(\Rightarrow ):$ Let $\Omega$ be a weakly S− prime hyperideal of $\Gamma$. Let $\left\{0\right\}\ne {\varrho }_1\diamond {\varrho }_2\diamond \dots \diamond {\varrho }_n\subseteq \Omega$ for some $i\in \left\{1,2,\dots ,n\right\}$ and ${\varrho }_1,{\varrho }_2,\dots ,{\varrho }_n\in \Gamma$. If $\left\{0\right\}\ne {\varrho }_1\diamond {\varrho }_2\diamond \dots \diamond {\varrho }_n\subseteq 〈\left\{0\right\}〉\ne 〈{\varrho }_1〉\diamond 〈{\varrho }_2〉\diamond \dots \diamond 〈{\varrho }_{n-1}〉\diamond 〈{\varrho }_n〉\subseteq 〈{\varrho }_1\diamond {\varrho }_2\diamond \dots \diamond {\varrho }_n〉\subseteq \Omega$ then there exists $s\in S$ with $s\diamond 〈{\varrho }_1〉\subseteq \Omega$ or $s\diamond 〈{\varrho }_2〉\subseteq \Omega$ or… or $s\diamond 〈{\varrho }_n〉\subseteq \Omega$. Therefore we get $s\diamond {\varrho }_i$ for some $i\in \left\{1,2,\dots ,n\right\}$.

$(\Leftarrow ):$ Let $\left\{0\right\}\ne {\varrho }_1\diamond {\varrho }_2\subseteq \Omega$ for ${\varrho }_1,{\varrho }_2\in \Gamma$ and for $n=2$. Since $s\in S$ such that $s\diamond {\varrho }_1\subseteq \Omega$ or $s\diamond {\varrho }_2\subseteq \Omega$. Hence $\Omega$ is a weakly S− prime hyperideal of $\Gamma$.

Corollary 3.26. 

Suppose that $\Gamma$ is a multiplicative hyperring, $S\subseteq \Gamma$ is a multiplicative closed subset of $\Gamma$ and $\Omega$ is a hyperideal of $\Gamma$ such that $\Lambda \cap S=\varnothing$. Then $\Lambda$ is a weakly prime hyperideal of $\Gamma$ if and only if for all hyperideals ${\Omega }_1,{\Omega }_2,\dots ,{\Omega }_n$ of $\Gamma$, if $\left\{0\right\}\ne {\Omega }_1{\Omega }_2\dots {\Omega }_n\subseteq \Lambda$ then ${\Omega }_1,{\Omega }_2,\dots ,{\Omega }_n\subseteq \Lambda$ for some $i\in \left\{1,2,\dots ,n\right\}$ .

Proof: 

Take $S=\left\{1\right\}$ in Corollary $3.24$.

According to Definition $2.16$, we know that the intersection of the prime hyperideals of $\Gamma$ containing the hyperideal $\Lambda$ of $\Gamma$ of the multiplicative hyperring $\Gamma$ is called the prime radical of $\Lambda$, and if $\Lambda$ is a C− hyperideal of $\Gamma$, then $Rad(\Lambda )=D(\Lambda )=\left\{\gamma \in \Gamma |{\gamma }^n\subseteq \Lambda \right\}$ for some $n\in \mathbb{N}$ . Therefore, if $\Lambda$ is a C− hyperideal of the multiplicative hyperring $\Gamma$, we will show in the following proposition and theorem, respectively, the property satisfied by Proposition $3.25$ for the set $Rad(\Lambda )$ and the property obtained in Proposition $3.16$ for many C− hyperideals of $\Gamma$.

Proposition 3.27. 

Assume that $\Gamma$ is a multiplicative hyperring, $S\subseteq \Gamma$ is a multiplicative closed subset of $\Gamma$ , $\Omega$ is a hyperideal of $\Gamma$ such that $\Omega \cap S=\varnothing$. If $\Omega$ is a weakly S− prime hyperideal of $\Gamma$ and $\Lambda$ is a C− hyperideal of $\Gamma$ with $\Lambda \subseteq \Omega$ then there exists $s\in S$ such that $s\diamond \left(Rad\right(\Lambda \left)\right)\subseteq \Omega$.

