#  (weakly) $(s,n)$-closed hyperideals

Mahdi Anbarloei  Department of Mathematics, Faculty of Sciences, Imam Khomeini
International University, Qazvin, Iran<EMAIL_ADDRESS>
###### Abstract.

A multiplicative hyperring is a well-known type of algebraic hyperstructures
which extend a ring to a structure in which the addition is an operation but
the multiplication is a hyperoperation. Let $G$ be a commutative
multiplicative hyperring and $s,n\in\mathbb{Z}^{+}$. A proper hyperideal $Q$
of $G$ is called (weakly) $(s,n)$-closed if ($0\neq a^{s}\subseteq Q$ )
$a^{s}\subseteq Q$ for $a\in G$ implies $a^{n}\subseteq Q$. In this paper, we
aim to investigate (weakly) $(s,n)$-closed hyperideals and give some results
explaining the structures of these notions.

###### Key words and phrases:

$(s,n)$-closed hyperideal, weakly $(s,n)$-closed hyperideal, $(s,n)$-tough-
zero element.

###### 2010 Mathematics Subject Classification:

Primary 20N20; Secondary 16Y99

##  1\. Introduction

Hyperstructures, which are a generalization of classical algebraic structures,
take an important place in both pure and applied mathematics. These structures
was first introduced by Marty in 1934 [20]. He published some notes on
hypergroups as a generalization of groups. Later on, many authors have worked
on this new topic of modern algebra and developed it [11, 12, 13, 14, 15, 21,
23, 29]. Similar to hypergroups, hyperrings are algebraic structures,
subsitutiting both or only one of the binary operations of addition and
multiplication by hyperoperations. A importan type of a hyperring called the
multiplicative hyperring was introduced by Rota in 1982 [25]. In this
hyperstructure, the multiplication is a hyperoperation, while the addition is
an operation. Many illustrations and results of the multiplicative hyperring
can be seen in [3, 4, 6, 7, 8, 22, 28].

A hyperoperation ”$\circ$” on nonempty set $A$ is a mapping of $A\times A$
into $P^{*}(A)$ where $P^{*}(A)$ is the family of all nonempty subsets of $A$.
If ”$\circ$” is a hyperoperation on $A$, then $(A,\circ)$ is called
hypergroupoid. The hyperoperation on $A$ can be extended to subsets of $A$.
Let $A_{1},A_{2}$ be two subsets of $A$ and $a\in A$, then $A_{1}\circ
A_{2}=\cup_{a_{1}\in A_{1},a_{2}\in A_{2}}a_{1}\circ a_{2},$ and $A_{1}\circ
a=A_{1}\circ\\{a\\}.$ A hypergroupoid $(A,\circ)$ is called a semihypergroup
if $\cup_{v\in b\circ c}a\circ v=\cup_{u\in a\circ b}u\circ c$ for all
$a,b,c\in A$ which means $\circ$ is associative. A semihypergroup is said to
be a hypergroup if $a\circ A=A=A\circ a$ for all $a\in A$. A nonempty subset
$B$ of a semihypergroup $(A,\circ)$ is called a subhypergroup if we have
$a\circ B=B=B\circ a$ for all $b\in B$. Recall from [18] that an algebraic
structure $(G,+,\circ)$ is said to be commutative multiplicative hyperring if
(1) $(G,+)$ is a commutative group; (2) $(G,\circ)$ is a semihypergroup; (3)
for all $a,b,c\in G$, we have $a\circ(b+c)\subseteq a\circ b+a\circ c$ and
$(b+c)\circ a\subseteq b\circ a+c\circ a$; (4) for all $a,b\in G$, we have
$a\circ(-b)=(-a)\circ b=-(a\circ b)$; (5) for all $a,b\in G$, $a\circ b=b\circ
a$ If in (3) the equality holds then we say that the multiplicative hyperring
is strongly distributive. Let $(\mathbb{Z},+,\cdot)$ be the ring of integers.
Corresponding to every subset $X\in P^{\star}(\mathbb{Z})$ with $|X|\geq 2$,
there exists a multiplicative hyperring $(\mathbb{Z}_{X},+,\circ)$ with
$\mathbb{Z}_{X}=\mathbb{Z}$ and for any $a,b\in\mathbb{Z}_{X}$, $a\circ
b=\\{a.x.b\ |\ x\in X\\}$. A non empty subset $H$ of a multiplicative
hyperring $R$ is a hyperideal of $G$ if (i) $a-b\in H$ for all $a,b\in H$;
(ii) $r\circ a\subseteq H$ for all $a\in H$ and $r\in G$. The notion of
primeness of hyperideal in a multiplicative hyperring was conceptualized by
Procesi and rota in [24]. Dasgupta extended the prime and primary hyperideals
in multiplicative hyperrings in [16]. Recall from [16] that a proper
hyperideal $P$ of multiplicative hyperring $G$ is called a prime hyperideal if
$xa\circ b\subseteq P$ for $a,b\in G$ implies that $a\in P$ or $b\in P$. The
intersection of all prime hyperideals of $R$ containing hyperideal $I$ is
called the prime radical of $I$, being denoted by $rad(I)$. If the
multiplicative hyperring $G$ does not have any prime hyperideal containing
$I$, we define $rad(I)=G$. Let C be the class of all finite products of
elements of $R$ i.e. ${\bf C}=\\{r_{1}\circ r_{2}\circ...\circ r_{n}\ :\
r_{i}\in R,n\in\mathbb{N}\\}\subseteq P^{\ast}(R)$. A hyperideal $I$ of $R$ is
said to be a C-hyperideal of $R$ if, for any $A\in{\bf C},A\cap
I\neq\varnothing$ implies $A\subseteq I$. Let I be a hyperideal of $R$. Then,
$D\subseteq rad(I)$ where $D=\\{r\in R:r^{n}\subseteq I\ for\ some\
n\in\mathbb{N}\\}$. The equality holds when $I$ is a C-hyperideal of $R$([16],
proposition 3.2). Recall that a hyperideal $I$ of $R$ is called a strong
C-hyperideal if for any $E\in\mathfrak{U}$, $E\cap I\neq\varnothing$, then
$E\subseteq I$, where $\mathfrak{U}=\\{\sum_{i=1}^{n}A_{i}\ :\ A_{i}\in{\bf
C},n\in\mathbb{N}\\}$ and ${\bf C}=\\{r_{1}\circ r_{2}\circ...\circ r_{n}\ :\
r_{i}\in R,n\in\mathbb{N}\\}$ (for more details see [17]). Recall from [1]
that a proper hyperideal $I$ of multiplicative hyperring $R$ is maximal in R
if for any hyperideal $J$ of $R$ with $I\subseteq J\subseteq R$ then $J=R$.
Also, we say that $R$ is a local multiplicative hyperring, if it has just one
maximal hyperideal. An element $e$ in $G$ is an identity element if $a\in
a\circ e$ for all $a\in G$. Moreover, an element $e$ in $G$ is a scalar
identity element if $a=a\circ e$ for all $a\in G$.

In this paper, we introduce the notions of $(s,n)$-closed hyperideals and
weakly $(s,n)$-closed hyperideals of a commutative multiplicative hyperring.
Many specific results are presented to explain the structures of these
notions. Additionally, we study the relationships between the new hyperideals
and classical hyperideals and explore some ways to connect them. Throughout
this paper, $G$ denotes a commutative multiplicative hyperring.

##  2\.  $(s,n)$-closed hyperideals

In this section, we define the concept of $(s,n)$-closed hyperideals and
present some basic properties of $(s,n)$-closed hyperideals.

######  Definition 2.1.

Let $s,n\in\mathbb{Z}^{+}$. A proper hyperideal $Q$ of $G$ is called
$(s,n)$-closed if $a^{s}\subseteq Q$ for $a\in G$ implies $a^{n}\subseteq Q$.

It is clear that every $(s,n)$-closed hyperideal of $G$ is
$(s^{\prime},n^{\prime})$-closed for all
$s,s^{\prime},n,n^{\prime}\in\mathbb{Z}^{+}$ with $s^{\prime}\leq s$ and
$n^{\prime}\geq n$.

