#  $J$-prime hyperideals and their generalizations

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

Let $R$ be a multiplicative hyperring with identity. In this paper, we define
the concept of $J$-prime hyperideals which is a generalization of
$n$-hyperideals and we will show some properties of them. Then, we extend the
notion of $J$-prime to quasi $J$-prime and 2-absorbing $J$-prime hyperideals.
Various characterizations of them are provided.

###### Key words and phrases:

$J$-prime hyperideal, Quasi $J$-prime hyperideal, 2-absorbing $J$-prime
hyperideal.

###### 2010 Mathematics Subject Classification:

20N20, 16Y99

##  1\. Introduction

The hyperstructure theory was introduced by Marty in 1934, at the 8th Congress
of Scandinavian Mathematicians [22], when he defined the hypergroups and began
to investigate their properties with applications to groups, algebraic
functions and rational fractions. Later on, many researchers have worked on
this new field of modern algebra and developed it. The multiplicative
hyperring, as one important class of hyperrings, was introduced by Rota in
1982 [30] and was subsequently investigated by many authors [23, 24, 25, 27].
In the hyperring, the multiplication is a hyperoperation, while the addition
is an operation. In 1990, the strongly distributive multiplicative hyperrings
were characterized by Rota [29]. The polynomials over multiplicative
hyperrings were studied by Procesi and Rota in [26]. Ameri and Kordi
introduced the notions of clean multiplicative hyperring and regular
multiplicative hyperring, as two generalizations of classical rings, in [3]
and [4]. The concept of derivation on multiplicative hyperrings was introduced
by Ardekani and Davvaz in [19]. Ameri and et. al. [5] have studied notion of
hyperring of fractions generated by a multiplicative hyperring. Soltani and
et. al. introduced zero-divisor graphs of a commutative multiplicative
hyperring, as a generalization of commutative rings [31]. The codes over
multiplicative hyperrings were studied by Akbiyik [1].

In the theory of rings, the key role of the notion of prime ideal as a
generalization of the notion of prime number in the ring $\mathbb{Z}$ is
undeniable. The notion of primeness of hyperideal in a multiplicative
hyperring was conceptualized by Procesi and Rota in [27]. The notions of prime
and primary hyperideals in multiplicative hyperrings were fully studied by
Dasgupta in [14]. The concept of $S$-prime hyperideals in multiplicative
hyperrings which is a generalization of prime hyperideals was studied in [17].
Badawi [11] introduced and studied a generalization of prime ideals called
2-absorbing ideals and this notion is further generalized by Anderson and
Badawi [9, 10]. In [16], Ghiasvand introduced the concept of 2-absorbing
hyperideal in a multiplicative hyperring which is a generalisation of prime
hyperideals. Several authors have extended and generalized this concept in
several ways [6, 18, 20, 21]. In [32], Ulucak defined the notion of
$\delta$-primary hyperideals in multiplicative hyperrings, which unifies the
prime and primary hyperideals under one frame. Recently, we introduce the
notions of $n$-hyperideals and $r$-hyperideal in a multiplicative hyperring
[8].

Our aim in this paper is to introduce and study the concept of $J$-prime
hyperideals which is a generalization of $n$-hyperideals. Furthermore, we
defined two generalizations of them in a multiplicative hyperring. The paper
is orgnized as follows. In Section 2, we have given some basic definitions and
results of multiplicative hyperrings which we need to develop our paper. In
Section 3, we introduce the concept of $J$-prime hyperideals and discuss their
relations with some other types of hyperideals. Furthermore, we investigate
the behaviour of $J$-prime hyperideals under a good homomorphism. In Section
4, we study a generalization of the $J$-prime hyperideals which is called
quasi $J$-prime hyperideals . We give a characterization of local
multiplicative hyperrings in terms of quasi $J$-prime hyperideals. In Section
5, we extend the notion of $J$-prime to 2-absorbing $J$-prime hyperideals and
give some properties of them.

##  2\. Preliminaries

Recall first the basic terms and definitions from the hyperring theory. A
hyperoperation on a non-empty set $G$ is a map $"\circ":G\times
G\longrightarrow P^{*}(G)$, where $P^{*}(G)$ is the set of all the nonempty
subsets of $G$. An algebraic system $(G,\circ)$ is called a hypergroupoid. A
hypergroupoid $(G,\circ)$ is called a hypergroup if it satisfies the
following:

(1) $a\circ(b\circ c)=(a\circ b)\circ c$, for all $a,b,c\in G$.

(2) $a\circ G=G\circ a=G$, for all $a\in G$.

A hypergroupoid with the associative hyperoperation is called a
semihypergroup.

A non-empty set $R$ with a operation + and a hyperoperation $\circ$ is called
a multiplicative hyperring if it satisfies the following:

(1)$(R,+)$ is an abelian group;

(2)$(R,\circ)$ is a semihypergroup;

(3) for all $a,b,c\in R$, we have $a\circ(b+c)\subseteq a\circ b+a\circ c$ and
$(b+c)\circ a\subseteq b\circ a+c\circ a$;

(4) for all $a,b\in R$, we have $a\circ(-b)=(-a)\circ b=-(a\circ b)$.

Let $A$ and $B$ be two nonempty subsets of $R$ and $r\in R$. Then we define

$A\circ B=\bigcup_{x\in A,\ y\in B}x\circ y,\ \ \ \ A\circ r=A\circ\\{r\\}$

A non empty subset $I$ of a multiplicative hyperring $R$ is a hyperideal if

  * (i)

If $a,b\in I$, then $a-b\in I$;

  * (iii)

If $x\in I$ and $r\in R$, then $r\circ x\subseteq I$.

######  Definition 2.1.

[15] Let $(R_{1},+_{1},\circ_{1})$ and $(R_{2},+_{2},\circ_{2})$ be two
multiplicative hyperrings. A mapping from $R_{1}$ into $R_{2}$ is said to be a
good homomorphism if for all $x,y\in R_{1}$,
$\phi(x+_{1}y)=\phi(x)+_{2}\phi(y)$ and
$\phi(x\circ_{1}y)=\phi(x)\circ_{2}\phi(y)$.

######  Definition 2.2.

[14] A proper hyperideal $P$ of $R$ is called a prime hyperideal if $x\circ
y\subseteq P$ for $x,y\in R$ implies that $x\in P$ or $y\in P$. The
intersection of all prime hyperideals of $R$ containing $I$ is called the
prime radical of $I$, being denoted by $\sqrt{I}$. If the multiplicative
hyperring $R$ does not have any prime hyperideal containing $I$, we define
$\sqrt{I}=R$.

######  Definition 2.3.

[5] A proper hyperideal $I$ of $R$ is maximal in $R$ if for any hyperideal $J$
of $R$ with $I\subseteq J\subseteq R$ then $J=R$. Also, we say that $R$ is a
local multiplicative hyperring, if it has just one maximal hyperideal. For a
hyperring $R$ we deﬁne the Jacobson radical $J(R)$ of $R$ as the intersection
of all maximal hyperideals of $R$.
Moreover, if $I$ is a proper hyperideal of $R$, then the Jacobson radical
$J(I)$ is defined as the intersection of all maximal hyperideals of $R$
containing $I$.

Let C be the class of all finite products of elements of $R$ i.e. ${\bf
C}=\\{r_{1}\circ r_{2}\circ...\circ r_{n}\ :\ r_{i}\in
R,n\in\mathbb{N}\\}\subseteq P^{\ast}(R)$. A hyperideal $I$ of $R$ is said to
be a C-hyperideal of $R$ if, for any $A\in{\bf C},A\cap I\neq\varnothing$
implies $A\subseteq I$. Let I be a hyperideal of $R$. Then,
$D\subseteq\sqrt{I}$ where $D=\\{r\in R:r^{n}\subseteq I\ for\ some\
n\in\mathbb{N}\\}$. The equality holds when $I$ is a C-hyperideal of $R$([14],
proposition 3.2). In this paper, we assume that all hyperideals are
C-hyperideal.

######  Definition 2.4.

[5] Let $R$ be commutative multiplicative hyperring and $1$ be an identity (i.
e., for all $a\in R$, $a\in a\circ 1$). An element $x\in R$ in is called unit,
if there exists $y\in R$, such that $1\in x\circ y$. Denote the set of all
unit elements in $R$ by $U(R)$.

######  Definition 2.5.

[2] An element $x\in R$ is said to be a zero divizor, if there exists $0\neq
y\in R$ such that $\\{0\\}=x\circ y$. The set of all zero divizor elements in
$R$ is denoted by $Z(R)$.