Proof: 

Let $\varrho \in Rad(\Lambda )$. As $\Lambda$ is a C− hyperideal of $\Gamma$ so $Rad(\Lambda )=D(\Lambda )=\left\{\gamma \in \Gamma |{\gamma }^n\subseteq \Lambda ,n\in \mathbb{N}\right\}$ We get ${\varrho }^n\subseteq \Lambda \subseteq \Omega$ $(\exists n\in \mathbb{N})$ and if $\left\{0\right\}\ne \varrho \diamond \varrho \diamond \dots \diamond \varrho \subseteq \Omega$ then there exists $s\in S$ such that $s\diamond {\varrho }^n\subseteq \Omega$ and hence $s\diamond \varrho \subseteq \Omega$ Therefore $s\diamond \left(Rad\right(\Lambda \left)\right)\subseteq \Omega$.

Theorem 3.28. 

Suppose that $\Gamma$ is multiplicative hyperring, $S\subseteq \Gamma$ is a multiplicative closed subset of $\Gamma$, $\Lambda$ is a hyperideal of $\Gamma$ and ${\Omega }_1,{\Omega }_2,\dots ,{\Omega }_n$ are weakly S− prime C− hyperideals of such that $\Lambda \subseteq {\bigcup }_{i=1}^n{\Omega }_i$ thus there exists an $s\in S$ such that $s\diamond \Lambda \subseteq {\Omega }_i$ for $i\in \left\{1,2,\dots ,n\right\}$.

Proof: 

Let ${\Omega }_i$ be weakly S− prime hyperideals of $\Gamma$ for all $1\le i\le n$ and let ${\bigcup }_{i=1}^n{\Omega }_i$. If ${\Omega }_i$ are weakly S− prime hyperideals then $({\Omega }_i:s_i)$ are weakly prime hyperideal for $s_i\in S$. If $\Lambda \subseteq {\bigcup }_{i=1}^n{\Omega }_i\subseteq {\bigcup }_{i=1}^n({\Omega }_i:s_i)$ by[7],there exists $i\in \left\{1,2,\dots ,n\right\}$ such that $\Lambda \subseteq ({\Omega }_i:s_i)$ and hence $s_i\diamond \Lambda \subseteq {\Omega }_i$.

Theorem 3.29. 

Let $\Gamma ={\Gamma }_1\times {\Gamma }_2$ be a multiplicative hyperring and let $S_1,{\Gamma }_1$ and $S_2,{\Gamma }_2$ respectively be multiplicative closed subsets of each other and $S=S_1\times S_2$. Let $0\ne {\Omega }_1$ and $0\ne {\Omega }_2$ be proper hyperideals of ${\Gamma }_1$ and ${\Gamma }_2$, respectively. The following statements are equivalent.

(i)$I={\Omega }_1\times {\Omega }_2$ be a weakly S− prime hyperideal of $\Gamma$.

(ii)${\Omega }_1$ be a weakly S− of weakly S− prime hyperideal of ${\Gamma }_1$ and $S_2\cap {\Omega }_2\ne \varnothing$ or ${\Omega }_2$ be a weakly S− prime hyperideal of ${\Gamma }_2$ and $S_1\cap {\Omega }_1\ne \varnothing$.

(iii)$I={\Omega }_1\times {\Omega }_2$ be a S− prime hyperideal.

Proof: 