######  Example 2.2.

Consider the set of all integers $\mathbb{Z}$ with ordinary addition and the
hyperoperation $\circ$ defined as
$\alpha\circ\beta=\\{2\alpha\beta,4\alpha\beta\\}$ for all
$\alpha,\beta\in\mathbb{Z}$. Then $105\mathbb{Z}$ is an $(s,3)$-closed
hyperideal of the multiplicative hyperring $(\mathbb{Z},+,\circ)$ for all
$s\in\mathbb{Z}^{+}$.

A proper hyperideal $I$ of $G$ refers to an $n$-absorbing hyperideal of $G$ if
$x_{1}\circ\cdots\circ x_{n+1}\subseteq I$ for $x_{1},\cdots,x_{n+1}\in G$
implies that there exist $n$ of the $x_{i}^{,}$s whose product is a subset of
$I$ [5].

######  Theorem 2.3.

Let $Q$ be an $n$-absorbing ${\bf C}$-hyperideal of $G$. Then $Q$ is an
$(s,n)$-closed hyper for each $s\in\mathbb{Z}^{+}$.

###### Proof.

Assume that $Q$ is an $n$-absorbing hyperideal of $G$ and $a^{s}=a^{n}\circ
a^{s-n}\subseteq Q$ for $a\in G$ and $s,n\in\mathbb{Z}^{+}$ with $n<s$. Then
we have $a^{n}\circ b\subseteq Q$ for every $b\in a^{s-n}$. Since $Q$ is an
$n$-absorbing hyperideal of $G$, we get $a^{n}\in Q$ or $a^{n-1}\circ b\in Q$.
In the first case, we are done. In the second case, we have $a^{s-1}\subseteq
Q$. Consider $a^{s-1}=a^{n}\circ a^{s-n-1}\subseteq Q$ and pick $c\in
a^{s-n-1}$. By continuing the process mentioned, we obtain $a^{n}\subseteq Q$.
Hence $Q$ is $(s,n)$-closed for $s>n$. If $s\leq n$, then $Q$ is
$(s,n)$-closed clearly. Thus $Q$ is an $(s,n)$-closed hyperideal for each
$s\in\mathbb{Z}^{+}$. ∎

######  Theorem 2.4.

Let $Q_{1},\cdots,Q_{t}$ be some prime hyperideals of $G$. Then
$Q_{1}\circ\cdots\circ Q_{t}$ is an $(s,n)$-closed hyperideal of $G$ for all
$s,n\in\mathbb{Z}^{+}$ with $1\leq s$ and $\min\\{s,t\\}\leq n$.

###### Proof.

Assume that $a^{s}\subseteq Q_{1}\circ\cdots\circ Q_{t}$ for $a\in G$. This
means $a^{s}\subseteq Q_{i}$ for each $i\in\\{1,\cdots,t\\}$. Since $Q_{i}$ is
a prime hyperideal of $G$, we have $a\in Q_{i}$. Hence $a^{t}\subseteq
Q_{1}\circ\cdots\circ Q_{t}$. This implies that $a^{n}\subseteq
Q_{1}\circ\cdots\circ Q_{t}$ for all $\min\\{s,t\\}\leq n$. ∎

######  Theorem 2.5.

Let $Q_{1},\cdots,Q_{t}$ be some hyperideals of $G$ such that for each
$i\in\\{1,\cdots,t\\}$, $Q_{i}$ is $(s_{i},n_{i})$-closed with
$s_{i},n_{i}\in\mathbb{Z}^{+}$. Then

  * (i)

$Q_{1}\circ\cdots\circ Q_{t}$ is $(s,n)$-closed for $s,n\in\mathbb{Z}^{+}$
such that $s\leq\min\\{s_{1},\cdots,s_{t}\\}$ and
$n\geq\min\\{s,n_{1}+\cdots+n_{t}\\}.$

  * (ii)

$Q_{1}\cap\cdots\cap Q_{t}$ is $(s,n)$-closed for $s,n\in\mathbb{Z}^{+}$ such
that $s\leq\min\\{s_{1},\cdots,s_{t}\\}$ and
$n\geq\min\\{s,\max\\{n_{1},\cdots,n_{t}\\}\\}.$

###### Proof.

$(i)$ Assume that $a^{s}\subseteq Q_{1}\circ\cdots\circ Q_{t}$ for
$s\leq\min\\{s_{1},\cdots,s_{t}\\}$ and $a\in G$. This means $a^{s}\subseteq
Q_{i}$ for all $i\in\\{1,\cdots,t\\}$ and so $a^{s_{i}}\subseteq Q_{i}$. Since
$Q_{i}$ is $(s_{i},n_{i})$-closed, we get $a^{n_{i}}\subseteq Q_{i}$. Then we
conclude that $a^{n_{1}}\circ\cdots\circ
a^{n_{t}}=a^{n_{1}+\cdots+n_{t}}\subseteq Q_{1}\circ\cdots\circ Q_{t}$. This
implies that $a^{n}\subseteq Q_{1}\circ\cdots\circ Q_{t}$ for
$n_{1}\cdots+n_{t}\leq n$. Consequently, we have $a^{n}\subseteq
Q_{1}\circ\cdots\circ Q_{t}$ for all $n\geq\min\\{s,n_{1}+\cdots+n_{t}\\}$.
$(ii)$ Suppose that $a^{s}\subseteq Q_{1}\cap\cdots Q_{t}$ for
$s\leq\min\\{s_{1},\cdots,s_{t}\\}$ and $a\in G$. Hence $a^{s}\subseteq Q_{i}$
for all $i\in\\{1,\cdots,t\\}$ and then $a^{s_{i}}\subseteq Q_{i}$. Since
$Q_{i}$ is $(s_{i},n_{i})$-closed, we obtain $a^{n_{i}}\subseteq Q_{i}$. Thus
we have $a^{n}\subseteq Q_{i}$ for every $i\in\\{1,\cdots,t\\}$ and
$n\geq\max\\{n_{1},\cdots,n_{t}\\}.$ Hence $a^{n}\subseteq Q_{1}\cap\cdots\cap
Q_{t}$ for all $n\geq\min\\{s,\max\\{n_{1},\cdots,n_{t}\\}\\}$. ∎

######  Corollary 2.6.

Let $Q_{1},\cdots,Q_{t}$ be some $(s,n)$-closed hyperideals of $G$ for some
$s,n\in\mathbb{Z}^{+}$. Then $Q_{1}\cap\cdots\cap Q_{t}$ is an $(s,n)$-closed
hyperideal of $G$.

Recall from [1] that two hyperideals $P$ and $Q$ of $G$ are coprime if
$P+Q=G$.

######  Corollary 2.7.

Let $Q_{1},\cdots,Q_{t}$ be some $(s,n)$-closed hyperideals of $G$ for some
$s,n\in\mathbb{Z}^{+}$ such that $Q_{1},\cdots,Q_{t}$ are pairwise coprime.
Then $Q_{1}\circ\cdots\circ Q_{t}$ is an $(s,n)$-closed hyperideal of $G$.

######  Theorem 2.8.

Let $P$ and $Q$ be two hyperideals of $G$ such that $Q$ is an $(s,2)$-closed
strong C-hyperideal for $s\in\mathbb{Z}^{+}$. If $P^{s}\subseteq Q$, then
$P^{2}+P^{2}\subseteq Q$.