######  Definition 2.6.

Let $I,J$ be two hyperideals of $R$ and $x\in R$. Then define:

$(I:a)=\\{r\in R:r\circ a\subseteq I\\}$ $ann(x)=\\{y\in R:x\circ
y=\\{0\\}\\}$

##  3\. $J$-prime hyperideals

######  Definition 3.1.

Let $I$ be a proper hyperideal of $R$. We call $I$ a $J$-prime hyperideal if
whenever $x,y\in R$ with $x\circ y\subseteq I$ , then either $x\in J(R)$ or
$y\in I$.

######  Example 3.2.

Let $\mathbb{Z}$ be the ring of integers. For all $a,b\in\mathbb{Z}$, we
define the hyperoperation $a\circ b=\\{a.x.b\ |\ x\in A\\}$ where
$A=\\{2,3\\}$. Then $(\mathbb{Z},+,\circ)$ is a multiplicative hyperring. In
the hyperring, every principal hyperideal generated by prime integer is a
$J$-prime hyperideal.

######  Example 3.3.

In Example 3.2, let $A=\\{14,21\\}$. Then the principal hyperideal $\langle
7\rangle$ is not a $J$-prime hyperideal. Because, $1\circ
1=\\{14,21\\}\subseteq\langle 7\rangle$, but neither $1\in\langle 7\rangle$
nor $1\in J(\mathbb{Z})$.

######  Theorem 3.4.

Let $I$ be a $J$-prime hyperideal of $R$. Then $I$ is in $J(R)$.

###### Proof.

Assume that $I$ is a $J$-prime hyperideal of $R$ such that it is not in
$J(R)$. Let $x\in I$ but $x\notin J(R)$. Since $I$ is a C-hyperideal and $x\in
x\circ 1$, then $x\circ 1\subseteq I$. Since $I$ is a $J$-prime hyperideal of
$R$ and $x\notin J(R)$, then $1\in I$ which is a contradiction. Thus, $I$ is
in $J(R)$. ∎

######  Theorem 3.5.

For a ring R, the following statements are equivalent:

  * (i)

$R$ is a local hyperring.

  * (ii)

Every proper hyperideal of $R$ is $J$-prime.

  * (iii)

Every proper principal hyperideal of $R$ is $J$-prime.

###### Proof.

$(i)\Longrightarrow(ii)$ Suppose that $I$ is a proper hyperideal of $R$ and
$M$ is the only maximal hyperideal of $R$. So $J(R)=M$. Let $x\circ y\subseteq
I$ for some $x,y\in R$ such that $x\notin M$. Therefore, $x\in U(R)$. Hence,
we have $y\in 1\circ y\subseteq(x^{-1}\circ x)\circ y=x^{-1}\circ(x\circ
y)\subseteq I$. Thus $I$ is a $J$-prime hyperideal of $R$.

$(ii)\Longrightarrow(iii)$ Obvious.

$(iii)\Longrightarrow(i)$ Let every proper principal hyperideal of $R$ is
$J$-prime. Suppose that the hyperideal $M$ of $R$ is maximal. Let $x\in M$.
Since the principal hyperideal $\langle x\rangle$ is $J$-prime and $x\circ
1\subseteq\langle x\rangle$, then we get $x\in J(R)$ or $1\in\langle
x\rangle$. In the second case, we have a contradiction. Then $x\in J(R)$ which
means $J(R)=M$. Thus, $R$ is a local hyperring. ∎

######  Lemma 3.6.

Each maximal hyperideal $M$ of a multiplicative hyperring $R$ is a prime
hyperideal.

###### Proof.

Let $M$ be a maximal hyperideal of $R$. Suppose that $x\circ y\subseteq M$ for
some $x,y\in R$ but $y\notin M$. So, $\langle y,M\rangle=R$ which means there
exists $m\in M$ such that $1\in\langle y\rangle+m$. Then there exist $r_{i}\in
R$ such that $1\in\sum_{i=1}^{n}r_{i}\circ y+m$. Hence, we get $x\in x\circ
1\subseteq x\circ(\sum_{i=1}^{n}r_{i}\circ y+m)\subseteq(\sum_{i=1}^{n}x\circ
r_{i}\circ y)+x\circ m\subseteq M$. Then $x\in M$ as needed. ∎

Recall from [8] that a proper hyperideal $I$ of $R$ is said to be an
$n$-hyperideal if $x\circ y\subseteq I$ implies $x\in\sqrt{0}$ or $y\in I$ for
any $x,y\in R$.

######  Theorem 3.7.

If $I$ is an n-hyperideal of $R$, then it is $J$-prime.

###### Proof.

Let $x\circ y\subseteq I$ for some $x,y\in R$ such that $x\notin J(R)$. By
Lemma 3.6, we have $\sqrt{0}\subseteq J(R)$. Therefore, $x\notin\sqrt{0}$.
Since $I$ is an n-hyperideal of $R$, then $y\in I$. Thus, $I$ is a $J$-prime
hyperideal of $R$. ∎

######  Theorem 3.8.

Let $R$ be a local multiplicative hyperring such that $\sqrt{0}\subsetneqq
J(R)$. Then $J(R)$ is a $J$-prime hyperideal of $R$ which is not an
n-hyperideal.

###### Proof.

Let $R$ be a local multiplicative hyperring. By Theorem 3.5, the hyperideal
$J(R)$ of $R$ is $J$-prime. Let $x\in J(R)-\sqrt{0}$. Then we get $x\in x\circ
1\subseteq J(R)$ such that $x\notin\sqrt{0}$ and $1\notin J(R)$. This implies
that $J(R)$ is not an n-hyperideal of $R$. ∎

Recall from [8] that a proper hyperideal $I$ of $R$ is said to be a
$r$-hyperideal if for all $x,y\in R$, $x\circ y\subseteq I$ and
$ann(x)=\\{0\\}$, then $y\in I$ .

######  Theorem 3.9.

Let $R$ be a multiplicative hyperring with $Z(R)\subseteq J(R)$. If $I$ is a
r-hyperideal of $R$, then $I$ is a $J$-prime hyperideal.

###### Proof.

Let $I$ be a r-hyperideal of $R$. Suppose that $x\circ y\subseteq I$ for some
$x,y\in R$ such that $x\notin J(R)$. Since $Z(R)\subseteq J(R)$, then $x\notin
Z(R)$ which implies $ann(x)=\\{0\\}$. Since $I$ is a r-hyperideal of $R$, we
get $y\in I$. Thus, $I$ is a $J$-prime hyperideal of $R$. ∎

######  Theorem 3.10.

Let $\\{I_{i}\\}_{i\in\Delta}$ be a nonmepty set of $J$-prime hyperideals of
$R$. Then $\bigcap_{i\in\Delta}I_{i}$ is a $J$-prime hyperideal of $R$.

###### Proof.

Let $x\circ y\subseteq\bigcap_{i\in\Delta}I_{i}$ for some $x,y\in R$ such that
$x\notin J(R)$. Then $x\circ y\subseteq I_{i}$ for every $i\in\Delta$. Since
$I_{i}$ is a $J$-prime hyperideal of $R$, we get the result that $y\in I_{i}$
and so $y\in\bigcap_{i\in\Delta}I_{i}$. ∎

######  Theorem 3.11.

Let $I$ be a proper hyperideal of $R$.Then the following statements are
equivalent:

  * (i)

$I$ is a $J$-prime hyperideal of $R$.

  * (ii)

$I=(I:x)$ for every $x\notin J(R)$.

  * (iii)

$I_{1}\circ I_{2}\subseteq I$ for some hyperideals $I_{1}$ and $I_{2}$ of $R$
implies that $I_{1}\subseteq J(R)$ or $I_{2}\subseteq I$.

###### Proof.

$(i)\Longrightarrow(ii)$ Let $I$ be a $J$-prime hyperideal of $R$. It is clear
that $I\subseteq(I:x)$ for all $x\in R$. Assume that $y\in(I:x)$ such that
$x\notin J(R)$. This means $x\circ y\subseteq I$. Since $I$ is a $J$-prime
hyperideal of $R$ and $x\notin J(R)$, then $y\in I$. Thus, we have $I=(I:x)$.

$(ii)\Longrightarrow(iii)$ Let $I_{1}\circ I_{2}\subseteq I$ for some
hyperideals $I_{1}$ and $I_{2}$ of $R$ such that $I_{1}\nsubseteq J(R)$.
Therefore, we get $x\in I_{1}$ such that $x\notin J(R)$. Hence, $x\circ
I_{2}\subseteq I$ which means $I_{2}\subseteq(I:x)$. Since $I=(I:x)$ for every
$x\notin J(R)$, then $I_{2}\subseteq I$.

