#  some hyperideals defined on the basis of the intersection of all $n$-ary
prime hyperideals

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

Various classes of hyperideals have been studied in many papers in order to
let us fully understand the structures of hyperrings in general. The purpose
of this paper is the study of some hyperideals whose concept is created on the
basis of the intersection of all $n$-ary prime hyperideals in category of
Krasner $(m,n)$-hyperrings. In this regard we introduce notions of
$N$-hyperideals, $\delta$-$N$-hyperideals, $(k,n)$-absorbing
$\delta$-$N$-hyperideals and $S$-$N$-hyperideals. The overall framework of
these structures is then explained, providing a number of major conclusions.

###### Key words and phrases:

$N$-hyperideal, $\delta$-$N$-hyperideal, $(k,n)$-absorbing
$\delta$-$N$-hyperideal, $S$-$N$-hyperideal, Krasner $(m,n)$-hyperring.

###### 2010 Mathematics Subject Classification:

16Y99

##  1\. Introduction

The notion of prime ideal as an extension of the notion of prime number in the
ring of integers plays a highly important role in the theory of rings. In
2017, Terkir et al. [13] proposed a new classe of ideals called $n$-ideals and
investigated some properties of them analogous with prime ideals. Let $I$ be a
ideal of ring $R$. $I$ refers to an $n$-ideal if the condition $xy\in I$ with
$x\notin\sqrt{0}$ implies $y\in I$ for all $x,y\in I$. Afterward, the concept
of $(2,n)$-ideals in a commutative ring was defined by Tamekhante and Bouba
[14]. They give many results to show the relations between this new notion and
others that already exist. Indeed, the notion is a generalization of
$n$-ideals. A proper ideal $I$ of $R$ is said to be a $(2,n)$-ideal if $xyz\in
I$ for $x,y,z\in R$ implies that $xy\in I$ or $yz\in\sqrt{0}$ or
$xz\in\sqrt{0}$.

Algebraic hyperstructures arise as natural extensions of classical algebraic
structures when the composition operator is multivalued. The pioneer of this
theory was the French mathematician F. Marty [8], who introduced the concept
of hypergroup in 1934 on the occasion of the 8th Congress of Scandinavian
Mathematicians in Stockholm. Since the 1970s, hyperstructure theory has
experienced a surge of interest , when its research field was greatly extended
by the introduction of other useful notions. Nowadays, hypercompositional
algebra has a variety of relationships with other areas of mathematics. One
significant type of hyperrings was introduced by Krasner, when the addition is
a hyperoperation, while the multiplication is an ordinary binary operation,
which is said to be Krasner hyperring. An extension of the Krasner hyperrings,
which is a subclass of $(m,n)$-hyperrings, was presented by Mirvakili and
Davvaz which is called Krasner $(m,n)$-hyperring, in [9]. Ameri and Norouzi
defined some substancial classes of hyperideals in Krasner $(m,n)$-hyperrings
[1]. Later, in [7] the concepts of $(k,n)$-absorbing hyperideals and
$(k,n)$-absorbing primary hyperideals were studied by Hila et al.. Norouzi and
his colleagues illustrated a new defnition for normal hyperideals in Krasner
$(m,n)$-hyperrings, with respect to that one given in [9] and they proposed in
[10] that these hyperideals correspond to strongly regular relations . In [5],
Asadi and Ameri showed direct limit of a direct system in the category of
Krasner $(m,n)$-hyperrigs. Ozel Ay et al. presented the idea of
$\delta$-primary on Krasner hyperrings [12]. The notion of $\delta$-primary
hyperideals in Krasner $(m,n)$-hyperrings, which unifies the prime and primary
hyperideals under one frame, was defined in [3]. For a Krasner hyperring, the
concept of $n$-hyperideals was briefly introduced by Omidi et al in [11].

For a commutative Krasner $(m,n)$-hyperring $H$, $\sqrt{0}^{(m,n)}$ is the
intersection is taken over all $n$-ary prime hyperideals.

$\sqrt{0}^{(m,n)}=\biggm{\\{}a\in H\ |\
\biggm{\\{}\begin{array}[]{lr}g(a^{(s)},1_{H}^{(n-s)})=0,&s\leq n\\\
g_{(l)}(a^{(s)})=0&s>n,s=l(n-1)+1\end{array}\biggm{\\}}\biggm{\\}}$

is an alternative definition of $\sqrt{0}^{(m,n)}$. In this paper, we aim to
analyze some notions of hyperideals established on basis of the intersection
of all $n$-ary prime hyperideals in a commutative Krasner $(m,n)$-hyperring.
The paper is orgnized as follows. In Section 2, we have given some basic
definitions and results of commutative Krasner $(m,n)$-hyperrings which we
need to develop our paper. In Section 3, we introduce the idea of $n$-ary
$N$-hyperiudeals in a commutative Krasner $(m,n)$-hyperring and give several
characterizations of them. In Section 4, we extend the notion of n-ary
$N$-hyperideals to $n$-ary $\delta$-$N$-hyperideals. We obtain many specific
results explaining the structures. Afterward, in Section 5, we study an
expansion of the previous concept called $(k,n)$-absorbing
$\delta$-$N$-hyperideals. Some properties of them are provided. The last
section is devoted for introducing the $n$-ary $S$-$N$-hyperideals.

##  2\. Preliminaries

For a non-empty set $H$, the mapping $f:H^{n}\longrightarrow P^{*}(H)$ is
called an $n$-ary hyperoperation where $P^{*}(H)$ is the family of all the
nonempty subsets of $H$. An $n$-ary hypergroupoid is an algebraic system
$(H,f)$. Let $H_{1},...,H_{n}$ be non-empty subsets of $H$. Then
$f(H^{n}_{1})=f(H_{1},...,H_{n})$ is defined by $\bigcup\\{f(a^{n}_{1})\ |\
a_{i}\in H_{i},1\leq i\leq n\\}$. The sequence $a_{i},a_{i+1},...,a_{j}$ is
denoted by $a^{j}_{i}$ and so we write

$f(a_{1},...,a_{i},b_{i+1},...,b_{j},c_{j+1},...,c_{n})=f(a^{i}_{1},b^{j}_{i+1},c^{n}_{j+1}).$

Moreover, if $b_{i+1}=...=b_{j}=b$, then we write
$f(a^{i}_{1},b^{(j-i)},c^{n}_{j+1})$. $a^{j}_{i}$ is the empty symbol if
$j<i$. Let $f$ be an $n$-ary hyperoperation. Then $(l(n-1)+1)$-ary
hyperoperation $f_{(l)}$ is given by
$f_{(l)}(a_{1}^{l(n-1)+1})=f(f(...,f(f(a^{n}_{1}),a_{n+1}^{2n-1}),...),a_{(l-1)(n-1)+1}^{l(n-1)+1})$.
Recall from [9] that an algebraic hyperstructure $(H,f,g)$ , or simply $H$, is
called a commutative Krasner $(m,n)$-hyperring if

  * (1)

$(H,f$) is a canonical $m$-ary hypergroup, i.e., for every $1\leq i<j\leq n$
and all $a_{1}^{2n-1}\in H$,
$f(a^{i-1}_{1},f(a_{i}^{n+i-1}),a^{2n-1}_{n+i})=f(a^{j-1}_{1},f(a_{j}^{n+j-1}),a_{n+j}^{2n-1})$
and ${}^{-1}:H\longrightarrow H$ is a unitary operation such that

    * (i)

there exists a unique $e\in H$, such that $f(a,e^{(m-1)})=a$ for every $a\in
H$ and $e^{-1}=e$;

    * (ii)

for each $a\in H$ there exists a unique $a^{-1}\in H$ with $e\in
f(a,a^{-1},e^{(m-2)})$;

    * (iii)

if $a\in f(a^{m}_{1})$, then $a_{i}\in
f(a,a^{-1},...,a^{-1}_{i-1},a^{-1}_{i+1},...,a^{-1}_{m})$ for all $i$,

    * (iv)

$f(a_{1}^{m})=f(a_{\sigma(1)}^{\sigma(m)})$ for all $\sigma\in\mathbb{S}_{n}$
and for all $a_{1}^{m}\in H$,

  * (2)

$(H,g)$ is a $n$-ary semigroup,

  * (3)

$g(a^{i-1}_{1},f(x^{m}_{1}),a^{n}_{i+1})=f(g(a^{i-1}_{1},x_{1},a^{n}_{i+1}),...,g(a^{i-1}_{1},x_{m},a^{n}_{i+1}))$
for every $a^{i-1}_{1},a^{n}_{i+1},x^{m}_{1}\in H$, and $1\leq i\leq n$;

  * (4)

$g(0,a^{n}_{2})=g(a_{2},0,a^{n}_{3})=...=g(a^{n}_{2},0)=0$ for every
$a^{n}_{2}\in H$,

  * (5)

$g(a_{1}^{n})=g(a_{\sigma(1)}^{\sigma(n)})$ for all $\sigma\in\mathbb{S}_{n}$
and for all $a_{1}^{n}\in H$.

For a non-empty subset $S$ of $H$, if $(S,f,g)$ is a Krasner $(m,n)$-hyperrin,
then $S$ is called a subhyperring of $H$ . Let $I$ be a non-empty subset of
$H$. $I$ refers to a hyperideal of $(R,f,g)$ if $(I,f)$ is an $m$-ary
subhypergroup of $(H,f)$ and for every $x^{n}_{1}\in R$ and $1\leq i\leq n$,
$g(a^{i-1}_{1},I,a_{i+1}^{n})\subseteq I$.

######  Definition 2.1.

[9] Suppose that $(H_{1},f_{1},g_{1})$ and $(H_{2},f_{2},g_{2})$ are two
Krasner $(m,n)$-hyperrings. A mapping $d:H_{1}\longrightarrow H_{2}$ is called
a homomorphism if

$d(f_{1}(a_{1},...,a_{m}))=f_{2}(d(a_{1}),...,d(a_{m}))$
$d(g_{1}(b_{1},...,b_{n}))=g_{2}(d(b_{1}),...,d(b_{n}))$

for all $a^{m}_{1}\in H_{1}$ and $b^{n}_{1}\in H_{1}$.

Now, We recall some definitions from [1]. Let $I$ be a hyperideal in a
commutative Krasner $(m,n)$-hyperring $R$ with scalar identity.
${\sqrt{I}}^{(m,n)}$ is the intersection is taken over all prime hyperideals
of $H$ which contain $I$. If the set of all prime hyperideals containing $I$
is empty, then ${\sqrt{I}}^{(m,n)}=H$. Theorem 4.23 in [1] shows that if
$a\in{\sqrt{I}}^{(m,n)}$ then there exists $t\in\mathbb{N}$ such that
$g(a^{(t)},1_{H}^{(n-t)})\in I$ for $t\leq n$, or $g_{(l)}(a^{(t)})\in I$ for
$t=l(n-1)+1$.

######  Definition 2.2.

Let $I$ be a proper hyperideal of a commutative Krasner $(m,n)$-hyperring $H$
with the scalar identity $1_{R}$. $I$ refers to a

  * (1)

maximal hyperideal of $H$ if for every hyperideal $I^{\prime}$ of $H$,
$I\subseteq I^{\prime}\subseteq H$ implies that $I^{\prime}=M$ or
$I^{\prime}=H$. The intersection of all maximal hyperideals of $H$ is called
the Jacobson radical of $H$ which is denoted by $J_{(m,n)}(H)$. If $H$ does
not have any maximal hyperideal, we define $J_{(m,n)}(H)=H$.

  * (2)

$n$-ary prime hyperideal if for hyperideals $I_{1},...,I_{n}$ of $H$,
$g(I_{1}^{n})\subseteq I$ implies that $I_{i}\subseteq I$ for some $1\leq
i\leq n$. Lemma 4.5 in [1] shows that if for all $a^{n}_{1}\in H$,
$g(a^{n}_{1})\in I$ implies that $a_{i}\in I$ for some $1\leq i\leq n$, then
$I$ is a prime hyperideal.

  * (2)

$n$-ary primary hyperideal if $g(a^{n}_{1})\in I$ and $a_{i}\notin I$, then
$g(a_{1}^{i-1},1_{H},a_{i+1}^{n})\in{\sqrt{I}}^{(m,n)}$ for some $1\leq i\leq
n$. Theorem 4.28 in [1] shows that if $I$ is a primary hyperideal of $H$, then
${\sqrt{I}}^{(m,n)}$ is prime.

######  Definition 2.3.

Let $a$ be an element in a commutative Krasner $(m,n)$-hyperring $H$. The
hyperideal generated by $a$ is denoted by $<a>$ and defined by
$g(H,a,1^{(n-2)})=\\{g(r,a,1^{(n-2)})\ |\ r\in H\\}$

######  Definition 2.4.

Let $a$ be an element in a commutative Krasner $(m,n)$-hyperring $H$. It is
invertible if there exists $b\in R$ with $1_{H}=g(a,b,1_{H}^{(n-2)})$. Let $U$
be a subset of $H$. Then $U$ is invertible if and only if every element of $U$
is invertible.

##  3\. $n$-ary $N$-hyperideals

We begin the section with the following definition.

######  Definition 3.1.

Let $I$ be a proper hyperideal of a commutative Krasner $(m,n)$-hyperring $H$.
$I$ refers to an $n$-ary $N$-hyperideal if whenever $x_{1}^{n}\in H$ with
$g(x_{1}^{n})\in I$ and $x_{i}\notin\sqrt{0}^{(m,n)}$ for some $1\leq i\leq n$
imply that $g(x_{1}^{i-1},1_{H},x_{i+1}^{n})\in I$.

######  Example 3.2.

Let $x\geq 1$. Consider $n$-ary hyperintegral domain
$([x,\infty)\cup\\{0\\},\boxplus,\cdot)$ where $\boxplus$ is defined by

$a\boxplus b=\begin{cases}b\boxplus a=\\{a\\}&\text{if $b=0$},\\\
\\{min\\{a,b\\}\\}&\text{if $a\neq 0,b\neq 0$ and $a\neq b$,}\\\
[a,\infty)\cup\\{0\\}&\text{if $a=b\neq 0$.}\end{cases}$

and $"\cdot"$ is the usual multiplication. The hyperideal $0$ is the only
$n$-ary $N$-hyperideal of $([x,\infty)\cup\\{0\\},\boxplus,\cdot)$.

######  Theorem 3.3.

For an $n$-ary $N$-hyperideal $I$ of a commutative Krasner $(m,n)$-hyperring
$H$, $I\subseteq\sqrt{0}^{(m,n)}$.

