#  merging the $N$-hyperideals and $J$-hyperideals in one frame

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

The notions of $N$-hyperideals and $J$-hyperideals as two new classes of
hyperideals were recently defined in the context of Krasner
$(m,n)$-hyperrings. These concepts are created on the basis of the
intersection of all $n$-ary prime hyperideals and the intersection of all
maximal hyperideals, respectively. Despite being vastly different in many
aspects, they share numerous similar properties. The aim of this research work
is to merge them under one frame called $n$-ary $\delta(0)$-hyperideals where
the function $\delta$ assigns to each hyperideal of a Krasner
$(m,n)$-hyperring a hyperideal of the same hyperring. We give various
properties of $n$-ary $\delta(0)$-hyperideals and use them to characterize
certain classes of hyperrings such as hyperintegral domains and local
hyperrings. Moreover, we introduce the notions of $(s,n)$-absorbing
$\delta(0)$-hypereideals and weakly $(s,n)$-absorbing
$\delta(0)$-hypereideals.

###### Key words and phrases:

$n$-ary $\delta(0)$-hyperideal, $(s,n)$-absorbing $\delta(0)$-hyperideal,
weakly $(s,n)$-absorbing $\delta(0)$-hyperideal.

###### 2010 Mathematics Subject Classification:

16Y20, 16Y99, 20N20

##  1\. Introduction

Although two of the most important structures in commutative algebra, namely
prime and primary ideals, differ significantly in various aspects, they share
several similar properties. The question was whether a unified approach to
studying these two structures is possible. In [14], Dongsheng presented the
concept of $\delta$-primary ideals unifing the prime and primary ideals under
one frame in a commutative ring. Furthermore, Fahid and Dongsheng put two
concepts of 2-absorbing ideals and 2-absorbing primary ideals in one frame
called 2-absorbing $\delta$-primary ideals in [15].

Hyperstructures are algebraic structures that have at least one multi-valued
operation, known as a hyperoperation. The idea of the algebraic
hyperstructures goes back to Marty,s research work on hypergroups presented
during the $8^{th}$ Congress of the Scandinavian Mathematicians in 1934. Since
then, numerous papers and books concerning this topic have been written. They
can be seen in [8, 9, 10, 25, 27, 29]. $n$-ary hypergroups as an extension of
the notion of a hypergroup in the sense of Marty were defined in [11]. For
more study on $n$-ary structures refer to [17, 18, 19]. The notion of
$(m,n)$-hyperrings was presented by Mirvakili and Davvaz in [23]. Krasner
$(m,n)$-hyperrings as a subclass of $(m,n)$-hyperrings were given in [22].
Some review of the hyperrings can be found in [7, 20, 22, 24, 26]. Ameri and
Norouzi defined $n$-ary prime and $n$-ary primary hyperideals in Krasner
$(m,n)$-hyperring in [1] and Hila et al. generalized these concepts and
studied $(k,n)$-absorbing and $(k,n)$-absorbing primary hyperideals in [16].
Also, Davvaz et al. defined weakly $(k,n)$-absorbing and weakly
$(k,n)$-absorbing primary hyperideals in [13]. As a recent study, the concept
of $N$-hyperideals was introduced in [5] and some properties of them have been
investigated analogous with prime hyperideals. Let $G$ be a commutative
Krasner $(m,n)$-hyperring. A hyperideal $A$ of $G$ is called an $n$-ary
$N$-hyperideal if $g(x_{1}^{n})$ is in $A$ for $x_{1}^{n}\in G$ such that
$x_{i}$ is not in the intersection of the all $n$-ary prime hyperideals of $G$
for some $i\in\\{1,\cdots,n\\}$, then $g(x_{1}^{i-1},1_{G},x_{i+1}^{n})$ is in
$A$. The notion of $J$-hyperideals as an extension of $N$-hyperideals was
defined in [3]. A hyperideal $A$ of $G$ refers to an $n$-ary $J$-hyperideal if
$g(x_{1}^{n})\in A$ for $x_{1}^{n}\in G$ such that $x_{i}$ is not in the
intersection of the all maximal hyperideals of $G$ for some
$i\in\\{1,\cdots,n\\}$, then $g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\in A$. The two
structures have been treated differently and their properties were
independently proven in the mentioned references. Despite their differences in
many aspects, they share several similar properties. It is therefore essential
to explore the possibility of a unified approach to studying these two
structures.

In this paper, we aim to introduce the notion of $\delta(0)$-hyperideals
unifying the $N$-hyperideals and $J$-hyperideals under one frame. This method
effectively demonstrates the similarities and relationships between the two
structures. The paper is orgnized as follows. In Section 2, we recall the
necessary background, and we fix notation. In Section 3, we define $n$-ary
$\delta(0)$-hyperideals as a frame containing the concepts of $N$-hyperideals
and $J$-hyperideals. However, Example 3.3 shows that an $n$-ary
$\delta(0)$-hyperideal may not be an $n$-ary $N$-hypeideal. We give equivalent
characterizations for an $n$-ary $\delta(0)$-hyperideals in Theorem 3.4. It is
shown that an $n$-ary $\delta(0)$-hyperideal of a Krasner $(m,n)$-hyperring is
contained in $\delta(0)$ in Theorem 3.10. We conclude that if $G$ admits an
$n$-ary $\delta(0)$-hyperideal, then there exists an $n$-ary
$\delta(0)$-hyperideal of $G$ such that there is no $\delta(0)$-hyperideal
containing it in Theorem 3.21. Moreover, we examine $n$-ary
$\delta(0)$-hyperideals under various contexts of constructions such as
cartesian products, homomorphic images, localizations. The penultimate section
is dedicated to the notion of $(s,n)$-absorbing $\delta(0)$-hyperideals. Some
properties concerning it are investigated. The final section deals with issues
surrounding the notion of weakly $(s,n)$-absorbing $\delta(0)$-hyperideals as
an extension of the $(s,n)$-absorbing $\delta(0)$-hyperideals. We finish this
section with a copy of Nakayama,s lemma for a kind of weakly $(s,n)$-absorbing
$\delta(0)$-hypereideals. In this paper, all the hyperrings used are
commutative Krasner $(m,n)$-hyperrings with scalar identity.

##  2\. Preliminaries

In this section, we give deﬁnitions and notations that will be used in this
paper.

An $n$-ary hyperoperation $``f"$ on nonempty set $G$ is a mapping of $G^{n}$
into the family of all non-empty subsets of $G$. If $``f"$ is an $n$-ary
hyperoperation on $G$, then $(G,f)$ is called an $n$-ary hypergroupoid. We can
extend the $n$-ary hyperoperation on $G$ to non-empty subsets of $G$ as
follows. For $G_{1},\cdots,G_{n}\subseteq G$, then

$f(G^{n}_{1})=f(G_{1},...,G_{n})=\bigcup\\{f(a_{1},\cdots,a_{n})\ |\ a_{i}\in
G_{i},i=1,...,n\\}.$

Let us use the notation $a^{j}_{i}$ instead of the sequence
$a_{i},a_{i+1},...,a_{j}$. Then we have
$f(a_{1},\cdots,a_{i},b_{i+1},\cdots,b_{j},c_{j+1},\cdots,c_{n})=f(a^{i}_{1},b^{j}_{i+1},c^{n}_{j+1})$
and $f(a_{1},\cdots,a_{i},\underbrace{b,\cdots,b}_{j-i},\break
c_{j+1},\cdots,c_{n})=f(a^{i}_{1},b^{(j-i)},c^{n}_{j+1})$ The above notation
is the empty symbol where $j<i$. Let $f$ be an $n$-ary hyperoperation. Then
$r$-ary hyperoperation $f_{(l)}$ for $r=l(n-1)+1$ is given by
$f_{(l)}(a_{1}^{r})=\underbrace{f(f(...,f(f}_{l}(a^{n}_{1}),a_{n+1}^{2n-1}),...),a_{r-n+1}^{r})$.
An $n$-ary hypergroupoid $(G,f)$ is commutative if
$f(a_{1}^{n})=f(a_{\sigma(1)}^{\sigma(n)})$ for all $a_{1}^{n}\in G$ and
$\sigma\in\mathbb{S}_{n}$. An $n$-ary semihypergroup is an $n$-ary
hypergroupoid $(G,f)$, which is associative, that is
$f(a^{i-1}_{1},f(a_{i}^{n+i-1}),a^{2n-1}_{n+i})=f(a^{j-1}_{1},f(a_{j}^{n+j-1}),a_{n+j}^{2n-1})$
for $1\leq i<j\leq n$ and $a_{1}^{2n-1}\in G$. If the equation $a\in
f(a_{1}^{i-1},x_{i},a_{i+1}^{n})$ in an $n$-ary hypergroupoid $(G,f)$ has a
solution $x_{i}\in G$ for all $a_{1}^{i-1},a_{i+1}^{n},a\in G$ and
$i\in\\{1,\cdots,n\\}$, then $(G,f)$ is called an $n$-ary quasihypergroup. An
$n$-ary hypergroup is an $n$-ary semihypergroup that is an $n$-ary
quasihypergroup. A non-empty subset $H$ of an $n$-ary hypergroup $G$ is called
an $n$-ary subhypergroup of $G$ if $(H,f)$ is an $n$-ary hypergroup [22].

######  Definition 2.1.

[22] An algebraic hyperstructure $(G,f,g)$, or simply $G$, is said to be a
Krasner $(m,n)$-hyperring if it satisfies the following axioms:

  * (1)

$(G,f$) is a canonical $m$-ary hypergroup, that is

    * (i)

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

    * (ii)

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

    * (iii)

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

  * (2)

$(G,g)$ is a $n$-ary semigroup;

  * (3)

$g(a^{i-1}_{1},f(b^{m}_{1}),a^{n}_{i+1})=f(g(a^{i-1}_{1},b_{1},a^{n}_{i+1}),...,g(a^{i-1}_{1},b_{m},a^{n}_{i+1}))$
for every $a^{i-1}_{1},a^{n}_{i+1},b^{m}_{1}\in G$, and
$i\in\\{1,\cdots,n\\}$;

  * (4)

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

We assume throughout this paper that $G$ is a commutative Krasner
$(m,n)$-hyperring with scalar identity $1_{G}$, that is $g(a,1_{G}^{(n-2)})=a$
for all $a\in G$.