$(iii)\Longrightarrow(i)$ Let $x\circ y\subseteq I$ for some $x,y\in R$ such
that $x\notin J(R)$. By Proposition 2.15 in [14], we have $\langle
x\rangle\circ\langle y\rangle\subseteq\langle x\circ y\rangle\subseteq I$ but
$\langle x\rangle\nsubseteq J(R)$. Then we get $\langle y\rangle\subseteq I$
which implies $y\in I$. Thus, $I$ is a $J$-prime hyperideal of $R$. ∎

######  Theorem 3.12.

Let $I$ be a proper hyperideal of $R$. Then $I$ is a $J$-prime hyperideal of
$R$ if and only if $(I:y)\subseteq J(R)$ for every $y\notin I$.

###### Proof.

$\Longrightarrow$ Let $x\in(I:y)$ such that $y\notin I$. So, $x\circ
y\subseteq I$. Since $I$ is a $J$-prime hyperideal of $R$, then $x\in J(R)$.

$\Longleftarrow$ Let $x\circ y\subseteq I$ for some $x,y\in R$ such that
$x\notin J(R)$. If $y\notin I$, then $x\in(I:y)\subseteq J(R)$, by the
hypothesis. This is a contradiction. Therefore, $y\in I$. Thus, $I$ is a
$J$-prime hyperideal of $R$ ∎

######  Theorem 3.13.

Let $I$ be a hyperideal of $R$ and let $T$ be a nonempty subset of $R$ such
that $T\nsubseteq I$. If $I$ is a $J$-prime hyperideal of $R$, then $(I:T)$ is
a $J$-prime hyperideal of $R$.

###### Proof.

Let $(I:T)=R$. Then $1\in(I:T)$ which means $T\subseteq I$. This is a
contradiction. Hence, $(I:T)$ is a proper hyperideal of $R$. Suppose that
$x\circ y\subseteq(I:T)$ for some $x,y\in R$ such that $x\notin J(R)$. This
implies that $x\circ y\circ t\subseteq I$ for all $t\in T$. Then we get
$y\circ t\subseteq I$ for all $t\in T$ as $I$ is a $J$-prime hyperideal of
$R$. Thus $y\in(I:T)$. ∎

######  Theorem 3.14.

Suppose that $I$ is a $J$-prime hyperideal of $R$ such that there is no
$J$-prime hyperideal which contains $I$ properly. Then $I$ is a prime
hyperideal.

###### Proof.

Suppose that $I$ is a $J$-prime hyperideal of $R$ such that there is no
$J$-prime hyperideal which contains $I$ properly. Let $x\circ y\subseteq I$
for some $x,y\in R$ such that $x\notin I$. By Theorem 3.13, $(I:x)$ is a
$J$-prime hyperideal of $R$. Since $I\subseteq(I:x)$, we conclude that
$y\in(I:x)=I$, by the hypothesis. Thus, $I$ is a prime hyperideal. ∎

The next Theorem shows that the inverse of Theorem 3.14 is true if $I=J(R)$.

######  Theorem 3.15.

Let the hyperideal $J(R)$ of $R$ be prime. Then $J(R)$ is a $J$-prime
hyperideal of $R$ such that there is no $J$-prime hyperideal which contains
$J(R)$ properly.

###### Proof.

Suppose that $I=J(R)$. Let $x\circ y\subseteq I$ for some $x,y\in R$ such that
$x\notin J(R)$. Since $I$ is a prime hyperideal of $R$, then $y\in I=J(R)$
which means the hyperideal $J(R)$ of $R$ is $J$-prime. By Theorem 3.4, we
conclude that there is no $J$-prime hyperideal which contains $I$ properly. ∎

######  Theorem 3.16.

Let $H$ be a hyperideal of $R$ such that $H\nsubseteq J(R)$. Then

  * (i)

If the hyperideals $A_{1}$ and $A_{2}$ of $R$ are $J$-prime such that
$A_{1}\circ H=A_{2}\circ H$, then $A_{1}=A_{2}$.

  * (ii)

If the hyperideal $I\circ H$ of $R$ is $J$-prime for some hyperideal $I$ of
$R$, then $I\circ H=I$.

###### Proof.

$(i)$ It is clear that $A_{1}\circ H=A_{2}\circ H\subseteq A_{2}$. By Theorem
3.11, we obtain $A_{1}\subseteq A_{2}$ as the hyperideal $A_{2}$ is $J$-prime.
By a similar argument we get $A_{2}\subseteq A_{1}$. Thus $A_{1}=A_{2}$.

$(ii)$ Since $I$ is a hyperideal of $R$, then $I\circ H\subseteq I$. Let the
hyperideal $I\circ H$ of $R$ be $J$-prime. Since $I\circ H\subseteq I\circ H$
and $H\nsubseteq J(R)$, then $I\subseteq I\circ H$, by Theorem 3.11. Thus,
$I\circ H=I$. ∎

######  Theorem 3.17.

Let $(R_{1},+_{1},\circ_{1})$ and $(R_{2},+_{2},\circ_{2})$ be multiplicative
hyperrings and $\phi:R_{1}\longrightarrow R_{2}$ be a good homomorphism. If
$I_{2}$ is a $J$-prime hyperideal of $R_{2}$ such that $Ker\phi\subseteq
J(R_{1})$, then $\phi^{-1}(I_{2})$ is a $J$-prime hyperideal of $R_{1}$.

###### Proof.

Let $I_{2}$ is a $J$-prime hyperideal of $R_{2}$. Suppose that
$x\circ_{1}y\subseteq\phi^{-1}(I_{2})$ for some $x,y\in R_{1}$ such that
$x\notin J(R_{1})$. This implies that
$\phi(x)\circ_{2}\phi(y)=\phi(x\circ_{1}y)\subseteq I_{2}$. Let $M$ be a
maximal hyperideal of $R_{1}$ and $\phi(x)\in J(R_{2})$. Then $\phi(M)$ is a
maximal hyperideal of $R_{2}$ which implies $\phi(x)\in\phi(M)$. Since
$Ker\phi\subseteq M$, then we have $x\in M$ which means $x\in J(R_{1})$, a
contradiction. Therefore $\phi(x)\notin J(R_{2})$. Now, we have $\phi(y)\in
I_{2}$ as $I_{2}$ is a $J$-prime hyperideal of $R_{2}$. Then we conclude that
$y\in\phi^{-1}(I_{2})$. Thus, $\phi^{-1}(I_{2})$ is a $J$-prime hyperideal of
$R_{1}$. ∎

######  Theorem 3.18.

Let $(R_{1},+_{1},\circ_{1})$ and $(R_{2},+_{2},\circ_{2})$ be multiplicative
hyperrings and $\phi:R_{1}\longrightarrow R_{2}$ be a good homomorphism. If
$I_{1}$ is a $J$-prime hyperideal of $R_{1}$ such that $Ker\phi\subseteq
I_{1}$, then $\phi(I_{1})$ is a $J$-prime hyperideal of $R_{2}$.

###### Proof.

Let $x_{2}\circ_{2}y_{2}\subseteq\phi(I_{1})$ for some $x_{2},y_{2}\in R_{2}$
such that $x_{2}\notin J(R_{2})$. Then for some $x_{1},y_{1}\in R_{1}$ we have
$\phi(x_{1})=x_{2}$ and $\phi(y_{1})=y_{2}$. So
$\phi(x_{1})\circ_{2}\phi(y_{1})=\phi(x_{1}\circ_{1}y_{1})\subseteq\phi(I_{1})$.
Now, take any $u\in ax_{1}\circ y_{1}$. Then $f(u)\in f(x_{1}\circ
y_{1})\subseteq\phi(I_{1})$ and so there exists $w\in I_{1}$ such that
$\phi(u)=\phi(w)$. This means $\phi(u-w)=0$, that is, $u-w\in Ker\phi\subseteq
I_{1}$ and then $u\in I_{1}$. Since $I_{1}$ is a C-hyperideal of $R_{1}$, then
we get $x_{1}\circ_{1}y_{1}\subseteq I_{1}$. Since $\phi(J(R_{1}))\subseteq
J(R_{2})$, then $x_{1}\notin J(R_{1})$. Hence, we have $y_{1}\in I_{1}$ as
$I_{1}$ is a $J$-prime hyperideal of $R_{1}$. Thus,
$y_{2}=\phi(y_{1})\in\phi(I_{1})$. It follows that $\phi(I_{1})$ is a
$J$-prime hyperideal of $R_{2}$. ∎

######  Corollary 3.19.

Let $H$ and $I$ be two hyperideals of $R$ such that $H\subseteq I$. If $I$ is
a $J$-prime hyperideal of $R$, then the hyperideal $I/H$ of $R/H$ is
$J$-prime.