###### Proof.

Assume that $a,b\in P$. Since $P^{s}\subseteq Q$, we have
$a^{s},b^{s},(a+b)^{s}\subseteq Q$. Since $Q$ is an $(s,2)$-closed hyperideal
of $G$, we get $a^{2},b^{2},(a+b)^{2}\subseteq Q$. Pick $c\in(a+b)^{2}$. Since
$(a+b)^{2}\subseteq a^{2}+a\circ b+a\circ b+b^{2}$, we have $c\in
a^{2}+d+b^{2}$ for some $d\in a\circ b+a\circ b$ and so $c-d\in
a^{2}+b^{2}\subseteq Q$. Then we get $d\in Q$ as $c\in(a+b)^{2}\subseteq Q$.
Since $d\in a\circ b+a\circ b$ and $Q$ is a strong C-hyperideal of $G$, we
obtain $a\circ b+a\circ b\subseteq Q$. Hence $P^{2}+P^{2}\subseteq Q$. ∎

We define the relation $\gamma$ on a multiplicative hyperring $(G,+,\circ)$ as
follows:
$x\gamma y$ if and only if $\\{x,y\\}\subseteq A$ where $A$ is a finite sum of
finite products of elements of $G$.
This means $x\gamma y$ if and only if there exist $z_{1},...,z_{n}\in G$ that
$\\{x,y\\}\subseteq\sum_{j\in J}\prod_{i\in I_{j}}z_{i}$ and
$I_{j},J\subseteq\\{1,...,n\\}$. The transitive closure of $\gamma$ is denoted
by $\gamma^{\ast}$. The relation $\gamma^{\ast}$ is the smallest equivalence
relation on a multiplicative hyperring $G$ such that the quotient
$G/\gamma^{\ast}$, the set of all equivalence classes, is a fundamental ring.
Let $\mathbb{A}$ be the set of all finite sums of products of elements of $G$.
The definition of $\gamma^{\ast}$ can be rewritten on $G$, i.e.,
$x\gamma^{\ast}y$ if and only if there exist $z_{1},...,z_{n}\in G$ with
$z_{1}=x,z_{n+1}=y$ and $u_{1},...,u_{n}\in\mathbb{A}$ such that
$\\{z_{i},z_{i+1}\\}\subseteq u_{i}$ for $i\in\\{1,...,n\\}$. Suppose that
$\gamma^{\ast}(x)$ is the equivalence class containing $x\in G$. Define
$\gamma^{\ast}(x)\oplus\gamma^{\ast}(y)=\gamma^{\ast}(z)$ for all
$z\in\gamma^{\ast}(x)+\gamma^{\ast}(y)$ and
$\gamma^{\ast}(x)\odot\gamma^{\ast}(y)=\gamma^{\ast}(w)$ for all
$w\in\gamma^{\ast}(x)\circ\gamma^{\ast}(y)$. Then $(G/\gamma^{\ast},+,\odot)$
is a ring, which is called a fundamental ring of $G$ [30].

######  Theorem 2.9.

Let $Q$ be a hyperideal of $G$. Then $Q$ is an $(s,n)$-closed hyperideal of
$(G,+,\circ)$ if and only if $Q/\gamma^{\ast}$ is an $(s,n)$-closed ideal of
$(G/\gamma^{\ast},\oplus,\odot)$.

###### Proof.

($\Longrightarrow$) Let $\underbrace{a\odot\cdots\odot a}_{s}\in
Q/\gamma^{\ast}$ for some $a\in G/\gamma^{\ast}$. Therefore there exists $x\in
G$ such that $a=\gamma^{\ast}(x)$ and $\underbrace{a\odot\cdots\odot
a}_{s}=\underbrace{\gamma^{\ast}(x)\odot\cdots\odot\gamma^{\ast}(x)}_{s}=\gamma^{\ast}(x^{s})$.
Since $\gamma^{\ast}(x^{s})\in Q/\gamma^{\ast}$, we obtain $x^{s}\subseteq Q$.
Since $Q$ is an $(s,n)$-closed hyperideal of $G$, we conclude that
$x^{n}\subseteq Q$. Thus $\underbrace{a\odot\cdots\odot
a}_{n}=\underbrace{\gamma^{\ast}(x)\odot\cdots\odot\gamma^{\ast}(x)}_{n}=\gamma^{\ast}(x^{n})\in
Q/\gamma^{\ast}$ which implies $Q/\gamma^{\ast}$ is an $(s,n)$-closed ideal of
$G/\gamma^{\ast}$.
($\Longleftarrow$) Assume that $x^{s}\subseteq Q$ for some $x\in G$. Therefore
$\gamma^{\ast}(x)\in G/\gamma^{\ast}$ and
$\underbrace{\gamma^{\ast}(x)\odot\cdots\odot\gamma^{\ast}(x)}_{s}=\gamma^{\ast}(x^{s})\in
Q/\gamma^{\ast}$. Since $Q/\gamma^{\ast}$ is an $(s,n)$-closed ideal of
$G/\gamma^{\ast}$, we get
$\underbrace{\gamma^{\ast}(x)\odot\cdots\odot\gamma^{\ast}(x)}_{n}=\gamma^{\ast}(x^{n})\in
Q/\gamma^{\ast}$ which means $x^{n}\subseteq Q$. Consequently $Q$ is an
$(s,n)$-closed hyperideal of $G$. ∎

Let us define $\mathfrak{C}(Q)=\\{(s,n)\in\mathbb{Z}^{+}\times\mathbb{Z}^{+}\
|\ Q\text{ is (s,n)-closed }\\}$ for some proper hyperideal $Q$ of $G$. Then
we have $\\{(s,n)\in\mathbb{Z}^{+}\times\mathbb{Z}^{+}\ |\ 1\leq s\leq
n\\}\subseteq\mathfrak{C}(Q)\subseteq\mathbb{Z}^{+}\times\mathbb{Z}^{+}$.
Moreover, $rad(Q)=Q$ if and only if
$\mathfrak{C}(Q)=\mathbb{Z}^{+}\times\mathbb{Z}^{+}$.

######  Theorem 2.10.

Let $Q$ be a proper hyperideal of $G$ and $s,n\in\mathbb{Z}^{+}$. If
$(s,n),(s+1,n+1)\in\mathfrak{C}(Q)$ with $s\neq n$, then
$(s+1,n)\in\mathfrak{C}(Q)$.

###### Proof.

Let $s<n$. Clearly, we have $(s+1,n)\in\mathfrak{C}(Q)$. Now, let us conseder
$s>n$. Assume that $a^{s+1}\subseteq Q$ for some $a\in G$. Since
$(s+1,n+1)\in\mathfrak{C}(Q)$, we get $a^{n+1}\subseteq Q$. Since $n+1\leq s$,
we obtain $a^{s}\subseteq Q$ which implies $a^{n}\subseteq Q$ as
$(s,n)\in\mathfrak{C}(Q)$. Hence $(s+1,n)\in\mathfrak{C}(Q)$. ∎

######  Lemma 2.11.

Let $Q$ be a proper hyperideal of $G$ and $s,n\in\mathbb{Z}^{+}$. If
$(s,n)\in\mathfrak{C}(Q)$, then $(s^{\prime},n^{\prime})\in\mathfrak{C}(Q)$
for $s^{\prime},n^{\prime}\in\mathbb{Z}^{+}$ such that $1\leq s^{\prime}\leq
s$ and $n\leq n^{\prime}$.

###### Proof.

Straightforward. ∎

######  Theorem 2.12.

Let $Q$ be a proper ${\bf C}-$hyperideal of $G$.

  * (i)

If $(n,2),(n+1,2)\in\mathfrak{C}(Q)$ for $n\in\mathbb{Z}^{+}$ such that $n\geq
3$, then $(n+2,2)\in\mathfrak{C}(Q)$ and then $(t,2)\in\mathfrak{C}(Q)$ for
all $t\in\mathbb{Z}^{+}$.

  * (ii)

If $(s,n)\in\mathfrak{C}(Q)$ for $s,n\in\mathbb{Z}^{+}$ such that
$n\leq\frac{s}{2}$, then $(s+1,n)\in\mathfrak{C}(Q)$ and then
$(t,n)\in\mathfrak{C}(Q)$ for all $t\in\mathbb{Z}^{+}$.

###### Proof.

(i) Let $a^{n+2}\subseteq Q$ for some $a\in G$. Since $2n\geq n+2$, we have
$(a^{2})^{n}=a^{2n}\subseteq Q$. Pick $x\in a^{2}$. Since $x^{n}\subseteq Q$
and $(n,2)\in\mathfrak{C}(Q)$, we get $x^{2}\subseteq Q$. From $x\in a^{2}$ it
follows that $x^{2}\subseteq a^{4}$. Since $Q$ is a ${\bf C}-$hyperideal of
$G$, we conclude that $a^{4}\subseteq Q$. Since $(n+1,2)\in\mathfrak{C}(Q)$
and $n\geq 3$, we obtain $a^{2}\subseteq Q$. Consequently,
$(n+2,2)\in\mathfrak{C}(Q)$. By a similar argument, we conclude that
$(t,2)\in\mathfrak{C}(Q)$ for all $t\in\mathbb{Z}^{+}$ such that $t\geq n+3$.
Thus, by Lemma 2.11, $(t,2)\in\mathfrak{C}(Q)$ for all $t\in\mathbb{Z}^{+}$.