If $(R,f,g)$ is a Krasner $(m,n)$-hyperring such that $\varnothing\neq
R\subseteq G$, then $R$ is called a subhyperring of $G$. If $\varnothing\neq
A\subseteq G$ such that $(A,f)$ is an $m$-ary subhypergroup of $(G,f)$ and
$g(a^{i-1}_{1},A,a_{i+1}^{n})\subseteq I$ for $a^{n}_{1}\in R$,
$i\in\\{1,\cdots,n\\}$, then $A$ is said to be a hyperideal of $G$. Let $A$ be
a hyperideal of $G$. Then the set $G/A=\\{f(a_{1}^{i-1},A,a_{i+1}^{n})\ |\
a_{1}^{i-1},a_{i+1}^{n}\in G\\}$ is a Krasner $(m,n)$-hyperring with $m$-ary
hyperoperation and $n$-hyperoperation $f$ and $g$, respectively [1].

######  Definition 2.2.

[1] Let $a\in G$. Then $\langle a\rangle$ denotes the hyperideal generated by
$a$ and defined by $\langle a\rangle=g(G,a,1^{(n-2)})=\\{g(r,a,1_{G}^{(n-2)})\
|\ r\in G\\}$.

######  Definition 2.3.

[1] A hyperideal $A$ of $G$ refers to a maximal hyperideal if $A\subseteq
B\subseteq G$ for every hyperideal $B$ of $G$ implies that $B=A$ or $B=G$.

$Max(G)$ denotes the set of all maximal hyperideal of $G$. Jacobson radical of
$G$, being denoted by $J(G)$, is the intersection of all maximal hyperideals
of $G$. If $G$ does not have any maximal hyperideal, then we define $J(G)=G$.
Moreover, if $G$ has just one maximal hyperideal, then $G$ is called local.

######  Definition 2.4.

[1] Let $a\in G$. The element is invertible if there exists $b\in G$ with
$1_{G}=g(a,b,1_{G}^{(n-2)})$.

######  Definition 2.5.

[22] Let $(G_{1},f_{1},g_{1})$ and $(G_{2},f_{2},g_{2})$ be two Krasner
$(m,n)$-hyperrings. A mapping $\psi:G_{1}\longrightarrow G_{2}$ is called a
homomorphism if for all $x^{m}_{1}\in G_{1}$ and $y^{n}_{1}\in G_{1}$ we have

  * (i)

$\psi(f_{1}(x_{1},...,x_{m}))=f_{2}(\psi(x_{1}),...,\psi(x_{m})),$

  * (ii)

$\psi(g_{1}(y_{1},...,y_{n}))=g_{2}(\psi(y_{1}),...,\psi(y_{n})),$

  * (iii)

$\psi(1_{G_{1}})=1_{G_{2}}.$

######  Definition 2.6.

[6] Assume that $M\neq\varnothing$. Then $(M,h,k)$ is an $(m,n)$-hypermodule
over $G$, if

  * (i)

$(M,h)$ is a canonical $m$-ary hypergroup.

  * (ii)

The map

$\hskip 85.35826ptk:\underbrace{G\times...\times G}_{n-1}\times
M\longrightarrow P^{*}(M)$
statisfied the following conditions:

  * (1)

$k(a_{1}^{n-1},h(x_{1}^{m}))=h(k(a_{1}^{n-1},x_{1}),...,k(a_{1}^{n-1},x_{m}))$

  * (2)

$k(a_{1}^{i-1},f(b_{1}^{m}),a_{i+1}^{n-1},x)=h(k(a_{1}^{i-1},b_{1},r_{i+1}^{n-1},x),...,k(r_{1}^{i-1}b_{m},r_{i+1}^{n-1},x))$

  * (3)

$k(a_{1}^{i-1},g(a_{i}^{i+n-1}),a_{i+m}^{n+m-2},x)=k(a_{1}^{n-1},k(a_{m}^{n+m-2},x))$

  * (4)

$0=k(r_{1}^{i-1},0,r_{i+1}^{n-1},x)$.

##  3\. $n$-ary $\delta(0)$-hyperideals

A hyperideal expansion $\delta$ of $G$ is a fuction which assigns to each
hyperideal $A$ of $G$ a hyperideal $\delta(A)$ of $G$ such that
$A\subseteq\delta(A)$, and $B\subseteq C$ for hyperideals $B,C$ of $G$ implies
$\delta(B)\subseteq\delta(C)$. For example, the functions
$\delta_{0},\delta_{1},\delta_{G}$ and $\delta_{M}$ from the set of all
hyperideals of $G$ to the same set, defined by
$\delta_{0}(A)=A,\delta_{1}(A)=rad(A),\delta_{G}(A)=G$ and
$\delta_{M}(A)=\cap_{A\subseteq B\in Max(G)}B$, are hyperideal expansions of
$G$. In [2], the $n$-ary prime and $n$-ary primary hyperideals were put in a
frame called $n$-ary $\delta$-primary hyperideals. In this section, we
introduce and study the notion of $n$-ary $\delta(0)$-hyperideals in a
commutative Krasner $(m,n)$-hyperring $G$ which unify the $n$-ary
$N$-hyperideals and $n$-ary $J$-hyperideals.

######  Definition 3.1.

Let $A$ be a proper hyperideal of $G$ and $\delta$ a hyperideal expansion of
$G$. $A$ refers to an $n$-ary $\delta(0)$-hyperideal if whenever $x_{1}^{n}\in
G$ with $g(x_{1}^{n})\in A$ and
$g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\notin\delta(0)$ for some
$i\in\\{1,\cdots,n\\}$, then $x_{i}\in A$.

######  Example 3.2.

Consider the hyperring $([\alpha,\infty)\cup\\{0\\},+,\cdot)$ where
$\alpha\geq 1$, $``\cdot"$ is the usual multiplication and $``+"$ is defined
by

$\hskip 85.35826pta+b=\begin{cases}[a,\infty)\cup\\{0\\}&\text{if $a=b\neq
0$.}\\\ b+a=\\{a\\}&\text{if $b=0$},\\\ \\{\min\\{a,b\\}\\}&\text{if $a\neq
0,b\neq 0$ and $a\neq b$.}\\\ \end{cases}$

Note that $([\alpha,\infty)\cup\\{0\\},+,\cdot)$ is a Krasner
$(m,n)$-hyperring with $f(a_{1}^{m})=\Sigma_{i=1}^{m}a_{i}$ and
$g(b_{1}^{n})=\Pi_{i=1}^{n}b_{i}$ for all
$a_{1}^{m},b_{1}^{n}\in[\alpha,\infty)\cup\\{0\\}$. The hyperideal $0$ is an
$n$-ary $\delta_{1}(0)$-hyperideal of $[\alpha,\infty)\cup\\{0\\}$.

The next example shows that $n$-ary $\delta(0)$-hyperideals and $n$-ary
$N$-hyperideals are different notions.

######  Example 3.3.

Note that $(\mathbb{Z}/6\mathbb{Z},f,g)$ is a commutative Krasner
$(m,n)$-hyperring in which $f$ and $g$ are usual addition and multiplication,
respectively. Consider $\delta(I)=\\{x\in\mathbb{Z}/6\mathbb{Z}\ |\
g(x,3\mathbb{Z}/6\mathbb{Z},1_{\mathbb{Z}/6\mathbb{Z}}^{(n-2)})\subseteq
I\\}$. Let $A=2\mathbb{Z}/6\mathbb{Z}$ for every hyperideal $I$ of $G$. Then
$A$ is an $n$-ary $\delta(0_{\mathbb{Z}/6\mathbb{Z}})$-hyperideal. However,
$A$ is not an $n$-ary $N$-hyperideal.

Our first theorem gives a characterization of $\delta(0)$-hyperideals.

######  Theorem 3.4.

Let $A$ be a proper hyperideal of $G$. Then we have the following equivalent
statements:

  * (i)

$A$ is an $n$-ary $\delta(0)$-hyperideal of $G$;

  * (ii)

$A=E_{x}$ where $E_{x}=\\{y\in G\ |\ g(x,y,1^{(n-2)})\in A\\}$ for each
$x\notin\delta(0)$;

  * (iii)

$g(A_{1}^{n})\subseteq A$ for hyperideals $A_{1}^{n}$ of $G$ such that
$g(A_{1}^{i-1},1_{G},A_{i+1}^{n})\bigcap(G-\delta(0))\neq\varnothing$ for some
$i\in\\{1,\cdots,n\\}$ implies that $A_{i}\subseteq A$

###### Proof.

(i) $\Longrightarrow$ (ii) Let $A$ be an $n$-ary $\delta(0)$-hyperideal of
$G$. We always have $A\subseteq E_{x}$ for each $x\in G$. Suppose that $y\in
E_{x}$ for $x\notin\delta(0)$. This means $g(x,y,1^{(n-2)})\in A$. Since $A$
is an $n$-ary $\delta(0)$-hyperideal of $G$ and
$x=g(x,1^{(n-1)})\notin\delta(0)$, we get the result that $y\in A$ and so
$A=E_{x}$.

(ii) $\Longrightarrow$ (iii) Let $g(A_{1}^{n})\subseteq A$ for hyperideals
$A_{1}^{n}$ of $G$ and
$g(A_{1}^{i-1},1_{G},A_{i+1}^{n})\bigcap(G-\delta(0))\neq\varnothing$ for some
$i\in\\{1,\cdots,n\\}$. Since
$g(A_{1}^{i-1},1_{G},A_{i+1}^{n})\bigcap(G-\delta(0))\neq\varnothing$ for some
$i\in\\{1,\cdots,n\\}$, there exists $x_{j}\in A_{j}$ for all
$j\in\\{1,\cdots,n\\}$ and $j\neq i$ such that
$g(x_{1}^{i-1},1_{G},1_{i+1`}^{n})\notin\delta(0)$. Since
$g(g(x_{1}^{i-1},1_{G},x_{i+1}^{n}),A_{i},1^{(n-2)})=g(x_{1}^{i-1},A_{i},x_{i+1}^{n})\subseteq
A$, we have $A_{i}\subseteq E_{g(x_{1}^{i-1},1_{G},x_{i+1}^{n})}$. Now, by
(ii) we get $A_{i}\subseteq A$.