###### Proof.

Define the natural epimorphism $\pi:R\longrightarrow R/H$ by $\pi(x)=x+H$.
Since $Ker\pi\subseteq I$, we conclude that $\pi(I)=I/H$ is a $J$-prime
hyperideal of $R/H$, by Theorem 3.18. ∎

######  Corollary 3.20.

Let $H$ and $I$ be two hyperideals of $R$ such that $H\subseteq I\cap J(R)$.
If the hyperideal $I/H$ of $R/H$ is $J$-prime, then $I$ is a $J$-prime
hyperideal of $R$.

###### Proof.

Consider the natural epimorphism $\pi$ defined in the proof part of Corollary
3.19. Now, the claim follows by Theorem 3.17. ∎

######  Corollary 3.21.

Let $H$ and $I$ be two hyperideals of $R$ with $H\subseteq I$ such that $H$ is
$J$-prime. If the hyperideal $I/H$ of $R/H$ is $J$-prime, then $I$ is a
$J$-prime hyperideal of $R$.

###### Proof.

This can be proved by using Corollary 3.20 and Theorem 3.4. ∎

######  Definition 3.22.

A nonempty subset $S$ of $R$ containing $R-J(R)$ is called a
$J$-multiplicatively closed subset if $x\circ y\subseteq S$ for every $x\in
R-J(R)$ and every $y\in S$.

######  Theorem 3.23.

Let $I$ be a proper hyperideal of $R$. Then $I$ is $J$-prime if and only if
$R-I$ is a $J$-multiplicatively closed subset of $R$.

###### Proof.

$\Longrightarrow$ Let $I$ be a proper hyperideal of $R$. By Theorem 3.4, we
conclude that $R-J(R)\subseteq R-I$. Suppose that $x\in R-J(R)$ and $y\in
R-I$. To establish the claim, suppose, on the contrary, that $x\circ
y\subseteq I$. From $x\in R-J(R)$ it follows that $y\in I$ as $I$ is a
$J$-prime of $R$. This is a contradiction. Hence $x\circ y\subseteq R-I$ as
needed.

$\Longleftarrow$ Let $x\circ y\subseteq I$ for some $x,y\in R$ such that
$x\notin J(R)$. If $y\notin I$, then we conclude that $x\circ y\subseteq R-I$.
Thus, we arrive at a contradiction. Therefore $y\in I$ which means $I$ is
$J$-prime. ∎

######  Theorem 3.24.

Let $S$ be a multiplicatively subset of $R$ and $I$ be a hyperideal of $R$
disjoint from $S$. Then there exists a hyperideal $Q$ which is maximal in the
set of all hyperideals of $R$ disjoint from $S$, containing $I$. Any such
hyperideal $Q$ is a $J$-prime hyperideal of $R$.

###### Proof.

Let $\Upsilon$ be the set of all hyperideals of $R$ disjoint from $S$,
containing $I$. Then $\Upsilon\neq\varnothing$, since $I\in\Upsilon$. So
$\Upsilon$ is a partially ordered set with respect to set inclusion relation.
By Zorn’s lemma, there is a hyperideal $Q$ which is maximal in $\Upsilon$.
Suppose that $Q$ is not a $J$-prime hyperideal of $R$. Let $x\circ y\subseteq
Q$ such that $x\in R-J(R)$ and $y\in R-Q$. Therefore we have
$Q\subsetneqq(Q:x)$. Since $Q$ is a maximal element of $\Upsilon$, then
$(Q:x)\cap S\neq\varnothing$. Let $t\in(Q:x)\cap S$. Then $t\circ x\subseteq
Q\cap S$ which means $Q\cap S\neq\varnothing$. Thus we arrive at a
contradiction. Consequently, $Q$ is a $J$-prime hyperideal of $R$. ∎

Let $(R_{1},+_{1},\circ_{1})$ and $(R_{2},+_{2},\circ_{2})$ be two
multiplicative hyperrings with non zero identity. [32] Recall $(R_{1}\times
R_{2},+,\circ)$ is a multiplicative hyperring with the operation $+$ and the
hyperoperation $\circ$ are defined respectively as

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

$(x_{1},x_{2})\circ(y_{1},y_{2})=\\{(x,y)\in R_{1}\times R_{2}\ |\ x\in
x_{1}\circ_{1}y_{1},y\in x_{2}\circ_{2}y_{2}\\}$.

######  Theorem 3.25.

Let $(R_{1},+_{1},\circ_{1})$ and $(R_{2},+_{2},\circ_{2})$ be two
multiplicative hyperrings with non zero identity. Then $R_{1}\times R_{2}$ has
no $J$-prime hyperideals.

###### Proof.

Let $I_{1}\times I_{2}$ is a $J$-prime hyperideal of $R_{1}\times R_{2}$ for
some hyperideals $I_{1},I_{2}$ of $R_{1}$, $R_{2}$. Since
$(0,0)\in(1,0)\circ(0,1)\cap I_{1}\times I_{2}$, then
$(1,0)\circ(0,1)\subseteq I_{1}\times I_{2}$. Let $(1,0),(0,1)\in
J(R_{1}\times R_{2})$. Then $(1,0)=(1,1)-(0,1)$ and $(0,1)=(1,1)-(1,0)$ are
unit elements of $R_{1}\times R_{2}$. Therefore we conclude that $(1,0),(0,1)$
are not in $J(R_{1}\times R_{2})$. Thus, $(1,0),(0,1)$ are in $I_{1}\times
I_{2}$. So, $(1,1)=(1,0)+(0,1)\in I_{1}\times I_{2}$ which means $I_{1}\times
I_{2}=R_{1}\times R_{2}$. ∎

######  Definition 3.26.

A proper hyperideal $I$ of $R$ is called $J$-primary if whenever elements
$x,y\in R$ and $x\circ y\subseteq I$, then $x\in J(I)$ or $y\in I$.

######  Theorem 3.27.

Let $I$ be a hyperideal of $R$ with $I\subseteq J(R)$. Then $I$ is a $J$-prime
hyperideal if and only if $I$ is $J$-primary.

###### Proof.

$\Longrightarrow$ Let $I$ be a $J$-prime hyperideal of $R$. Suppose that
$x\circ y\subseteq I$ for some $x,y\in R$ such that $x\notin J(I)$. This means
$x\notin J(R)$ as $J(R)\subseteq J(I)$. Since $I$ is a $J$-prime hyperideal of
$R$ and $x\notin J(R)$, we get $y\in I$. Consequently, the hyperideal $I$ is
$J$-primary.

$\Longleftarrow$ Suppose that the hyperideal $I$ of $R$ is $J$-primary. Let
$x\circ y\subseteq I$ for some $x,y\in R$ such that $x\notin J(R)$. By the
hypothesis, we have $J(I)\subseteq J(R)$. This implies that $x\notin J(I)$.
Since $I$ is a $J$-primary hyperideal of $R$ and $x\notin J(I)$, then we get
$y\in I$. Thus, $I$ is a $J$-prime hyperideal. ∎

##  4\. quasi $J$-prime hyperideals

In this section, we define the concept of quasi $J$-prime hyperideals as a
generalization of $J$-prime hyperideals.

######  Definition 4.1.

A proper hyperideal $I$ of $R$ is called a quasi $J$-prime hyperideal if
$\sqrt{I}$ is $J$-prime.

######  Example 4.2.

In Example 3.2, the hyperideals $\langle 2\rangle$ and $\langle 3\rangle$ are
quasi $J$-prime. The hyperideal $\langle 12\rangle$ is not a quasi $J$-prime
hyperideal. In fact, $3\circ 4=\\{24,36\\}\subseteq\langle 12\rangle$, but
$3,4\notin\sqrt{\langle 12\rangle}=\langle 2\rangle\cap\langle 3\rangle$ and
$3,4\notin J(\mathbb{Z})$.

######  Theorem 4.3.

Let $I$ be a quasi $J$-prime hyperideal of $R$. If $x\circ H\subseteq I$ for
some hyperideal $H$ of $R$ and some element $x\in R$, then $x\in J(R)$ or
$H\subseteq\sqrt{I}$.

###### Proof.

Let $I$ be a quasi $J$-prime hyperideal of $R$. Suppose that $x\circ
H\subseteq I$ for some hyperideal $H$ of $R$ and some element $a\in R$ such
that $x\notin J(R)$. By Theorem 3.11, we get $\sqrt{I}=(\sqrt{I}:x)$. Since
$(I:x)\subseteq(\sqrt{I}:x)$, then $H\subseteq\sqrt{I}$. ∎

######  Theorem 4.4.

Let $I$ be a proper hyperideal of $R$. Then the following are equivalent:

  * (i)

$I$ is a quasi $J$-prime hyperideal of $R$.