$\left(i\right)\Rightarrow \left(ii\right):$ Let $\Omega ={\Omega }_1\times {\Omega }_2$ be a weakly S− prime hyperideal of $\Gamma$. Let $\left\{0\right\}\ne (\varrho ,\sigma )=(\varrho ,1)\diamond (1,\sigma )\subseteq \Omega$ for $\varrho \in {\Gamma }_1$ and $\sigma \in {\Gamma }_2$ Then there exists $s^{\prime }=(s_1,s_2)\in S=S_1\times S_2$ either $(s_1,s_2)\diamond (\varrho ,1)=(s_1\diamond \varrho ,s_2)\subseteq \Omega$ or $(s_1,s_2)\diamond (1,\sigma )=(s_1,s_2\diamond \sigma )\subseteq \Omega$, we have $S_1\cap {\Omega }_1=\varnothing$ and $S_2\cap {\Omega }_2=\varnothing$. Now let $S_2\cap {\Omega }_2=\varnothing$ $(S_1\cap {\Omega }_1=\varnothing )$ as $\Omega \cap S=\varnothing$ so $S_1\cap {\Omega }_1=\varnothing$ We will show that ${\Omega }_1$ be a weakly S− prime hyperideal of ${\Gamma }_1$. Let $\varrho \diamond {\varrho }_1\subseteq \Omega$ for $\varrho ,{\varrho }_1\in {\Gamma }_1$. Since $S_2\cap {\Gamma }_2=\varnothing$. We find $\left\{0\right\}=(\varrho ,\sigma )\diamond ({\varrho }_1,1)\subseteq \Omega$ for $0\ne \sigma \in S_2\cap {\Omega }_2$ thus there exists $s^{\prime }=(s_1,s_2)\in S$ either $(s_1,s_2)\diamond (\varrho ,\sigma )=(s_1\diamond \varrho ,s_2\diamond \sigma )\subseteq \Omega$ or $(s_1,s_2)\diamond ({\varrho }_1,1)=(s_1\diamond {\varrho }_1,s_2)\subseteq \Omega$. Since either $s_1\diamond \varrho \subseteq \Omega$ or $s_1\diamond {\varrho }_1\subseteq {\Omega }_1$ then ${\Omega }_1$ is a weakly $S_1$− prime hyperideal of ${\Gamma }_1$. Similarly, if $S_1\cap {\Omega }_1=\varnothing$ then ${\Omega }_2$ is a weakly $S_2$− prime hyperideal of ${\Gamma }_2$.

$\left(ii\right)\Rightarrow \left(iii\right):$ Let ${\Omega }_1$ be a weakly $S_1$− prime hyperideal of ${\Gamma }_1$ and $S_2\cap {\Omega }_2\ne \varnothing$. Let $({\varrho }_1,{\sigma }_1)\diamond ({\varrho }_2,{\sigma }_2)=({\varrho }_1\diamond {\varrho }_2,{\sigma }_1\diamond {\sigma }_2)\subseteq \Omega$ for ${\varrho }_1,{\varrho }_2\in {\Gamma }_1,{\sigma }_1,{\sigma }_2\in {\Gamma }_2$ and $s_2\in S_2\cap {\Omega }_2$. If ${\varrho }_1\diamond {\varrho }_2\subseteq \Omega$ then there exists $s_1\in S_1$ such that $s_1\diamond {\varrho }_1\subseteq {\Omega }_1$ or $s_1\diamond {\varrho }_2\subseteq {\Omega }_1$ and if $s=(s_1,s_2)\in S$ then we get $(s_1,s_2)\diamond ({\varrho }_1,{\sigma }_1)=(s_1\diamond {\varrho }_1,s_2\diamond {\sigma }_1)\subseteq \Omega$.

Consequently; $\Omega$ is a $S_1$− prime hyperideal of ${\Gamma }_1$. Similarly, if $S_1\cap {\Omega }_1\ne \varnothing$ then ${\Omega }_2$ is a $S_2$− prime hyperideal of ${\Gamma }_2$.

$\left(iii\right)\Rightarrow \left(i\right):$ Every S− prime hyperideal is a weakly S− prime hyperideal.

Theorem 3.30. 

Let $\Gamma ={\Gamma }_1\times {\Gamma }_2\times \dots \times {\Gamma }_m$ be a multiplicative hyperring and $S=S_1\times S_2\times \dots \times S_m$ for $m\ge 2$. Let $S_k$ respectively be a multiplicative closed subset of ${\Gamma }_k$ for each $k=1,2,\dots ,m$ and $0\ne {\Omega }_k$ be a proper hyperideal of ${\Gamma }_k$. The following statements are equivalent.

$\left(i\right)$$\Omega ={\Omega }_1\times {\Omega }_2\times \dots \times {\Omega }_m$ be a weakly S− prime hyperideal of $\Gamma$.

$\left(ii\right)$ For all $k\in \left\{1,2,\dots ,m\right\}-\left\{n\right\}$, $S_k\cap {\Omega }_k\ne \varnothing$ and ${\Omega }_n$ be $S_n$− prime hyperideal of ${\Gamma }_n$ for some $n\in \left\{1,2,\dots ,m\right\}$.

$\left(iii\right)$$\Omega ={\Omega }_1\times \dots \times {\Omega }_m$ be an S− prime hyperideal of $\Gamma$.