(iii) $\Longrightarrow$ (i) Let $g(x_{1}^{n})\in A$ for $x_{1}^{n}\in G$ such
that $g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\notin\delta(0)$. Put $A_{i}=\langle
x_{i}\rangle$ for each $i\in\\{1,\cdots,n\\}$. Since $g(x_{1}^{n})\in
g(A_{1}^{n})\subseteq A$ and $g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\in
g(A_{1}^{i-1},1_{G},A_{i+1}^{n})\bigcap(G-\delta(0))$, we get the result that
$A_{i}\subseteq A$. This means that $x_{i}\in A$. Hence $A$ is an $n$-ary
$\delta(0)$-hyperideals, as desired. ∎

######  Theorem 3.5.

Assume that $A_{1}^{n-1},A$ and $B$ are some hyperideals of $G$ such that
$g(A_{1}^{n-1},1_{G})\bigcap(G-\delta(0))\neq\varnothing$. Then:

  * (i)

If $A$ and $B$ are $n$-ary $\delta(0)$-hyperideals of $G$ that
$g(A_{1}^{n-1},A)=g(A_{1}^{n-1},B)$, then $A=B$.

  * (ii)

If $g(A_{1}^{n-1},A)$ is an $n$-ary $\delta(0)$-hyperideal of $G$, then
$g(A_{1}^{n-1},A)=A$.

###### Proof.

(i) Suppose that $g(A_{1}^{n-1},A)=g(A_{1}^{n-1},B)$ for hyperideals
$A_{1}^{n-1},A$ and $B$ of $G$. Then we have $g(A_{1}^{n-1},B)\subseteq A$.
Since $A$ is an $n$-ary $\delta(0)$-hyperideal of $G$ and
$g(A_{1}^{n-1},1_{G})\bigcap(G-\delta(0))\neq\varnothing$, we get the result
that $B\subseteq A$ by Theorem 3.4. Similarly, we obtain $A\subseteq B$ and so
$A=B$.

(ii) Let $g(A_{1}^{n-1},A)$ be an $n$-ary $\delta(0)$-hyperideal of $G$ for
hyperideals $A_{1}^{n-1},A$ of $G$. Since $g(A_{1}^{n-1},A)\subseteq
g(A_{1}^{n-1},A)$ and
$g(A_{1}^{n-1},1_{G})\bigcap(G-\delta(0))\neq\varnothing$, we have $A\subseteq
g(A_{1}^{n-1},A)$. Since $g(A_{1}^{n-1},A)\subseteq A$, we get
$g(A_{1}^{n-1},A)=A$. ∎

######  Proposition 3.6.

Let $\delta$ and $\gamma$ be two hyperideal expansions of $G$ such that
$\gamma(0)\subseteq\delta(0)$. Then every $n$-ary $\gamma(0)$-hyperideal of
$G$ is an $n$-ary $\delta(0)$-hyperideal.

######  Theorem 3.7.

Assume that $R$ is a non-empty subset of $G$. If $A$ is an $n$-ary
$\delta(0)$-hyperideal of $G$ such that $R\nsubseteq A$, then $E_{R}=\\{x\in
G\ |\ g(x,R,1_{G}^{(n-2)})\subseteq A\\}$ is an $n$-ary $\delta(0)$-hyperideal
of $G$.

###### Proof.

It is clear that $E_{R}\neq G$. Suppose that $g(x_{1}^{n})\in E_{R}$ for
$x_{1}^{n}\in G$ such that $g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\notin\delta(0)$
for some $i\in\\{1,\cdots,n\\}$. Then we conclude that
$g(g(x_{i},r,1_{G}^{(n-2)}),g(x_{1}^{i-1},1_{G},x_{i+1}^{n}),1^{(n-2)})=g(g(x_{1}^{n}),r,1^{(n-2)})\in
A$ for each $r\in R$. Since $A$ is an $n$-ary $\delta(0)$-hyperideal and
$g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\notin\delta(0)$, we obtain
$g(x_{i},r,1^{(n-2)})\in A$ which implies $x_{i}\in E_{R}$. Thus $E_{R}$ is an
$n$-ary $\delta(0)$-hyperideal of $G$. ∎

######  Theorem 3.8.

If $\\{A_{j}\\}_{j\in J}$ is a non-empty set of $n$-ary
$\delta(0)$-hyperideals of $G$, then so is $\bigcap_{j\in J}A_{j}.$

###### Proof.

Assume that $g(x_{1}^{n})\in\bigcap_{j\in J}A_{j}$ for $x_{1}^{n}\in G$ such
that $g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\notin\delta(0)$ for some
$i\in\\{1,\cdots,n\\}$. Since $A_{j}$ is an $n$-ary $\delta(0)$-hyperideal of
$G$ and $g(x_{1}^{n})\in A_{j}$ for all $j\in J$, we obtain $x_{i}\in A_{j}$
which means $x_{i}\in\bigcap_{j\in J}A_{j}$. ∎

Recall from [3] that a proper hyperideal $A$ of $G$ is an $n$-ary
$J$-hyperideal if $g(x_{1}^{n})\in A$ for $x_{1}^{n}$ and $x_{i}\notin J(G)$
imply that $g(x_{1}^{i-1},1_{G},x_{i-1}^{n})\in A$.

######  Theorem 3.9.

Let $\delta(0)$ be a maximal hyperideal of $G$ and $A$ be an $n$-ary
$J$-hyperideal of $G$. Then $A$ is an $n$-ary $\delta(0)$-hyperideal.

###### Proof.

Assume that $g(x_{1}^{n})\in A$ for $x_{1}^{n}\in G$ such that $x_{i}\notin A$
for some $i\in\\{1,\cdots,n\\}$. Since
$g(g(x_{1}^{i-1},1_{G},x_{i+1}^{n}),x_{i},1_{G}^{(n-2)})\in A$ and
$g(x_{i},1_{G}^{(n-1)})\notin A$, we conclude that
$g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\in J(G)$. Since $\delta(0)$ is a maximal
hyperideal of $G$, $J(G)\subseteq\delta(0)$ and so
$g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\in\delta(0)$. Thus $A$ is an $n$-ary
$\delta(0)$\- hyperideal of $G$. ∎

######  Theorem 3.10.

Let $A$ be an $n$-ary $\delta(0)$-hyperideal of $G$. Then $A$ is contained in
$\delta(0)$.

###### Proof.

Suppose that $A$ is an $n$-ary $\delta(0)$-hyperideal of $G$ such that it is
not contained in $\delta(0)$. Let $x\in A$ but $x\notin\delta(0)$. Since $A$
is an $n$-ary $\delta(0)$-hyperideal of $G$, $g(x,1_{G}^{(n-1)})\in A$ and
$g(x,1_{G}^{(n-1)})\notin\delta(0)$, we get $1\in A$ which is a contradiction.
Thus $A$ is contained in $\delta(0)$. ∎

Recall from [1] that a proper hyperideal $A$ of $G$ is an $n$-ary prime
hyperideal if for hyperideals $A_{1}^{n}$ of $G$, $g(A_{1}^{n})\subseteq A$
implies that $A_{i}\subseteq A$ for some $i\in\\{1,\cdots,n\\}$. In Lemma 4.5
in [1], it was shown that a proper hyperideal $A$ of $G$ is an $n$-ary prime
hyperideal if for $x^{n}_{1}\in G$, $g(x^{n}_{1})\in A$ implies that $x_{i}\in
A$ for some $i\in\\{1,\cdots,n\\}$.

######  Remark 3.11.

$\delta(0)$ is an $n$-ary prime hyperideal of $G$ if and only if $\delta(0)$
is an $n$-ary $\delta(0)$-hyperideal of $G$.

######  Theorem 3.12.

Assume that $A$ is an $n$-ary prime hyperideal of $G$ with $\delta(A)=A$. Then
$A$ is an $n$-ary $\delta(0)$-hyperideal of $G$ if and only if $A=\delta(0)$.

###### Proof.

($\Longrightarrow$) Since $0\in A$, we have $\delta(0)\subseteq\delta(A)=A$.
Since $A$ is an $n$-ary $\delta(0)$-hyperideal of $G$, the inclusion
$A\subseteq\delta(0)$ holds by 3.10. Thus $A=\delta(0)$.

($\Longleftarrow$) Let $A=\delta(0)$ and $g(x_{1}^{n})\in A$ for $x_{1}^{n}\in
G$ such that $g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\notin\delta(0)$. Since $A$ is
an $n$-ary prime hyperideal of $G$ and $g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\notin
A$, we get $x_{i}\in A$. Consequently, $A$ is an $n$-ary
$\delta(0)$-hyperideal of $G$. ∎

######  Theorem 3.13.

Let $A$ be an $n$-ary $\delta(0)$-hyperideal of $G$ such that $\delta(A)=A$.
If there is no $\delta(0)$-hyperideal containing $A$ properly, then
$A=\delta(0)$.

###### Proof.

It is sufficient to show the hyperideal $A$ is prime. Assume that
$g(x_{1}^{n})\in A$ for $x_{1}^{n}\in G$ but $x_{i}\notin A$ for some
$i\in\\{1,\cdots,n\\}$. By Theorem 3.7, $E_{x_{i}}=\\{x\in G\ |\
g(x,x_{i},1_{G}^{(n-2)})\subseteq A\\}$ is an $n$-ary $\delta(0)$-hyperideal
of $G$ as $A$ is an $n$-ary $\delta(0)$-hyperideal and $x_{i}\notin A$. Since
there is no $\delta(0)$-hyperideal containing $A$ properly, we get
$g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\in E_{x_{i}}=A$. Since $A$ is an $n$-ary
$\delta(0)$-hyperideal of $G$, we can continue the process and get $x_{j}\in
A$ for some $j\in\\{1,\cdots,n\\}$. Then $A$ is an $n$-ary prime hyperideal of
$G$. Now, by Theorem 3.12, we get the result that $A=\delta(0)$. ∎

A commutative Krasner $(m,n)$-hyperring $G$ is said to be an $n$-ary
hyperintegral domain if $g(x_{1}^{n})=0$ for $x_{1}^{n}\in G$ implies that
$x_{i}=0$ for some $i\in\\{1,\cdots,n\\}$ [1].