  * (ii)

If $H\circ T\subseteq I$ for some hyperideals $H$ and $T$ of $R$, then
$H\subseteq J(R)$ or $T\subseteq\sqrt{I}$.

  * (iii)

If $x\circ y\subseteq I$ for some $x,y\in R$, then $x\in J(R)$ or
$y\in\sqrt{I}$.

###### Proof.

$(i)\Longrightarrow(ii)$ Let $H\circ T\subseteq I$ for some hyperideals $H$
and $T$ of $R$ such that $H\nsubseteq J(R)$. Take $x\in H$ such that $x\notin
J(R)$. Clearly, $x\circ T\subseteq I$. Since $I$ is a quasi $J$-prime
hyperideal of $R$ and $x\notin J(R)$, then $T\subseteq\sqrt{I}$, by Theorem
4.3.

$(ii)\Longrightarrow(iii)$ Let $x\circ y\subseteq I$ for some $x,y\in R$. Put
$H=\langle x\rangle$ and $T=\langle y\rangle$. Hence, $\langle
x\rangle\circ\langle y\rangle\subseteq\langle x\circ y\rangle\subseteq I$. By
the assumption, we get $x\in\langle x\rangle\subseteq J(R)$ or $y\in\langle
y\rangle\subseteq\sqrt{I}$.

$(iii)\Longrightarrow(i)$ Let $x\circ y\subseteq\sqrt{I}$ for some $x,y\in R$
such that $x\notin J(R)$. This means we have $(x\circ y)^{n}=x^{n}\circ
y^{n}\subseteq I$ for some $n\in\mathbb{N}$. Since $x\notin J(R)$, then
$x^{n}\notin J(R)$, by Theorem 3.6. Now, by the assumption, we obtain
$y^{n}\subseteq\sqrt{I}$ which implies $y\in\sqrt{\sqrt{I}}$. Thus
$y\in\sqrt{I}$ as needed. ∎

######  Theorem 4.5.

Let $I$ be a proper hyperideal of $R$. Then $I$ is a quasi $J$-prime
hyperideal of $R$ if and only if $I\subseteq J(R)$ and for $x,y\in R$, $x\circ
y\subseteq I$ implies that $x\in J(I)$ or $y\in\sqrt{I}$.

###### Proof.

$\Longrightarrow$ Let $\sqrt{I}$ be a $J$-prime hyperideal of $R$. By Theorem
3.4, we have $\sqrt{I}\subseteq J(R)$. Since $I\subseteq\sqrt{I}$, then we get
$I\subseteq J(R)$. Let $x\circ y\subseteq I$ for some $x,y\in R$. Then we have
$x\circ y\subseteq\sqrt{I}$. Since $\sqrt{I}$ is a $J$-prime hyperideal of
$R$, then we obtain $x\in J(R)\subseteq J(I)$ or $y\in\sqrt{I}$.

$\Longleftarrow$ Let $x\circ y\subseteq I$ such that $x\notin J(R)$. By the
hypothesis, we get $J(I)\subseteq J(R)$ which implies $x\notin J(I)$. Hence,
we have $y\in\sqrt{I}$. We conclude that the hyperideal $I$ of $R$ is a quasi
$J$-prime hyperideal. ∎

######  Lemma 4.6.

Let $I$ be a hyperideal of $R$ and $S$ be a subset of $R$ with $S\nsubseteq
J(R)$. If $I$ is a quasi $J$-prime hyperideal of $R$, then
$(\sqrt{I}:S)\subseteq\sqrt{(I:S)}$.

###### Proof.

Let $x\in(\sqrt{I}:S)$. Then $x\circ S=\sqrt{I}$. Since $x\circ
S=\bigcup_{y\in S}x\circ y$, we have $x\circ y\subseteq\sqrt{I}$ for every
$y\in S$. Since $S\nsubseteq J(R)$, then there exists $z\in S-J(R)$ such that
$x\circ z\subseteq\sqrt{I}$. Since $\sqrt{I}$ is a $J$-prime hyperideal of
$R$, we get $x\in\sqrt{I}$. Therefore for some $n\in\mathbb{N}$,
$x^{n}\subseteq I$. This implies that $x^{n}\subseteq(I:S)$ which means
$x\in\sqrt{(I:S)}$. Consequently, $(\sqrt{I}:S)\subseteq\sqrt{(I:S)}$. ∎

######  Theorem 4.7.

Let $I$ be a hyperideal of $R$ and $S$ be a subset of $R$ such that
$S\nsubseteq J(R)$. If the hyperideal $I$ of $R$ is quasi $J$-prime, then so
is $(I:S)$.

###### Proof.

It is clear that $(I:S)$ is a proper hyperideal of $R$. Let $x\circ
y\subseteq(I:S)$ for some $x,y\in R$ such that $x\notin J(R)$. Then $x\circ
y\circ S\subseteq I$. Now we have $x\circ r\subseteq I$ for all $r\in y\circ
S$. By Theorem 4.4, we get $r\in\sqrt{I}$. Since $y\circ
S\cap\sqrt{I}\neq\varnothing$ and $I$ is a C-hyperideal, then we conclude that
$y\circ S\subseteq\sqrt{I}$. This means $y\in(\sqrt{I}:S)$ which implies
$y\in\sqrt{(I:S)}$, by Lemma 4.6. Thus, $(I:S)$ is a quasi $J$-prime
hyperideal of $R$. ∎

######  Theorem 4.8.

Suppose that $I$ is a quasi $J$-prime hyperideal of $R$ such that there is no
quasi $J$-prime hyperideal which contains $I$ properly. Then the hyperideal
$I$ of $R$ is $J$-prime.

###### Proof.

Suppose that $I$ is a quasi $J$-prime hyperideal of $R$ such that there is no
quasi $J$-prime hyperideal which contains $I$ properly. Let $x\circ y\subseteq
I$ for some $x,y\in R$ such that $x\notin J(R)$. By Theorem 4.7, we conclude
that $(I:x)$ is a quasi $J$-prime hyperideal of $R$. Since $I\subseteq(I:x)$,
then $I=(I:x)$, by the hypothesis. Thus, we get $y\in I$. Consequently, the
hyperideal $I$ of $R$ is $J$-prime. ∎

######  Corollary 4.9.

The hyperideal $J(R)$ of $R$ is $J$-prime if and only if $J(R)$ is quasi
$J$-prime.

######  Theorem 4.10.

Let $M$ be a maximal hyperideal of $R$. If $\langle a\circ b\rangle$ is a
quasi $J$-prime hyperideal of $R$ for each $a,b\in R$ is quasi $J$-prime, then
so is $M$.

###### Proof.

Let $M$ be a maximal hyperideal of $R$. Let $x\circ y\subseteq M$ and
$x\notin\sqrt{M}$. Since $\sqrt{\langle x\circ y\rangle}\subseteq\sqrt{M}$ and
$x\notin\sqrt{M}$, then $x\notin\sqrt{\langle x\circ y\rangle}$. Since
$\langle x\circ y\rangle$ is a quasi $J$-prime hyperideal of $R$ and $x\circ
y\subseteq\langle x\circ y\rangle$, then we obtain $y\in J(R)$ which means $M$
is a quasi $J$-prime hyperideal of $R$. ∎

######  Theorem 4.11.

Every maximal hyperideal of $R$ is a quasi $J$-prime hyperideal if and only if
$R$ is a local hyperring.

###### Proof.

$\Longrightarrow$ Let the maximal hyperideal $M$ of $R$ be a quasi $J$-prime
hyperideal. By Theorem 3.4, we get $M\subseteq J(R)$. Since $M$ is maximal,
then $M$ is a prime hyperideal of $R$. Hence $M=\sqrt{M}$ and so $M\subseteq
J(R)$.

$\Longleftarrow$ Let $R$ ibe a local hyperring. By Theorem 3.5, every proper
hyperideal of $R$ is $J$-prime and so every proper hyperideal of $R$ is quasi
$J$-prime. Now, the claim follows by Theorem 4.10. ∎

Recall from [7] that a proper hyperideal $I$ of $R$ is said to be a quasi
primary hyperideal if $\sqrt{I}$ is prime.

######  Theorem 4.12.

Let every prime hyperideal of $R$ be maximal. If the hyperideal $I$ of $R$ is
quasi $J$-prime and $I\subseteq J(R)$, then $I$ is a quasi primary hyperideal
of $R$.