Proof: 

$\left(i\right)\Rightarrow \left(ii\right):$ Suppose that $\Omega ={\Omega }_1\times \dots \times {\Omega }_m$ is a weakly S− prime hyperideal of $\Gamma$. We will continue by induction on m. Let ${\Gamma }^{\prime }={\Gamma }_1\times \dots \times {\Gamma }_{m-1}$, $S^{\prime }=S_1\times \dots \times S_{m-1}$ and ${\Omega }^{\prime }=I_1\times \dots \times {\Omega }_{m-1}$ be true for $t<m$. Thus $\left\{0\right\}\ne {\Omega }^{\prime }={\Omega }_1\times \dots \times {\Omega }_{m-1}$ is a weakly $(S^{\prime }=S_1\times \dots \times S_{m-1})$− prime hyperideal of ${\Gamma }^{\prime }={\Gamma }_1\times \dots \times {\Gamma }_{m-1}$. We have ${\Omega }_m$ is an $S_m$− prime hyperideal of ${\Gamma }_m$ with ${\Omega }^{\prime }\cap S^{\prime }\ne \varnothing$ or ${\Omega }^{\prime }$ is an $S^{\prime }$− prime hyperideal of ${\Gamma }^{\prime }$ such that ${\Omega }_m\cap S_m\ne \varnothing$. If ${\Omega }^{\prime }\cap S^{\prime }=\varnothing$ and ${\Omega }_m$ be $S_m$− prime hyperideal of ${\Gamma }_m$, then the proof is complete. Let ${\Omega }_m\cap S_m\ne \varnothing$ and weakly S− prime hyperideal of ${\Gamma }^{\prime }$ to ${\Omega }^{\prime }$. If we apply induction theorem for $k=m-1$ then ${\Omega }_i$ is $S_i$− prime hyperideal of ${\Gamma }_i$ for some $i\in \left\{1,2,\dots ,m\right\}$ and for ${\Omega }_i\cap S_i\ne \varnothing$.

$\left(ii\right)\Rightarrow \left(iii\right):$ Let we apply induction theorem for m. Let we assume our claim is true for $t\le m$ and let ${\Gamma }^{\prime }={\Gamma }_1\times \dots \times {\Gamma }_{m-1}$, $S^{\prime }=S_1\times \dots \times S_{m-1}$ and ${\Omega }^{\prime }={\Omega }_1\times \dots \times {\Omega }_{m-1}$ If $\Gamma ={\Gamma }^{\prime }\times {\Gamma }_m$, $S=S^{\prime }\times S_m$ and $\Omega ={\Omega }^{\prime }\times {\Omega }_m$ then ${\Omega }_m$ is $S_m$− prime hyperideal of ${\Gamma }_m$ and ${\Omega }^{\prime }\cap S^{\prime }\ne \varnothing$ or if ${\Gamma }^{\prime }$ is a $S^{\prime }$− prime hyperideal of ${\Gamma }^{\prime }$ then ${\Omega }_m\cap S_m\ne \varnothing$. If ${\Omega }_m$ be $S_m$− prime hyperideal of ${\Gamma }_m$ and ${\Omega }^{\prime }\cap S^{\prime }=\varnothing$, then the proof is complete. Now suppose that ${\Omega }^{\prime }\cap S^{\prime }=\varnothing$ and ${\Omega }^{\prime }$ is an $S^{\prime }$− prime hyperideal of ${\Gamma }^{\prime }$, and let us apply the induction theorem for $t=m-1$. Therefore ${\Omega }_i$ be $S_i$− prime hyperideal of ${\Gamma }_i$ for some $i\in \left\{1,2,\dots ,m\right\}-\left\{n\right\}$ and ${\Omega }_i\cap S_i\ne \varnothing$. Conversely, assume that ${\Omega }_1$ is a $S_1$− prime hyperideal of ${\Gamma }_1$ for $s_1\in S_1$ and for all $k\ne 1$, $s_k\in S_k\cap {\Omega }_k$. If $s=(t,s_2,\dots ,s_m)$ then $\Omega ={\Omega }_1\times \dots \times {\Omega }_m$ be S− prime hyperideal of $\Gamma ={\Gamma }_1\times \dots \times {\Gamma }_m$.

$\left(iii\right)\Rightarrow \left(i\right):$ Every S− prime hyperideal is a weakly S− prime hyperideal.

Proposition 3.31. 