######  Remark 3.14.

If $G$ is an $n$-ary hyperintegral domain, then $0$ is an $n$-ary
$\delta(0)$-hyperideal of $G$.

######  Proposition 3.15.

If $G$ is a commutative Krasner $(m,n)$-hypering which is not an $n$-ary
hyperintegral domain and $\delta(0)=0$, then $G$ has no
$\delta(0)$-hyperideal.

###### Proof.

Let $\delta(0)=0$. Assume that $g(x_{1}^{n})\in\delta(0)$ for $x_{1}^{n}\in
G$. Since $G$ is not an $n$-ary hyperintegral domain, we get $x_{i}\neq 0$ for
all $i\in\\{1,\cdots,n\\}$ which means $\delta(0)$ is not an $n$-ary prime
hyperideal. Therefore $\delta(0)$ is not an $n$-ary $\delta(0)$-hyperideal of
$G$ by Remark 3.11. Now, assume that $A\neq 0$ is an arbitrary hyperideal of
$G$. If $A$ is an $n$-ary $\delta(0)$-hyperideal of $G$, then $A$ is containd
in $\delta(0)$ by Theorem 3.10 which means $A=0$, a contradiction.
Consequently, $G$ has no $\delta(0)$-hyperideal. ∎

In view of Remark 3.14 and Proposition 3.15, we have the following result.

######  Corollary 3.16.

Let $\delta(0)=0$. Then $G$ is an $n$-ary hyperintegral domain if and only if
$0$ is an $n$-ary $\delta(0)$-hyperideal of $G$.

Recall from [1] that the radical of the hyperideal $A$ of $G$, denoted by
$rad(A)$, is the intersection of all prime hyperideals $G$ containing $A$. If
the set of all prime hyperideals containing $A$ is empty, then we define
$rad(A)=G$. Moreover, a proper hyperideal $A$ of $G$ is an $n$-ary primary
hyperideal if $x_{1}^{n}\in G$ and $g(x^{n}_{1})\in A$ imply either $x_{i}\in
A$ or $g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\in rad(A)$ for some
$i\in\\{1,\cdots,n\\}$.

######  Theorem 3.17.

Let $A$ be an $n$-ary primary hyperideal of $G$. If $rad(A)$ is containd in
$\delta(0)$, then $A$ is an $n$-ary $\delta(0)$-hyperideal.

###### Proof.

Assume that $A$ is an $n$-ary primary hyperideal of $G$. Let $g(x_{1}^{n})\in
A$ for $x_{1}^{n}\in G$ but $g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\notin\delta(0)$
for some $i\in\\{1,\cdots,n\\}$. This means
$g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\notin rad(A)$ and so $x_{i}\in A$ as $A$ is
an $n$-ary primary hyperideal of $G$. Hence $A$ is an $n$-ary
$\delta(0)$-hyperideal. ∎

As an immediate consequence of the previous theorem, we have the following
result.

######  Corollary 3.18.

If $A$ is an $n$-ary prime hyperideal of $G$ such that $A$ is contained in
$\delta(0)$, then $A$ is an $n$-ary $\delta(0)$-hyperideal.

In [1], it was shown that if $x\in rad(A)$ then there exists $r\in\mathbb{N}$
such that $g(x^{(r)},1_{G}^{(n-r)})\in A$ for $r\leq n$, or
$g_{(l)}(x^{(r)})\in A$ for $r=l(n-1)+1$.

######  Theorem 3.19.

Let $0$ be an $n$-ary $\delta(0)$-hyperideal of $G$. If
$rad(\delta(0))=\delta(0)$, then $rad(0)$ is an $n$-ary $\delta(0)$-hyperideal
of $G$.

###### Proof.

Let $g(x_{1}^{n})\in rad(0)$ for $x_{1}^{n}\in G$ such that
$g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\notin\delta(0)=rad(\delta(0))$ for some
$i\in\\{1,\cdots,n\\}$. Then we conclude that there exists $r\in\mathbb{N}$
such that $g(g(x_{1}^{n})^{(r)},1_{G}^{(n-r)})=0$ for $r\leq n$, or
$g_{(l)}(g(x_{1}^{n})^{(r)})=0$ for $r=l(n-1)+1$. Let
$g(g(x_{1}^{(r)},1_{G}^{(n-r)}),\cdots,g(x_{n}^{(r)},1_{G}^{(n-r)}))=g(g(x_{1}^{n})^{(r)},1_{G}^{(n-r)})=0$
for some $r\leq n$. Since $g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\notin
rad(\delta(0))$, we get the result that

$g(g(x_{1}^{(r)},1_{G}^{(n-r)}),\cdots,g(x_{i-1}^{(r)},1_{G}^{(n-r)}),1_{G},g(x_{i+1}^{(r)},1_{G}^{(n-r)}),\cdots,g(x_{n}^{(r)},1_{G}^{(n-r)})$

$\hskip 28.45274pt=g(g(x_{1}^{i-1},1_{G},1_{i+1}^{n})^{(r)},1_{G}^{(n-r)})$

$\hskip 28.45274pt\notin\delta(0)$
for all $r\leq n$ and
$g_{(l)}(g(x_{1}^{i-1},1_{G},x_{i+1}^{n})^{(r)})\notin\delta(0)$ for
$r=l(n-1)+1$. Since $0$ is an $n$-ary $\delta(0)$-hyperideal of $G$, we have
$g(x_{i}^{(r)},1_{G}^{(n-r)})=0$ which means $x_{i}\in rad(0)$. By using a
similar argument, one can easily complete the proof where $r=l(n-1)+1$. ∎

A proper hyperideal $A$ of $G$ is said to be an $n$-ary $\delta$-primary
hyperideal if $g(x_{1}^{n})\in A$ for $x_{1}^{n}\in G$ implies that $x_{i}\in
A$ or $g(x_{1}^{i-1},1_{G},x_{i-1}^{n})\in\delta(A)$ for some
$i\in\\{1,\cdots,n\\}$ [2].

######  Theorem 3.20.

Let $A$ be a proper hyperideal of $G$.

  * (i)

If $A$ is an $n$-ary $\delta(0)$-hyperideal of $G$, then $A$ is an $n$-ary
$\delta$-primary hyperideal.

  * (ii)

Let $\delta^{2}(0)\subseteq\delta(0)$. Then $A$ is an $n$-ary
$\delta(0)$-hyperideal of $G$ if and only if $A$ is an $n$-ary
$\delta$-primary hyperideal and it is contained in $\delta(0)$.

###### Proof.

(i) Let $A$ be an $n$-ary $\delta(0)$-hyperideal of $G$. Assume that
$g(x_{1}^{n})\in A$ for $x_{1}^{n}\in G$ such that
$g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\notin\delta(A)$ for some
$i\in\\{1,\cdots,n\\}$. Therefore
$g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\notin\delta(0)$ as
$\delta(0)\subseteq\delta(A)$. Since $A$ is an $n$-ary $\delta(0)$-hyperideal,
we get $x_{i}\in A$. Thus $A$ is an $n$-ary $\delta$-primary hyperideal.

(ii) ($\Longrightarrow$) Let $A$ be an $n$-ary $\delta(0)$-hyperideal of $G$.
By (i) and Theorem 3.10, we are done.

($\Longleftarrow$) Assume that $A$ is an $n$-ary $\delta$-primary hyperideal
and it is contained in $\delta(0)$. Suppose that $g(x_{1}^{n})\in A$ for
$x_{1}^{n}\in G$ such that $x_{i}\notin A$ for some $i\in\\{1,\cdots,n\\}$.
Since $A$ is an $n$-ary $\delta$-primary hyperideal and $A\subseteq\delta(0)$,
we conclude that
$g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\in\delta(A)\subseteq\delta^{2}(0)$ which
means $g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\in\delta(0)$. Consequently, $A$ is an
$n$-ary $\delta(0)$-hyperideal of $G$. ∎

######  Theorem 3.21.

If $G$ admits an $n$-ary $\delta(0)$-hyperideal, then there exists an $n$-ary
$\delta(0)$-hyperideal of $G$ such that there is no $\delta(0)$-hyperideal
containing it. Moreover, if $\delta(A)=A$ for every $n$-ary
$\delta(0)$-hyperideal $A$ that there is no $\delta(0)$-hyperideal containing
it, then $\delta(0)$ is an $n$-ary prime hyperideal of $G$.

###### Proof.

Let $A$ be an $n$-ary $\delta(0)$-hyperideal of $G$. Assume that $\Sigma$ is
the set of all $n$-ary $\delta(0)$-hyperideals of $G$. From $A\in\Sigma$ it
follows that $\Sigma\neq\varnothing$. $\Sigma$ is a parially ordered set with
respect to set inclusion relation. Assume that $A_{1}\subseteq
A_{2}\subseteq\cdots$ is some chain in $\Sigma$. Clearly,
$\cup_{i=1}^{\infty}A_{i}$ is a hyperideal of $G$. Assume that
$g(x_{1}^{n})\in\cup_{i=1}^{\infty}A_{i}$ for $x_{1}^{n}\in G$ such that
$g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\notin\delta(0)$ for some
$i\in\\{1,\cdots,n\\}$. This means that there exists $t\in\mathbb{N}$ such
that $g(x_{1}^{n})\in A_{t}$. Since $A_{t}$ is an $n$-ary
$\delta(0)$-hyperideal of $G$ and
$g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\notin\delta(0)$, we have $x_{i}\in
A_{t}\subseteq\cup_{i=1}^{\infty}A_{i}$. Therefore $\cup_{i=1}^{\infty}A_{i}$
is an upper bound of the mentioned chain. By Zorn,s lemma, there is an $n$-ary
$\delta(0)$-hyperideal $\mathbb{A}$ which is maximal in $\Sigma$. Hence we
conclude that $\mathbb{A}=\delta(0)$ by Theorem 3.13. Thus $\delta(0)$ is an
$n$-ary prime hyperideal of $G$ by Remark 3.11. ∎

The concept of Krasner $(m,n)$-hyperring of fractions was defined in [4]. Let
$S$ a non-empty subset of $G$. $S$ refers to an $n$-ary multiplicative subset
of $G$, if $1\in S$ and $g(x_{1}^{n})\in S$ for $x_{1},\cdots,x_{n}\in S$.
Suppose that $\delta$ is a hyperideal expansion of a $G$, $S$ is an $n$-ary
multiplicative subset of $G$ and $A$ a hyperideal of $G$. Then $\delta_{S}$ is
a hyperideal expansion of $S^{-1}G$ with
$\delta_{S}(S^{-1}A)=S^{-1}\Big{(}\delta(A)\Big{)}$ [5]. $S$ is said to be an
$n$-ary $\delta(0)$-multiplicative subset of $G$ if $G-\delta(0)\subseteq S$
and $g(x_{1}^{n-1},x)\in S$ for all $x\in S$ and $g(x_{1}^{n-1},1_{G})\in
G-\delta(0)$.