###### Proof.

Let $I$ be an $n$-ary $N$-hyperideal of a commutative Krasner
$(m,n)$-hyperring $H$ however $I\nsubseteq\sqrt{0}^{(m,n)}$. Assume that $x\in
I$ but $x\notin\sqrt{0}^{(m,n)}$. Since $I$ is an $n$-ary $N$-hyperideal of
$H$ and $g(x,1_{H}^{(n-1)})\in I$, we get $g(1_{H}^{(n)})\in I$, yielding a
contradiction. Consequently, $I\subseteq\sqrt{0}^{(m,n)}$. ∎

The next Theorem gives a characterization of $n$-ary $N$-hyperideals.

######  Theorem 3.4.

Let $I$ be an $n$-ary prime hyperideal of a commutative Krasner
$(m,n)$-hyperring $H$. Then $I$ is an $n$-ary $N$-hyperideal if and only if
$I=\sqrt{0}^{(m,n)}$.

###### Proof.

$\Longrightarrow$ Let $I$ be an $n$-ary prime hyperideal of $H$. Clearly,
$\sqrt{0}^{(m,n)}\subseteq I$. Suppose that $I$ is an $n$-ary $N$-hyperideal
of $H$. Hence we obtain $I\subseteq\sqrt{0}^{(m,n)}$, by Theorem 3.3. Then we
conclude that $I=\sqrt{0}^{(m,n)}$.
$\Longleftarrow$ Let $I=\sqrt{0}^{(m,n)}$. We presume $x_{1}^{n}\in H$ with
$g(x_{1}^{n})\in I$ such that $x_{i}\notin\sqrt{0}^{(m,n)}$ for some $1\leq
i\leq n$. Since $I$ is an $n$-ary prime hyperideal of $H$ and $x_{i}\notin I$,
then there exists $1\leq j\leq i-1$ or $i+1\leq j\leq n$ such that $x_{j}\in
I$. This implies that $g(x_{1}^{i-1},1_{H},x_{i+1}^{n})\in I$. Thus, $I$ is an
$n$-ary $N$-hyperideal of $H$. ∎

In view of Theorem 3.4, we have the following result.

######  Corollary 3.5.

Let $H$ be a commutative Krasner $(m,n)$-hyperring. Then $\sqrt{0}^{(m,n)}$ is
an $n$-ary prime hyperideal of $H$ if and only if it is an $n$-ary
$N$-hyperideal of $H$.

###### Proof.

$\Longrightarrow$ Suppose that $\sqrt{0}^{(m,n)}$ is an $n$-ary prime
hyperideal of $H$. By Theorem 3.4, $\sqrt{0}^{(m,n)}$ is an $n$-ary
$N$-hyperideal of $H$.
$\Longleftarrow$ Let $\sqrt{0}^{(m,n)}$ be an $n$-ary $N$-hyperideal of $H$.
Assume that $g(x_{1}^{n})\in\sqrt{0}^{(m,n)}$ for some $x_{1}^{n}\in H$ such
that $x_{i}\notin\sqrt{0}^{(m,n)}$ for some $1\leq i\leq n$. Then we get
$g(x_{1}^{i-1},1_{H},x_{i+1}^{n})\in\sqrt{0}^{(m,n)}$. Since
$\sqrt{0}^{(m,n)}$ is an $n$-ary $N$-hyperideal of $H$, we can continue the
process and obtain $x_{j}\in\sqrt{0}^{(m,n)}$ for some $1\leq j\leq n$. This
means that $\sqrt{0}^{(m,n)}$ is an $n$-ary prime hyperideal of $H$. ∎

######  Theorem 3.6.

Let $I$ be a proper hyperideal of a commutative Krasner $(m,n)$-hyperring $H$.
If every proper principal hyperideal is an $n$-ary $N$-hyperideal, then so is
$I$.

###### Proof.

Assume that $I$ is a proper hyperideal of $H$. Let $g(x_{1}^{n})\in I$ for
some $x_{1}^{n}\in H$ such that $x_{i}\notin\sqrt{0}^{(m,n)}$ for some $1\leq
i\leq n$. Since every proper principal hyperideal is an $n$-ary $N$-hyperideal
and $g(x_{1}^{n})\in\langle g(x_{1}^{n})\rangle$, we get
$g(x_{1}^{i-1},1_{H},x_{i+1}^{n})\in\langle g(x_{1}^{n})\rangle\subseteq I$.
This means that $I$ is a an $n$-ary $N$-hyperideal of $H$. ∎

######  Theorem 3.7.

Let $I$ be a proper hyperideal of a commutative Krasner $(m,n)$-hyperring $H$.
Then the following statements are equivalent:

  * (1)

$I$ is an $n$-ary $N$-hyperideal of $H$.

  * (2)

$I=E_{x}$ where $E_{x}=\\{y\in H\ |\ g(x,y,1_{H}^{(n-2)})\in I\\}$ for every
$x\notin\sqrt{0}^{(m,n)}$.

  * (3)

$g(I_{1}^{n})\subseteq I$ for some hyperideals $I_{1}^{n}$ of $H$ such that
$I_{i}\cap(H-\sqrt{0}^{(m,n)})\neq\varnothing$ for some $1\leq i\leq n$ imply
that $g(I_{1}^{i-1},1_{H},I_{i+1}^{n})\subseteq I$.

###### Proof.

$(1)\Longrightarrow(2)$ Let $I$ be an $n$-ary $N$-hyperideal of $H$. We always
have $I\subseteq E_{x}$ for every $x\in H$. Assume that $y\in E_{x}$ but
$x\notin\sqrt{0}^{(m,n)}$. This implies that $g(x,y,1_{H}^{(n-2)})\in I$.
Since $I$ is an $n$-ary $N$-hyperideal of $H$ and $x\notin\sqrt{0}^{(m,n)}$,
then $y=g(y,1_{H}^{(n-2)})\in I$. Therefore, we have $I=E_{x}$.

$(2)\Longrightarrow(3)$ Let $g(I_{1}^{n})\subseteq I$ for some hyperideals
$I_{1}^{n}$ of $H$ such that $I_{i}\cap(H-\sqrt{0}^{(m,n)})\neq\varnothing$
for some $1\leq i\leq n$. Then we get $x_{i}\in I_{i}$ such that
$x_{i}\notin\sqrt{0}^{(m,n)}$. Hence,
$g(I_{1}^{i-1},x_{i},I_{I+1}^{n})\subseteq I$ which implies
$g(I_{1}^{i-1},1_{H},I_{i+1}^{n})\subseteq E_{x_{i}}$. Since $I=E_{x_{i}}$ for
every $x_{i}\notin\sqrt{0}^{(m,n)}$, then we get
$g(I_{1}^{i-1},1_{H},I_{i+1}^{n})\subseteq I$.

$(3)\Longrightarrow(1)$ Assume that $g(x_{1}^{n})\in I$ for some $x_{1}^{n}\in
H$ such that $x_{i}\notin\sqrt{0}^{(m,n)}$ for some $1\leq i\leq n$. We get
$g(\langle x_{1}\rangle,...,\langle x_{n}\rangle)\subseteq\langle
g(x_{1}^{n})\rangle\subseteq I$ and $\langle
x_{i}\rangle\nsubseteq\sqrt{0}^{(m,n)}$. Then we conclude that $g(\langle
x_{1}\rangle,...,\langle x_{i-1}\rangle,1_{H},\langle
x_{i+1}\rangle,...,\langle x_{n}\rangle)\subseteq I$ which means
$g(x_{1}^{i-1},1_{H},x_{i+1}^{n})\in I$. Thus, $I$ is an $n$-ary
$N$-hyperideal of $H$. ∎

######  Theorem 3.8.

Let $T$ be a non-empty subset of a commutative Krasner $(m,n)$-hyperring $H$.
If $I$ is an $n$-ary $N$-hyperideal of $H$ such that $T\nsubseteq I$, then
$E_{T}=\\{x\in H\ |\ g(x,T,1_{H}^{(n-2)})\subseteq I\\}$ is an $n$-ary
$N$-hyperideal of $H$.

###### Proof.

Clealy, $E_{T}\neq H$. Assume that $g(x_{1}^{n})\in E_{T}$ for some
$x_{1}^{n}\in H$ such that $x_{i}\notin\sqrt{0}^{(m,n)}$ for some $1\leq i\leq
n$. This means that $g(g(x_{1}^{n}),t,1_{H}^{(n-2)})\in I$ for each $t\in T$
and so $g(x_{i},g(x_{1}^{i-1},t,x_{i+1}^{n}),1_{H}^{(n-2)})\in I$. Since $I$
is an $n$-ary $N$-hyperideal of $H$ and $x_{i}\notin\sqrt{0}^{(m,n)}$, we get
$g(g(x_{1}^{i-1},1_{H},x_{i+1}^{n}),t,1_{H}^{(n-2)})=g(x_{1}^{i-1},t,x_{i+1}^{n})\in
I$ which means $g(x_{1}^{i-1},1_{H},x_{i+1}^{n})\in E_{T}$. Consequently,
$E_{T}$ is an $n$-ary $N$-hyperideal of $H$. ∎

######  Theorem 3.9.

For a maximal $N$-hyperideal $I$ of a commutative Krasner $(m,n)$-hyperring
$H$, $I=\sqrt{0}^{(m,n)}$.

###### Proof.

Suppose that $I$ is a maximal $N$-hyperideal of $H$. Let us consider
$g(x_{1}^{n})\in I$ for some $x_{1}^{n}\in H$. We may assume that $x_{n}\notin
I$. By Theorem 3.8, we conclude that $E_{x_{n}}$ is an $n$-ary $N$-hyperideal
of $H$ with $I\subseteq E_{x_{n}}$. Since $I$ is a maximal $N$-hyperideal of
$H$, we have $E_{x_{n}}=I$ and so $g(x_{1}^{n-1},1_{H})\in I$. Now, we assume
that $x_{n-1}\notin I$. In a similar way, we get
$g(x_{1}^{n-2},1_{H}^{(2)})\in I$. By continuing the argument, we obtain
$x_{1}\in I$ which implies $I$ is an $n$-ary prime hyperideal of $H$. Then, we
conclude that $I=\sqrt{0}^{(m,n)}$, by Theorem 3.4. ∎

The next result characterizes hyperrings admitting $n$-ary $N$-hyperideals.

######  Theorem 3.10.

Let $H$ be a commutative Krasner $(m,n)$-hyperring. Then $H$ admits an $n$-ary
$N$-hyperideal if and only if $\sqrt{0}^{(m,n)}$ is an $n$-ary prime
hyperideal of $H$.

###### Proof.

$\Longrightarrow$ Let $I$ be an $n$-ary $N$-hyperideal of $H$ and let $\Sigma$
be the set of all $n$-ary $N$-hyperideals of $H$. Then
$\Sigma\neq\varnothing$, since $I\in\Sigma$. So $\Sigma$ is a partially
ordered set with respect to set inclusion relation. Now, we take the chain
$I_{1}\subseteq I_{2}\subseteq\cdots\subseteq I_{n}\subseteq\cdots$ in
$\Sigma$. Put $P=\bigcup_{i=1}^{\infty}I_{i}$. We prove $P$ is an $n$-ary
$N$-hyperideal of $H$. Assume that $g(x_{1}^{n})\in P$ for some $x_{1}^{n}\in
H$ such that $x_{i}\notin\sqrt{0}^{(m,n)}$. Then there exists $s\in\mathbb{N}$
such that $g(x_{1}^{n})\in I_{s}$. Then we get
$g(x_{1}^{i-1},1_{H},x_{i+1}^{n})\in I_{s}\subseteq P$, as $I_{s}$ is an
$n$-ary $N$-hyperideal of $H$ and $x_{i}\notin\sqrt{0}^{(m,n)}$. This means
that $P$ is a upper bound of the mentioned chain. By Zorn’s lemma, there is a
hyperideal $Q$ which is maximal in $\Sigma$. Hence $Q=\sqrt{0}^{(m,n)}$ is an
$n$-ary prime hyperideal of $H$, by Theorem 3.9.
$\Longleftarrow$ Let $\sqrt{0}^{(m,n)}$ be an $n$-ary prime hyperideal of $H$.
Then it is an $n$-ary $N$-hyperideal of $H$, by Corollary 3.5. ∎

######  Corollary 3.11.

Let $I$ be a hyperideal of a commutative Krasner $(m,n)$-hyperring $H$ with
$I\subseteq\sqrt{0}^{(m,n)}$. Then $I$ is an $n$-ary $N$-hyperideal of $H$ if
and only if $I$ is an $n$-ary primary hyperideal of $H$.

###### Proof.

Straightforward. ∎

Now, we are interested in the hyperrings over which $0$ is the only $n$-ary
$N$-hyperideal of $H$.

######  Theorem 3.12.

Let $H$ be a commutative Krasner $(m,n)$-hyperring. Then $0$ is the only
$n$-ary $N$-hyperideal of $H$ if and only if $H$ is an $n$-ary hyperintegral
domain.

###### Proof.

$\Longrightarrow$ Let $0$ be the only $n$-ary $N$-hyperideal of $H$. Then by
Theorem 3.10, $\sqrt{0}^{(m,n)}$ is an $n$-ary prime hyperideal of $H$.
Therefore $\sqrt{0}^{(m,n)}$ is an $n$-ary $N$-hyperideal of $H$, by Corollary
3.5. Thus $0=\sqrt{0}^{(m,n)}$ is an $n$-ary prime hyperideal of $H$ which
means $H$ is an $n$-ary hyperintegral domain.
$\Longleftarrow$ Let $H$ be an $n$-ary hyperintegral domain. Suppose that $I$
is an $n$-ary $N$-hyperideal of $H$. Then $I\subseteq\sqrt{0}^{(m,n)}$, by
Theorem 3.3. Since $H$ is an $n$-ary hyperintegral domain, we get
$0=\sqrt{0}^{(m,n)}$. Hence $I=0$ is the only $n$-ary $N$-hyperideal of $H$. ∎

Let $(H_{1},f_{1},g_{1})$ and $(H_{2},f_{2},g_{2})$ be two commutative Krasner
$(m,n)$-hyperrings such that $1_{H_{1}}$ and $1_{H_{2}}$ be scalar identitis
of $H_{1}$ and $H_{2}$, respectively. Then the $(m,n)$-hyperring $(H_{1}\times
H_{2},f_{1}\times f_{2},g_{1}\times g_{2})$ is defined by $m$-ary
hyperoperation $f=f_{1}\times f_{2}$ and n-ary operation $g=g_{1}\times
g_{2}$, as follows:

$f_{1}\times f_{2}((x_{1},y_{1}),\cdots,(x_{m},y_{m}))=\\{(x,y)\ |\ \ x\in
f_{1}(x_{1}^{m}),y\in f_{2}(y_{1}^{m})\\}$

$\hskip 34.14322ptg_{1}\times
g_{2}((a_{1},b_{1}),\cdots,(a_{n},b_{n}))=(g_{1}(a_{1}^{n}),g_{2}(b_{1}^{n}))$,
for all $x_{1}^{m},a_{1}^{n}\in H_{1}$ and $y_{1}^{m},b_{1}^{n}\in H_{2}$ [2].