(ii) Assume that $a^{s+1}\subseteq Q$ for some $a\in G$. Therefore we have
$(a^{2})^{s}=a^{2s}\subseteq Q$. Choose $x\in a^{2}$. So $x^{s}\subseteq Q$.
Since $(s,n)\in\mathfrak{C}(Q)$, we get $x^{n}\subseteq Q$. From $x\in a^{2}$
it follows that $x^{n}\subseteq a^{2n}$. Then we have $a^{2n}\subseteq Q$ as
$Q$ is a ${\bf C}-$hyperideal of $G$. By the hypothesis, we have
$a^{s}\subseteq Q$. Since $(s,n)\in\mathfrak{C}(Q)$, we get $a^{n}\subseteq Q$
which implies $(s+1,n)\in\mathfrak{C}(Q)$. By a similar argument, we have
$(t,n)\in\mathfrak{C}(Q)$ for all $t\in\mathbb{Z}^{+}$ such that $t\geq n$.
Now, by Lemma 2.11, we conclude that $(t,n)\in\mathfrak{C}(Q)$ for all
$t\in\mathbb{Z}^{+}$. ∎

Assume that $Q$ is a proper hyperideal of $G$ and $s,n\in\mathbb{Z}^{+}$. We
consider the mappings $\omega:\mathbb{Z}^{+}\longmapsto\mathbb{Z}^{+}$,
defined by $\omega_{Q}(s)=\min\\{n\ |\ Q\text{ is
(s,n)-closed}\\}\in\\{1,\cdots,s\\}$, and
$\Omega:\mathbb{Z}^{+}\longmapsto\mathbb{Z}^{+}\cup\\{\infty\\}$, defined by
$\Omega_{Q}(n)=\sup\\{s\ |\ Q\text{ is
(s,n)-closed}\\}\in\\{n,n+1,\cdots\\}\cup\\{\infty\\}$. The rows of
$\mathfrak{C}(Q)$ determine $\omega_{Q}$ and the columns of $\mathfrak{C}(Q)$
determine $\Omega_{Q}$. Since $(n,n)\in\mathfrak{C}(Q)$ for all
$n\in\mathbb{Z}^{+}$, we have $1\leq\omega_{Q}(s)\leq s$. Moreover, by Lemma
2.11, we conclude that $\omega_{Q}(s)\leq\omega_{Q}(s+1)$ and
$\Omega_{Q}(n)\leq\Omega_{Q}(n+1)$.

######  Remark 2.13.

Let $P$ and $Q$ be two proper hyperideals of $G$. It can be easily seen that
$\mathfrak{C}(P)\subseteq\mathfrak{C}(Q)$ for if and only if
$\omega_{P}(s)\leq\omega_{Q}(s)$ for all $s\in\mathbb{Z}^{+}$ if and only if
$\Omega_{P}(n)\leq\Omega_{Q}(n)$ for all $n\in\mathbb{Z}^{+}$.

######  Theorem 2.14.

Let $Q$ be a proper hyperideal of $G$ and $s,n\in\mathbb{Z}^{+}$. If
$\omega_{Q}(s)<s$, then either $\omega_{Q}(s+1)=\omega_{Q}(s)$ or
$\omega_{Q}(s+1)\geq\omega_{Q}(s)+2$.

###### Proof.

Let $\omega_{Q}(s+1)=\omega_{Q}(s)+1$ and $\omega_{Q}(s)=n$. Then $n<s$ and
$\omega_{Q}(s+1)=n+1$. Hence we have $(s+1,n+1),(s,n)\in\mathfrak{C}(Q)$. By
Theorem 2.10, we conclude that $(s+1,n)\in\mathfrak{C}(Q)$ which implies
$\omega_{Q}(s+1)\leq n$ which is a contradiction. ∎

######  Theorem 2.15.

Let $Q$ be a proper hyperideal of $G$ and $s,n\in\mathbb{Z}^{+}$. If
$\Omega_{Q}(n)>n$, then either $\Omega_{Q}(n+1)=\Omega_{Q}(n)$ or
$\Omega_{Q}(n+1)\geq\Omega_{Q}(n)+2$.

###### Proof.

By using an argument similar to that in the proof of Theorem 2.14, one can
easily complete the proof. ∎

######  Theorem 2.16.

Let $P$ and $Q$ be proper hyperideals of $G$. Then

  * (i)

$\omega_{P\cap Q}\leq\omega_{P}\vee\omega_{Q}$.

  * (ii)

$\Omega_{P}\wedge\Omega_{Q}\leq\Omega_{P\cap Q}$.

###### Proof.

Put $\omega_{P}(s)=n^{\prime}$ and $\omega_{Q}(s)=n^{\prime\prime}$ for
$s\in\mathbb{Z}^{+}$ such that $n=\max\\{n^{\prime},n^{\prime\prime}\\}.$ By
Lemma 2.11 and Corollary 2.6, we have
$(s,n)\in\mathfrak{C}(P)\cap\mathfrak{C}(Q)\subseteq\mathfrak{C}(P\cap Q)$.
Therefore we conclude that $\omega_{P\cap Q}(s)\leq n$. Since
$n=(\omega_{P}\vee\omega_{Q})(s)$, we have $\omega_{P\cap
Q}(s)\leq(\omega_{P}\vee\omega_{Q})(s)$.

(ii) This can be proved in a very similar manner to the way in which (i) was
proved. ∎

######  Theorem 2.17.

Let $P$ and $Q$ be proper hyperideals of $G$. Then $\omega_{P\cap
Q}=\omega_{P}\vee\omega_{Q}$ if and only if
$\mathfrak{C}(P)\cap\mathfrak{C}(Q)=\mathfrak{C}(P\cap Q)$.

###### Proof.

$(\Longrightarrow)$Let $\omega_{P\cap Q}=\omega_{P}\vee\omega_{Q}$. This means
that $\omega_{P}\leq\omega_{P\cap Q}$ and $\omega_{Q}\leq\omega_{P\cap Q}$. By
Remark 2.13, we get $\mathfrak{C}(P\cap
Q)\subseteq\mathfrak{C}(P)\cap\mathfrak{C}(Q)$. On the other hand, we have
$\mathfrak{C}(P)\cap\mathfrak{C}(Q)\subseteq\mathfrak{C}(P\cap Q)$ by
Corollary 2.6. Consequently,
$\mathfrak{C}(P)\cap\mathfrak{C}(Q)=\mathfrak{C}(P\cap Q)$.

$(\Longleftarrow)$ Let $\mathfrak{C}(P)\cap\mathfrak{C}(Q)=\mathfrak{C}(P\cap
Q)$. This implies that $\omega_{P}\leq\omega_{P\cap Q}$ and
$\omega_{Q}\leq\omega_{P\cap Q}$ by Remark 2.13. Therefore we obtain
$\omega_{P}\vee\omega_{Q}\leq\omega_{P\cap Q}$. On the other hand, by Theorem
2.16, we get $\omega_{P\cap Q}\leq\omega_{P}\vee\omega_{Q}$. Consequently,
$\omega_{P\cap Q}=\omega_{P}\vee\omega_{Q}$. ∎

######  Theorem 2.18.

Let $P$ and $Q$ be proper hyperideals of $G$. Then $\Omega_{P\cap
Q}=\Omega_{P}\wedge\Omega_{Q}$ if and only if
$\mathfrak{C}(P)\cap\mathfrak{C}(Q)=\mathfrak{C}(P\cap Q)$.

###### Proof.

By an argument similar to that one given in Theorem 2.17, we are done. ∎

In view of Theorem 2.17 and Theorem 2.18, the following result is obtained.