######  Theorem 3.22.

Let $A$ be an $n$-ary $\delta(0_{G})$-hyperideal of $(G,f,g)$ and $S$ an
$n$-ary multiplicative subset of $G$ such that $S\cap A=\varnothing$. Then
$S^{-1}A$ is an $n$-ary $\delta_{S}(0_{S^{-1}G})$-hyperideals of
$(S^{-1}G,H,K)$.

###### Proof.

Suppose that $K(\frac{x_{1}}{s_{1}},\cdots,\frac{x_{n}}{s_{n}})\in S^{-1}A$
for $\frac{x_{1}}{s_{1}},\cdots,\frac{x_{n}}{s_{n}}\in S^{-1}G$ such that
$K(\frac{x_{1}}{s_{1}},\cdots,\frac{x_{i-1}}{s_{i-1}},\frac{1_{G}}{1_{G}},\frac{x_{i+1}}{s_{i+1}},\cdots,\frac{x_{n}}{s_{n}})\notin\delta_{S}(0_{S^{-1}G})$
for some $i\in\\{1,\cdots,n\\}$. Therefore
$\frac{g(x_{1}^{n})}{g(s_{1}^{n})}\in S^{-1}A$ which implies
$g(x_{1}^{i-1},g(s,x_{i},1_{G}^{(n-2)}),x_{i+1}^{n})=g(s,g(x_{1}^{n}),1_{G}^{(n-2)})\in
A$ for some $s\in S$. Since $A$ is an $n$-ary $\delta(0_{G})$-hyperideal of
$G$ and $g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\notin\delta(0_{G})$, we get the
result that $g(s,x_{i},1_{G}^{(n-2)})\in A$. Hence we have
$\frac{x_{i}}{s_{i}}=\frac{g(x_{i},s,1_{G}^{(n-2)})}{g(s_{i},s,1_{G}^{(n-2)})}=K(\frac{x_{i}}{s_{i}},\frac{s}{s},\frac{1_{G}}{1_{G}}^{(n-2)})\in
S^{-1}A$. Thus we conclude that $S^{-1}A$ is an $n$-ary
$\delta_{S}(0_{S^{-1}G})$-hyperideals of $(S^{-1}G,H,K)$. ∎

######  Theorem 3.23.

Let $A$ be a hyperideal of $G$. Then $A$ is an $n$-ary $\delta(0)$-hyperideal
of $G$ if and only if $G-A$ is an $n$-ary $\delta(0)$-multiplicative subset of
$G$.

###### Proof.

($\Longrightarrow$)Let $A$ be an $n$-ary $\delta(0)$-hyperideal of $G$.
Therefore $A$ is contained in $\delta(0)$ by Theorem 3.10. Then we have
$G-\delta(0)\subseteq G-A$. Let $x\in G-A$ and $g(x_{1}^{n-1},1_{G})\in
G-\delta(0)$. Suppose that $g(x_{1}^{n-1},x)\in A$. Hence we get $x\in A$ as
$A$ is an $n$-ary $\delta(0)$-hyperideal of $G$ and
$g(x_{1}^{n-1},1_{G})\notin\delta(0)$. This is a contradiction. Therefore
$g(x_{1}^{n-1},x)\in G-A$ which means $G-A$ is an $n$-ary
$\delta(0)$-multiplicative subset of $G$.

($\Longleftarrow$) Let $G-A$ be an $n$-ary $\delta(0)$-multiplicative subset
of $G$ for some hyperideal $A$ of $G$. Assume that $g(x_{1}^{n})\in A$ for
$x_{1}^{n}\in G$ such that $g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\notin\delta(0)$
for some $i\in\\{1,\cdots,n\\}$. If $x_{i}\notin A$, then we get
$g(x_{1}^{i-1},x_{i},x_{i+1}^{n})\notin A$ as $G-A$ is an $n$-ary
$\delta(0)$-multiplicative subset of $G$ and
$g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\in G-\delta(0)$. This is a contradiction.
Thus $x_{i}\in A$. Consequently, $A$ is an $n$-ary $\delta(0)$-hyperideal of
$G$. ∎

######  Theorem 3.24.

Let $A$ be a hyperideal of $G$ and $S$ an $n$-ary $\delta(0)$-multiplicative
subset of $G$ such that $S\cap A=\varnothing$. Then there exists an $n$-ary
$\delta(0)$-hyperideal $B$ such that $A\subseteq B$ and $B\cap S=\varnothing$.

###### Proof.

Let $\Phi$ be the set of all hyperideals of $G$ that any hyperideal has an
empty intersection with $S$. By the hypothesis, $A\in\Phi$ and so
$\Phi\neq\varnothing$. $\Phi$ is a partially ordered set with respect to set
inclusion relation. Then there exists a maximal element $B$ in $\Phi$ by
Zorn,s Lemma. We assume that $B$ is not an $n$-ary $\delta(0)$-hyperideal of
$G$ and look for a contradiction. The assumption means that $g(x_{1}^{n})\in
B$ for $x_{1}^{n}\in G$ and neither $x_{i}\in B$ nor
$g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\in\delta(0)$ for all $i\in\\{1,\cdots,n\\}$.
Put $E_{g(x_{1}^{i-1},1_{G},x_{i+1}^{n})}=\\{x\in G\ |\
g(x,g(x_{1}^{i-1},1_{G},x_{i-1}^{n}),1_{G}^{(n-2)})\in B\\}$ for
$i\in\\{1,\cdots,n\\}$. Since $B\subsetneq
E_{g(x_{1}^{i-1},1_{G},x_{i+1}^{n})}$, we conclude that
$E_{g(x_{1}^{i-1},1_{G},x_{i+1}^{n})}\cap S\neq\varnothing$ by the maximality
$B$. Let $t\in E_{g(x_{1}^{i-1},1_{G},x_{i+1}^{n})}\cap S$. Then
$g(x_{1}^{i-1},t,x_{i+1}^{n})=g(t,g(x_{1}^{i-1},1_{G},x_{i+1}^{n}),1_{G}^{(n-2)})\in
B$. On the other hand, since $S$ is an $n$-ary $\delta(0)$-multiplicative
subsete of $G$, $t\in S$ and $g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\in
G-\delta(0)$, we get the result
$g(x_{1}^{i-1},t,x_{i+1`}^{n})=g(t,g(x_{1}^{i-1},1_{G},x_{i+1}^{n}),1_{G}^{(n-1)})\in
S$ . This implies that $g(x_{1}^{i-1},t,x_{i-1}^{n})\in B\cap S$ and so $B\cap
S\neq\varnothing$. This is a contradiction because $B\in\Phi$. Consequently,
$B$ is an $n$-ary $\delta(0)$-hyperideal of $G$. ∎

In following theorem, we discuss hyperrings of which every proper hyperideal
is an $n$-ary $\delta(0)$-hyperideal.

######  Theorem 3.25.

We have the following equivalent statements:

  * (i)

$G$ is a local Krasner $(m,n)$-hyperring with maximal hyperideal $\delta(0)$.

  * (ii)

Every proper principal hyperideal of $G$ is an $n$-ary $\delta(0)$-hyperideal.

  * (iii)

Every proper hyperideal of $G$ is an $n$-ary $\delta(0)$-hyperideal.

###### Proof.

(i) $\Longrightarrow$ (ii) Let $\langle x\rangle$ be a proper hyperideal of
$G$ for $x\in G$. Assume that $g(x_{1}^{n})\in\langle x\rangle$ for
$x_{1}^{n}\in G$ such that $g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\notin\delta(0)$
for some $i\in\\{1,\cdots,n\\}$. Since $\delta(0)$ is the only maximal
hyperideal of $G$ and $g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\notin\delta(0)$, we
conclude that $g(x_{1}^{i-1},1_{G},x_{i+1}^{n})$ is invertible. Then there
exists $y\in R$ such that
$1_{G}=g(y,g(x_{1}^{i-1},1_{G},x_{i+1}^{n}),1_{G}^{(n-2)})$. Hence we have
$x_{i}=g(x_{i},1_{G}^{(n-1)})=g(x_{i},g(y,g(x_{1}^{i-1},1_{G},x_{i+1}^{n}),1_{G}^{(n-2)}),1_{G}^{(n-2)})=g(y,g(x_{1}^{n}),1_{G}^{(n-2)})\in\langle
x\rangle$. This shows that $\langle x\rangle$ is an $n$-ary
$\delta(0)$-hyperideal.

(ii) $\Longrightarrow$ (iii) Assume that $A$ is a hyperideal of $G$. Let
$g(x_{1}^{n})\in A$ for $x_{1}^{n}\in G$ such that
$g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\notin\delta(0)$. By (ii), $\langle
g(x_{1}^{n})\rangle$ is an $n$-ary $\delta(0)$-hyperideal of $G$. From
$g(x_{1}^{n})\in\langle g(x_{1}^{n})\rangle$ it follows that $x_{i}\in\langle
g(x_{1}^{n})\rangle$ as $g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\notin\delta(0)$.
Since $\langle g(x_{1}^{n})\rangle\subseteq A$, we obtain $x_{i}\in A$ which
implies $A$ is an $n$-ary $\delta(0)$-hyperideal of $G$.