###### Proof.

We show that $\sqrt{I}$ is prime. Let $x\circ y\subseteq\sqrt{I}$ for some
$x,y\in R$ such that $x\notin\sqrt{I}$. Hence we have $x^{n}\circ
y^{n}\subseteq I$ for some $n\in\mathbb{N}$. By the hypothesis, we conclude
that $\sqrt{I}=J(R)$. Since $x^{n}\notin J(R)$ and the hyperideal $I$ of $R$
is quasi $J$-prime, then $y^{n}\subseteq\sqrt{I}$, by Theorem 4.4. This means
$y\in\sqrt{I}$. Thus $I$ is a quasi primary hyperideal of $R$. ∎

######  Theorem 4.13.

Let every prime hyperideal of $R$ be maximal and $I$ be a hyperideal of $R$
such that $I\subseteq J(R)$. Then $I$ is quasi $J$-prime if and only if the
multiplicative hyperring $R$ is local with the maximal hyperideal $\sqrt{I}$.

###### Proof.

$\Longrightarrow$ Let $I$ be a quasi $J$-prime hyperideal of $R$. Hence
$\sqrt{I}$ is a prime hyperideal of $R$, by Theorem 4.12. By the hypothesis,
$\sqrt{I}$ is a maximal hyperideal of $R$. Then there exists some prime
hyperideal $Q$ such that $I=Q^{n}$ for some $n\in\mathbb{N}$. This means
$\sqrt{I}=Q$ is maximal and so $J(R)\subseteq\sqrt{I}=Q$. Since $I\subseteq
J(R)$, then we conclude that $\sqrt{I}=J(R)$. Thus the multiplicative
hyperring $R$ is local with the maximal hyperideal $\sqrt{I}$.

$\Longleftarrow$ The claim follows by Theorem 4.11. ∎

######  Theorem 4.14.

  * (i)

If $I_{i}$ is a quasi $J$-prime hyperideal of $R$ for each $1\leq i\leq n$,
then is so $\bigcap_{i=1}^{n}I_{i}$.

  * (ii)

If $I_{i}$ is a quasi $J$-prime hyperideal of $R$ for each $1\leq i\leq n$,
then is so $\Pi_{i=1}^{n}I_{i}$.

###### Proof.

(i) By Proposition 3.3. in [14], we have
$\sqrt{\bigcap_{i=1}^{n}I_{i}}=\bigcap_{i=1}^{n}\sqrt{I_{i}}$. Since
$\sqrt{I_{i}}$ is a $J$-prime hyperideal of $R$ for each $1\leq i\leq n$, we
conclude that $\bigcap_{i=1}^{n}\sqrt{I_{i}}$ is a $J$-prime hyperideal of
$R$, by Theorem 3.10. Thus, $\sqrt{\bigcap_{i=1}^{n}I_{i}}$ is a $J$-prime
hyperideal of $R$ which means $\bigcap_{i=1}^{n}I_{i}$ is a quasi $J$-prime
hyperideal of $R$.

(ii) Let $x\circ y\subseteq\Pi_{i=1}^{n}I_{i}$ for some $x,y\in R$ such that
$x\notin J(R)$. Since $I_{i}$ is a quasi $J$-prime hyperideal of $R$ for each
$1\leq i\leq n$ and $\Pi_{i=1}^{n}I_{i}\subseteq\bigcap_{i=1}^{n}I_{i}$, we
conclude that $y\in\sqrt{I_{i}}$ for each $1\leq i\leq n$. This means
$y^{t_{i}}\subseteq I_{i}$ for each $1\leq i\leq n$. Put
$t=t_{1}+t_{2}+...+t_{n}$. Therefore, $y^{t}=y^{t_{1}}\circ
y^{t_{2}}\circ...\circ y^{t_{n}}\subseteq\Pi_{i=1}^{n}I_{i}$ which implies
$y\in\sqrt{\Pi_{i=1}^{n}I_{i}}$. Consequently, the hyperideal
$\Pi_{i=1}^{n}I_{i}$ of $R$ is quasi $J$-prime. ∎

##  5\. 2-absorbing $J$-prime hyperideals

In this section , we extend the notion of $J$-prime to 2-absorbing $J$-prime
hyperideals and give some properties of them.

######  Definition 5.1.

Let $I$ be a proper hyperideal of $R$. $I$ is called a 2-absorbing $J$-prime
hyperideal of $R$ if whenever $x,y,z\in R$ with $x\circ y\circ z\subseteq I$,
then $x\circ y\subseteq I$ or $x\circ z\subseteq J(R)$ or $y\circ z\subseteq
J(R)$.

######  Example 5.2.

In Example 3.2, let $A=\\{2,4\\}$. Then the principal hyperideal $\langle
15\rangle$ is a 2-absorbing $J$-prime hyperideal.

######  Example 5.3.

Consider the ring
$(\\{\bar{0},\bar{1},\bar{2},\bar{4},\bar{5}\\}=\mathbb{Z}_{6},\oplus,\star)$
that for all $\bar{x},\bar{y}\in\mathbb{Z}_{6}$, $\bar{x}\oplus\bar{y}$ and
$\bar{x}\star\bar{y}$ are the remainder of $\frac{x+y}{6}$ and $\frac{x\cdot
y}{6}$, respectively, which $"+"$ and $"\cdot"$ are ordinary addition and
multiplication, and $x,y\in\mathbb{Z}$. We define the hyperoperation
$\bar{x}\odot\bar{y}=\\{\bar{xy},\bar{2xy},\bar{3xy},\bar{4xy},\bar{5xy}\\}$,
for all $\bar{x},\bar{y}\in\mathbb{Z}_{6}$. In the commutative multiplicative
hyperring $(\mathbb{Z},\oplus,\odot)$, hyperideal $\\{\bar{0}\\}$ is
2-absorbing $J$-prime.

######  Theorem 5.4.

If $I$ is a $J$-prime hyperideal of $R$, then $I$ is a 2-absorbing $J$-prime
hyperideal of $R$.

###### Proof.

Let $I$ be a $J$-prime hyperideal of $R$. Suppose that $x\circ y\circ
z\subseteq I$ for some $x,y,z\in R$. Choose $u\in x\circ z$. Since $y\circ
u\subseteq I$ and $I$ is a $J$-prime hyperideal of $R$, we get $y\in I$ or
$u\in J(R)$. In the former case, we have $y\circ z\subseteq I$ which means
$y\circ z\subseteq J(R)$, by Theorem 3.4. In the second case, we obtain
$x\circ z\subseteq J(R)$ as $J(R)$ is a C-hyperideal of $R$. ∎

######  Theorem 5.5.

If $I$ is a 2-absorbing $J$-prime hyperideal of $R$, then $I\subseteq J(I)$.

###### Proof.

Let $I$ be a 2-absorbing $J$-prime hyperideal of $R$. We suppose that
$I\nsubseteq J(I)$ and look for a contradiction. From $I\nsubseteq J(I)$ it
follows that there exists $x\in I$ such that $x\notin J(R)$. Since $1\circ
1\circ x\subseteq I$ and $I$ is a 2-absorbing $J$-prime hyperideal of $R$, we
have $1\circ 1\subseteq I$, a contradiction or $x\in 1\circ x\subseteq J(R)$,
a contradiction. Thus, $I\subseteq J(I)$. ∎

Recall from [6] that a proper hyperideal $I$ of $R$ is said to be an
2-absorbing primary hyperideal if $x\circ y\circ z\subseteq I$ implies $x\circ
y\subseteq I$ or $x\circ z\subseteq\sqrt{I}$ or $y\circ z\subseteq\sqrt{I}$
for any $x,y,z\in R$.
The next Theorem shows that the inverse of Theorem 5.5 is true if $I$ is a
2-absorbing primary hyperideal of $R$

######  Theorem 5.6.

Let $I$ be a 2-absorbing primary hyperideal of $R$ and $I\subseteq J(R)$. Then
$I$ is a 2-absorbing $J$-prime hyperideal of $R$.

###### Proof.

Suppose that $I$ is a 2-absorbing primary hyperideal of $R$ such that
$I\subseteq J(R)$. Let $x\circ y\circ z\subseteq I$ for some $x,y,z\in R$ such
that $x\circ z,y\circ z\nsubseteq J(R)$. Then we conclude that $x\circ
z,y\circ z\nsubseteq\sqrt{I}$. This implies that $x\circ y\subseteq I$ as $I$
is a 2-absorbing primary hyperideal of $R$. Consequently, $I$ is a 2-absorbing
$J$-prime hyperideal of $R$. ∎

######  Theorem 5.7.

Let the multiplicative hyperring $R$ has at most two maximal hyperideals.
Suppose that the hyperideal $I$ of $R$ is 2-absorbing primary that is not
quasi primary. Then $I$ is a 2-absorbing $J$-prime hyperideal of $R$.