Suppose that $\Gamma$ is a multiplicative hyperring, $S\subseteq \Gamma$ is a multiplicative closed subset of $\Gamma$ and if $\Omega$ is a weakly S− prime hyperideal of $\Gamma$ such that $\Omega \cap S=\varnothing$ then $S^{-1}\Omega$ is a weakly prime hyperideal of $S^{-1}\Gamma$.

Proof: 

Since $S\cap \Omega =\varnothing$ we have $S^{-1}\Omega \ne S^{-1}\Gamma$. Let $\left\{0\right\}\ne {\displaystyle \frac{\varrho }{s_1}}\otimes {\displaystyle \frac{\sigma }{s_2}}\subseteq S^{-1}\Omega$ for $\varrho ,\sigma \in \Gamma$ and for $s_1,s_2\in S$. If ${\displaystyle \frac{\varrho }{s_1}}\otimes {\displaystyle \frac{\sigma }{s_2}}={\displaystyle \frac{\vartheta }{s_3}}$ then there exists $t\in S$ such that $\left\{0\right\}\ne t\diamond s_3\diamond \varrho \diamond \sigma =t\diamond s_1\diamond s_2\diamond \vartheta \subseteq \Omega$ for some $\vartheta \in \Omega$ and $s_3\in S$. Because $S\cap \Omega =\varnothing$ we get $s\diamond (\varrho \diamond \sigma )\subseteq \Omega$. Therefore $s\diamond (s\diamond \varrho )\subseteq \Omega$ or $s\diamond \sigma \subseteq \Omega$ then we get $s\diamond \varrho \subseteq \Omega$ or $s\diamond \sigma \subseteq \Omega$ and hence ${\displaystyle \frac{\varrho }{s_1}}={\displaystyle \frac{s\diamond \varrho }{s\diamond s_1}}\subseteq S^{-1}\Omega$ or ${\displaystyle \frac{\sigma }{s_2}}={\displaystyle \frac{s\diamond \sigma }{s\diamond s_2}}\subseteq S^{-1}\Gamma$ as required.

Proposition 3.32. 

Assume that $\Gamma$ is a multiplicative hyperring, $S,T\subseteq \Gamma$ are multiplicative closed subsets of $\Gamma$ and $\Omega$ is a weakly T− prime hyperideal of $\Gamma$ with $\Omega \cap S=\varnothing$ and $\Omega \cap T=\varnothing$. If $\Omega$ is a weakly T− prime hyperideal of $\Gamma$ then $S^{-1}\Omega$ is an $S^{-1}T$− prime hyperideal of $S^{-1}\Gamma$.

Proof: 

If T is a multiplicative closed subset of $\Gamma$ then $S^{-1}T$ is a multiplicative closed subset of $S^{-1}\Gamma$ by [10]. Therefore $T\cap \Omega =\varnothing$ we have $(S^{-1}\Omega )\cap \left(S^{-1}T\right)=\varnothing$. Let $\left\{0\right\}\ne {\displaystyle \frac{\varrho }{s_1}}\otimes {\displaystyle \frac{\sigma }{s_2}}={\displaystyle \frac{\varrho \diamond \sigma }{s_1\diamond s_2}}\subseteq S^{-1}\Omega$ for $\varrho ,\sigma \in \Gamma$ and $s_1,s_2\in S$. Therefore there exists $s\in S$ with $\left\{0\right\}\ne s\diamond (\varrho \diamond \sigma )\subseteq \Omega$. According to our hypothesis, there exists $t\in T$ either $t\diamond (s\diamond \varrho )\subseteq \Omega$ or $t\diamond \sigma \subseteq \Omega$ thus we find ${\displaystyle \frac{t}{1}}\otimes \left({\displaystyle \frac{\varrho }{s_1}}\right)={\displaystyle \frac{t}{1}}\otimes {\displaystyle \frac{s\diamond \varrho }{s\diamond s_1}}\subseteq S^{-1}\Omega$ or ${\displaystyle \frac{t}{1}}\otimes \left({\displaystyle \frac{\sigma }{s_2}}\right)\subseteq S^{-1}\Omega$ Hence $S^{-1}\Omega$ is a $S^{-1}T$− prime hyperideal of $S^{-1}\Gamma$.

Let us define the concept of a saturated multiplicative set, as defined in classical algebra, for the localization of multiplicative closed hyperrings.