(iii) $\Longrightarrow$ (i) Assume that $A$ is an arbitrary hyperideal of $G$.
Hence $A$ is an $n$-ary $\delta(0)$-hyperideal of $G$ by (iii). Therefore we
conclude that $A$ is contained in $\delta(0)$ by Theorem 3.10. Thus
$\delta(0)$ is the only maximal hyperideal of $G$ and so $G$ is local. ∎

Suppose that $(G_{1},f_{1},g_{1})$ and $(G_{2},f_{2},g_{2})$ are two Krasner
$(m,n)$-hyperrings. Let $\delta$ and $\gamma$ be hyperideal expansions of
$G_{1}$ and $G_{2}$, respectively. Then the hyperring homomorphism
$\psi:G_{1}\longrightarrow G_{2}$ is said to be a $\delta\gamma$-homomorphism
if $\delta(\psi^{-1}(A_{2}))=\psi^{-1}(\gamma(A_{2}))$ for the hyperideal
$A_{2}$ of $G_{2}$. It is remarkable that
$\gamma(\psi(A_{1})=\psi(\delta(A_{1}))$ for $\delta\gamma$-epimorphism $\psi$
and for hyperideal $A_{1}$ of $G_{1}$ with $Ker(\psi)\subseteq A_{1}$ [2].
Every homomorphism $\psi:G_{1}\longrightarrow G_{2}$ is a
$\delta_{1}\gamma_{1}$-homomorphism.

######  Theorem 3.26.

Assume that $G_{1}$ and $G_{2}$ are two Krasner $(m,n)$-hyperrings and
$\delta$ and $\gamma$ are hyperideal expansions of $G_{1}$ and $G_{1}$,
respectively, such that $\psi:G_{1}\longrightarrow G_{2}$ is a
$\delta\gamma$-homomorphism.

  * (i)

If $\psi$ is a monomorphism and $A_{2}$ is an $n$-ary
$\gamma(0_{G_{2}})$-hyperideal of $G_{2}$, then $\psi^{-1}(A_{2})$ is an
$n$-ary $\delta(0_{G_{1}})$-hyperideal of $G_{1}$.

  * (ii)

If $\psi$ is an epimorphism and $A_{1}$ is an $n$-ary
$\delta(0_{G_{1}})$-hyperideal of $G_{1}$ such that $Ker(\psi)\subseteq
A_{1}$, then $\psi(A_{1})$ is an n-ary $\gamma(0_{G_{2}})$-hyperideal of
$G_{2}$.

###### Proof.

(i) Let $g_{1}(x_{1}^{n})\in\psi^{-1}(A_{2})$ for $x_{1}^{n}\in G_{1}$ such
that $g_{1}(x_{1}^{i-1},1_{G_{1}},x_{i+1}^{n})\notin\delta(0_{G_{1}})$ for
some $i\in\\{1,\cdots,n\\}$. This means that
$\psi(g_{1}(x_{1}^{n}))=g_{2}(\psi(x_{1}),\cdots,\psi(x_{n}))\in A_{2}$. Since
$\psi$ is a monomorphism and
$g_{1}(x_{1}^{i-1},1_{G_{1}},x_{i+1}^{n})\notin\delta(0_{G_{1}})$, we conclude
that
$g_{2}(\psi(x_{1}),\cdots,\psi(x_{i-1}),1_{G_{2}},\psi(x_{i+1}),\cdots,\psi(x_{n}))=\psi(g_{1}(x_{1}^{i-1},1_{G_{1}},x_{i+1}^{n}))\notin\gamma(0_{G_{2}})$.
Since $A_{2}$ is an $n$-ary $\gamma(0_{G_{2}})$-hyperideal of $G_{2}$, we get
$\psi(x_{i})\in A_{2}$ which means $x_{i}\in\psi^{-1}(A_{2})$. Thus
$\psi^{-1}(A_{2})$ is an $n$-ary $\delta(0_{G_{1}})$-hyperideal of $G_{1}$.

(ii) Let $g_{2}(y_{1}^{n})\in\psi(A_{1})$ for $y_{1}^{n}\in G_{2}$ such that
$g_{2}(y_{1}^{i-1},1_{G_{2}},y_{i+1}^{n})\notin\gamma(0_{G_{2}})$ for some
$i\in\\{1,\cdots,n\\}$. Since $\psi$ is an epimorphism, then there exist
$x_{1}^{n}\in G_{1}$ such that $\psi(x_{1})=y_{1},...,\psi(x_{n})=y_{n}$. Thus
$\psi(g_{1}(x_{1}^{n}))=g_{2}(\psi(x_{1}),...,\psi(x_{n}))=g_{2}(y_{1}^{n})\in\psi(A_{1})$.
Since $Ker(\psi)$ is contained in $A_{1}$, we have $g_{1}(x_{1}^{n})\in
A_{1}$. Since $A_{1}$ is an $n$-ary $\delta(0_{G_{1}})$-hyperideal of $G_{1}$
and $g_{1}(x_{1}^{i-1},1_{G_{1}},x_{i+1}^{n})\notin\delta(0_{G_{1}})$, we get
the result that $x_{i}\in A_{1}$ which implies
$y_{i}=\psi(x_{i})\in\psi(A_{1})$. This shows that $\psi(A_{1})$ is an n-ary
$\gamma(0_{G_{2}})$-hyperideal of $G_{2}$. ∎

Let $A$ be a hyperideal of $G$. The projection map $\pi:G\longrightarrow G/A$,
defined by $a\mapsto f(a,A,0^{(m-2)})$ is a homomorphism by Theorem 3.2 in
[1].

######  Corollary 3.27.

Let $A$ and $B$ be hyperideals of $G$ such that $A\subseteq B$. If $B$ is an
$n$-ary $\delta(0)$-hyperideal of $G$, then $B/A$ is an $n$-ary
$\delta_{q}(0)$-hyperideal of $G/A$.

###### Proof.

Follows by applying Theorem 3.26 (ii) to the epimorphism $\pi:G\longrightarrow
G/A$. ∎

######  Corollary 3.28.

Assume that $B/A$ is an $n$-ary $\delta_{q}(0)$-hyperideal of $G/A$ where $A$
and $B$ are hyperideals of $G$ with $A\subseteq B$.

  * (i)

If $A\subseteq\delta(0)$, then $B$ is an $n$-ary $\delta(0)$-hyperideal of
$G$.

  * (ii)

If $A$ is an $n$-ary $\delta(0)$-hyperideal of $G$, then so is $B$.

###### Proof.

(i) Let $g(x_{1}^{n})\in B$ for $x_{1}^{n}\in G$ such that
$g(x_{1}^{i-1},1_{G},x_{i+1}^{n})\notin\delta(0)$ for some
$i\in\\{1,\cdots,n\\}$. So
$g(f(x_{1},A,0^{(m-2)}),\cdots,f(x_{n},A,0^{(m-2)}))=f(g(x_{1}^{n}),A,0^{(m-2)})\in
B/A$. Since $B/A$ is an $n$-ary $\delta_{q}(0)$-hyperideal of $G/A$ and

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

$\hskip 82.51282pt\cdots,f(x_{n},A,0^{(m-2)}))$

$\hskip 42.67912pt=f(g(x_{1}^{i-1},1_{G},x_{i+1}^{n}),A,0^{(m-2)})$

$\hskip 42.67912pt\notin\delta_{q}(0),$
we get the result that $f(x_{i},A,0^{(m-2)})\in B/A$ which means $x_{i}\in B$.
This implies that $B$ is an $n$-ary $\delta(0)$-hyperideal of $G$.

(ii) Since $A$ is an $n$-ary $\delta(0)$-hyperideal of $G$, we conclude that
$A$ is contained in $\delta(0)$ by Theorem 3.10. Therefore the claim is clear
by (i). ∎

Assume that $(G_{1},f_{1},g_{1})$ and $(G_{2},f_{2},g_{2})$ are two
commutative Krasner $(m,n)$-hyperrings such that $1_{G_{1}}$ and $1_{G_{2}}$
are scalar identities of $G_{1}$ and $G_{2}$, respectively. Then $(G_{1}\times
G_{2},f_{1}\times f_{2},g_{1}\times g_{2})$ endowed with the following $m$-ary
hyperoperation $f_{1}\times f_{2}$ and $n$-ary operation $g_{1}\times g_{2}$
is a Krasner $(m,n)$-hyperring.

$f_{1}\times
f_{2}\Big{(}(a_{1},b_{1}),\cdots,(a_{m},b_{m})\Big{)}=\Big{\\{}(a,b)\ |\ \
a\in f_{1}(a_{1}^{m}),b\in f_{2}(b_{1}^{m})\Big{\\}}$

$\hskip 34.14322ptg_{1}\times
g_{2}\Big{(}(x_{1},y_{1}),\cdots,(x_{n},y_{n})\Big{)}=\Big{(}g_{1}(x_{1}^{n}),g_{2}(y_{1}^{n})\Big{)}$,
for all $a_{1}^{n},x_{1}^{m}\in G_{1}$ and $b_{1}^{n},y_{1}^{m}\in G_{2}$
[12].

######  Theorem 3.29.

Let $\delta_{1}$ and $\delta_{2}$ be hyperideal expansions of the Krasner
$(m,n)$-hyperring $G_{1}$ and $G_{2}$, respectively, such that
$\delta(0_{G_{1}\times
G_{2}})=\delta_{1}(0_{G_{1}})\times\delta_{2}(0_{G_{2}})$ and $A_{1}$ a proper
hyperideal of $G_{1}$. If $A_{1}\times G_{2}$ is an $n$-ary
$\delta(0_{G_{1}\times G_{2}})$-hyperideal of $G_{1}\times G_{2}$, then
$A_{1}$ is an $n$-ary $\delta_{1}(0_{G_{1}})$-hyperideal of $G_{1}$.