######  Theorem 3.13.

Let $(H_{1},f_{1},g_{1})$ and $(H_{2},f_{2},g_{2})$ be two commutative Krasner
$(m,n)$-hyperrings such that $1_{H_{1}}$ and $1_{H_{2}}$ be scalar identitis
of $H_{1}$ and $H_{2}$, respectively. Then $H_{1}\times H_{2}$ has no $n$-ary
$N$-hyperideals.

###### Proof.

Let $I_{1}$ and $I_{2}$ be hyperideals of $H_{1}$ and $H_{2}$, respectively,
and let $I_{1}\times I_{2}$ be an $n$-ary $N$-hyperideal of $H_{1}\times
H_{2}$. It is clear that
$(1_{H_{1}},0_{H_{2}}),(0_{H_{1}},1_{H_{2}}),(1_{H_{1}},1_{H_{2}})\notin\sqrt{0_{H_{1}\times
H_{2}}}^{(m,n)}$ but $g_{1}\times
g_{2}((1_{H_{1}},0_{H_{2}}),(0_{H_{1}},1_{H_{2}}),(1_{H_{1}},1_{H_{2}})^{(n-2)})\in
I_{1}\times I_{2}$. Therefore we obtain $(1_{H_{1}},0_{H_{2}})=g_{1}\times
g_{2}((1_{H_{1}},0_{H_{2}}),(1_{H_{1}},1_{H_{2}})^{(n-1)})\in I_{1}\times
I_{2}$ and $(0_{H_{1}},1_{H_{2}})=g_{1}\times
g_{2}((1_{H_{1}},1_{H_{2}}),(0_{H_{1}},1_{H_{2}}),(1_{H_{1}},1_{H_{2}})^{(n-2)})\in
I_{1}\times I_{2}$. This implies that $(1_{H_{1}},1_{H_{2}})=f_{1}\times
f_{2}((1_{H_{1}},0_{H_{2}}),(0_{H_{1}},1_{H_{2}}),(0_{H_{1}},0_{H_{2}})^{(m-2)}))\in
I_{1}\times I_{2}$ which means $I_{1}\times I_{2}=H_{1}\times H_{2}$. ∎

######  Theorem 3.14.

Suppose that $(H_{1},f_{1},g_{1})$ and $(H_{2},f_{2},g_{2})$ are two
commutative Krasner $(m,n)$-hyperrings and $h:H_{1}\longrightarrow H_{2}$ is a
homomorphism. Then:

$(1)$ If $h$ is a monomorphism and $I_{2}$ is an $n$-ary $N$-hyperideal of
$H_{2}$, then $h^{-1}(I_{2})$ is an $n$-ary $N$-hyperideal of $H_{1}$.

$(2)$ Let $h$ be an epimorphism and $I_{1}$ be a hyperideal of $H_{1}$ with
$Ker(h)\subseteq I_{1}$. If $I_{1}$ is an $n$-ary $N$-hyperideal of $H_{1}$,
then $h(I_{1})$ is an $n$-ary $N$-hyperideal of $H_{2}$.

###### Proof.

$(1)$ Let $g_{1}(x_{1}^{n})\in h^{-1}(I_{2})$ for $x_{1}^{n}\in H_{1}$ such
that $x_{i}\notin\sqrt{0_{H_{1}}}^{(m,n)}$ for some $1\leq i\leq n$. Then we
have $h(g_{1}(x_{1}^{n}))=g_{2}(h(x_{1}^{n}))\in I_{2}$. Since $h$ is a
monomorphism and $x_{i}\notin\sqrt{0_{H_{1}}}^{(m,n)}$, then
$h(x_{i})\notin\sqrt{0_{H_{2}}}^{(m,n)}$. Since $I_{2}$ is a $N$-hyperideal of
$H_{2}$, we get the result that

$g_{2}(h(x_{1}),...,h(x_{i-1}),1_{H_{2}},h(x_{i+1}),...,h(x_{n}))$

$\hskip 85.35826pt=h(g_{1}(x_{1}^{i-1},1_{H_{1}},x_{i+1}^{n}))$

$\hskip 85.35826pt\in I_{2}$
which follows $g_{1}(x_{1}^{i-1},1_{H_{1}},x_{i+1}^{n})\in h^{-1}(I_{2})$.
Thus $h^{-1}(I_{2})$ is an $n$-ary $N$-hyperideal of $H_{1}$.

$(2)$ Let $g_{2}(y_{1}^{n})\in h(I_{1})$ for some $y_{1}^{n}\in H_{2}$ such
that $y_{i}\notin\sqrt{0_{H_{2}}}^{(m,n)}$. Since $h$ is an epimorphism, then
there exist $x_{1}^{n}\in R_{1}$ with $h(x_{1})=y_{1},...,h(x_{n})=y_{n}$.
Therefore
$h(g_{1}(x_{1}^{n}))=g_{2}(h(x_{1}),...,h(x_{n}))=g_{2}(y_{1}^{n})\in
h(I_{1})$. Since $Ker(h)\subseteq I_{1}$, then we have $g_{1}(x_{1}^{n})\in
I_{1}$. Since $y_{i}\notin\sqrt{0_{H_{2}}}^{(m,n)}$, then
$x_{i}\notin\sqrt{0_{H_{1}}}^{(m,n)}$. Since $I_{1}$ is an n-ary
$N$-hyperideal of $R_{1}$ and $x_{i}\notin\sqrt{0_{H_{1}}}^{(m,n)}$, we obtain
$g_{1}(x_{1}^{i-1},1_{H_{1}},x_{i+1}^{n})\in I_{1}$ which means

$h(g_{1}(x_{1}^{i-1},1_{H_{1}},x_{i+1}^{n}))=g_{2}(h(x_{1}),...,h(x_{i-1}),1_{H_{2}},h(x_{i+1}),...,h(x_{n}))$

$\hskip 128.0374pt=g_{2}(y_{1}^{i-1},1_{H_{2}},y_{i+1}^{n})\in h(I_{1})$
Consequently, $h(I_{1})$ is an $n$-ary $N$-hyperideal of $H_{2}$. ∎

######  Corollary 3.15.

Let $H^{\prime}$ be a subhyperring of a commutative Krasner $(m,n)$-hyperring
$H$. If $I$ is an $n$-ary $N$-hyperideal of $H$ such that
$H^{\prime}\nsubseteq I$, then $H^{\prime}\cap I$ is an $n$-ary $N$-hyperideal
of $H^{\prime}.$

###### Proof.

Let us consider the identity map $j$ from $H^{\prime}$ into $H$. We conclude
that $H^{\prime}\cap I$ is an $n$-ary $N$-hyperideal of $H^{\prime}$, by
Theorem 3.14 (2). ∎

Let $J$ be a hyperideal of a commutative Krasner $(m,n)$-hyperring $(H,f,g)$.
Then the set

$H/J=\\{f(a^{i-1}_{1},J,a^{m}_{i+1})\ |\ a^{i-1}_{1},a^{m}_{i+1}\in H\\}$

endowed with m-ary hyperoperation $f$ which for all
$a_{11}^{1m},...,a_{m1}^{mm}\in H$

$f(f(a_{11}^{1(i-1)},J,a^{1m}_{1(i+1)}),...,f(a_{m1}^{m(i-1)},J,a^{mm}_{m(i+1)}))$

$\hskip
85.35826pt=f(f(a^{m1}_{11}),...,f(a^{m(i-1)}_{1(i-1)}),J,f(a^{m(i+1)}_{1(i+1)}),...,f(a^{mm}_{1m}))$
and with $n$-ary hyperoperation g which for all
$a_{11}^{1m},...,a_{n1}^{nm}\in H$

$g(f(a_{11}^{1(i-1)},J,a^{1m}_{1(i+1)}),...,f(a_{n1}^{n(i-1)},J,a^{nm}_{n(i+1)}))$

$\hskip
85.35826pt=f(g(a^{n1}_{11}),...,g(a^{n(i-1)}_{1(i-1)}),J,g(a^{n(i+1)}_{1(i+1)}),...,f(a^{nm}_{1m}))$
construct a commutative Krasner $(m,n)$-hyperring, and $(H/J,f,g)$ is said to
be the quotient Krasner $(m,n)$-hyperring of $H$ by $J$ [1].

Now, we determine when the hyperideal $I/J$ is n-ary $N$-hyperideal in $H/J$.

######  Theorem 3.16.

Let $I$ and $J$ be two hyperideals of a commutative Krasner $(m,n)$-hyperring
$H$ with $J\subseteq I$. If $I$ is an $n$-ary $N$-hyperideal of $H$, then
$I/J$ is an $n$-ary $N$-hyperideal of $H/I$.

###### Proof.

Consider the projection map of $H$ of $J$, that is, $\theta:H\longrightarrow
H/J$, defined by $a\longrightarrow f(a,I,0^{(m-2)})$. By using Theorem 3.14
(2), we are done. ∎

######  Theorem 3.17.

Let $I$ and $J$ be two hyperideals of a commutative Krasner $(m,n)$-hyperring
$H$ with $J\subseteq I$. If $I/J$ is an $n$-ary $N$-hyperideal of $H/J$ with
$J\subseteq\sqrt{0_{H}}^{(m,n)}$, then $I$ is an $n$-ary $N$-hyperideal of
$H$.

###### Proof.

Let $g(x_{1}^{n})\in I$ for $x_{1}^{n}\in H$ such that
$x_{i}\notin\sqrt{0_{H}}^{(m,n)}$ for some $1\leq i\leq n$. Then we get
$g(f(x_{1},J,0_{H}^{(m-2)}),\cdots,f(x_{n},J,0_{H}^{(m-2)}))=f(g(x_{1}^{n}),J,0_{H/J}^{(m-2)})\in
I/J$. Note that $f(x_{i},J,0_{H}^{(m-2)})\notin\sqrt{0_{H/J}}^{(m,n)}$. Since
$I/J$ is an $n$-ary $N$-hyperideal of $H/J$ and
$f(x_{i},J,0_{H}^{(m-2)})\notin\sqrt{0_{H/J}}^{(m,n)}$, we get the result that

$g(f(x_{1},J,0_{H}^{(m-2)}),\cdots,f(x_{i-1},J,0_{H}^{(m-2)}),1_{H/J},f(x_{i+1},J,0_{H}^{(m-2)}),$

$\hskip 56.9055pt\cdots,f(x_{n},J,0_{H}^{(m-2)}))\in I/J$
which implies $f(g(x_{1}^{i-1},1_{H},x_{i+1}^{n}),J,0_{H/J}^{(m-2)})\in I/J$.
Therefore $g(x_{1}^{i-1},1_{H},x_{i+1}^{n})\in I$. Thus $I$ is an n-ary
$N$-hyperideal of $H$. ∎

Using Theorem 3.3 and Theorem 3.17, we have the next corollary.

######  Corollary 3.18.

Let $I$ and $J$ be two hyperideals of a commutative Krasner $(m,n)$-hyperring
$H$ with $J\subseteq I$. If $I/J$ is an $n$-ary $N$-hyperideal of $H/J$ such
that $J$ is an $n$-ary $N$-hyperideal of $H$, then $I$ is an $n$-ary
$N$-hyperideal of $H$.

###### Proof.

The proof follows from Theorem 3.3 and Theorem 3.17. ∎

##  4\. $n$-ary $\delta$-$N$-hyperideals

Recall from [3] that a function $\delta$ is a hyperideal expansion of a
commutative Krasner $(m,n)$-hyperring $H$ if it assigns to each hyperideal $I$
of $H$ a hyperideal $\delta(I)$ of $H$ such that $I\subseteq\delta(I)$ and if
$I\subseteq J$ for any hyperideals $I,J$ of $H$, then
$\delta(I)\subseteq\delta(J)$. For instance, $\delta_{0}(I)=I$,
$\delta_{1}(I)=\sqrt{I}^{(m,n)}$ and $\delta_{H}(I)=H$ for all hyperideals $I$
of $H$ are hyperideal expansions of $H$. Moreover,
$\delta_{q}(I/J)=\delta(I)/J$ for expansion function $\delta$ of $H$ and for
all hyperideals $I$ of $H$ containing hyperideal $J$ is an hyperideal
expansion of $H$. By using a hyperideal expansion $\delta$ of $H$, we present
the following definition.

######  Definition 4.1.

Assume that $\delta$ is a hyperideal expansion of a commutative Krasner
$(m,n)$-hyperring $H$. A proper hyperideal $I$ of $H$ is said to be an $n$-ary
$\delta$-$N$-hyperideal if $g(x_{1}^{n})\in I$ for $x_{1}^{n}\in H$ and
$x_{i}\notin\sqrt{0}^{(m,n)}$ for some $1\leq i\leq n$ imply that
$g(x_{1}^{i-1},1_{H},x_{i+1}^{n})\in\delta(I).$

######  Example 4.2.

Assume that $\mathbb{Z}_{12}=\\{0,1,2,3,\cdots,11\\}$ is the set of all
congruence classes of integers modulo $12$ and
$\mathbb{Z}^{\star}_{12}=\\{1,5,7,11\\}$ is multiplicative subgroup of units
$\mathbb{Z}_{12}$. Construct $H$ as $\mathbb{Z}_{12}/\mathbb{Z}_{12}^{\star}$.
Then we have $H=\\{\bar{0},\bar{1},\bar{2},\bar{3},\bar{4},\bar{6}\\}$ in
which $\bar{0}=\\{0\\}$, $\bar{1}=\\{1,5,7,11\\}$,
$\bar{2}=\bar{10}=\\{2,10\\}$, $\bar{3}=\bar{9}=\\{3,9\\}$,
$\bar{4}=\bar{8}=\\{4,8\\}$, $\bar{6}=\\{6\\}$. Consider Krasner hyperring
$(H,\boxplus,\circ)$ that for all $\bar{x},\bar{y}\in H$,
$\bar{x}\circ\bar{y}=\overline{xy}$ and 2-ary hyperoperation $\boxplus$ is
defined as follows

$\boxplus$ | $\bar{0}$ | $\bar{1}$ | $\bar{2}$ | $\bar{3}$ | $\bar{4}$ | $\bar{6}$
---|---|---|---|---|---|---
$\bar{0}$ | $\bar{0}$ | $\bar{1}$ | $\bar{2}$ | $\bar{3}$ | $\bar{4}$ | $\bar{6}$
$\bar{1}$ | $\bar{1}$ | $\bar{0},\bar{2},\bar{\bar{4}},\bar{6}$ | $\bar{1},\bar{3}$ | $\bar{2},\bar{4}$ | $\bar{1},\bar{3}$ | $\bar{1}$
$\bar{2}$ | $\bar{2}$ | $\bar{1},\bar{3}$ | $\bar{0},\bar{4}$ | $\bar{1}$ | $\bar{2},\bar{6}$ | $\bar{4}$
$\bar{3}$ | $\bar{3}$ | $\bar{2},\bar{4}$ | $\bar{1}$ | $\bar{0},\bar{6}$ | $\bar{1}$ | $\bar{3}$
$\bar{4}$ | $\bar{4}$ | $\bar{1},\bar{3}$ | $\bar{2},\bar{6}$ | $\bar{1}$ | $\bar{0},\bar{4}$ | $\bar{2}$
$\bar{6}$ | $\bar{6}$ | $\bar{1}$ | $\bar{4}$ | $\bar{3}$ | $\bar{2}$ | $\bar{0}$

In the hyperring, $\sqrt{0}^{(2,2)}=\\{\bar{0},\bar{6}\\}$ and
$I=\\{\bar{0},\bar{2},\bar{4},\bar{6}\\}$ is a 2-ary
$\delta_{1}$-$N$-hyperring.