###### Proof.

Assume that the multiplicative hyperring $R$ has at most two maximal
hyperideals. Let the hyperideal $I$ of $R$ be 2-absorbing primary. Then
$\sqrt{I}=P$ or $\sqrt{I}=P_{1}\cap P_{2}$ for some prime hyperideals of $R$,
by Theorem 4.5 in [6]. In the former case, we have a contradiction since $I$
is not a quasi primary hyperideal of $R$. Therefore, we obtain
$I\subseteq\sqrt{I}\subseteq J(R)$ as the multiplicative hyperring $R$ has at
most two maximal hyperideals. Thus, $I$ is a 2-absorbing $J$-prime hyperideal
of $R$, by Theorem 5.6. ∎

######  Corollary 5.8.

Let $R_{1}$ and $R_{2}$ be two local multiplicative hyperrings. Then every
2-absorbing primary hyperideal of $R_{1}\times R_{2}$ that is not quasi
primary is 2-absorbing $J$-prime.

######  Theorem 5.9.

Let $I$ be a 2-absorbing $J$-prime of $R$. Then for each $a,b\in I$ with
$a\circ b\nsubseteq I$, $(I:a\circ b)\subseteq(J(R):a)$ or $(I:a\circ
b)\subseteq(J(R):b)$.

###### Proof.

Let $I$ be a 2-absorbing $J$-prime of $R$. Let $c\in(I:a\circ b)$ for some
$a,b\in R$ with $a\circ b\nsubseteq I$. This means $a\circ b\circ c\subseteq
I$. Since the hyperideal $I$ of $R$ is 2-absorbing $J$-prime and $a\circ
b\nsubseteq I$, then $a\circ c\subseteq J(R)$ of $b\circ c\subseteq J(R)$
which means $c\in(J(R):a)$ or $c\in(J(R):b)$. This implies that $(I:a\circ
b)\subseteq(J(R):a)$ or $(I:a\circ b)\subseteq(J(R):b)$. ∎

Recall that a hyperideal $I$ of $R$ is called a strong C-hyperideal if for any
$E\in\mathfrak{U}$, $E\cap I\neq\varnothing$, then $E\subseteq I$, where
$\mathfrak{U}=\\{\sum_{i=1}^{n}A_{i}\ :\ A_{i}\in{\bf C},n\in\mathbb{N}\\}$
and ${\bf C}=\\{r_{1}\circ r_{2}\circ...\circ r_{n}\ :\ r_{i}\in
R,n\in\mathbb{N}\\}$.

######  Theorem 5.10.

Let $I$ be a proper strong C-hyperideal of $R$. Then the following are
equivalent:

  * (i)

$I$ is a 2-absorbing $J$-prime hyperideal of $R$.

  * (ii)

If $a\circ b\circ H\subseteq I$ for some $a,b\in R$ and some hyperideal $H$ of
$R$, then $a\circ b\subseteq I$ or $a\circ H\subseteq J(R)$ or $b\circ
H\subseteq J(R)$.

  * (iii)

If $a\circ H\circ T\subseteq I$ for some $a\in R$ and some hyperideals $H,T$
of $R$, then $a\circ H\subseteq I$ or $a\circ T\subseteq J(R)$ or $H\circ
T\subseteq J(R)$.

  * (iv)

If $K\circ H\circ T\subseteq I$ for some hyperideals $K,H,T$ of $R$, then
$K\circ H\subseteq I$ or $K\circ T\subseteq J(R)$ or $H\circ T\subseteq J(R)$.

###### Proof.

$(i)\Longrightarrow(ii)$ Let $a\circ b\circ H\subseteq I$ for some $a,b\in R$
and some hyperideal $H$ of $R$ such that $a\circ b\nsubseteq I$. Hence, we get
$H\subseteq(J(R):a)$ or $H\subseteq(J(R):b)$, by Theorem 5.9. This implies
that $a\circ H\subseteq J(R)$ or $b\circ H\subseteq J(R)$.

$(ii)\Longrightarrow(iii)$ Let $a\circ H\circ T\subseteq I$ for some $a\in R$
and some hyperideals $H,T$ of $R$ such that $a\circ H\nsubseteq I$ and $a\circ
T\nsubseteq J(R)$. Therefore, we get $a\circ h_{1}\nsubseteq I$ and $a\circ
t_{1}\nsubseteq J(R)$ for some $h_{1}\in H$ and $t_{1}\in T$. Since $a\circ
h_{1}\circ T\subseteq I$, then $h_{1}\circ T\subseteq J(R)$, by (i). Let
$h_{2}\in H$. Suppose that $h_{2}\circ T\nsubseteq J(R)$. By (i), we get
$a\circ h_{2}\subseteq I$ as $a\circ h_{2}\circ T\subseteq I$. Hence
$a\circ(h_{1}+h_{2})\nsubseteq I$. Again by (i), we obtain $(h_{1}+h_{2})\circ
T\subseteq J(R)$ as $a\circ(h_{1}+h_{2})\circ T\subseteq I$. As $h_{1}\circ
T\subseteq J(R)$, $h_{2}\circ T\subseteq J(R)$ which means $H\circ T\subseteq
J(R)$.

$(iii)\Longrightarrow(iv)$ Let $K\circ H\circ T\subseteq I$ for some
hyperideals $K,H,T$ of $R$ such that $K\circ H\nsubseteq I$ and $H\circ
T\nsubseteq J(R)$. Then we have $k\circ H\nsubseteq I$ for some $k\in K$. It
is clear that $k\circ H\circ T\subseteq I$. Hence we have $k\circ T\subseteq
J(R)$, by (iii). Assume that $x\in K$. By (iii), we conclude that $x\circ
H\subseteq I$ or $x\circ T\subseteq J(R)$ as $x\circ H\circ T\subseteq I$. If
$x\circ H\subseteq I$, then $(k+x)\circ H\subseteq(k\circ H)+(x\circ
H)\nsubseteq I$. Clearly, $(k+x)\circ H\circ T\subseteq I$. Then $(k+x)\circ
T\subseteq J(R)$. Since $I$ is a strong C-hyperideal of $R$, then $(k\circ
T)+(x\circ T)\subseteq J(R)$. Hence, $x\circ T\subseteq J(R)$ as $k\circ
T\subseteq J(R)$. Now, let $x\circ H\nsubseteq I$. It is obvious that $x\circ
H\circ T\subseteq I$. Therefore, $x\circ T\subseteq J(R)$, by (iii).
Consequently, $K\circ T\subseteq J(R)$.

$(iv)\Longrightarrow(i)$ Let $x\circ y\circ z\subseteq I$ for some $x,y,z\in
R$ such that $x\circ z\nsubseteq J(R)$ and $y\circ z\nsubseteq J(R)$. We
consider $K=\langle x\rangle$, $H=\langle y\rangle$ and $T=\langle z\rangle$.
Thus, we conclude that $K\circ T=\langle x\rangle\circ\langle
z\rangle\subseteq\langle x\circ z\rangle\nsubseteq J(R)$ and $H\circ T=\langle
y\rangle\circ\langle z\rangle\subseteq\langle y\circ z\rangle\nsubseteq J(R)$.
Since $K\circ H\circ T\subseteq I$, by (iv), we get $K\circ H\subseteq I$
which means $x\circ y\subseteq I$ and that completes the proof. ∎

######  Theorem 5.11.

If every proper hyperideal of $R$ is 2-absorbing $J$-prime, then $R$ is a
local multiplicative hyperring.

###### Proof.

Let every proper hyperideal of $R$ be 2-absorbing $J$-prime. Suppose that $M$
is a maximal hyperideal of $R$. We show that $M\subseteq J(R)$. Let $a\in M$
and $I=\langle a\rangle$. Since $I$ is a 2-absorbing $J$-prime of $R$, then
$I\subseteq J(R)$ by Theorem 5.5. Therefore, $a\in J(R)$ which means
$M\subseteq J(R)$. Thus $J(R)=M$ as $J(R)\subseteq M$. Consequently, $R$ is a
local multiplicative hyperring. ∎

######  Theorem 5.12.

Let $\\{I_{i}\\}_{i\in\Delta}$ be a nonmepty set of 2-absorbing $J$-prime
hyperideals of $R$. Then $\bigcap_{i\in\Delta}I_{i}$ is a 2-absorbing
$J$-prime hyperideal of $R$.