###### Proof.

Assume that $g_{1}(x_{1}^{n})\in A_{1}$ for $x_{1}^{n}\in G_{1}$. This means
that

$g_{1}\times
g_{2}\Big{(}(x_{1},1_{G_{2}}),\cdots,(x_{n},1_{G_{2}})\Big{)}=\Big{(}g_{1}(x_{1}^{n}),1_{G_{2}}\Big{)}\in
A_{1}\times G_{2}.$

Since $A_{1}\times G_{2}$ is an $n$-ary $\delta(0_{G_{1}\times
G_{2}})$-hyperideal of $G_{1}\times G_{2}$, we conclude that either
$(x_{i},1_{G_{2}})\in A_{1}\times G_{2}$ or

$g_{1}\times
g_{2}\Big{(}(x_{1},1_{G_{2}}),\cdots,(x_{i-1},1_{G_{2}}),(1_{G_{1}},1_{G_{2}})(x_{i+1},1_{G_{2}}),\cdots,(x_{n},1_{G_{2}})\Big{)}$

$\hskip
22.76228pt=\Big{(}g_{1}(x_{1}^{i-1},1_{G_{1}},x_{i+1}^{n}),1_{G_{2}}\Big{)}$

$\hskip 22.76228pt\in\delta(0_{G_{1}\times G_{2}})$
for some $i\in\\{1,\cdots,n\\}$. This implies that $x_{i}\in A_{1}$ or
$g(x_{1}^{i-1},1_{G_{1}},x_{i+1}^{n})\in\delta_{1}(0_{G_{1}})$. Consequently,
$A_{1}$ is an $n$-ary $\delta_{1}(0_{G_{1}})$-hyperideal of $G_{1}$. ∎

##  4\. $(s,n)$-absorbing $\delta(0)$-hyperideals

In this section, we present the concept of $(s,n)$-absorbing
$\delta(0)$-hyperideals as a generalization of the $n$-ary
$\delta(0)$-hyperideals and a subclass of the $(s,n)$-absorbing
$\delta$-primary hyperideals.

######  Definition 4.1.

Suppose that $\delta$ is a hyperideal expansion of $G$ and
$s\in\mathbb{Z}^{+}$. A proper hyperideal $A$ of $G$ is said to be an
$(s,n)$-absorbing $\delta(0)$-hyperideal if $g(x_{1}^{sn-s+1})\in A$ for
$x_{1}^{sn-s+1}\in G$, then either $g(x_{1}^{(s-1)n-s+2})\in A$ or a
$g$-product of $(s-1)n-s+2$ of $x_{i}^{,}$s, other than
$g(x_{1}^{(s-1)n-s+2})$, is in $\delta(0)$.

Obviously, every $n$-ary $\delta(0)$-hyperideal is an $(s,n)$-absorbing
$\delta(0)$-hyperideal and every $(s,n)$-absorbing $\delta(0)$-hyperideal is
an $(s,n)$-absorbing $\delta$-primary hyperideal. However, these are different
notions.

######  Example 4.2.

Let us consider the hyperring $(G=[0,a],+,\cdot)$ where $0<a\leq 1$, $``+"$ is
defined by

$p+q=\begin{cases}$[0,p]$,&\text{if $p=q$}\\\ \\{\max\\{p,q\\}\\}&\text{if
$p\neq q$}\end{cases}.$

and $``\cdot"$ is the usual multiplication on real numbers. Then $([0,a],f,g)$
is a Krasner $(m,n)$-hyperrin such that $f(p_{1}^{m})=p_{1}+\cdots+p_{m}$ and
$g(q_{1}^{n})=q_{1}\cdots q_{n}$ for all $p_{1}^{m},q_{1}^{n}\in[0,a]$. In the
hyperring $0$ is an $(s,n)$-absorbing $\delta_{1}$-hyperideal. Now, let $a=1$.
Then $A=[0,0.5]$ is a $(2,2)$-absorbing $\delta_{1}$-primary hyperideal.
However, it is not $(2,2)$-absorbing $\delta_{1}(0)$-hyperideals. Take
$0.6,0.9\in G$. We have $g(0.6,0.9,0.9)\in A$, but neither
$g(0.9^{(2)},1_{G})$ and $g(0.6,0.9,1,1_{G})$ are in $A$ nor in
$\delta_{1}(0)$.

######  Theorem 4.3.

Let $A$ be an $n$-ary $\delta(0)$-hyperideal of $G$. Then $A$ is a
$(2,n)$-absorbing $\delta(0)$-hyperideal of $G$.

###### Proof.

Let $g(x_{1}^{2n-1})\in A$ for $x_{1}^{2n-1}\in G$. Since $A$ is an $n$-ary
$\delta(0)$-hyperideal of $G$, we conclude that $g(x_{1}^{n})\in A$ or
$g(x_{n+1}^{2n-1})\in\delta(0)$. In the second case, we get
$g(x_{i},x_{n+1}^{2n-1})\in\delta(0)$, for $i\in\\{1,\cdots,n\\}$ because
$\delta(0)$ is a hyperideal of $G$. Thus $A$ is $(2,n)$-absorbing
$\delta(0)$-hyperideal of $G$. ∎

######  Theorem 4.4.

Let $A$ be an $(s,n)$-absorbing $\delta(0)$-hyperideal of $G$. Then $A$ is a
$(k,n)$-absorbing $\delta(0)$-hyperideal for $k>n$.

###### Proof.

Let $g(x_{1}^{(s+1)n-(s+1)+1})\in A$ for $x_{1}^{(s+1)n-(s+1)+1}\in G$. Let us
consider $g(x_{1}^{n+2})=x$. Since $A$ is $(s,n)$-absorbing
$\delta(0)$-hyperideal, we get the result that
$g(x,\cdots,x_{(s+1)n-(s+1)+1})\in A$ or a $g$-product of $sn-s+1$ of the
$x_{i}^{,}$s except $g(x,\cdots,x_{(s+1)n-(s+1)+1})$ is in $\delta(0)$. This
means that $g(x_{i},x_{n+3}^{(s+1)n-(s+1)+1})\in\delta(0)$ for
$i\in\\{1,\cdots n+2\\}$ which means $A$ is an $(s+1,n)$-absorbing
$\delta(0)$-hyperideal. This shows that $A$ is a $(k,n)$-absorbing
$\delta(0)$-hyperideal for $k>n$. ∎

##  5\. weakly $(s,n)$-absorbing $\delta(0)$-hyperideals

In this section, we aim to discuss some of the fundamental results concerning
weakly $(s,n)$-absorbing $\delta(0)$-hyperideals as an expansion of the
$(s,n)$-absorbing $\delta(0)$-hyperideals.

######  Definition 5.1.

Assume that $\delta$ is a hyperideal expansion of $G$ and
$s\in\mathbb{Z}^{+}$. A proper hyperideal $A$ of $G$ is called a weakly
$(s,n)$-absorbing $\delta(0)$-hyperideal if $x_{1}^{sn-t+1}\in G$ and $0\neq
g(x_{1}^{sn-s+1})\in A$ imply either $g(x_{1}^{(s-1)n-s+2})\in A$ or a
$g$-product of $(s-1)n-s+2$ of $x_{i}^{,}$s, other than
$g(x_{1}^{(s-1)n-s+2})$, is in $\delta(0)$.

######  Example 5.2.

Cosider Krasner $(m,n)$-hyperring $(G/B,f,g)$ where
$G={\mathbb{Z}_{3}[X,Y,Z]}$, $B=\langle X^{3}Y^{3}Z^{3}\rangle$ and $f,g$ are
ordinary addition and ordinary multiplication, respectively. In the hyperring,
$A=rad(B/B)$ is not a weakly $(1,3)$-absorbing
$\delta_{1}(0_{G/B})$-hyperideal of $G/B$ because $0\neq
2XYZ+B=(2X+B)(Y+B)(Z+B)\in A$ but none of the elements $(2X+B),(Y+B)$ and
$(Z+B)$ are not in $\delta_{1}(0_{G/B})$.

######  Proposition 5.3.

Assume that $\delta$ and $\gamma$ are hyperideal expansions of $G$ and $A$ is
an $(s,n)$-absorbing $\gamma(0)$-hyperideal of $G$. Then $A$ is a weakly
$(s,n)$-absorbing $\gamma o\delta(0)$-hyperideal if and only if $A$ is an
$(s,n)$-absorbing $\gamma o\delta(0)$-hyperideal.

###### Proof.

($\Longrightarrow$) Let $A$ be a weakly $(s,n)$-absorbing $\gamma
o\delta(0)$-hyperideal of $G$. Suppose that $g(x_{1}^{sn-s+1})\in A$ for
$x_{1}^{sn-s+1}\in G$. If $g(x_{1}^{sn-s+1})\neq 0$, we have
$g(x_{1}^{(s-1)n-s+2})\in A$ or a $g$-product of $(s-1)n-s+2$ of $x_{i}^{,}$s,
other than $g(x_{1}^{(s-1)n-s+2})$, is in $\gamma o\delta(0)$ as $A$ is a
weakly $(s,n)$-absorbing $\gamma\circ\delta(0)$-hyperideal of $G$. Let us
assume that $g(x_{1}^{sn-s+1})=0\in A$ such that $g(x_{1}^{(s-1)n-s+2})\notin
A$. Since $A$ is an $(s,n)$-absorbing $\gamma(0)$-hyperideal of $G$, we get
the result that a $g$-product of $(s-1)n-s+2$ of $x_{i}^{,}$s, other than
$g(x_{1}^{(s-1)n-s+2})$, is in $\gamma(0)$. Since $\gamma(0)\subseteq\gamma
o\delta(0)$, we conclude that the $g$-product of $(s-1)n-s+2$ of $x_{i}^{,}$s
is in $\gamma o\delta(0)$. Thus $A$ is an $(s,n)$-absorbing $\gamma
o\delta(0)$-hyperideal.

($\Longleftarrow$) It is straightforward. ∎

We say that a proper hyperideal $A$ of $G$ is called a strongly weakly
$(s,n)$-absorbing $\delta(0)$-hyperideal if $0\neq g(A_{1}^{sn-s+1})\subseteq
A$ for any hyperideal $A_{1}^{sn-n+1}$ of $G$ implies
$g(A_{1}^{(s-1)n-s+2})\subseteq A$ or a $g$-product of $(s-1)n-s+2$ of
$A_{i}^{,}$s, other than $g(A_{1}^{(s-1)n-s+2})$, is a subset of $\delta(0)$.
Moreover, we say that $(x_{1}^{s(n-1)+1})$ is an $(s,n)$-$\delta(0)$-zero of a
weakly $(s,n)$-absorbing $\delta(0)$-hyperideal $A$ if
$g(x_{1}^{s(n-1)+1})=0$, $g(x_{1}^{(s-1)n-s+2})\notin A$ and all $g$-products
of $(s-1)n-s+2$ of $x_{i}^{,}$s, except $g(x_{1}^{(s-1)n-s+2})$ are not in
$\delta(0)$. Note that the notation
$x_{1},\cdots,\widehat{x_{i}},\cdots,x_{sn-s+1}$ indicates that $i$-th term is
excluded from the sequence $x_{1}^{sn-s+1}$.