######  Theorem 4.3.

Let $I$ be an $n$-ary $\delta$-primary hyperideal of a commutative Krasner
$(m,n)$-hyperring $H$ such that $I\subseteq\sqrt{0}^{(m,n)}$. Then $I$ is an
$n$-ary $\delta$-$N$-hyperideal of $H$.

###### Proof.

Let $g(x_{1}^{n})\in I$ for $x_{1}^{n}\in H$ such that
$x_{i}\notin\sqrt{0}^{(m,n)}$ for some $1\leq i\leq n$. By the assumption,
$x_{i}\notin I$. Since $I$ is an $n$-ary $\delta$-primary hyperideal of $H$,
we conclude that $g(x_{1}^{i-1},1_{H},x_{i+1}^{n})\in\delta(I)$. Thus $I$ is
an $n$-ary $\delta$-$N$-hyperideal of $H$. ∎

The next Theorem shows that the inverse of Theorem 4.3 is true if
$I=\sqrt{0}^{(m,n)}$.

######  Theorem 4.4.

Let $\sqrt{0}^{(m,n)}$ be an $n$-ary $\delta$-$N$-hyperideal of a commutative
Krasner $(m,n)$-hyperring $H$. Then $\sqrt{0}^{(m,n)}$ is an $n$-ary
$\delta$-primary hyperideal of $H$.

###### Proof.

Let $g(x_{1}^{n})\in\sqrt{0}^{(m,n)}$ for some $x_{1}^{n}\in H$ such that
$x_{i}\notin\sqrt{0}^{(m,n)}$ for some $1\leq i\leq n$. Since
$\sqrt{0}^{(m,n)}$ is an $n$-ary $\delta$-$N$-hyperideal of $H$, we get the
result that $g(x_{1}^{i-1},1_{H},x_{i+1}^{n})\in\delta(\sqrt{0}^{(m,n)})$, as
needed. ∎

######  Theorem 4.5.

Let $I$ be a proper hyperideal of a commutative Krasner $(m,n)$-hyperring $H$.
Then the followings are equivalent:

  * (1)

$I$ is an $n$-ary $\delta$-$N$-hyperideal of $H$.

  * (2)

$E_{x}\subseteq\sqrt{0}^{(m,n)}$ for all $x\notin\delta(I)$ where
$E_{x}=\\{y\in H\ |\ g(x,y,1_{H}^{(n-2)})\in I\\}$.

  * (3)

If $g(x,I_{1}^{n-1})\subseteq I$ for some hyperideals $I_{1}^{n-1}$ of $H$ and
for some $x\in H$ implies $x\in\sqrt{0}^{(m,n)}$ or
$g(1_{H},I_{1}^{n-1})\subseteq\delta(I)$.

  * (4)

If $g(I_{1}^{n})\subseteq I$ for some hyperideals $I_{1}^{n-1}$ of $H$, then
$I_{i}\cap(H-\sqrt{0}^{(m,n)})=\varnothing$ for some $1\leq i\leq n$ or
$g(I_{1}^{i-1},1_{H},I_{i+1}^{n})\subseteq\delta(I)$.

###### Proof.

$(1)\Longrightarrow(2)$ Assume that $y\in E_{x}$. So $g(y,x,1_{H}^{(n-2)})\in
I$. Since $I$ is an $n$-ary $\delta$-$N$-hyperideal of $H$ and
$x=g(x,1_{H}^{(n-2)})\notin\delta(I)$, we obtain $y\in\sqrt{0}^{(m,n)}$ which
means $E_{x}\subseteq\sqrt{0}^{(m,n)}$.
$(2)\Longrightarrow(3)$ Let $g(x,I_{1}^{n-1})\subseteq I$ for some hyperideals
$I_{1}^{n-1}$ of $H$ and for some $x\in H$ such that
$g(1_{H},I_{1}^{n-1})\nsubseteq\delta(I)$. Hence there exist $a_{1}\in
I_{1},\cdots,a_{n-1}\in I_{n-1}$ such that
$g(1_{H},a_{1}^{n-1})\notin\delta(I)$. Since $x\in E_{g(1_{H},a_{1}^{n-1})}$,
we conclude that $x\in\sqrt{0}^{(m,n)}$.
$(3)\Longrightarrow(4)$ Let $g(I_{1}^{n})\subseteq I$ for some hyperideals
$I_{1}^{n-1}$ of $H$ such that $I_{i}\cap(H-\sqrt{0}^{(m,n)})\neq\varnothing$
for some $1\leq i\leq n$. Therefore we have some $x\in
I_{i}\cap(H-\sqrt{0}^{(m,n)})$. Since $g(I_{1}^{i-1},x,I_{i+1}^{n})\subseteq
I$, we get $g(I_{1}^{i-1},1_{H},I_{i+1}^{n})\subseteq\delta(I)$, as
$x\notin\sqrt{0}^{(m,n)}$.
$(4)\Longrightarrow(1)$ Suppose that $g(x_{1}^{n})\in I$ for $x_{1}^{n}\in H$
such that $x_{i}\notin\sqrt{0}^{(m,n)}$ for some $1\leq i\leq n$. Let us
consider $I_{1}=\langle x_{1}\rangle,\cdots,I_{n}=\langle x_{n}\rangle$. Since
$I_{i}\cap(H-\sqrt{0}^{(m,n)})\neq\varnothing$, we conclude that
$g(x_{1}^{i-1},1_{H},x_{i+1}^{n})\in
g(I_{1}^{i-1},1_{H},I_{i+1}^{n})\subseteq\delta(I)$. Thus $I$ is an $n$-ary
$\delta$-$N$-hyperideal of $H$. ∎

######  Theorem 4.6.

Let $I$ be an $n$-ary $\delta$-$N$-hyperideal of a commutative Krasner
$(m,n)$-hyperring $H$ and let $x\notin\delta(I).$ If
$F_{x}\subseteq\delta(E_{x})$ where $F_{x}=\\{y\in H\ |\
g(y,x,1_{H}^{(n-2)})\in\delta(I)\\}$ and $E_{x}=\\{y\in H\ |\
g(y,x,1_{H}^{(n-2)})\in I\\}$, then $E_{x}$ is an $n$-ary
$\delta$-$N$-hyperideal of $H$.

###### Proof.

Let $g(x_{1}^{n})\in E_{x}$ for $x_{1}^{n}\in H$ such that
$x_{i}\notin\sqrt{0}^{(m,n)}$ for some $1\leq i\leq n$. Hence
$g(g(x_{1}^{n}),x,1_{H}^{(n-2)})=g(x_{i},g(x_{1}^{i-1},x,x_{i+1}^{n}),1_{H}^{(n-2)})\in
I$. Since $I$ is an $n$-ary $\delta$-$N$-hyperideal of $H$, we have
$g(x_{1}^{i-1},x,x_{i+1}^{n})\in\delta(I)$. Hence
$g(x_{1}^{i-1},1_{H},x_{i+1}^{n})\in F_{x}\subseteq\delta(E_{x})$, as needed.
∎

Recall that a hyperideal expansion $\delta$ of $H$ is intersection preserving
if it satisfies $\delta(I_{1}\cap I_{2})=\delta(I_{1})\cap\delta(I_{2})$ for
each hyperideals $I_{1}$ and $I_{2}$ of a commutative Krasner
$(m,n)$-hyperring $H$. For instance, the hyperideal expansion $\delta_{1}$ of
$H$ is intersection preserving.

######  Theorem 4.7.

Let $I_{1}^{n}$ be some $n$-ary $\delta$-$N$-hyperideals of a Krasner
$(m,n)$-hyperring $H$ and let the hyperideal expansion $\delta$ of $H$ be
intersection preserving. Then $\bigcap_{i=1}^{n}I_{i}$ is an $n$-ary
$\delta$-$N$-hyperideal of $H$.

###### Proof.

Put $I=\bigcap_{i=1}^{n}I_{i}$. Suppose that $g(x_{1}^{n})\in I$ for
$x_{1}^{n}\in H$ with $g(x_{1}^{i-1},1_{H},x_{i+1}^{n})\notin\delta(I)$. Since
the hyperideal expansion $\delta$ of $H$ is intersection preserving, then
there exists $1\leq t\leq n$ with
$g(x_{1}^{i-1},1_{H},x_{i+1}^{n})\notin\delta(I_{t})$. As $I_{t}$ ia an
$n$-ary $\delta$-$N$-hyperideal of $H$, we have $x_{i}\in\sqrt{0}^{(m,n)}$.
Therefore $I=\bigcap_{i=1}^{n}I_{i}$ is an $n$-ary $\delta$-$N$-hyperideal of
$H$. ∎

######  Theorem 4.8.

Suppose that $I_{1},I_{2}$ and $I_{3}$ are proper hyperideals of a commutative
Krasner $(m,n)$-hyperring $H$ with $I_{1}\subseteq I_{2}\subseteq I_{3}$ and
$\delta(I_{1})=\delta(I_{3})$. If $I_{3}$ is an $n$-ary
$\delta$-$N$-hyperideal of $H$, then $I_{2}$ is an $n$-ary
$\delta$-$N$-hyperideal of $H$.

###### Proof.

Let $g(x_{1}^{n})\in I_{2}$ for some $x_{1}^{n}\in H$ such that
$x_{i}\notin\sqrt{0}^{(m,n)}$. Since $I_{3}$ is an $n$-ary
$\delta$-$N$-hyperideal of $H$ and $I_{2}\subseteq I_{3}$, then we conclude
that $g(x_{1}^{i-1},1_{H},x_{i+1}^{n})\in\delta(I_{3})$. Then we get
$g(x_{1}^{i-1},1_{H},x_{i+1}^{n})\in\delta(I_{1})$. Therefore
$g(x_{1}^{i-1},1_{H},x_{i+1}^{n})\in\delta(I_{2})$, as $I_{1}\subseteq I_{2}$.
Thus $I_{2}$ is an $n$-ary $\delta$-$N$-hyperideal of $H$. ∎

######  Theorem 4.9.

Suppose that $I$ is a proper hyperideal of a commutative Krasner
$(m,n)$-hyperring $H$ such that $\delta(I)$ is an $n$-ary $N$-hyperideal of
$H$. Then $I$ is an $n$-ary $\delta$-$N$-hyperideal of $H$.

###### Proof.

Let $\delta(I)$ be an $n$-ary $N$-hyperideal of $H$. Let $g(x_{1}^{n})\in I$
for $x_{1}^{n}\in H$ with $x_{i}\notin\sqrt{0}^{(m,n)}$ for some $1\leq i\leq
n$. Since $I\subseteq\delta(I)$ and $\delta(I)$ is an $n$-ary $N$-hyperideal
of $H$, then $g(x_{1}^{i-1},1_{H},x_{i+1}^{n})\in\delta(I)$. It follows that
$I$ is an $n$-ary $\delta$-$N$-hyperideal of $H$. ∎

The inverse of the previous theorem is true if $\delta=\delta_{1}$. See the
next Theorem.

######  Theorem 4.10.

If $I$ is an $n$-ary $\delta_{1}$-$N$-hyperideal of a commutative Krasner
$(m,n)$-hyperring $H$, then $\delta_{1}(I)$ is an $n$-ary $N$-hyperideal of
$H$.

###### Proof.

Let $g(x_{1}^{n})\in\delta_{1}(I)$ for $x_{1}^{n}\in H$ such that
$x_{i}\notin\sqrt{0}^{(m,n)}$. It implies that there exists $t\in\mathbb{N}$
such that if $t\leq n$, then $g(g(x_{1}^{n})^{(t)},1_{H}^{(n-t)})\in I$.
Therefore we get

$g(x_{i}^{(t)},g(x_{i}^{i-1},1_{H},x_{i+1}^{n})^{(t)},1_{H}^{(n-2t)})$

$\hskip
96.73918pt=g(x_{i}^{(t)},g(x_{i}^{i-1},1_{H},x_{i+1}^{n})^{(t)},g(1_{R}^{(n)}),1_{H}^{(n-2t-1)})$

$\hskip
96.73918pt=g(g(x_{i}^{(t)},1_{H}^{(n-t)}),g(x_{i}^{i-1},1_{H},x_{i+1}^{n})^{(t)},1_{H}^{(n-t-1)})$

$\hskip 96.73918pt\subseteq I$.
Since $I$ is an $n$-ary $\delta_{1}$-$N$-hyperideal of $H$ and
$g(x_{i}^{(t)},1^{(n-t)})\notin\sqrt{0}^{(m,n)}$, we get the result that
$g(g(x_{i}^{i-1},1_{H},x_{i+1}^{n})^{(t)},1_{H}^{(n-t)})\in\delta_{1}(I)$ and
so $g(x_{i}^{i-1},1_{H},x_{i+1}^{n})\in\delta_{1}(I)$. It follows that
$\delta_{1}(I)$ is an $n$-ary $N$-hyperideal of $H$. By using a similar
argument, one can easily complete the proof where $t=l(n-1)+1$. ∎

######  Theorem 4.11.

Suppose that $\sqrt{0}^{(m,n)}$ is the only maximal hyperideal of a
commutative Krasner $(m,n)$-hyperring $H$. Then for all $a\in H$, $\langle
a\rangle$ is an $n$-ary hyperideal of $H$.