Definition 3.33. 

Let $\Gamma$ be a multiplicative hyperring and $S\subseteq \Gamma$ be a multiplicative closed subset of $\Gamma$. A set S is called a saturated multiplicative closed subset if $\varrho \diamond \sigma \subseteq S$ then both $\varrho \in S$ and $\sigma \in S$ for all $\varrho ,\sigma \in \Gamma$.

Example 3.34. 

Let $(\Gamma ,+,*)$ be a commutative ring with identity 1. Let us define a hyperoperation $\varrho \diamond \sigma =\left\{\varrho .\sigma ,2.\varrho .\sigma ,3.\varrho .\sigma ,\dots \right\}$. Then $(\Gamma ,{+}^{\prime }\diamond )$ is a commutative with identity multiplicative ring. Let $\varrho ,\sigma \in \Gamma$ such that $\varrho ,\sigma \notin nil(\Gamma )$ and let $S=\Gamma -nil(\Gamma )$. Then S is a saturated multiplicative closed subset of $\Gamma$.

Suppose that S is a multiplicative closed subset of $\Gamma$, $S^{*}=\left\{\gamma \in \Gamma |{\displaystyle \frac{\gamma }{1}}\in S^{-1}\Gamma \right\}$ which is ${\displaystyle \frac{\gamma }{1}}$ is unit, denotes the saturation of S. Also, $S^{*}$ is a multiplicative closed subset containing S. A multiplicative closed subset S of $\Gamma$ is called saturated if $S=S^{*}$. Hence $S^{*}$ is always a saturated multiplicative closed subset of $\Gamma$.

Proposition 3.35. 

Let $\Gamma$ be a multiplicative hyperring , $S\subseteq \Gamma$ be a multiplicative closed subset of $\Gamma$ and $\Omega$ be a hyperideal of $\Gamma$ which is $\Omega \cap S=\varnothing$. $\Omega$ is a weakly S− prime hyperideal of $\Gamma$ if and only if $\Omega$ is a weakly $S^{*}$− prime hyperideal.

Proof: 

Since $S^{*}$ is a saturation of S such that $S\subseteq S^{*}$ we have $\Omega \cap S^{*}=\varnothing$ If $r\in S^{*}$ then there exists $r^{\prime }\in S^{*}$ such that $r\diamond r^{\prime }\in S$. Let $r\in S^{*}$ we get ${\displaystyle \frac{r}{1}}\diamond {\displaystyle \frac{\kappa }{s}}=1$ and there exists $t\in \Gamma$ which is $t\diamond (r\diamond s)=t\diamond \kappa \subseteq S$ where $\kappa \in \Gamma$ and $s\in S$. If we take $r^{\prime }=t\diamond \kappa$ we get $r^{\prime }\diamond r=t\diamond (r\diamond s)=t\diamond \kappa \subseteq S$ and let $r^{\prime }=t\diamond \kappa$ we find $r^{\prime }\diamond r=(t\diamond \kappa )\diamond r=1$. Hence $\Omega$ is a weakly $S^{*}$− prime hyperideal of $\Gamma$.

The following proposition and theorem will characterize weakly S− prime hyperideals.

Proposition 3.36. 

Assume that $\Gamma$ is a multiplicative hyperring containing regular elements, $S\subseteq \Gamma$ is a multiplicative closed subset of $\Gamma$ and $\Omega$ is a hyperideal of $\Gamma$ such that $S\cap \Omega =\varnothing$. The following statements are equivalent.

$\left(i\right)$ $\Omega$ is a weakly S− prime hyperideal of $\Gamma$.

$\left(ii\right)$ $(\Omega :s)$ is a weakly prime hyperideal of $\Gamma$.

$\left(iii\right)$ Let $S^{-1}\Omega$ be weakly prime hyperideal of $S^{-1}\Gamma$ and for all $t\in S$ there exists $s\in S$ such that $(\Omega :t)\subseteq (\Omega :s)$ .

$\left(iv\right)$ $S^{-1}\Omega$ be a weakly prime hyperideal of $S^{-1}\Gamma$ and $(S^{-1}\Omega )\cap \Gamma =(\Omega :s)$ for some $s\in S$.

Proof: 

$\left(i\right)\iff \left(ii\right):$ $\Omega$ is a weakly S− prime hyperideal if and only if $(\Omega :s)$ is a weakly prime hyperideal of $\Gamma$ thus by Proposition $3.16$.