######  Corollary 2.19.

Let $P$ and $Q$ be proper hyperideals of $G$. Then $\omega_{P\cap
Q}=\omega_{P}\vee\omega_{Q}$ in and only if $\Omega_{P\cap
Q}=\Omega_{P}\wedge\Omega_{Q}$

##  3\. weakly $(s,n)$-closed hyperideals

######  Definition 3.1.

Let $Q$ be a proper hyperideal of $G$ and $s,n\in\mathbb{Z}^{+}$. $Q$ refers
to a weakly $(s,n)$-closed hyperideal if $0\notin x^{s}\subseteq Q$ for $x\in
G$ implies that $x^{n}\subseteq Q$.

######  Example 3.2.

Consider the multiplicative hyperring $(\mathbb{Z}_{A},+,\circ)$ where
$\mathbb{Z}_{A}=\mathbb{Z}$, $A=\\{7,11\\}$, $a\circ b=\\{a\cdot x\cdot b\ |\
x\in A\\}$ for all $a,b\in\mathbb{Z}_{A}$ and $+$ is ordinary addition. Then
$\langle 390\rangle$ is an weakly $(s,4)$-closed hyperideal of $\mathbb{Z}$
for all $s\in\mathbb{Z}^{+}$.

It can be easily verified that an intersection of weakly $(s,n)$-closed
hyperideals of $G$ is weakly $(s,n)$-closed. Furthermore, every weakly
$(s,n)$-closed hyperideal of $G$ is weakly $(s,n^{\prime})$-closed for all
$n^{\prime}\in\mathbb{Z}^{+}$ such that $n^{\prime}\geq n$. Let $Q$ be a
weakly $(s,n)$-closed ${\bf C}$-hyperideal of $G$. We define $x\in G$ to be an
$(s,n)$-tough-zero element of $Q$ if $0\in x^{s}$ and $x^{n}\nsubseteq Q$.
Indeed, it means that $Q$ is not $(s,t)$-closed if and only if $Q$ has an
$(s,n)$-tough-zero element.

######  Theorem 3.3.

Let $Q$ be a weakly $(s,n)$-closed strong ${\bf C}$-hyperideal of $G$. If $x$
is an $(s,n)$-tough-zero element of $Q$, then $0\in(x+a)^{s}$ for all $a\in
Q$.

###### Proof.

Assume that $a\in Q$. So $\sum_{i=1}^{s}\tbinom{s}{i}x^{s-i}a^{i}\subseteq Q$.
On the other hand, $0+\sum_{i=1}^{s}\tbinom{s}{i}x^{s-i}a^{i}\subseteq
x^{s}+\sum_{i=1}^{s}\tbinom{s}{i}x^{m-i}a^{i}$. Since $Q$ is a strong ${\bf
C}$-hyperideal of $G$, we have
$x^{s}+\sum_{i=1}^{s}\tbinom{s}{i}x^{s-i}a^{i}\subseteq Q$ which implies
$(x+a)^{s}\subseteq Q$ as $(x+a)^{s}\subseteq
x^{s}+\sum_{i=1}^{s}\tbinom{s}{i}x^{m-i}a^{i}$. If $(x+a)^{n}\subseteq Q$,
then $x^{n}+\sum_{i=1}^{n}\tbinom{n}{i}x^{n-i}a^{i}\subseteq Q$ because $Q$ is
a strong ${\bf C}$-hyperideal of $G$. Since
$\sum_{i=1}^{n}\tbinom{n}{i}x^{n-i}a^{i}\subseteq Q$, we get $x^{n}\subseteq
Q$, a contradiction. Then $(x+a)^{n}\nsubseteq Q$. Since $Q$ is a weakly
$(s,n)$-closed of $G$, we obtain $0\in(x+a)^{s}$. ∎

Recall from [1] that an element $x\in G$ is nilpotent if there exists an
integer $t$ such that $0\in x^{t}$. The set of all nilpotent elements of $G$
is denoted by $\Upsilon$.

######  Theorem 3.4.

If $Q$ is a weakly $(s,n)$-closed strong ${\bf C}$-hyperideal of $G$ such that
is not $(s,t)$-closed, then $Q\subseteq\Upsilon$.

###### Proof.

Suppose that $Q$ is a weakly $(s,n)$-closed strong ${\bf C}$-hyperideal of
$G$. If $Q$ is not $(s,t)$-closed, then there exists $x\in G$ such that $x$ is
an $(s,n)$-tough-zero element of $Q$. Assume that $a\in Q$. We have $0\in
x^{s}$ which implies $0\in(x+a)^{s}$ by theorem 3.3. This means that
$x,x+a\in\Upsilon$. Since by Theorem 3.5 in [1] $\Upsilon$ is a hyperideal of
$G$ , we conclude that $a=(x+a)-x\in\Upsilon$. This means that
$Q\subseteq\Upsilon$. ∎

######  Theorem 3.5.

Let $G$ be a strongly distributive multiplicative hyperring with scalar
identity $e$ such that has an $i$-set and $s,n\in\mathbb{Z}^{+}$ with $s>n$.
Then every hyperideal of $G$ contained in $\Upsilon$ is weakly $(s,n)$-closed
if and only if $0\in x^{s}$ for all $x\in\Upsilon$.

###### Proof.

$(\Longrightarrow)$ Let $0\notin x^{s}$ for some $x\in\Upsilon$. Put
$Q=\langle x^{s}\rangle$. So $\langle x^{s}\rangle\subseteq\Upsilon$. Since
every hyperideal of $G$ contained in $\Upsilon$ is weakly $(s,n)$-closed, $Q$
is a weakly $(s,n)$-closed hyperideal of $G$. Therefore $0\notin
x^{s}\subseteq Q$ which implies $x^{n}\subseteq Q$ and $0\notin x^{n}$. Then
for every $a\in x^{n}$ there exist $a_{1},\cdots,a_{t}\in G$ and
$x_{1},\cdots,x_{t}\in x^{s}$ such that $a\in\sum_{i=1}^{t}a_{i}\circ x_{i}$.
Hence $0\in\sum_{i=1}^{t}x_{i}\circ a_{i}-a\subseteq\sum_{i=1}^{t}x^{s}\circ
a_{i}-x^{n}=x^{n}\circ(\sum_{i=1}^{t}x^{s-n}\circ a_{i}-e)$. This means that
$0\in x^{n}\circ c-e$ for some $c\in\sum_{i=1}^{t}x^{s-n}\circ a_{i}$. Since
$x\in\Upsilon$, we have $\sum_{i=1}^{t}x^{s-n}\circ a_{i}\in\Upsilon$. Hence
$c-e$ is invertible by Theorem 3.20 in [2]. Thus we have $0\in x^{n}$ which is
a contradiction.

$\Longleftarrow$ Let $0\in x^{s}$ for all $x\in\Upsilon$. Suppose that $Q$ is
a hyperideal of $G$ such that $Q\subseteq\Upsilon$ and $0\notin a^{s}\subseteq
Q$ for some $a\in G$ but $a^{n}\nsubseteq Q$. Then there exist $c\in a^{n}$
such that $c\notin Q$. If $c\in\Upsilon$, then $0\in c^{s}\subseteq a^{sn}$
and so $a\in\Upsilon$. Hence $0\in a^{s}$, a contradiction. If
$c\notin\Upsilon$, then $0\notin c^{s}$. Since $c^{s}\subseteq a^{ns}\subseteq
Q$, we have $c^{s}\subseteq\Upsilon$. Let $d\in c^{s}$. Then $0\in
d^{s}\subseteq c^{s^{2}}$ which means $c\in\Upsilon$, a contradiction. ∎

######  Definition 3.6.

Let $s,n\in\mathbb{Z}^{+}$. An element $a\in G$ is $(s,n)$-regular if
$a^{n}\subseteq a^{s}\circ b$ for some $b\in G$. An element $a\in G$ is
$(s,n)$-Regular if $a^{n}\subseteq a^{s}\circ B$ for some subset $B$ of $G$.

Recall from [2] that an element $a\in G$ is called weak zero divisor if for
$0\notin b\in G$ we have $0\in a\circ b$. Denote the set of weak zero divisor
by $Z_{w}(G)$. Also, recall from [1] that an element $x\in G$ is said to be
invertible if there exists $y\in G$, such that $e\in x\circ y$. The set of all
invertible elements in $G$ is denoted by $U(G)$.