###### Proof.

Let $x\circ y\circ z\subseteq\bigcap_{i\in\Delta}I_{i}$ for some $x,y,z\in R$
such that $x\circ z\nsubseteq J(R)$ and $y\circ z\nsubseteq J(R)$. This
implies that $x\circ y\circ z\subseteq I_{i}$ for every $i\in\Delta$. Since
$I_{i}$ is a 2-absorbing $J$-prime hyperideal of $R$ for every $i\in\Delta$,
we get the result that $x\circ y\subseteq I_{i}$ and so $x\circ
y\subseteq\bigcap_{i\in\Delta}I_{i}$. Thus $\bigcap_{i\in\Delta}I_{i}$ is a
2-absorbing $J$-prime hyperideal of $R$. ∎

######  Definition 5.13.

A proper hyperideal $I$ of $R$ is called 2-absorbing $J$-primary if whenever
elements $x,y,z\in R$ and $x\circ y\circ z\subseteq I$, then $x\circ
y\subseteq I$ or $x\circ z\in J(I)$ or $y\circ z\in J(I)$.

######  Theorem 5.14.

Let $I$ be a hyperideal of $R$ . Then $I$ is a 2 -absorbing $J$-prime
hyperideal of $R$ if and only if $I$ is a 2-absorbing $J$-primary hyperideal
of $R$ with $J(I)=J(R)$.

###### Proof.

$\Longrightarrow$ Let $I$ be a 2 -absorbing $J$-prime hyperideal of $R$.
Suppose that $x,y,z\subseteq I$ for some $x,y,z\in R$. This implies that
$x\circ y\subseteq I$ or $x\circ z\subseteq J(R)$ or $y\circ z\subseteq J(R)$
as $I$ is a 2 -absorbing $J$-prime hyperideal of $R$. Since $J(R)\subseteq
J(I)$, we have $x\circ y\subseteq I$ or $x\circ z\subseteq J(I)$ or $y\circ
z\subseteq J(I)$. Thus $I$ is a 2-absorbing $J$-primary hyperideal of $R$. It
is clear that $J(R)\subseteq J(I)$. For the reverse inclusion, we have
$I\subseteq J(R)$ by Theorem 5.5. Thus $J(I)\subseteq J(R)$ and so
$J(I)=J(R)$.

$\Longleftarrow$ Let $I$ be a 2-absorbing $J$-primary hyperideal of $R$ such
that $J(I)=J(R)$. Suppose that $x,y,z\subseteq I$ for some $x,y,z\in R$. This
implies that $x\circ y\subseteq I$ or $x\circ z\subseteq J(I)$ or $y\circ
z\subseteq J(I)$ as $I$ is a 2-absorbing $J$-primary hyperideal of $R$. Since
$J(I)=J(R)$, we get $x\circ y\subseteq I$ or $x\circ z\subseteq J(R)$ or
$y\circ z\subseteq J(R)$. This means $I$ is a 2 -absorbing $J$-prime
hyperideal of $R$. ∎

######  Theorem 5.15.

Let $(R_{1},+_{1},\circ_{1})$ and $(R_{2},+_{2},\circ_{2})$ be two
multiplicative hyperrings and let $I_{1}$ and $I_{2}$ be hyperideals of
$R_{1}$ and $R_{2}$, respectively. Then $I_{1}\times I_{2}$ is a 2-absorbing
$J$-prime hyperideal of $R_{1}\times R_{2}$ if and only if $I_{1}$ is a
$J$-prime hyperideal of $R_{1}$ and $I_{2}$ is a $J$-prime hyperideal of
$R_{2}$.

###### Proof.

$\Longrightarrow$ Let $I_{1}\times I_{2}$ is a 2-absorbing $J$-prime
hyperideal of $R_{1}\times R_{2}$. Suppose that $a\circ_{1}b\subseteq I_{1}$
such that $b\notin J(R_{1})$. Therefore $(a,1)\circ(1,0)\circ(b,1)=\\{(x,y)\
|\ x\in a\circ_{1}1\circ_{1}b,y\in 1\circ_{2}0\circ_{2}1\\}\subseteq
I_{1}\times I_{2}$, $(a,1)\circ(b,1)=\\{(x^{\prime},y^{\prime})\ |\
x^{\prime}\in a\circ_{1}b,y^{\prime}\in 1\circ_{2}1\\}\nsubseteq J(R_{1}\times
R_{2})$ and $(b,0)\in(1,0)\circ(b,1)\nsubseteq J(R_{1}\times R_{2})$. Hence we
get $(a,0)\in(a,1)\circ(1,0)\subseteq I_{1}\times I_{2}$ and so $a\in I_{1}$.
Consequently, $I_{1}$ is a $J$-prime hyperideal of $R_{1}$. By a similar
argument, we can prove that $I_{2}$ is a $J$-prime hyperideal of $R_{2}$.

$\Longleftarrow$ Let $I_{1}$ be a $J$-prime hyperideal of $R_{1}$ and $I_{2}$
is a $J$-prime hyperideal of $R_{2}$. Suppose that
$(a_{1},b_{1})\circ(a_{2},b_{2})\circ(a_{3},b_{3})\subseteq I_{1}\times I_{2}$
such that $(a_{2},b_{2})\circ(a_{3},b_{3})\nsubseteq J(R_{1}\times R_{2})$.
Take $a\in a_{2}\circ_{1}a_{3}$. Then $a_{1}\circ_{1}a\subseteq
a_{1}\circ_{1}a_{2}\circ_{1}a_{3}\subseteq I_{1}$. We assume that
$a_{2}\circ_{1}a_{3}\nsubseteq J(R_{1})$ then $a\notin J(R_{1})$ as $J(R_{1})$
is a C-hyperideal. Since $I_{1}$ is a $J$-prime hyperideal of $R_{1}$,
$a_{1}\in I_{1}$ and so $a_{1}\circ_{1}a_{2}\subseteq I_{1}$. On the other
hand, if $b_{1}\circ_{2}b_{3}\subseteq J(R_{2})$, then
$(a_{1},b_{1})\circ(a_{3},b_{3})\subseteq I_{1}\times J(R_{2})\subseteq
J(R_{1}\times R_{2})$, as needed. Then we may assume $b_{1}\circ
b_{3}\nsubseteq J(R_{2})$. Take $b\in b_{1}\circ_{2}b_{3}$. Then
$b_{2}\circ_{2}b\subseteq I_{2}$. Since $I_{2}$ is a $J$-prime hyperideal of
$R_{2}$ and $b\notin J(R_{2})$, we have $b_{2}\in I_{2}$ and so
$b_{1}\circ_{2}b_{2}\subseteq I_{2}$. This
$(a_{1},b_{1})\circ(a_{2},b_{2})\subseteq I_{1}\times I_{2}$. This implies
that $I_{1}\times I_{2}$ is a 2-absorbing $J$-prime hyperideal of $R_{1}\times
R_{2}$. ∎

######  Theorem 5.16.

Let $(R_{1},+_{1},\circ_{1})$, $(R_{2},+_{2},\circ_{2})$ and
$(R_{3},+_{3},\circ_{3})$ be three multiplicative hyperrings with non zero
identity. Then, $R_{1}\times R_{2}\times R_{3}$ has no 2-absorbing $J$-prime
hyperideals.

###### Proof.

Let $I_{1}\times I_{2}\times I_{3}$ is a 2-absorbing $J$-prime hyperideal of
$R_{1}\times R_{2}\times R_{2}$ for some hyperideals $I_{1},I_{2}$ and $I_{3}$
of $R_{1}$, $R_{2}$ and $R_{3}$, respectively. Since
$(0,0,0)\in(1,1,0)\circ(0,1,1)\circ(1,0,1)\cap I_{1}\times I_{2}\times I_{3}$,
then $(1,1,0)\circ(0,1,1)\circ(1,0,1)\subseteq I_{1}\times I_{2}\times I_{3}$.
Since $(1,1,0)\circ(0,1,1)\nsubseteq J(R_{1}\times R_{2}\times R_{3})$ and
$(1,1,0)\circ(1,0,1)\nsubseteq J(R_{1}\times R_{2}\times R_{3})$, then
$(0,0,1)\in(0,0,1)\circ(1,0,1)\subseteq I_{1}\times I_{2}\times I_{3}$.
Moreover, we can get $(0,1,0),(1,0,0)\in I_{1}\times I_{2}\times I_{3}$. Thus
$(1,1,1)=(1,0,0)+(0,1,0)+(0,0,1)\in I_{1}\times I_{2}\times I_{3}$ which is a
contradiction. ∎

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