$\left(i\right)\Rightarrow \left(iii\right):$ If $\Omega$ is a weakly S− prime hyperideal of $\Gamma$ then $S^{-1}\Omega$ is a weakly prime hyperideal of $S^{-1}\Gamma$. Let $s\in S$ which is associated with $\Omega$ and assume that $\left\{0\right\}\ne \varrho \subseteq (\Omega :t)$ for $t\in S$ then we have $\left\{0\right\}\ne t\diamond \varrho \subseteq \Omega$. Since $\Omega$ is a weakly S− prime hyperideal of $\Gamma$ then there exists $s\in S$ either $s\diamond t\subseteq \Gamma$ or $s\diamond \varrho \subseteq \Omega$. Also, since $\Omega \cap S=\varnothing$ we get $s\diamond t\subseteq \Omega$, that is $s\diamond \varrho \subseteq \Omega$. Hence $\varrho \in (\Omega :s)$.

$\left(iii\right)\Rightarrow \left(i\right):$ We may assume $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq \Omega$ where $\varrho ,\sigma \in \Gamma$. Therefore we have either ${\displaystyle \frac{\varrho }{1}}\subseteq S^{-1}\Gamma$ or ${\displaystyle \frac{\sigma }{1}}\subseteq S^{-1}\Omega$ with $\left\{0\right\}\ne {\displaystyle \frac{\varrho }{1}}\otimes {\displaystyle \frac{\sigma }{1}}\subseteq S^{-1}\Omega$. If ${\displaystyle \frac{\varrho }{1}}\subseteq S^{-1}\Omega$ then there exists $\mu \in \Omega$ and $t\in S$ such that ${\displaystyle \frac{\varrho }{1}}={\displaystyle \frac{\mu }{t}}$ thus $t\diamond \varrho \subseteq \Omega$ so we get $\varrho \subseteq (\Omega :t)\subseteq (\Omega :s)$.

Similarly, it is obtained for ${\displaystyle \frac{\sigma }{1}}\subseteq S^{-1}\Omega$, thus the desired result is achieved.

$\left(iv\right)\Rightarrow \left(i\right):$ Let $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq \Omega$ where $\varrho ,\sigma \in \Gamma$ so if $\left\{0\right\}\ne {\displaystyle \frac{\varrho }{1}}\diamond {\displaystyle \frac{\sigma }{1}}\subseteq S^{-1}\Omega$ then there exists either ${\displaystyle \frac{\varrho }{1}}\subseteq S^{-1}\Omega$ or ${\displaystyle \frac{\sigma }{1}}\subseteq S^{-1}\Omega$. If ${\displaystyle \frac{\varrho }{1}}\subseteq S^{-1}\Omega$ then there exists $\mu \in \Omega$ or $\nu \in \Omega$ such that ${\displaystyle \frac{\varrho }{1}}={\displaystyle \frac{\mu }{\nu }}$. Therefore we find $\nu \diamond \varrho \subseteq \Omega$ and $\varrho ={\displaystyle \frac{\nu \diamond \varrho }{\nu }}\subseteq (S^{-1}\Omega )\cap \Gamma$. According to our assumption $\varrho \subseteq (\Omega :s)$ then $s\diamond \varrho \subseteq \Omega$. Similarly it is obtained for ${\displaystyle \frac{\sigma }{1}}\subseteq S^{-1}\Omega$. Hence $\Omega$ is a weakly S− prime of $\Gamma$.

Theorem 3.37. 

Suppose that $\Gamma$ is a multiplicative hyperring $S\subseteq \Gamma$ is a multiplicative closed subset of $\Gamma$ and $\Omega$ is a hyperideal of $\Gamma$ such that $\Omega \cap S=\varnothing$. Then the following assertions are equivalent.

$\left(i\right)$ $\Omega$ is a weakly S− prime hyperideal of $\Gamma$.

$\left(ii\right)$ There exists $s\in S$ such that $(\Omega :\varrho )\subseteq (\Omega :s)$ or $(\Omega :\varrho )=(0:\varrho )$ for each $\varrho \notin (\Omega :s)$

$\left(iii\right)$ There is an $s\in S$, such that if $\left\{0\right\}\ne \Phi \Lambda \subseteq \Omega$ then $s\diamond \Phi \subseteq \Omega$ or $s\diamond \Lambda \subseteq \Omega$ for each $\Phi ,\Lambda$ two hyperideals of $\Gamma$.