######  Theorem 3.7.

Assume that $G$ is a strongly distributive multiplicative hyperring with
scalar identity $e$, $a\in G\backslash(Z_{w}(G)\cup U(G))$ and
$s,n\in\mathbb{Z}^{+}$. Then $a$ is $(s,n)$-regular if and only if $s\leq n$.

###### Proof.

Let $a$ be $(s,n)$-regular and $s>n$. From $a^{n}\subseteq a^{s}\circ b$ for
some $b\in G$, it follows that $0\in a^{s}\circ
b-a^{n}=a^{n}\circ(a^{s-n}\circ b-e)$. By the hypothesis, we have $0\in
a^{s-n}\circ b-e$ which implies $e\in a^{s-n}\circ b$. This means $e\in a\circ
d$ for some $d\in a^{s-n-1}\circ b$. Then $a\in U(G)$ which is a
contradiction. The converse is clear. ∎

######  Theorem 3.8.

Let $s,n\in\mathbb{Z}^{+}$ with $s>n$. If $a$ is $(s,n)$-regular, then $a$ is
$(s+1,n)$-Regular.

###### Proof.

Assume that $a$ is $(s,t)$-regular such that $s>n$. From $a^{n}\subseteq
a^{s}\circ b$ for some $b\in G$, it follows that $a^{n}\subseteq a^{s}\circ
b=a^{n}\circ(a^{s-n}\circ b)\subseteq(a^{s}\circ b)\circ(a^{s-n}\circ
b)=a^{s+1}\circ(a^{s-n-1}\circ b^{2})$. Put $B=a^{s-n-1}\circ b^{2}$. Thus we
have $a^{n}\subseteq a^{s+1}\circ B$ which means $a$ is $(s+1,n)$-Regular. ∎

######  Theorem 3.9.

If $a\in U(G)$, then $a$ is $(s,n)$-Regular for all $s,n\in\mathbb{Z}^{+}$.

###### Proof.

Let $a\in U(G)$. Then there exists $b\in G$ such that $e\in a\circ b$. So we
have $e\in a^{s-n}\circ b^{s-n}$. Therefore $a^{n}\circ e\subseteq a^{s}\circ
b^{s-n}$. If $x\in a^{n}$, then $x\in x\circ e\subseteq a^{n}\circ e$ which
implies $a^{n}\subseteq a^{n}\circ e$. Hence we get $a^{n}\subseteq a^{s}\circ
b^{s-n}$. Put $B=b^{s-n}$. Then $a^{n}\subseteq a^{s}\circ B$ which means $a$
is $(s,n)$-Regular. ∎

A non-empty finite subset $\xi=\\{e_{1},\cdots,e_{n}\\}$ of a multiplicative
hyperring $G$ is said to be $i$-set if $e_{i}\neq 0$ for at least one $1\leq
i\leq n$ and for every $x\in G$, $x\in\sum_{i=1}^{n}x\circ e_{i}$ [27].

######  Theorem 3.10.

Let $G$ be a strongly distributive multiplicative hyperring that has an
$i$-set and $s,n\in\mathbb{Z}^{+}$ with $s>n$. Then every proper hyperideal of
$G$ is weakly $(s,n)$-closed if and only if every non-nilpotent element of $G$
is $(s,n)$-Regular and $0\in a^{s}$ for all $a\in\Upsilon$.

###### Proof.

$(\Longrightarrow)$ Since every hyperideal of $G$ contained in $\Upsilon$ is
weakly $(s,n)$-closed, we have $0\in a^{s}$ for all $a\in\Upsilon$ by Theorem
3.5. Let $a\in G\backslash\Upsilon.$ Assume that $a\in U(G)$. Therefore $a$ is
$(s,n)$-Regular by Theorem 3.9. Now, assume that $a\notin U(R)$. Put
$Q=\langle a^{s}\rangle$. By the hypothesis, $Q$ is a weakly $(s,n)$-closed
hyperideal of $G$. Therefore $0\notin a^{s}\subseteq Q$ which implies
$a^{n}\subseteq Q$. Then for every $x\in a^{n}$ there exist
$b_{1_{x}},\cdots,b_{t_{a}}\in a^{s}$ and $c_{1_{x}},\cdots,c_{t_{x}}\in G$
such that $x\in\sum_{i=1}^{t}b_{i_{x}}\circ
c_{i_{x}}\subseteq\sum_{i=1}^{t}a^{s}\circ
c_{i_{x}}=a^{s}\circ\sum_{i=1}^{t}c_{i_{x}}$. Put
$c_{x}=\sum_{i=1}^{t}c_{i_{x}}$. Then we obtain $a^{n}=\cup_{x\in
a^{n}}\\{x\\}\subseteq\cup_{x\in a^{n}}a^{s}\circ c_{x}=a^{s}\circ\cup_{x\in
a^{n}}c_{x}$. Now, put $B=\cup_{x\in a^{n}}c_{x}$. Thus, $a^{n}\subseteq
a^{s}\circ B$ which means $a$ is $(s,n)$-Regular.

$(\Longleftarrow)$ Assume that $Q$ is a proper hyperideal of $G$ and $0\notin
a^{s}\subseteq Q$ for some $a\in G$. So $a\notin\Upsilon$. By the hypothesis,
$a$ is $(s,n)$-Regular. This means $a^{n}\subseteq a^{s}\circ B$ for some
subset $B$ of $G$. Since $a^{s}\circ B\subseteq Q$, we have $a^{n}\subseteq
Q$. Consequently, $Q$ is weakly $(s,n)$-closed. ∎

Assume that $(G_{1},+_{1},\circ_{1})$ and $(G_{2},+_{2},\circ_{2})$ are two
multiplicative hyperrings. Recall from [18] that a mapping $\psi$ from $G_{1}$
into $G_{2}$ is said to be a hyperring good homomorphism if for all $x,y\in
G_{1}$, $\psi(x+_{1}y)=\psi(x)+_{2}\psi(y)$ and
$\psi(x\circ_{1}y)=\psi(x)\circ_{2}\psi(y)$.

######  Theorem 3.11.

Let $G_{1}$ and $G_{2}$ be two multiplicative hyperrins,
$\psi:G_{1}\longrightarrow G_{2}$ a hyperring good homomorphism and
$s,n\in\mathbb{Z}^{+}$.

  * (1)

If $Q_{2}$ is a weakly $(s,n)$-closed hyperideal of $G_{2}$ and $\psi$ is
injective, then $\psi^{-1}(Q_{2})$ is a weakly $(s,n)$-closed hyperideal of
$G_{1}$.

  * (2)

If $\psi$ is surjective and $Q_{1}$ is a weakly $(s,n)$-closed hyperideal of
$G_{1}$ with $Ker(\psi)\subseteq Q_{1}$ , then $\psi(Q_{1})$ is a weakly
$(s,n)$-closed hyperideal of $G_{2}$.

###### Proof.

(1) Let $0\notin a^{s}\subseteq\psi^{-1}(Q_{2})$ for some $a\in G_{1}$. Then
we get $0\notin\psi(a^{s})=\psi(a)^{s}\subseteq Q_{2}$ as $\psi$ is injective.
Since $Q_{2}$ is a weakly $(s,n)$-closed hyperideal of $G_{2}$, we obtain
$(\psi(a))^{n}\subseteq Q_{2}$ which implies $\psi(a^{n})\subseteq Q_{2}$
which means $a^{n}\subseteq\psi^{-1}(Q_{2})$. Thus $\psi^{-1}(Q_{2})$ is a
weakly $(s,n)$-closed hyperideal of $G_{1}$.