###### Proof.

Assume that $\sqrt{0}^{(m,n)}$ is the only maximal hyperideal of a commutative
Krasner $(m,n)$-hyperring $H$. Let $a\in H$ and $g(a_{1}^{n})\in\langle
a\rangle$ for $a_{1}^{n}\in H$ such that $a_{i}\notin\langle a\rangle$ for
some $1\leq i\leq n$. From $a_{i}\notin\langle a\rangle$, it follows that
$a_{i}$ is an invertible element. Then we have
$g(a_{1}^{i-1},1_{H},a_{i+1}^{n})\in\langle a\rangle\subseteq\delta(\langle
a\rangle)$ which means that $\langle a\rangle$ is an $n$-ary hyperideal of
$H$. ∎

######  Theorem 4.12.

Let $I$ be a hyperideal of a commutative Krasner $(m,n)$-hyperring $H$ and
$\delta$ be a a hyperideal expansion of $H$ such that
$\delta(\delta(I))=\delta(I)$. If $I$ is an $n$-ary $\delta$-$N$-hyperideal of
$H$ and $x\notin\sqrt{0}^{(m,n)}$, then $\delta(E_{x})=\delta(I)$ where
$E_{x}=\\{y\in H\ |\ g(y,x,1_{H}^{(n-2)})\in I\\}$.

###### Proof.

Assume that $I$ is an $n$-ary $\delta$-$N$-hyperideal of $H$ and
$x\notin\sqrt{0}^{(m,n)}$. Let $y\in E_{x}$. This means
$g(y,x,1_{H}^{(n-2)})\in I$. Since $I$ is an $n$-ary hyperideal of $H$ and
$x\notin\sqrt{0}^{(m,n)}$, we get $y=(y,1^{(n-1)})\in\delta(I)$ which implies
$E_{x}\subseteq\delta(I)$ and so $\delta(E_{x})\subseteq\delta(\delta(I))$. By
the assumpption we have $\delta(E_{x})\subseteq\delta(I)$. On the other hand,
we have $I\subseteq E_{x}$ and so $\delta(I)\subseteq\delta(E_{x})$.
Consequently, $\delta(E_{x})\subseteq\delta(I)$. ∎

######  Theorem 4.13.

Suppose that $I$ is a maximal $n$-ary $\delta$-$N$-hyperideal of a commutative
Krasner $(m,n)$-hyperring $H$ and $x\notin\delta(I).$ If
$F_{x}\subseteq\delta(E_{x})$ where $F_{x}=\\{y\in H\ |\
g(y,x,1_{H}^{(n-2)})\in\delta(I)\\}$ and $E_{x}=\\{y\in H\ |\
g(y,x,1_{H}^{(n-2)})\in I\\}$, then $I=\sqrt{0}^{(m,n)}$ is an $n$-ary prime
hyperideal of $H$.

###### Proof.

Let $I$ be a maximal $n$-ary $\delta$-$N$-hyperideal of $H$. Assume that
$g(a_{1}^{n})\in I$ for $a_{1}^{n}\in H$ such that $a_{i}\notin I$ for some
$1\leq i\leq n$. Then, by Theorem 4.6, $E_{x}$ is an $n$-ary
$\delta$-$N$-hyperideal of $H$. Therefore $E_{x}=I$ by the maximality of $I$.
Thus we have $g(a_{1}^{i-1}1_{H},a_{i+1}^{n})\in I$. Since $I$ is a maximal
$n$-ary $\delta$-$N$-hyperideal of $H$, we can continue the process and so get
$a_{j}\in I$ for some $1\leq j\leq n$ which implies $I$ is an $n$-ary prime
hyperideal of $H$. This means that $\sqrt{0}^{(m,n)}\subseteq I$. Now, suppose
that $I\nsubseteq\sqrt{0}^{(m,n)}$. Then there exist some $x\in I$ such that
$x\notin\sqrt{0}^{(m,n)}$. Since $g(x,1_{H}^{(n-1)})\in I$ and $I$ is an
$n$-ary $\delta$-$N$-hyperideal of $H$, then we get the result that
$g(1_{H}^{(n)})\in I$ which is a contradiction. Then we conclude that
$I=\sqrt{0}^{(m,n)}$. ∎

######  Theorem 4.14.

Let $\gamma$ and $\delta$ be two hyperideal expansions of a commutative
Krasner $(m,n)$-hyperring $H$. If $I$ is an $n$-ary $\gamma$-$N$-hyperideal of
$H$ and $\gamma(J)\subseteq\delta(J)$ for all hyperideals $J$ of $H$, then $I$
is an $n$-ary $\delta$-$N$-hyperideal of $H$. Moreover, if $\delta(I)$ is an
$n$-ary $\gamma$-$N$-hyperideal of $H$, then $I$ is an $n$-ary
$\gamma\circ\delta$-$N$-hyperideal of $H$.

###### Proof.

The proof of the first assertion is straightforward. For the second assertion,
suppose that $g(x_{1}^{n})\in I$ for $x_{1}^{n}\in H$ such that
$x_{i}\notin\sqrt{0}^{(m,n)}$. Therefore we get $g(x_{1}^{n})\in\delta(I)$, as
$I\subseteq\delta(I)$. By the assumption, we conclude that
$g(x_{1}^{i-1},1_{H},x_{i+1}^{n})\in\gamma(\delta(I))=\gamma\circ\delta(I)$
which means $I$ is an $n$-ary $\gamma\circ\delta$-$N$-hyperideal of $H$. ∎

Assume that $(H_{1},f_{1},g_{1})$ and $(H_{2},f_{2},g_{2})$ are two
commutative Krasner $(m,n)$-hyperrings and $h:H_{1}\longrightarrow H_{2}$ a
hyperring homomorphism. Let $\delta$ and $\gamma$ be two hyperideal expansions
of $H_{1}$ and $H_{2}$, respectively. Recall from [3] that $h$ is a
$\delta\gamma$-homomorphism if $\delta(h^{-1}(I_{2}))=h^{-1}(\gamma(I_{2}))$
for the hyperideal $I_{2}$ of $H_{2}$. Note that
$\gamma(h(I_{1})=h(\delta(I_{1})$ for $\delta\gamma$-epimorphism $h$ and for
hyperideal $I_{1}$ of $H_{1}$ such that $Ker(h)\subseteq I_{1}$.
For instance, suppose that $(H_{1},f_{1},g_{1})$ and $(H_{2},f_{2},g_{2})$ are
two Krasner $(m,n)$-hyperrings. If $\delta_{1}$ of $H_{1}$ and $\gamma_{1}$ of
$H_{2}$ are the hyperideal expansions defined in Example 3.2 of [3], then all
homomorphism $h:H_{1}\longrightarrow H_{2}$ is a
$\delta_{1}\gamma_{1}$-homomorphism.

######  Theorem 4.15.

Let $\delta$ and $\gamma$ be two hyperideal expansions of commutative Krasner
$(m,n)$-hyperrings $(H_{1},f_{1},g_{1})$ and $(H_{2},f_{2},g_{2})$,
respectively, and let $h:H_{1}\longrightarrow H_{2}$ be a
$\delta\gamma$-homomorphism. Then the followings hold :

  * (1)

If $I_{2}$ is an $n$-ary $\gamma$-$N$-hyperideal of $H_{2}$ and $h$ is a
monomorphism, then $h^{-1}(I_{2})$ is an $n$-ary $\delta$-$N$-hyperideal of
$H_{1}$.

  * (2)

If $h$ is an epimorphism and $I_{1}$ is an $n$-ary $\delta$-$N$-hyperideal of
$H_{1}$ containing $Ker(h)$, then $h(I_{1})$ is an $n$-ary
$\gamma$-$N$-hyperideal of $H_{2}$.

###### Proof.

$(1)$ Let $g_{1}(x_{1}^{n})\in h^{-1}(I_{2})$ for $x_{1}^{n}\in H_{1}$. It
follows that $g_{2}(h(x_{1}),\cdots,h(x_{n}))=h(g_{1}(x_{1}^{n}))\in I_{2}$.
Since $I_{2}$ is an $n$-ary $\gamma$-$N$-hyperideal of $H_{2}$, then
$h(x_{i})\in\sqrt{0_{H_{2}}}^{(m,n)}$ for some $1\leq i\leq n$ or

$g_{2}(h(x_{1}),...,h(x_{i-1}),1_{H_{2}},h(x_{i+1}),...,h(x_{n}))$

$\hskip 96.73918pt=h(g_{1}(x_{1}^{i-1},1_{H_{1}},x_{i+1}^{n}))$

$\hskip 85.35826pt\in\gamma(I_{2})$.
In the first possibility, since $Ker(h)=\\{0_{H_{1}}\\}$, we obtain
$x_{i}\in\sqrt{0_{H_{1}}}^{(m,n)}$. In the second possibility, we get the
result that $g_{1}(x_{1}^{i-1},1_{H_{1}},x_{i+1}^{n})\in
h^{-1}(\gamma(I_{2}))=\delta(h^{-1}(I_{2})$. Consequently, $h^{-1}(I_{2})$ is
a $\delta$-$N$-hyperideal of $H_{1}$.

$(2)$ Assume that $g_{2}(y_{1}^{n})\in h(I_{1})$ for $y_{1}^{n}\in H_{2}$ with
$y_{i}\notin\sqrt{0_{H_{2}}}^{(m,n)}$ for some $1\leq i\leq n$. Then there
exist $x_{1}^{n}\in H_{1}$ such that $h(x_{1})=y_{1},...,h(x_{n})=y_{n}$ as
$h$ is an epimorphism. Therefor
$h(g_{1}(x_{1}^{n}))=g_{2}(h(x_{1}),...,h(x_{n}))=g_{2}(y_{1}^{n})\in
h(I_{1})$.
Since $I_{1}$ contains $Ker(h)$, we have $g_{1}(x_{1}^{n})\in I_{1}$. Since
$y_{i}\notin\sqrt{0_{H_{2}}}^{(m,n)}$, then
$x_{i}\notin\sqrt{0_{H_{1}}}^{(m,n)}$. Since $I_{1}$ is a
$\delta$-$N$-hyperideal of $H_{1}$ and $x_{i}\notin\sqrt{0_{H_{1}}}^{(m,n)}$,
it follows that $g_{1}(x_{1}^{i-1},1_{H_{1}},x_{i+1}^{n})\in\delta(I_{1})$
which implies

$h(g_{1}(x_{1}^{i-1},1_{H_{1}},x_{i+1}^{n}))=g_{2}(h(x_{1}),...,h(x_{i-1}),1_{H_{2}},h(x_{i+1}),...,h(x_{n}))$

$\hskip 116.65646pt=g_{2}(y_{1}^{i-1},1_{H_{2}},y_{i+1}^{n})$

$\hskip 116.65646pt\in h(\delta(I_{1})).$
By the assumption, we have $h(\delta(I_{1}))=\gamma(h(I_{1}))$. So
$g_{2}(y_{1}^{i-1},1_{H_{2}},y_{i+1}^{n})\in\gamma(h(I_{1}))$. Hence
$h(I_{1})$ is an $n$-ary $\gamma$-$N$-hyperideal of $H_{2}$. ∎

######  Corollary 4.16.

Suppose that $I$ and $J$ are two hyperideals of a commutative Krasner
$(m,n)$-hyperring $H$ with $J\subseteq I$. If $I$ is an $n$-ary
$\delta$-$N$-hyperideal of $H$, then $I/J$ is an $n$-ary
$\delta_{q}$-$J$-hyperideal of $H/J$.

###### Proof.

The claim follows by using Theorem 4.15 (2) and by a similar argument to that
of 4.18. ∎

The next theorem shows that for an $n$-ary $\delta$-$N$-hyperideal $I$ of $H$
if $\delta(\sqrt{I}^{(m,n)})$ contains $\sqrt{\delta(I)}^{(m,n)}$, then
$\sqrt{I}^{(m,n)}$ is an $n$-ary $\delta$-$N$-hyperideal of $H$.

######  Theorem 4.17.

Suppose that $I$ is an $n$-ary $\delta$-$N$-hyperideal of a commutative
Krasner $(m,n)$-hyperring $H$ with
$\sqrt{\delta(I)}^{(m,n)}\subseteq\delta(\sqrt{I}^{(m,n)})$. Then
$\sqrt{I}^{(m,n)}$ is an $n$-ary $\delta$-$N$-hyperideal of $H$.

###### Proof.

Assume that $g(x_{1}^{n})\in\sqrt{I}^{(m,n)}$ for $x_{1}^{n}\in H$ with
$x_{i}\notin\sqrt{0}^{(m,n)}$ for some $1\leq i\leq n$. Then there exists
$t\in\mathbb{N}$ such that if $t\leq n$ implies that
$g(g(x_{1}^{n})^{(t)},1_{H}^{(n-t)})\in I$. Then we get the result that

$g(x_{i}^{(t)},g(x_{1}^{i-1},1_{H},x_{i+1}^{n})^{(t)},1_{H}^{(n-2t)})=g(g(x_{i}^{(t)},1_{H}^{(n-t)}),g(x_{1}^{i-1},1_{H},x_{i+1}^{n})^{(t)},$

$\hskip 14.22636pt1_{H}^{(n-t-1)})\subseteq I$.
Since $g(x_{i}^{(t)},1^{(n-t)})\notin\sqrt{0}^{(m,n)}$ and $I$ is an $n$-ary
$\delta$-$N$-hyperideal of $H$, then we have
$g(g(x_{1}^{i-1},1_{H},x_{i+1}^{n})^{(t)},1_{H}^{(n-t)})\in\delta(I)$. It
follows that $g(x_{1}^{i-1},1_{R},x_{i+1}^{n})\in\sqrt{\delta(I)}^{(m,n)}$.
Since $\sqrt{\delta(I)}^{(m,n)}\subseteq\delta(\sqrt{I}^{(m,n)})$, then
$g(x_{1}^{i-1},1_{H},x_{i+1}^{n})\in\delta(\sqrt{I}^{(m,n)})$, as needed. A
similar argument will show that if $t=l(n-1)+1$, then $\sqrt{I}^{(m,n)}$ is an
$n$-ary $\delta$-$N$-hyperideal of $H$. ∎

Recall from [1] that a non-empty subset $S$ of a Krasner $(m,n)$-hyperring $H$
is said to be an $n$-ary multiplicative, if $g(t_{1}^{n})\in S$ for
$t_{1},...,t_{n}\in S$. The notion of Krasner $(m,n)$-hyperring of fractions
was introduced in [4]. Assume that $\delta$ is a hyperideal expansion of a
commutative Krasner $(m,n)$-hyperring $H$ and $S$ is an $n$-ary multiplicative
subset of $H$ with $1\in S$. Then $\delta_{S}$ is a hyperideal expansion of
$S^{-1}H$ with $\delta_{S}(S^{-1}I)=S^{-1}(\delta(I))$.