Proof: 

$\left(i\right)\Rightarrow \left(ii\right):$ Since $\Omega$ is a weakly S− prime hyperideal of $\Gamma$, there exists an $s\in S$ such that if $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq \Omega$ then $s\diamond \varrho \subseteq \Omega$ for $\varrho ,\sigma \in \Gamma$. Let $\varrho \in \Omega /(\Omega :s)$ and $(\Omega :\varrho )\ne (0:\varrho )$ We will show that $(\Omega :\varrho )\subseteq (\Omega :s)$. Since $(0:\varrho )\subseteq (\Omega :\varrho )$ there exists $\omega \in (\Omega :\varrho )$ with $\left\{0\right\}\ne \omega \diamond \varrho \subseteq \Omega$. Since $s\diamond \varrho ⊈\Omega$ we have $s\diamond \omega \subseteq \Omega$. Now, let $\sigma \in (\Omega :\varrho )$. We have $\varrho \diamond \sigma \subseteq \Omega$ If $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq \Omega$ then $s\diamond \sigma \subseteq \Omega$ so we get $\sigma \in (\Omega :s)$. Suppose that $\left\{0\right\}\ne \varrho \diamond \sigma$. We have $\left\{0\right\}\ne \varrho \diamond \left(\omega {+}^{\prime }\sigma \right)=\varrho \diamond \omega \subseteq \Omega$ Since $\Omega$ is a weakly S− prime hyperideal then we have $s\diamond \left(\omega {+}^{\prime }\sigma \right)\subseteq \Omega$. Hence $s\diamond \sigma \subseteq \Omega$ so we have $\sigma \in (\Omega :s)$. Consequently $(\Omega :\varrho )\subseteq (\Omega :s)$.

$\left(ii\right)\Rightarrow \left(i\right):$ Let $\left\{0\right\}\ne \varrho \diamond \sigma \subseteq \Omega$ and $s\diamond \varrho ⊈\Omega$ for $\varrho ,\sigma \in \Gamma$. Thus $\varrho \notin (\Omega :s)$. We have $\sigma \in (\Omega :\varrho )$ and $\left\{0\right\}\ne \varrho \diamond \sigma$. We will show that $(\Omega :\varrho )\subseteq (\Omega :s)$. If $\sigma \in (\Omega :s)$ then $s\diamond \sigma \subseteq \Omega$. We get $(\Omega :\varrho )\subseteq (\Omega :s)$. Therefore $\Omega$ is a weakly S− prime hyperideal of $\Gamma$.

$\left(ii\right)\Rightarrow \left(iii\right):$ Assume that $\Phi$ and $\Lambda$ are two hyperideals of $\Gamma$ then $\Phi \Lambda \subseteq \Omega$. There is an exists an $s\in S$ either $(\Omega :\varrho )\subseteq (\Omega :s)$ or $(\Omega :\varrho )=(0:\varrho )$ for all $\varrho \notin (\Omega :s)$ so we get $s\diamond \Lambda \subseteq \Omega$ and $s\diamond \Phi \subseteq \Omega$ Now let show that $\left\{0\right\}=\Omega \Lambda$. Consider $\varrho \in \Phi /(\Omega :s)$ such that $\Lambda \subseteq (\Omega :\varrho )$ for $\varrho \diamond \Lambda \subseteq \Omega$. Because $\Lambda ⊈(\Omega :s)$ so we get $\Lambda \subseteq (\Omega :\varrho )=(0:\varrho )$. Hence $\left\{0\right\}=\varrho \diamond \Lambda$.

4. Conclusions

The generalization of prime hyperideals was first undertaken by Farzalipour and colleagues. This study included a definition and basic properties of the weakly S− prime hyperideal, a generalization of the S−prime hyperideal. Furthermore, as shown in the diagram below, the relationship between prime hyperideals, weakly prime hyperideals, S− prime, and weakly S− prime hyperideals is illustrated in this study. We also find the properties of weakly S− prime hyperideals under homomorphism of multiplicative hyperrings, Cartesian product, in factor hyperrings, and the fundamental relation in the condition of multiplicative hyperrings with related results.