######  Theorem 5.4.

Let $(x_{1}^{s(n-1)+1})$ be an $(s,n)$-$\delta(0)$-zero of a strongly weakly
$(s,n)$-absorbing $\delta(0)$-hyperideal $A$. Then for
$u\in\\{1,\cdots,(s-1)n-s+2\\}$ and
$i_{1},\cdots,i_{u}\in\\{1,\cdots,s(n-1)+1\\}$,
$g(x_{1},\cdots,\widehat{x_{i_{1}}},\cdots,\widehat{x_{i_{2}}},\cdots,\widehat{x_{i_{u}}},\cdots,x_{s(n-1)+1},A^{(u)})=0$

###### Proof.

We use the induction on $u$. Suppose that $u=1$. Now we show that
$g(x_{1},\cdots,\widehat{x_{i_{1}}}\cdots,x_{s(n-1)+1},A)=0$. Let
$g(x_{1},\cdots,\widehat{x_{i_{1}}}\cdots,x_{s(n-1)+1},A)\neq 0$. Without loss
of generality, we may assume that $g(x_{2}^{s(n-1)+1},A)\neq 0$. Then we
obtain $g(x_{2}^{s(n-1)+1},a)\neq 0$ for some $a\in A$. Since $0\neq
g(x_{2}^{sn-s+1},f(x_{1},a,0^{(m-2)}))=f(g(x_{1}^{s(n-1)+1}),g(x_{2}^{s(n-1)+1},a),0^{(m-2)})\subseteq
A$ and $A$ is a strongly weakly $(s,n)$-absorbing $\delta(0)$-hyperideal of
$G$, we get the result that $\delta(0)$ contains a $g$-product containing $a$
of the terms $(s-1)n-s+2$ of $x_{i}^{,}$s. Then we may assume that
$f(g(x_{1},x_{3}^{s(n-1)+1}),g(a,x_{3}^{s(n-1)+1}),0^{(m-2)})=g(x_{3}^{s(n-1)+1},f(x_{1},a,0^{(m-2)}))\subseteq\delta(0)$.
Thus $g(x_{1},x_{3}^{s(n-1)+1})\in\delta(0)$ as $g(a,x_{3}^{s(n-1)+1})\in
A\subseteq\delta(0)$. This is a contradiction. Therefore,
$g(x_{1},\cdots,\widehat{x_{i_{1}}}\cdots,x_{s(n-1)+1},A)=0$. We suppose that
the claim holds for all positive integers which are less than $u>1$ and show
that it holds for $u$. Suppose that
$g(x_{1},\cdots,\widehat{x_{i_{1}}},\cdots,\widehat{x_{i_{2}}},\cdots,\widehat{x_{i_{u}}},\cdots,x_{s(n-1)+1},A^{(u)})\neq
0$. Let us assume that $g(x_{u+1}^{s(n-1)+1},A^{(u)})\neq 0$, namely, we omit
$x_{1}^{u}$. Hence we get $0\neq g(x_{u+1}^{s(n-1)+1},a_{1}^{u})\in A$ for
some $x_{1}^{u}\in A$. Then $0\neq
g(x_{u+1}^{s(n-1)+1},f(x_{1},a_{1},0^{(m-2)}),\cdots,f(x_{u},a_{u},0^{(m-2)}))\subseteq
A$, by induction hypothesis. Thus we obtain

$g(x_{u+1}^{s(n-1)+1},f(x_{1},a_{1},0^{(m-2)}),\cdots,\widehat{f(x_{1},a_{1},0^{(m-2)})_{i_{1}}},\cdots,\widehat{f(x_{2},a_{2},0^{(m-2)})_{i_{2}}},$

$\cdots,\widehat{f(x_{n-1},a_{n-1},0^{(m-2)})_{i_{n-1}}},\cdots,\widehat{f(x_{u},a_{u},0^{(m-2)})})\subseteq\delta(0)$
or

$g(x_{u+1},\cdots,\widehat{x_{i_{u+1}}},\cdots,\widehat{x_{i_{u+2}}},\cdots,\widehat{x_{i_{u+(n-1)}}},\cdots,x_{s(n-1)+1},f(x_{1},a_{1},0^{(m-2)}),$

$\cdots,f(x_{u},a_{u},0^{(m-2)})\subseteq\delta(0)$
for some $i\in\\{1,\cdots,u\\}$. Therefore we get the result that for some
$i\in\\{1,\cdots,u\\}$, $g(x_{u+1}^{s(n-1)+1},\cdots,x_{n}^{t})\in\delta(0)$
or $g(x_{u+n}^{s(n-1)+1},\cdots,x_{1}^{u})\in\delta(0)$ or

$g(x_{u+1},\cdots,\widehat{x_{i_{u+1}}},\cdots,\widehat{x_{i_{u+2}}},\cdots,\widehat{x_{i_{u+s}}},\cdots,x_{s(n-1)+1},f(x_{1},a_{1},0^{(m-2)}),$

$\cdots,\widehat{f(x_{1},a_{1},0^{(m-2)})_{i_{u+(s+1)}}},\widehat{f(x_{n-1-s},a_{n-1-s},0^{(m-2)})_{i_{u+(n-1-s)}}},$

$\cdots,f(x_{u},a_{u},0^{(m-2)}))\subseteq\delta(0),$
a contradiction because $(x_{1}^{s(n-1)+1})$ is an $(s,n)$-$\delta(0)$-zero of
$A$. Consequently,
$g(x_{1},\cdots,\widehat{x_{i_{1}}},\cdots,\widehat{x_{i_{2}}},\cdots,\widehat{x_{i_{u}}},\cdots,x_{s(n-1)+1},A^{(u)})=0$.
∎

Now, we give a copy of Nakayama,s lemma for strongly weakly $(s,n)$-absorbing
$\delta(0)$-hyperideals.

######  Theorem 5.5.

Assume that $A$ is a strongly weakly $(s,n)$-absorbing $\delta(0)$-hyperideal
but is not $(s,n)$-absorbing $\delta(0)$-hyperideal. Then the following hold:

  * (i)

$g(A^{(s(n-1)+1)})=0$

  * (ii)

If $M=k(A,1_{G}^{(n-2)},M)$ for some $(m,n)$-hypermodule $M$ over $G$ , then
$M=0$.

###### Proof.

(i) Since $A$ is not $(s,n)$-absorbing $\delta(0)$-hyperideal, we obtain an
$(s,n)$-$\delta(0)$-zero $(x_{1}^{s(n-1)+1})$ of $A$. Let us assume that
$g(A^{(s(n-1)+1)})\neq 0$. Hence we have $g(a_{1}^{s(n-1)+1})\neq 0$ for some
$a_{1}^{s(n-1)+1}\in A$. Therefore we get the result that

$\hskip 28.45274pt0\neq
g(f(x_{1},a_{1},0^{(m-2)}),\cdots,f(x_{s(n-1)+1},a_{s(n-1)+1},0^{(m-2)}))\subseteq
A$
by Theorem 5.4. By the hypothesis, we conclude that
$g(f(x_{1},a_{1},0^{(m-2)})^{(s-1)n-s+2})\subseteq A$ or $\delta(0)$ contains
a $g$-product of $(s-1)n-s+2$ of $f(x_{i},a_{i},0^{(m-2)})^{,}$s other that
$g(f(x_{1},a_{1},0^{(m-2)})^{(s-1)n-s+2})$. In the first possibilty, we
conclude that $A$ contains
$f(\underbrace{g(x_{1}^{(s-1)n-s+2}),g(x_{1}^{u_{1}},a_{1}^{u_{2}}),g(a_{1}^{(s-1)n-s+2})}_{u},0^{(m-u)})$
such that $u_{1}+u_{2}=(s-1)n-s+2$. It follows that $g(x_{1}^{(s-1)n-s+2})\in
A$. But this is a contradiction since $(x_{1}^{s(n-1)+1})$ is an
$(s,n)$-$\delta(0)$-zero of $A$. In the second possibility, we may assume that
$g(f(x_{2},a_{2},0^{(m-2)}),\cdots,f(x_{(s-1)n-s+3},a_{(s-1)n-s+3},0^{(m-2)}))$
is contained in $\delta(0)$. Thus we get the result that
$f(\underbrace{g(x_{2}^{(s-1)n-s+3}),g(x_{1}^{u_{1}},a_{2}^{u_{2}}),g(a_{1}^{(s-1)n-s+2})}_{u},0^{(m-u)})$
is contained in $\delta(0)$ such that $u_{1}+u_{2}=(s-1)n-s+2$ which implies
$g(x_{2}^{(s-1)n-s+3})\in\delta(0)$. It contradicts the fact that
$(x_{1}^{s(n-1)+1})$ is an $(s,n)$-$\delta(0)$-zero of $A$. Consequently,
$g(A^{(s(n-1)+1)})=0$.

(ii) Let $M$ be a $(m,n)$-hypermodule over $G$ such that
$M=k(A,1_{G}^{(n-1)},M)$. By (i), we have
$k(g(A^{(s(n-1)+1)}),1_{G}^{(n-1)},M)=0$. This implies that

$\hskip 71.13188pt0=k(g(A^{(s(n-1)+1)}),1_{G}^{(n-1)},M)$

$\hskip 79.6678pt=k(g(A^{(s(n-1))},1_{G}),A,1_{G}^{(n-2)},M)$

$\hskip 79.6678pt=k(g(A^{(s(n-1)}),1_{G}),1_{G}^{(n-1)},k(A,1_{G}^{(n-1)},M))$

$\hskip 79.6678pt=k(g(A^{(s(n-1))},1_{G}),1_{G}^{(n-1)},M)$

$\hskip 79.6678pt=\cdots$

$\hskip 79.6678pt=k(A,1_{G}^{(n-1)},k(A,1_{G}^{(n-1)},M))$

$\hskip 79.6678pt=k(A,1_{G}^{(n-1)},M)$

$\hskip 79.6678pt=M$ ∎

## References

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