######  Theorem 4.18.

Suppose that $S$ is an $n$-ary multiplicative subset of a commutative Krasner
$(m,n)$-hyperring $H$ with $1\in S$. If $I$ is an $n$-ary
$\delta$-$N$-hyperideal of $H$ and $I\cap S=\varnothing$, then $S^{-1}I$ is an
$n$-ary $\delta_{S}$-$N$-hyperideal of $S^{-1}H$.

###### Proof.

Let $I$ be an $n$-ary $\delta$-$N$-hyperideal of $H$. Assume that
$G(\frac{x_{1}}{s_{1}},...,\frac{x_{n}}{s_{n}})\in S^{-1}I$ for
$\frac{x_{1}}{s_{1}},...,\frac{x_{n}}{s_{n}}\in S^{-1}H$ such that
$\frac{x_{i}}{s_{i}}\notin\sqrt{0_{S^{-1}H}}^{(m,n)}$ for some $1\leq i\leq
n$. Hence $\frac{g(x_{1}^{n})}{g(s_{1}^{n})}\in S^{-1}I$. This means that
there exists $t\in S$ such that $g(t,g(x_{1}^{n}),1_{H}^{(n-2)})\in I$ and
then $g(x_{i},g(x_{1}^{i-1},t,x_{i+1}^{n}),1_{H}^{(n-2)})\in I$. Since
$x_{i}\notin\sqrt{0}^{(m,n)}$ and $I$ is an $n$-ary $\delta$-$N$-hyperideal of
$H$, then we get the result that $g(x_{1}^{i-1},t,x_{i+1}^{n})\in\delta(I)$.
Therefore
$G(\frac{x_{1}}{s_{1}},\cdots,\frac{x_{i-1}}{s_{i-1}},\frac{1_{H}}{1_{H}},\frac{x_{i+1}}{s_{i+1}},\cdots,\frac{x_{n}}{s_{n}})=\frac{g(x_{1}^{i-1},1_{H},x_{i+1}^{n})}{g(s_{1}^{i-1},1_{H},s_{i+1}^{n})}=\frac{g(x_{1}^{i-1},t,x_{i+1}^{n})}{g(s_{1}^{i-1},t,s_{i+1}^{n})}\in
S^{-1}(\delta(I))=\delta_{S}(S^{-1}(I))$. Consequently, $S^{-1}I$ is an
$n$-ary $\delta_{S}$-$N$-hyperideal of $S^{-1}H$. ∎

##  5\. $(k,n)$-absorbing $\delta$-$N$-hyperideals

In this section, we extend the concept of $\delta$-$N$-hyperideals to the
notion of $(k,n)$-absorbing $\delta$-$N$-hyperideals.

######  Definition 5.1.

Given a hyperideal expansion $\delta$, a proper hyperideal $I$ of a
commutative Krasner $(m,n)$-hyperring $H$ is called $(k,n)$-absorbing
$N$-hyperideal if $g(a_{1}^{kn-k+1})\in I$ for $a_{1}^{kn-k+1}\in H$ implies
that $g(a_{1}^{(k-1)n-k+2})\in\sqrt{0}^{(m,n)}$ or a $g$-product of
$(k-1)n-k+2$ of $a_{i}^{,}$ s except $g(a_{1}^{(k-1)n-k+2})$ is in
$\delta(I)$.

######  Example 5.2.

If we continue with Example 4.2 and use its notation, then $I=\\{0,3,6\\}$ is
a $(2,2)$-absorbing $\delta_{1}$-$N$-hyperring of
$\mathbb{Z}_{12}/\mathbb{Z}_{12}^{\star}$.

######  Theorem 5.3.

Let $I$ be a $(k,n)$-absorbing $N$-hyperideal of a commutative Krasner
$(m,n)$-hyperring $H$. Then $\sqrt{I}^{(m,n)}$ is a $(k,n)$-absorbing
$\delta$-$N$-hyperideal.

###### Proof.

Assume that $g(a_{1}^{kn-k+1})\in\sqrt{I}^{(m,n)}$ for $a_{1}^{kn-k+1}\in H$.
We presume none of the $g$-products of $(k-1)n-k+2$ of the $a_{i}^{,}$s other
than $g(a_{1}^{(k-1)n-k+2})$ are in $\delta(\sqrt{I}^{(m,n)})$. Since
$g(a_{1}^{kn-k+1})\in\sqrt{I}^{(m,n)}$, then for some $t\in\mathbb{N}$ we have
for $t\leq n$, $g(g(a_{1}^{kn-k+1})^{(t)},1_{H}^{(n-t)})\in I$ or for $t>n$
with $t=l(n-1)+1$, $g_{(l)}(g(a_{1}^{kn-k+1})^{(t)})\in I$. In the first
possibilty, since all $g$-products of the $a_{i}^{,}$s other than
$g(a_{1}^{(k-1)n-k+2})$ are not in $\delta(\sqrt{I}^{(m,n)})$, then they are
not in $I$. Since $I$ is a $(k,n)$-absorbing $J$-hyperideal of $H$, then we
have $g(g(x_{1}^{(k-1)n-k+2)})^{(l)},1_{H}^{(n-t)})\in\sqrt{0}^{(m,n)}$ which
means $g(x_{1}^{(k-1)n-k+2})\in\sqrt{0}^{(m,n)}$. In the second possibilty,
the claim follows by using a similar argument. ∎

######  Theorem 5.4.

Suppose that $I$ is a hyperideal of a commutative Krasner $(m,n)$-hyperring
$H$ such that $\delta(I)$ is a $(2,n)$-absorbing $N$-hyperideal. Then $I$ is a
$(3,n)$-absorbing $\delta$-$N$-hyperideal of $H$.

###### Proof.

Let $g(a_{1}^{3n-2})\in I$ for $a_{1}^{3n-2}\in H$ but
$g(a_{1}^{2n-1})\notin\sqrt{0}^{(m,n)}$. This means that
$g(g(a_{1},a_{2n}^{3n-2}),a_{2}^{2n-1})\in I\subseteq\delta(I)$. Since
$\delta(I)$ is a $(2,n)$-absorbing $N$-hyperideal of $H$ and
$g(a_{2}^{2n-1})\notin\sqrt{0}^{(m,n)}$, then we get the result that
$g(a_{1}^{n},a_{2n}^{3n-2})\in\delta(I)$ or
$g(a_{1},a_{n+1}^{2n-1},a_{2n}^{3n-2})\in\delta(I)$. Consequently, $I$ is a
$(3,n)$-absorbing $\delta$-$N$-hyperideal of $H$. ∎

######  Theorem 5.5.

Assume that $I$ is a hyperideal of a commutative Krasner $(m,n)$-hyperring $H$
such that $\delta(I)$ is a $(k+1,n)$-absorbing $\delta$-$N$-hyperideal of $H$.
Then $I$ is a $(k+1,n)$-absorbing $\delta$-$N$-hyperideal of $H$.

###### Proof.

Suppose that $g(a_{1}^{(k+1)n-(k+1)+1})\in I$ for $a_{1}^{(k+1)n-(k+1)+1}\in
H$ such that $g(a_{1}^{kn-k+1})\notin\sqrt{0}^{(m,n)}$. So
$g(a_{1}^{(k+1)n-(k+1)+1})=g(a_{1}^{kn-k},g(a_{kn-k+1}^{(k+1)n-(k+1)+1}))\in
I\subseteq\delta(I)$. Since $\delta(I)$ is a $(k+1,n)$-absorbing
$\delta$-$N$-hyperideal and $g(a_{1}^{kn-k+1})\notin\sqrt{0}^{(m,n)}$, we get
the result that
$g(a_{1}^{i-1},a_{i+1}^{kn-k},g(a_{kn-k+1}^{(k+1)n-(k+1)+1}))\in\delta(I)$ for
$1\leq i\leq n$. Thus $I$ is a $(k+1,n)$-absorbing $\delta$-$N$-hyperideal of
$H$. ∎

######  Theorem 5.6.

Let $I$ be a $\delta$-$N$-hyperideal of a commutative Krasner
$(m,n)$-hyperring $H$. Then $I$ is a $(2,n)$-absorbing $\delta$-$N$-hyperideal
of $H$.

###### Proof.

Let $g(a_{1}^{2n-1})\in I$ for $a_{1}^{2n-1}\in H$. Since $I$ is a
$\delta$-$N$-hyperideal of $H$, we get the result that
$g(a_{1}^{n})\in\sqrt{0}^{(m,n)}$ or $g(a_{n+1}^{2n-1})\in\delta(I)$.
Therefore we have $g(a_{i},x_{n+1}^{2n-1})\in\delta(I)$, for $1\leq i\leq n$,
as $\delta(I)$ is a hyperideal of $H$. Thus $I$ is $(2,n)$-absorbing
$\delta$-primary. ∎

Next, we determine all integers $k>n$.

######  Theorem 5.7.

Let $I$ is a $(k,n)$-absorbing $\delta$-$N$-hyperideal of a commutative
Krasner $(m,n)$-hyperring $H$. Then $I$ is $(s,n)$-absorbing
$\delta$-$N$-hyperideal for $s>n$.

###### Proof.

Let $g(a_{1}^{(k+1)n-(k+1)+1})\in I$ for $a_{1}^{(k+1)n-(k+1)+1}\in H$. Put
$g(a_{1}^{n+2})=a$. Since $I$ is $(k,n)$-absorbing $\delta$-$N$-hyperideal,
then we obtain $g(a,\cdots,a_{(k+1)n-(k+1)+1})\in\sqrt{0}^{(m,n)}$ or a
$g$-product of $kn-k+1$ of the $a_{i}^{,}$s except
$g(a,\cdots,a_{(k+1)n-(k+1)+1})$ is in $\delta(I)$. This implies that
$g(a_{i},a_{n+3}^{(k+1)n-(k+1)+1})\in\delta(I)$ for all $1\leq i\leq n+2$
which implies $I$ is a $(k+1,n)$-absorbing $\delta$-$N$-hyperideal.
Consequently, $I$ is an $(s,n)$-absorbing $\delta$-$N$-hyperideal for $s>n$. ∎

##  6\. $n$-ary $S$-$N$-hyperideals

######  Definition 6.1.

Assume that $S$ is an $n$-ary multiplicative subset of a commutative Krasner
$(m,n)$-hyperring $H$. A hyperideal $I$ of $R$ with $I\cap S=\varnothing$ is
said to be an $n$-ary $S$-$N$-hyperideal if there exists an $s\in S$ such that
for all $x_{1}^{n}\in H$ if $g(x_{1}^{n})\in I$ with
$g(s,x_{i},1_{H}^{(n-2)})\notin\sqrt{0}^{(m,n)}$ for some $1\leq i\leq n$,
then $g(x_{1}^{i-1},s,x_{i+1}^{n})\in I$. This element $s$ in $S$ is called an
$S$-element of $I$.

######  Example 6.2.

Let $H=\\{0,1,a\\}$. Consider commutative Krasner $(3,3)$-hyperring $(H,f,g)$
that 3-ary $g$ is defined as $g(1,1,1)=1$, $g(1,1,a)=g(1,a,a)=g(a,a,a)=a$ and
for all $x,y\in H$, $g(0,x,y)=0$ and 3-ary hyperoeration $f$ is defined as
$f(0,0,0)=0$, $f(0,0,1)=1$, $f(0,1,1)=1$, $f(1,1,1)=1$, $f(1,1,a)=H$,
$f(0,1,a)=H$, $f(0,0,a)=a$, $f(0,a,a)=a$, $f(1,a,a)=H$, $f(a,a,a)=a$. In the
Krasner $(3,3)$-hyperring, $I=\\{0,a\\}$ is an 3-ary $S$-$N$-hyperideal of $H$
such that 3-ary multiplicative subset $S$ is $\\{1,2\\}$.

Now we give a charactrization of an $n$-ary $S$-$N$-hyperideal.

######  Theorem 6.3.

Assume that $S$ is an $n$-ary multiplicative subset of a commutative Krasner
$(m,n)$-hyperring $H$ and $I$ is a hyperideal of $H$ disjoint with $S$. Then
$I$ is an $n$-ary $S$-$N$-hyperideal of $H$ if and only if there exists $s\in
S$, for all hyperideals $I_{1}^{n}$ of $H$, if $g(I_{1}^{n})\subseteq I$, then
$g(s,I_{i},1^{(n-1)})\subseteq\sqrt{0}^{(m,n)}$ for some $1\leq i\leq n$ or
$g(I_{1}^{i-1},s,I_{i+1}^{n})\subseteq I$.

###### Proof.

$\Longrightarrow$ Let $I$ be an $n$-ary $S$-$N$-hyperideal of $H$. Suppose
that $g(I_{1}^{n})\in I$ for some hyperideals $I_{1}^{n}$ of $H$ such that
$g(s,I_{i},1_{H}^{(n-2)})\nsubseteq\sqrt{0}^{(m,n)}$ and
$g(I_{1}^{i-1},s,I_{i+1}^{n})\nsubseteq I$ for all $s\in S$. Then there exists
$a_{i}\in I_{i}$ for each $1\leq i\leq n$ such that $g(a_{1}^{n})\in I$ but
$g(s,a_{i},1_{H}^{(n-2)})\notin\sqrt{0}^{(m,n)}$ and
$g(a_{1}^{i-1},s,a_{i+1}^{n})\notin I$, a contradiction.
$\Longleftarrow$ Suppose that $g(x_{1}^{n})\in I$ for some $x_{1}^{n}\in H$.
Then $g(\langle x_{1}\rangle,\cdots,\langle x_{n}\rangle)\subseteq I$. By the
assumption, we are done. ∎

######  Theorem 6.4.

Let $S$ be an $n$-ary multiplicative subset of a commutative Krasner
$(m,n)$-hyperring $H$. If $I_{1}^{n}$ are some $n$-ary $S$-$N$-hyperideals of
$H$, then $\bigcap_{t=1}^{n}I_{t}$ is an $n$-ary $S$-$N$-hyperideal of $H$.