(2) Let $0\notin b^{s}\subseteq\psi(Q_{1})$ for some $b\in G_{2}$. Then
$\psi(a)=b$ for some $a\in G_{1}$ as $\psi$ is surjective. Therefore
$\psi(a^{s})=\psi(a)^{s}\subseteq\psi(Q_{1})$. Now, pick any $x\in a^{s}$.
Then $\psi(x)\in\psi(a^{s})\subseteq\psi(Q_{1})$ and so there exists $y\in
Q_{1}$ such that $\psi(x)=\psi(y)$. Then we have $\psi(x-y)=0$ which means
$x-y\in Ker(\psi)\subseteq Q_{1}$ and so $x\in Q_{1}$. So $a^{s}\subseteq
Q_{1}$. Since $Q_{1}$ is weakly $(s,n)$-closed and $0\notin a^{s}$, we get
$a^{n}\subseteq Q_{1}$. This means that
$\psi(a^{n})=b^{n}\subseteq\psi(Q_{1})$. Consequently, $\psi(Q_{1})$ is a
weakly $(s,n)$-closed hyperideal of $G_{2}$. ∎

######  Corollary 3.12.

Let $P$ and $Q$ be two hyperideals of $G$ with $P\subseteq Q$ and
$s,n\in\mathbb{Z}^{+}$. If $Q$ is a weakly $(s,n)$-closed hyperideal of $G$,
then $Q/P$ is a weakly $(s,n)$-closed hyperideal of $G/P$.

###### Proof.

By Theorem 3.11 and using the mapping $\pi:G\longrightarrow G/P$ defined by
$\pi(a)=a+P$ we are done. ∎

Let $(G_{1},+_{1},\circ_{1})$ and $(G_{2},+_{2},\circ_{2})$ be two
multiplicative hyperrings with non zero identity. The set $G_{1}\times G_{2}$
with the operation $+$ and the hyperoperation $\circ$ defined as

$(x_{1},x_{2})+(y_{1},y_{2})=(x_{1}+_{1}y_{1},x_{2}+_{2}y_{2})$

$(x_{1},x_{2})\circ(y_{1},y_{2})=\\{(x,y)\in R_{1}\times R_{2}\ |\ x\in
x_{1}\circ_{1}y_{1},y\in x_{2}\circ_{2}y_{2}\\}$
is a multiplicative hyperring [28]. Now, we give some characterizations of
weakly $(s,n)$-closed hyperideals on cartesian product of commutative
multiplicative hyperring.

######  Theorem 3.13.

Let $(G_{1},+_{1},\circ_{1})$ and $(G_{2},+_{2},\circ_{2})$ be two
multiplicative hyperrings with scalar identities $e_{1}$ and $e_{2}$
respectively, $Q_{1}$ a ${\bf C}$-hyperideal of $G_{1}$ and
$s,n\in\mathbb{Z}^{+}$ . Then the followings are equivalent.

  * (i)

$Q_{1}\times G_{2}$ is a weakly $(s,n)$-closed hyperideal of $G_{1}\times
G_{2}$.

  * (ii)

$Q_{1}$ is an $(s,n)$-closed hyperideal of $G_{1}$.

  * (iii)

$Q_{1}\times G_{2}$ is an $(s,n)$-closed hyperideal of $G_{1}\times G_{2}$.

###### Proof.

(i) $\Longrightarrow$ (ii) Assume that $Q_{1}\times G_{2}$ is a weakly
$(s,n)$-closed hyperideal of $G_{1}\times G_{2}$. By Theorem 3.11 (2), we
conclude that $Q_{1}$ is a weakly $(s,n)$-closed hyperideal of $G_{1}$.
Suppose that $Q_{1}$ is not an $(s,n)$-closed hyperideal of $G_{1}$. This
implies that $Q_{1}$ has an $(s,n)$-tough-zero element $x$. Hence $0\in x^{s}$
and $x^{n}\nsubseteq Q_{1}$. This implies that
$(0,0)\notin(x,e_{2})^{s}\subseteq Q_{1}\times G_{2}$ and
$(x,e_{2})^{n}\nsubseteq Q_{1}\times G_{2}$ which is a contradiction since
$Q_{1}\times G_{2}$ is a weakly $(s,n)$-closed hyperideal of $G_{1}\times
G_{2}$. Thus $Q_{1}$ is an $(s,n)$-closed hyperideal of $G_{1}$.

(ii) $\Longrightarrow$ (iii) The claim follows by Theorem 2.12 in [9] and
Theorem 2.9.

(iii) $\Longrightarrow$ (i) This follows directly from the deﬁnitions. ∎

######  Lemma 3.14.

Let $(G_{1},+_{1},\circ_{1})$ and $(G_{2},+_{2},\circ_{2})$ be two
multiplicative hyperrings and $I_{1}$ and $I_{2}$ be hyperideals of $G_{1}$
and $G_{2}$, respectively. Then $I_{1}$ and $I_{2}$ are ${\bf C}$-hyperideals
if and only if $I_{1}\times I_{2}$ is ${\bf C}$-hyperideal of $G_{1}\times
G_{2}$.

###### Proof.

$\Longrightarrow$ Let $I_{1}$ and $I_{2}$ be ${\bf C}$-hyperideals of $G_{1}$
and $G_{2}$, respectively and $(a_{1},b_{1})\circ\cdots\circ(a_{n},b_{n})\cap
I_{1}\times I_{2}\neq\varnothing$ for some $a_{1},\cdots,a_{n}\in G_{1}$ and
$b_{1},\cdots,b_{n}\in G_{2}$. This means
$(a,b)\in(a_{1},b_{1})\circ\cdots\circ(a_{n},b_{n})$ for some $(a,b)\in
I_{1}\times I_{2}$. Therefore we have $a\in
a_{1}\circ_{1}\cdots\circ_{1}a_{n}$ and $b\in
b_{1}\circ_{2}\cdots\circ_{2}b_{n}$. Since $I_{1}$ and $I_{2}$ are ${\bf
C}$-hyperideals, we get $a_{1}\circ_{1}\cdots\circ_{1}a_{n}\subseteq I_{1}$
and $b_{1}\circ_{2}\cdots\circ_{2}b_{n}\subseteq I_{2}$. This implies that
$(a_{1},b_{1})\circ\cdots\circ(a_{n},b_{n})\subseteq I_{1}\times I_{2}$, as
needed.

$\Longleftarrow$ Let $a_{1}\circ_{1}\cdots\circ_{1}a_{n}\cap
I_{1}\neq\varnothing$ for some $a_{1},\cdots,a_{n}\in G_{1}$ and
$b_{1}\circ_{2}\cdots\circ_{2}b_{n}\cap I_{2}\neq\varnothing$ for some
$b_{1},\cdots,b_{n}\in G_{2}$. This means
$(a_{1},b_{1})\circ\cdots\circ(a_{n},b_{n})\cap I_{1}\times
I_{2}\neq\varnothing$. Since $I_{1}\times I_{2}$ is ${\bf C}$-hyperideal of
$G_{1}\times G_{2}$, we have
$(a_{1},b_{1})\circ\cdots\circ(a_{n},b_{n})\subseteq I_{1}\times I_{2}$ which
means $a_{1}\circ_{1}\cdots\circ_{1}a_{n}\subseteq I_{1}$ and
$b_{1}\circ_{2}\cdots\circ_{2}b_{n}\subseteq I_{2}$, as claimed. ∎

######  Theorem 3.15.

Let $(G_{1},+_{1},\circ 1)$ and $(G_{2},+_{2},\circ_{2})$ be two
multiplicative hyperrings with scalar identities $e_{1}$ and $e_{2}$,
repectively, and $s,n\in\mathbb{Z}^{+}$. Then the followings are equivalent.

  * (i)

$Q$ is a weakly $(s,n)$-closed ${\bf C}$-hyperideal of $G_{1}\times G_{2}$
that is not $(s,n)$-closed.

  * (ii)

$Q=Q_{1}\times Q_{2}$ for some ${\bf C}$-hyperideals $Q_{1}$ and $Q_{2}$ of
$G_{1}$ and $G_{2}$, respectively, such that either

    * (1)

$Q_{1}$ is a weakly $(s,n)$-closed hyperideal of $G_{1}$ that is not
$(s,n)$-closed, $0\in b^{s}$ whenever $b^{s}\subseteq Q_{2}$ for $b\in G_{2}$.
If $0\notin a^{s}\subseteq Q_{1}$ for $a\in G_{1}$, then $Q_{2}$ is an
$(s,n)$-closed hyperideal of $G_{2}$, or

    * (2)

$Q_{2}$ is a weakly $(s,n)$-closed hyperideal of $G_{2}$ that is not
$(s,n)$-closed, $0\in b^{s}$ whenever $b^{s}\subseteq Q_{1}$ for $b\in G_{1}$.
If $0\notin a^{s}\subseteq Q_{2}$ for $a\in G_{2}$, then $Q_{1}$ is an
$(s,n)$-closed hyperideal of $G_{1}$.