###### Proof.

Let $I_{1}^{n}$ be $n$-ary $S$-$N$-hyperideals of $H$. Suppose that for each
$1\leq t\leq n$, there exists $s_{t}\in S$ such that if $g(a_{1}^{n})\in
I_{t}$ for some $a_{1}^{n}\in H$, then
$g(s_{t},a_{i},1_{H}^{(n-2)})\in\sqrt{0}^{(m,n)}$ or
$g(a_{1}^{i-1},s_{t},a_{i+1}^{n})\in I_{t}$. Now, assume that
$g(a_{1}^{n})\in\bigcap_{t=1}^{n}I_{t}$ for some $a_{1}^{n}\in H$. This means
that $g(a_{1}^{n})\in I_{t}$ for each $1\leq t\leq n$. Put
$\Pi_{t=1}^{n}s_{t}\in S$. Then we get the result thet
$g(s,a_{i},1_{H}^{(n-2)})\in\sqrt{0}^{(m,n)}$ or
$g(a_{1}^{i-1},s,a_{i+1}^{n})\in\bigcap_{t=1}^{n}I_{t}$, as claimed. ∎

######  Theorem 6.5.

Assume that $S$ is an $n$-ary multiplicative subset of a commutative Krasner
$(m,n)$-hyperring $H$ and $I$ is a hyperideal of $H$ such that $I\cap
S=\varnothing$. If $E_{s}=\\{y\in H\ |\ g(y,s,1_{H}^{(n-2)})\in I\\}$ is an
$n$-ary $N$-hyperideal of $H$ for some $s\in S$, then $I$ is an $n$-ary
$S$-$N$-hyperideal of $H$.

###### Proof.

Let $E_{s}$ be an $n$-ary $N$-hyperideal of $H$ for some $s\in S$. Suppose
that $g(x_{1}^{n})\in I$ for $x_{1}^{n}\in H$ such that
$g(s,x_{i},1_{H}^{(n-2)})\notin\sqrt{0}^{(m,n)}$ for some $1\leq i\leq n$.
Therefore $g(g(x_{1}^{n}),s,1_{H}^{(n-2)}))\in I$ and so $g(x_{1}^{n})\in
E_{s}$. Since $E_{s}$ is an $n$-ary $N$-hyperideal of $H$ for some $s\in S$
and $x_{i}\notin\sqrt{0}^{(m,n)}$, we get the result that
$g(x_{1}^{i-1},1_{H},x_{i+1}^{n})\in E_{s}$ which means
$g(g(x_{1}^{i-1},1_{H},x_{i+1}^{n}),s,1_{H}^{(n-2)})=g(x_{1}^{i-1},s,x_{i+1}^{n})\in
I$. Consequently, $I$ is an $n$-ary $S$-$N$-hyperideal of $H$. ∎

In the following theorem, we determine a condition on $I$ when the converse
holds.

######  Theorem 6.6.

Assume that $S$ is an $n$-ary multiplicative subset of a commutative Krasner
$(m,n)$-hyperring $H$ and $I$ is a hyperideal of $H$ such that $I\cap
S=\varnothing$. If $I$ is an $n$-ary $S$-$N$-hyperideal of $H$ and
$F_{s}=\\{y\in H\ |\ g(y,s,1_{H}^{(n-2)})\in\sqrt{0}^{(m,n)}\\}$ is an $n$-ary
$N$-hyperideal for an $S$-element $s\in S$ of $I$, then $E_{s}=\\{y\in H\ |\
g(y,s,1_{H}^{(n-2)})\in I\\}$ is an $n$-ary $N$-hyperideal of $H$.

###### Proof.

Let $g(x_{1}^{n})\in E_{s}$ for $x_{1}^{n}\in H$. Then we have
$g(g(x_{1}^{n}),s,1_{H}^{(n-2)})\in I$ and so
$g(x_{i},g(x_{1}^{i-1},s,x_{i+1}^{n}),1_{H}^{(n-2)})\in I$. Since $I$ is an
$n$-ary $S$-$N$-hyperideal of $H$, we get the result that either
$g(s,x_{i},1_{H}^{(n-2)})\in\sqrt{0}^{(m,n)}$ or
$g(g(x_{1}^{i-1},s,x_{i+1}^{n}),s,1_{H}^{(n-2)})=g(g(x_{1}^{i-1},1_{H},x_{i+1}^{n}),g(s^{(2)},1_{H}^{(n-2)}),1_{H}^{(n-2)})\in
I$. In the former case, by Theorem 3.3 we conclude that
$F_{s}=\sqrt{0}^{(m,n)}$, as $F_{s}$ is an $n$-ary $N$-hyperideal. Therefore
$x_{i}\in\sqrt{0}^{(m,n)}$. In the second case, suppose that
$g(x_{1}^{i-1},s,x_{i+1}^{n})\notin I$. Then we obtain
$g(g(s^{(2},1_{H}^{(n-2)}),s,1_{H}^{(n-2)})=g(s^{(3)},1_{H}^{(n-3)})\in\sqrt{0}^{(m,n)}$.
It implies that $s\in\sqrt{0}^{(m,n)}$, a contradiction. Then we conclude that
$g(x_{1}^{i-1},s,x_{i+1}^{n})\in I$ which means
$g(x_{1}^{i-1},1_{H},x_{i+1}^{n})\in E_{s}$ which implies $E_{s}$ is an
$n$-ary $N$-hyperideal of $H$. ∎

######  Theorem 6.7.

Let $S\subseteq S^{\prime}$ be two $n$-ary multiplicative subsets of a
commutative Krasner $(m,n)$-hyperring $H$ and $I$ be an $n$-ary
$S^{\prime}$-$N$-hyperideal of $H$. If for each $s\in S^{\prime}$, there is an
element $s^{\prime}\in S^{\prime}$ with $g(s,s^{\prime},1_{H}^{(n-2)})\in S$,
then $I$ is an $n$-ary $S$-$N$-hyperideal of $H$.

###### Proof.

Let $g(a_{1}^{n})\in I$. Since $I$ is an $n$-ary $S^{\prime}$-$N$-hyperideal
of $H$, we have either $g(s,a_{i},1_{H}^{(n-2)})\in\sqrt{0}^{(m,n)}$ or
$g(a_{1}^{i-1},s,a_{i+1}^{n})\in I$ for a $S^{\prime}$-element $s\in
S^{\prime}$ of $I$. By the assumption, there exists $s^{\prime}\in S^{\prime}$
such that $s^{\prime\prime}=g(s,s^{\prime},1_{H}^{(n-2)})\in S$. From
$g(s,a_{i},1_{H}^{(n-2)})\in\sqrt{0}^{(m,n)}$ , it follows that
$g(s^{\prime},g(s,a_{i},1_{H}^{(n-2)}),1_{H}^{(n-2)})\in\sqrt{0}^{(m,n)}$.
Also, from $g(a_{1}^{i-1},s,a_{i+1}^{n})\in I$ it follows that
$g(s^{\prime},g(a_{1}^{i-1},s,a_{i+1}^{n}),1_{H}^{(n-2)})\in I$. Consequently,
we get the result that
$g(s^{\prime\prime},a_{i},1_{H}^{(n-2)})\in\sqrt{0}^{(m,n)}$ or
$g(a_{1}^{i-1},s^{\prime\prime},a_{i+1}^{n})\in I$, as needed. ∎

######  Theorem 6.8.

Let $S\subseteq S^{\prime}$ be two $n$-ary multiplicative subsets of a
commutative Krasner $(m,n)$-hyperring $H$ such that $1_{H}\in S$ and $I$ be a
hyperideal of $H$ with $I\cap S^{\prime}=\varnothing$ . If $I$ is an $n$-ary
$S$-$N$-hyperideal of $H$, then ${S^{\prime}}^{-1}I$ is an $n$-ary
${S^{\prime}}^{-1}S$-$N$-hyperideal of ${S^{\prime}}^{-1}H$ and
${S^{\prime}}^{-1}I\cap H=E_{s}$ where $E_{s}=\\{y\in H\ |\
g(y,s,1^{(n-2)})\in I\\}$ and $s$ is an $S$-element of $I$.

###### Proof.

Let $I$ be an $n$-ary $S$-$N$-hyperideal of $H$. It is easy to see that
${S^{\prime}}^{-1}S\cap{S^{\prime}}^{-1}I=\varnothing$ . Assume that
$\frac{s}{1_{H}}\in{S^{\prime}}^{-1}S$ for some $S$-element $s$ of $I$. Let
$G(\frac{a_{1}}{s_{1}},\cdots,\frac{a_{n}}{s_{n}})\in{S^{\prime}}^{-1}I$ for
$a_{1}^{n}\in H$ and $s_{1}^{n}\in S^{\prime}$ such that
$G(\frac{s}{1_{H}},\frac{a_{i}}{s_{i}},\frac{1_{H}}{1_{H}}^{(n-2)})\notin\sqrt{0_{{S^{\prime}}^{-1}H}}^{(m,n)}$
for some $1\leq i\leq n$. So
$\frac{g(a_{1}^{n})}{g(s_{1}^{n})}\in{S^{\prime}}^{-1}I$ which follows there
exists $t\in S^{\prime}$ such that $g(t,g(a_{1}^{n}),1_{H}^{(n-2)})\in I$ and
then $g(a_{i},g(a_{1}^{i-1},t,a_{i+1}^{n}),1_{H}^{(n-2)})\in I$. Since $I$ is
an $n$-ary $S$-$N$-hyperideal of $H$ and
$g(s,a_{i},1_{H}^{(n-2)})\notin\sqrt{0_{H}}^{(m,n)}$, we get the result that
$g(s,g(a_{1}^{i-1},t,a_{i+1}^{n}),1_{H}^{(n-2)})\in I$ which means
$G(\frac{a_{1}}{s_{1}},\cdots,\frac{a_{i-1}}{s_{i-1}},\frac{s}{1_{H}},\frac{a_{i+1}}{s_{i+1}},\cdots,\frac{a_{n}}{s_{n}})=\frac{g(a_{1}^{i-1},s,a_{i+1}^{n})}{g(s_{1}^{i-1},1_{H},s_{i+1}^{n})}=\frac{g(s,g(a_{1}^{i-1},t,a_{i+1}^{n}),1^{(n-2)})}{g(s_{1}^{i-1},t,s_{i+1}^{n})}\in{S^{\prime}}^{-1}I$.
Thus ${S^{\prime}}^{-1}I$ is an $n$-ary ${S^{\prime}}^{-1}S$-$N$-hyperideal of
${S^{\prime}}^{-1}H$. For the second assertion, suppose that
$x\in{S^{\prime}}^{-1}I\cap H$. Then there exists $a\in I$ such that
$\frac{x}{1_{H}}=\frac{a}{t}$ for some $t\in S^{\prime}$. Therefore there
exists $u\in S^{\prime}$ such that $g(u,a,1_{H}^{(n-2)})\in I$. Since $I$ is
an $n$-ary $S$-$N$-hyperideal of $H$, then there exists $s\in S\subseteq
S^{\prime}$ such that we have $g(s,u,1_{H}^{(n-2)})\in\sqrt{0_{H}}^{(m,n)}$ or
$g(s,a,1_{H}^{(n-2)})\in I$. In the former case, we have a contradiction since
$S^{\prime}\cap\sqrt{0_{H}}^{(m,n)}=\varnothing$. Hence
$g(s,a,1_{H}^{(n-2)})\in I$ which implies $a\in E_{s}$ which means
${S^{\prime}}^{-1}I\cap H\subseteq E_{s}$. Since the inclusion
$E_{s}\subseteq{S^{\prime}}^{-1}I\cap H$ holds, we get ${S^{\prime}}^{-1}I\cap
H=E_{s}$. ∎

We have the following corollary of Theorem 6.8.

######  Corollary 6.9.

Let $S$ be an $n$-ary multiplicative subset of a commutative Krasner
$(m,n)$-hyperring $H$ such that $1_{H}\in S$ and $I$ be a hyperideal of $H$
with $I\cap S=\varnothing$ . If $I$ is an $n$-ary $S$-$N$-hyperideal of $H$,
then $S^{-1}I$ is an $n$-ary $N$-hyperideal of $S^{-1}H$ and $S^{-1}I\cap
H=E_{s}$ where $E_{s}=\\{y\in H\ |\ g(y,s,1^{(n-2)})\in I\\}$ and $s$ is an
$S$-element of $I$.

###### Proof.

Let $I$ be an $n$-ary $S$-$N$-hyperideal of $H$. Then, by Theorem 6.8,
$S^{-1}I$ is an $n$-ary $S^{-1}S$-N-hyperideal of $S^{-1}H$. Suppose that
$G(\frac{a_{1}}{s_{1}},\cdots,\frac{a_{n}}{s_{n}})\in S^{-1}I$ for
$a_{1}^{n}\in H$ and $s_{1}^{n}\in S$. There there exist an $S^{-1}S$-element
$\frac{u}{v}$ of $S^{-1}I$ such that
$G(\frac{u}{v},\frac{a_{i}}{s_{i}},\frac{1_{H}}{1_{H}}^{(n-2)})\in\sqrt{0_{S^{-1}H}}^{(m,n)}$
or
$G(\frac{a_{1}}{s_{1}},\cdots,\frac{a_{i-1}}{s_{i-1}},\frac{u}{v},\frac{a_{i+1}}{s_{i+1}},\cdots,\frac{a_{n}}{s_{n}})\in
S^{-1}I$ for some $1\leq i\leq n$. Then we get the result that $S^{-1}I$ is an
$n$-ary $N$-hyperideal of $S^{-1}H$, as $\frac{u}{v}$ is invertible. For the
second assertion, we use a similar argument to that of Theorem 6.8. ∎

From Theorem 6.8 and Corollary 6.9, we can conclude the following.

######  Theorem 6.10.

Let $S$ be an $n$-ary multiplicative subset of a commutative Krasner
$(m,n)$-hyperring $H$ such that $1_{H}\in S$ and $I$ be a hyperideal of $H$
with $I\cap S=\varnothing$. Then $I$ is an $n$-ary $S$-$N$-hyperideal of $H$
if and only if $S^{-1}I$ is an $n$-ary $N$-hyperideal of $S^{-1}H$ and
$S^{-1}I\cap H=E_{s}$ and $S^{-1}\sqrt{0}^{(m,n)}\cap H=F_{s^{\prime}}$ where
$E_{s}=\\{y\in H\ |\ g(y,s,1^{(n-2)})\in I\\}$ and $F_{s^{\prime}}=\\{z\in H\
|\ g(z,s^{\prime},1^{(n-2)})\in\sqrt{0}^{(m,n)}\\}$ for some $s,s^{\prime}\in
S$.