###### Proof.

(i) $\Longrightarrow$ (ii) Assume that $Q$ is a weakly $(s,n)$-closed ${\bf
C}$-hyperideal of $G_{1}\times G_{2}$ that is not $(s,n)$-closed. Then
$Q=Q_{1}\times Q_{2}$ for some hyperideals $Q_{1}$ and $Q_{2}$ of $G_{1}$ and
$G_{2}$, respectively, such that one of them is weakly $(s,n)$-closed but is
not $(s,n)$-closed by Theorem 3.13. We may assume that $Q_{1}$ is a weakly
$(s,n)$-closed hyperideal of $G_{1}$ but is not $(s,n)$-closed. Then we
conclude that $Q_{1}$ has an $(s,n)$-tough-zero element $a_{1}$ which means
$0\in a_{1}^{s}$ and $a_{1}^{n}\nsubseteq Q_{1}$. Let $b^{s}\subseteq Q_{2}$
for some $b\in G_{2}$. Since $(a_{1},b)^{s}\subseteq Q$, we get
$(0,0)\in(a_{1},b)^{s}$ which means $0\in b^{s}$. Suppose that $0\notin
a^{s}\subseteq Q_{1}$ for $a\in G_{1}$. Let $b\in G_{2}$ with $b^{s}\subseteq
Q_{2}$. Therefore $(0,0)\notin(a,b)^{s}\subseteq Q$. This means that
$b^{n}\subseteq Q_{2}$ which implies $Q_{2}$ is an $(s,n)$-closed hyperideal
of $G_{2}$. Similiar for the other case.

(ii) $\Longrightarrow$ (i) Assume that $Q_{1}$ is a weakly $(s,n)$-closed
hyperideal of $G_{1}$ that is not $(s,n)$-closed. Suppose that $x$ is an
$(s,n)$-tough-zero element of $Q_{1}$. This means that $(x,0)$ is an
$(s,n)$-tough-zero element of $Q$. Hence $Q$ is not an $(s,n)$-closed. Let
$(0,0)\notin(a,b)^{s}\subseteq Q$ for $a\in G_{1}$ and $b\in G_{2}$. Since
$0\in b^{s}$, we get $0\notin a^{s}\subseteq Q_{1}$. Then $(a,b)^{n}\subseteq
Q$ as $Q_{1}$ is a weakly $(s,n)$-closed hyperideal of $G_{1}$ and $Q_{2}$ is
an $(s,n)$-closed hyperideal of $G_{2}$. Consequently, $Q$ is a weakly
$(s,n)$-closed ${\bf C}$-hyperideal of $G_{1}\times G_{2}$ that is not
$(s,n)$-closed. ∎

## References

  * [1] R. Ameri, A. Kordi, S. Hoskova-Mayerova, Multiplicative hyperring of fractions and coprime hyperideals, An. St. Univ. ovidious Constanta, 25(2017), No. 1, 5-23.
  * [2] R. Ameri, M. Norouzi, On commutative hyperrings, Int. Journal of Algebraic Hyperstructures and its Applications,  1(1) (2014) 45-58.
  * [3] R. Ameri, A. Kordi, On regular multiplicative hyperrings, European Journal of Pure and Applied Mathematics, 9(4)(2016) 402-418.
  * [4] R. Ameri, A. Kordi, Clean multiplicative hyperrings, Italian Journal of Pure and Applied Mathematics, (35) (2015) 625-636.
  * [5] M. Anbarloei, A generalization of prime hyperideals, Journal of Algebraic Systems, 8 (1) (2020) 113-127.
  * [6] M. Anbarloei, On 1-absorbing prime hyperideal and some of its generalizations, Journal of Mathematics, (2022), https://doi.org/10.1155/2022/4947019.
  * [7] M. Anbarloei, $\alpha$-prime hyperideals of a multiplicative hyperring, Journal of Mathematical Extension, 16(12) (2022) 1-18.
  * [8] M. Anbarloei, 2-prime hyperideals of multiplicative hyperrings, Journal of Mathematics, (2022), https://doi.org/10.1155/2022/5721946.
  * [9] D.F. Anderson, A. Badawi, On $(m,n)$-closed ideals of commutative rings, Journal of Algebra and Its Applications, 16(1)(2017) 1-21.
  * [10] E. Ay, G. Yesilot, D. Sonmez, $\delta$-Primary hyperideals on commutative hyperrings, International Journal Mathematics and Mathematical Sciences (2017): Article ID 5428160, 4 pages.
  * [11] P. Corsini, Prolegomena of hypergroup theory, Second ed., Aviani Editore, 1993.
  * [12] P. Corsini, V. Leoreanu, Applications of hyperstructures theory, Adv. Math., Kluwer Academic Publishers, 2003.
  * [13] J. Chvalina, J., S. Krehlik, M. Novak, Cartesian composition and the problem of generalizing the MAC condition to quasi- multiautomata, An. Stiint. Univ. Ovidius Constanta Ser. Mat., 24(3) (2016) 79-100.
  * [14] I. Cristea, S. Jancic-Rasovic, Compositions hyperrings, An. Stiint. Univ. Ovidius Constanta Ser. Mat., 21(2) (2013) 81-94.
  * [15] I. Cristea, Regularity of intuitionistic fuzzy relations on hypergroupoids, An. Stiint. Univ. Ovidius Constanta Ser. Mat., 22(1) (2014) 105-119.
  * [16] U. Dasgupta, On prime and primary hyperideals of a multiplicative hyperring, Annals of the Alexandru Ioan Cuza University-Mathematics, LVIII (1) (2012) 19-36.
  * [17] U. Dasgupta, On certain classes of hypersemirings, PhD Thsis, University of Calcutta, 2012.
  * [18] B. Davvaz, V. Leoreanu-Fotea, Hyperring theory and applications, International Academic Press, USA, 2007.
  * [19] F. Farzalipour, P. Ghiasvand, Semiprime hyperideals in multiplicative hyperrings, 51th Annual Iranian Mathematics conference, 1 (2021) 801-804.
  * [20] F. Marty, Sur une generalization de la notion de groupe, In: 8iem Congres des Mathematiciens Scandinaves, Stockholm, (1934) 45-49.
  * [21] C. G. Massouros, On the theory of hyperrings and hyperfields, Algebra i Logika, 24 (1985) 728-742.
  * [22] P. Ghiasvand, F. Farzalipour, On $S$-prime hyperideals in multiplicative hyperrings, journal of Algebraic Hyperstructures and Logical Algebras, 2 (2) (2021) 25-34.
  * [23] M. Novak, n-ary hyperstructures constructed from binary quasi-orderer semigroups, An. Stiint. Univ. Ovidius Constanta Ser. Mat., 22 (3) (2014) 147-168.
  * [24] R. Procesi, R. Rota, On some classes of hyperstructures, Combinatorics Discrete Math., 208/209 (1999) 485-497.
  * [25] R. Rota, Sugli iperanelli moltiplicativi, Rend. Di Math., Series VII, 4 (1982) 711-724.
  * [26] E. Sengelen Sevim, B.A. Ersoy, B. Davvaz, Primary hyperideals of multiplicative hyperrings, Eurasian Bulletin of Mathematics, 1(1) (2018) 43-49.
  * [27] M.K. Sen, U. Dasgupta, Some aspects of $G_{H}$-rings, An. Stiint. Univ. ”Al. I. Cuza” Iasi. Mat., 56 (2010) 253-272.
  * [28] G. Ulucak, On expansions of prime and 2-absorbing hyperideals in multiplicative hyperrings, Turkish Journal of Mathematics, 43 (2019) 1504-1517.
  * [29] T. Vougiouklis, Hyperstructures and their representations, Hadronic Press, Inc., Palm Harber, USA, 1994.
  * [30] T. Vougiouklis, Hyperstructures and their representations, Hadronic Press Inc., Florida, (1994).