###### Proof.

$\Longrightarrow$ Let $I$ be an $n$-ary $S$-$N$-hyperideal of $H$. Then, by
Corollary 6.9 we conclude that $S^{-1}I$ is an $n$-ary $N$-hyperideal of
$S^{-1}H$. The rest of the claim follows by a similar argument to that of
Theorem 6.8.
$\Longleftarrow$ Let $S^{-1}I$ be an $n$-ary $N$-hyperideal of $S^{-1}H$,
$S^{-1}I\cap H=E_{s}$ and $S^{-1}\sqrt{0}^{(m,n)}\cap H=F_{s^{\prime}}$ where
$E_{s}=\\{y\in H\ |\ g(y,s,1_{H}^{(n-2)})\in I\\}$ and $F_{s^{\prime}}=\\{z\in
H\ |\ g(z,s^{\prime},1_{H}^{(n-2)})\in\sqrt{0}^{(m,n)}\\}$ for some
$s,s^{\prime}\in S$. Put $g(s,s^{\prime},1_{H}^{(n-2)})=s^{\prime\prime}$.
Assume that $g(a_{1}^{n})\in I$ for $a_{1}^{n}\in H$. This implies that
$G(\frac{a_{1}}{1_{H}},\cdots,\frac{a_{n}}{1_{H}})\in S^{-1}I$. Then we get
the result that either
$\frac{a_{i}}{1_{H}}\in\sqrt{S^{-1}0}^{(m,n)}=S^{-1}\sqrt{0}^{(m,n)}$ or
$G(\frac{a_{1}}{1_{H}},\cdots,\frac{a_{i-1}}{1_{H}},\frac{1_{H}}{1_{H}},\frac{a_{i+1}}{1_{H}},\cdots,\frac{a_{n}}{1_{H}})\in
S^{-1}I$ for some $1\leq i\leq n$. In the first case, we get
$g(t,a_{i},1_{H}^{(n-2)})\in\sqrt{0}^{(m,n)}$ for some $t\in S$ which implies
$a_{i}=\frac{g(t,a_{i},1_{H}^{(n-2)})}{g(t,1_{H}^{(n-1)})}\in
S^{-1}\sqrt{0}^{(m,n)}\cap H=F_{s^{\prime}}$. This means that
$g(s^{\prime},a_{i},1_{H}^{(n-2)})\in\sqrt{0}^{(m,n)}$. Then
$g(s^{\prime\prime},a_{i},1_{H}^{(n-2)})=g(g(s,s^{\prime},1_{H}^{(n-2)}),a_{i},1_{H}^{(n-2)})\in\sqrt{0}^{(m,n)}$.
In the second case, we get $g(a_{1}^{i-1},t^{\prime},a_{i+1}^{n})\in I$ for
some $t^{\prime}\in S$. This implies that
$g(a_{1}^{i-1},1_{H},a_{i+1}^{n})=\frac{g(a_{1}^{i-1},t^{\prime},a_{i+1}^{n})}{g(t^{\prime},1_{H}^{(n-1)})}\in
S^{-1}I\cap H=E_{s}$. Then we conclude that
$g(a_{1}^{i-1},s^{\prime\prime},a_{i+1}^{n})=g(a_{1}^{i-1},g(s,s^{\prime},1_{H}^{(n-2)}),a_{i+1}^{n})\in
I$. Consequently, $I$ is an $n$-ary $S$-$N$-hyperideal of $H$. ∎

######  Theorem 6.11.

Let $S_{1}$ and $S_{2}$ be $n$-ary multiplicative subsets of commutative
Krasner $(m,n)$-hyperrings $(H_{1},f_{1},g_{1})$ and $(H_{2},f_{2},g_{2})$,
respectively, and let $1_{H_{1}}$ and $1_{H_{2}}$ be scalar identitis of
$H_{1}$ and $H_{2}$, respectively. Then the following statements hold.

  * (1)

$I_{1}\times H_{2}$ is an $n$-ary $S$-$N$-hyperideal of $H_{1}\times H_{2}$
where $S=S_{1}\times S_{2}$ if and only if $I_{1}$ is an $n$-ary
$S_{1}$-$N$-hyperideal of $H_{1}$ and
$S_{2}\cap\sqrt{0_{H_{2}}}\neq\varnothing$.

  * (2)

$H_{1}\times I_{2}$ is an $n$-ary $S$-$N$-hyperideal of $H_{1}\times H_{2}$
where $S=S_{1}\times S_{2}$ if and only if $I_{2}$ is an $n$-ary
$S_{2}$-$N$-hyperideal of $H_{2}$ and
$S_{1}\cap\sqrt{0_{H_{1}}}\neq\varnothing$.

###### Proof.

$\Longrightarrow$ Let $I_{1}\times H_{2}$ be an $n$-ary $S$-$N$-hyperideal of
$H_{1}\times H_{2}$ where $S=S_{1}\times S_{2}$ and let $(s_{1},s_{2})$ be an
$S$-element of $I_{1}\times H_{2}$. Suppose that $g_{1}(a_{1}^{n})\in I_{1}$
for some $a_{1}^{n}\in H_{1}$. So we consider the following cases.
Case 1: Let $g_{1}(a_{1}^{i-1},s_{1},a_{i+1}^{n})\notin I_{1}$. Then we get
$g_{1}\times g_{2}((a_{1},1_{H_{2}}),\cdots,(a_{n},1_{H_{2}}))\in I_{1}\times
H_{2}$ but $g_{1}\times
g_{2}((a_{1},1_{H_{2}}),\cdots,(a_{i-1},1_{H_{2}}),(s_{1},s_{2}),(a_{i+1},1_{H_{2}}),\cdots(a_{n},1_{H_{2}}))\notin
I_{1}\times H_{2}$. Then we get the result that $g_{1}\times
g_{2}((s_{1},s_{2}),(a_{i},1_{H_{2}}),(1_{H_{1}},1_{H_{2}})^{(n-2)})\in\sqrt{0_{H_{1}\times
H_{2}}}=\sqrt{0_{H_{1}}}\times\sqrt{0_{H_{2}}}$, as $I_{1}\times H_{2}$ is an
$n$-ary $S$-$N$-hyperideal of $H_{1}\times H_{2}$. Hence
$g(s_{1},a_{i},1_{H_{1}}^{(n-2)})\in\sqrt{0_{H_{1}}}$ and $s_{2}\in
S_{2}\cap\sqrt{0_{H_{2}}}$, as needed.
Case 2: In this case we suppose that $g_{1}(a_{1}^{i-1},s_{1},a_{i+1}^{n})\in
I_{1}$. Then we have $g_{1}\times
g_{2}((a_{1},1_{H_{2}}),\cdots,(a_{i-1},1_{H_{2}}),(s_{1},s_{2}),(a_{i+1},1_{H_{2}}),\cdots(a_{n},1_{H_{2}}))\in
I_{1}\times H_{2}$. So we get the result that $g_{1}\times
g_{2}((a_{1},1_{H_{2}}),\cdots,(a_{i-1},1_{H_{2}}),(s_{1},s_{2}),(a_{i+1},1_{H_{2}}),\cdots(a_{n},1_{H_{2}}))\in\sqrt{0_{H_{1}\times
H_{2}}}=\sqrt{0_{H_{1}}}\times\sqrt{0_{H_{2}}}$, since $g_{1}\times
g_{2}((s_{1},s_{2})^{(2)},(1_{H_{1}},1_{H_{2}})^{(n-2)})\notin I_{1}\times
H_{2}$.
$\Longleftarrow$ Let $I_{1}$ be an $n$-ary $S_{1}$-$N$-hyperideal of $H_{1}$
and let $s_{1}$ be an $S_{1}$-element of $I_{1}$ and $s_{2}\in
S_{2}\cap\sqrt{0_{H_{2}}}$. Now, assume that $g_{1}\times
g_{2}((a_{1},b_{1}),\cdots,(a_{n},b_{n}))\in I_{1}\times H_{2}$ for some
$a_{1}^{n}\in H_{1}$ and $b_{1}^{n}\in H_{2}$. Then we have
$g_{1}(a_{1}^{n})\in I_{1}$. Hence we get either
$g_{1}(s_{1},a_{i},1_{H}^{(n-2)})\in\sqrt{0_{H_{1}}}$ for some $1\leq i\leq n$
or $g(a_{1}^{i-1},s_{1},a_{i+1}^{n})\in I_{1}$. Therefore $g_{1}\times
g_{2}((s_{1},1_{H_{2}}),(a_{i},b_{i}),(1_{H_{1}},1_{H_{2}})^{(n-2)})\in\sqrt{0_{H_{1}}}\times\sqrt{0_{H_{2}}}$
or $g_{1}\times
g_{2}((a_{1},b_{1}),\cdots,(a_{i-1},b_{i-1}),(s_{1},s_{2}),(a_{i+1},b_{i+1}),\cdots,(a_{n},b_{n}))\in
I_{1}\times H_{2}$. So $(s_{1},s_{2})$ is an $S$-element of $I_{1}\times
H_{2}$.

(2) The proof is similar to (1). ∎

######  Theorem 6.12.

Let $S_{1}$ and $S_{2}$ be $n$-ary multiplicative subsets of commutative
Krasner $(m,n)$-hyperrings $(H_{1},f_{1},g_{1})$ and $(H_{2},f_{2},g_{2})$,
respectively, and let $I_{1}$ and $I_{2}$ be proper hyperideals of $H_{1}$ and
$H_{2}$, respectively. If one of the following cases holds,

  * (1)

$\sqrt{0_{H_{2}}}^{(m,n)}\cap S_{2}\neq\varnothing$ and $I_{1}$ is an $n$-ary
$S_{1}$-$N$-hyperideal of $H_{1}$.

  * (2)

$\sqrt{0_{H_{1}}}^{(m,n)}\cap S_{1}\neq\varnothing$ and $I_{2}$ is an $n$-ary
$S_{2}$-$N$-hyperideal of $H_{2}$.

then $I_{1}\times I_{2}$ is an $n$-ary $S$-$N$-hyperideal of $H_{1}\times
H_{2}$ where $S=S_{1}\times S_{2}$.

###### Proof.

Let $\sqrt{0_{H_{2}}}^{(m,n)}\cap S_{2}\neq\varnothing$ and $I_{1}$ be an
$n$-ary $S_{1}$-$N$-hyperideal of $H_{1}$. So $0_{H_{2}}\in I_{2}\cap
S_{2}\neq\varnothing$ and $I_{1}\cap S_{1}=\varnothing$. Suppose that
$g_{1}\times g_{2}((a_{1},b_{1}),\cdots,(a_{n},b_{n}))\in I_{1}\times I_{2}$
for some $a_{1}^{n}\in H_{1}$ and $b_{1}^{n}\in H_{2}$ and $s_{1}$ is an
$S_{1}$-element of $I_{1}$. Hence we have $g_{1}(a_{1}^{n})\in I_{1}$. Then
either $g_{1}(s_{1},a_{i},1_{H_{1}}^{(n-2)})\in\sqrt{0_{H_{1}}}^{(m,n)}$ for
some $1\leq i\leq n$ or $g_{1}(a_{1}^{i-1},s_{1},a_{i+1}^{n})\in I_{1}$. So
$g_{1}\times
g_{2}((s_{1},0),(a_{i},b_{i}),(1_{H_{1}},1_{H_{2}})^{(n-2)})\in\sqrt{0_{H_{1}}}^{(m,n)}\times\sqrt{0_{H_{2}}}^{(m,n)}$
or $g_{1}\times
g_{2}((a_{1},b_{1}),\cdots,(a_{i-1},b_{i-1}),(s_{1},0),(a_{i+1},b_{i+1}),\cdots,(a_{n},b_{n}))\in
I_{1}\times I_{2}$. Thus $(s_{1},0_{H_{2}})$ is an $S$-element of $I_{1}\times
I_{2}$.
Also, if Case 2 holds, then by using a similar argument, we can show that
$I_{1}\times I_{2}$ is an $n$-ary $S$-$N$-hyperideal of $H_{1}\times H_{2}$
where $S=S_{1}\times S_{2}$. ∎

## References

  * [1] R. Ameri, M. Norouzi, Prime and primary hyperideals in Krasner $(m,n)$-hyperrings, European Journal Of Combinatorics, (2013) 379-390.
  * [2] M. Anbarloei, $n$-ary 2-absorbing and 2-absorbing primary hyperideals in Krasner $(m,n)$-hyperrings, Matematicki Vesnik, 71 (3) (2019) 250-262.
  * [3] M. Anbarloei, Unifing the prime and primary hyperideals under one frame in a Krasner $(m,n)$-hyperring, Comm. Algebra, 49 (2021) 3432-3446.
  * [4] M. Anbarloei, Krasner (m, n)-hyperring of fractions, Jordan Journal of Mathematics and Statistic, In press.
  * [5] A. Asadi, R. Ameri, Direct limit of Krasner (m,n)-hyperrings,Journal of Sciences, 31 (1) (2020) 75-83.
  * [6] B. Fahid, Z. Dongsheng, 2-Absorbing $\delta$-primary ideals of commutative rings. Kyungpook Mathematical Journal, 57 (2017) 193-198.
  * [7] K. Hila, K. Naka, B. Davvaz, On $(k,n)$-absorbing hyperideals in Krasner $(m,n)$-hyperrings, Quarterly Journal of Mathematics, 69 (2018) 1035-1046.
  * [8] F. Marty, Sur une generalization de la notion de groupe, $8^{th}$ Congress Math. Scandenaves, Stockholm, (1934) 45-49.
  * [9] S. Mirvakili, B. Davvaz, Relations on Krasner $(m,n)$-hyperrings, European J. Combin., 31(2010) 790-802.
  * [10] M. Norouzi, R.Ameri, V. Leoreanu-Fotea, Normal hyperideals in Krasner $(m,n)$-hyperrings, An. St. Univ. Ovidius Constanta 26 (3) (2018) 197-211.
  * [11] S. Omidi, B. Davvaz, J. Zhan, Some properties of $n$-hyperideals in commutative hyperrings, Journal of Algebraic Hyperstructures and Logical Algebras, 1 (2) (2020) 23-30.
  * [12] E. Ozel Ay, G. Yesilot, D. Sonmez, $\delta$-Primary Hyperideals on Commutative Hyperrings, Int. J. Math. and Mathematical Sciences, (2017) Article ID 5428160, 4 pages.
  * [13] U. Tekir, S. Koc, K. H. Oral, $n$-ideals of commutative rings, Filomat, 31(10) (2017) 2933-2941.
  * [14] M. Tamekkante, E. M. Bouba, $(2,n)$-ideals of commutative rings, Journal of Algebra and Its Applications, 18 (6) (2019). DOI: 10.1142/S0219498819501032.
  * [15] M.M. Zahedi, R. Ameri, On the prime, primary and maximal subhypermodules, Ital. J. Pure Appl. Math., 5 (1999) 61-80.