#  $\alpha$-prime hyperideals in a multiplicative hyperring

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

The notion of multiplicative hyperrings is an important class of algebraic
hyperstructures which generalize rings where the multiplication is a
hyperoperation, while the addition is an operation . Let $R$ be a commutative
multiplicative hyperring and $\alpha\in End(R)$. A proper hyperideal $I$ of
$R$ is called $\alpha$-prime if $x\circ y\subseteq I$ for some $x,y\in R$ then
$x\in I$ or $\alpha(y)\in I$. Indeed, the $\alpha$-prime hyperideals are a new
generalization of prime hyperideals. In this paper, we aim to study
$\alpha$-prime hyperideals and give the basic properties of this new type of
hyperideals.

###### Key words and phrases:

$\alpha$-prime hyperideal, $\alpha$-radical, $\alpha$-nilpotent,
$\alpha$-nilradical.

###### 2010 Mathematics Subject Classification:

20N20

##  1\. Introduction

Algebraic hyperstructures are a suitable generalization of classical algebraic
structures. This theory has been introduced by Marty in 1934 during the
$8^{th}$ Congress of the Scandinavian Mathematicians [26]. He defined the
hypergroups as a generalization of groups. Afterwards, many researchers have
been worked on this new ﬁeld of modern algebra and developed it [12, 13, 14,
15, 16, 19, 27, 31, 36]. In an algebraic hyperstructure, the composition of
two elements is a set, while in a classical algebraic structure, the
composition of two elements is an element. Similar to hypergroups, hyperrings
are algebraic structures more general than rings, subsitutiting both or only
one of the binary operations of addition and multiplication by
hyperoperations. The hyperrings were introduced and studied by many authors
[3, 4, 30, 9]. Krasner introduced a type of the hyperring where addition is a
hyperoperation and multiplication is an ordinary binary operation. Such a
hyperring is called a Krasner hyperring [25]. Mirvakili and Davvaz introduced
$(m,n)$-hyperrings in [28] and they defined Krasner $(m,n)$-hyperrings as a
subclass of $(m,n)$-hyperrings and as a generalization of Krasner hyperrings
in [29]. The notion of multiplicative hyperrings is an important class of
algebraic hyperstructures that generalize rings, initiated the study by Rota
in 1982 which the multiplication is a hyperoperation, while the addition is an
operation [33]. There exists a general type of hyperrings that both the
addition and multiplication are hyperoperations [37]. Ameri and Kordi have
studied Von Neumann regularity in multiplicative hyperrings [5]. Moreover,
they introduced the concept of clean multiplicative hyperrings and studied
some topological concepts to realize clean elements of a multiplicative
hyperring by clopen subsets of its Zariski topology[6]. The notions such as
(weak)zero divisor, (weak)nilpotent and unit in an arbitrary commutative
hyperrings were introduced in [2]. Some equivalence relations - called
fundamental relations - play important roles in the the theory of algebraic
hyperstructures. The fundamental relations are one of the most important and
interesting concepts in algebraic hyperstructures that ordinary algebraic
structures are derived from algebraic hyperstructures by them. For more
details about hyperrings and fundamental relations on hyperrings see [18, 19,
21, 22, 24, 26, 34, 37]. Prime ideals and primary ideals are two of the most
important structures in commutative algebra. The notion of primeness of
hyperideal in a multiplicative hyperring was conceptualized by Procesi and
rota in [32]. Dasgupta extended the prime and primary hyperideals in
multiplicative hyperrings in [17]. Beddani and Messirdi [11] introduced a
generalization of prime ideals called 2-prime ideals and this idea is further
generalized by Ulucak and et. al. [35]. In [8], we investigated
$\delta$-primary hyperideals in a Krasner $(m,n)$-hyperring which unify prime
hyperideals and primary hyperideals. $\alpha$-prime ideals in a commutative
ring with nonzero identity have been introduced and studied by Akray and
Mohammad-Salih in [7].

In this paper we consider the class of multiplicative hyperring as a
hyperstructure $(R,+,\circ)$, where $(R,+)$ is an abelian group, $(R,\circ)$
is a semihypergroup and the hyperoperation ”$\circ$” is distributive with
respect to the operation $"+"$. In this paper we introduce and study the
notion of $\alpha$-prime hyperideals of a multiplicative hyperring which is a
new generalization to prime hyperideals. Several properties of them are
provided. For example we show (Theorem 3.17) that if $R$ is a multiplicative
hyperring such that it has zero absorbing property and $\langle 0\rangle$ is a
prime hyperideal of $R$, then

$Nil_{\alpha}(R)=\bigcap_{I{\text{\ }is\ \alpha-prime\ of\ R}}I.$

It is shown (Theorem 3.24) that the hyperideal $I$ of $R$ is $\alpha$-prime if
and only if $I/Ker\alpha$ is prime in $R/Ker\alpha$. We show (Theorem 3.28)
that thehyperideal $I$ is $\alpha$-prime if and only if $R/I$ is an
$\alpha$-integral hyperdomain. Also, we investigate the stability of
$\alpha$-prime hyperideals in some hyperring-theoric constructions.

##  2\. Preliminaries

In this section we give some defnitions and results which we need to develop
our paper.

A hyperoperation ”$\circ$” on nonempty set $G$ is a mapping of $G\times G$
into the family of all nonempty subsets of $G$. Assume that ”$\circ$” is a
hyperoperation on $G$. Then $(G,\circ)$ is called hypergroupoid. The
hyperoperation on $G$ can be extended to subsets of $G$ as follows. Let $X,Y$
be subsets of $G$ and $g\in G$, then

$X\circ Y=\cup_{x\in X,y\in Y}x\circ y,\ \ X\circ g=X\circ\\{g\\}.$

A hypergroupoid $(G,\circ)$ is called a semihypergroup if for all $x,y,z\in
G$, $(x\circ y)\circ z=x\circ(y\circ z)$, which is associative. A
semihypergroup is said to be a hypergroup if $g\circ G=G=G\circ g$ for all
$g\in G$. A nonempty subset $H$ of a semihypergroup $(G,\circ)$ is called a
subhypergroup if for all $x\in H$ we have $x\circ H=H=H\circ x$. A commutative
hypergroup $(G,\circ)$ is canonical if

  * (i)

there exists $e\in G$ with $e\circ x=\\{x\\}$, for every $x\in G$.

  * (ii)

for every $x\in G$ there exists a unique $x^{-1}\in G$ with $e\in x\circ
x^{-1}$.

  * (iii)

$x\in y\circ z$ implies $y\in x\circ z^{-1}$.

A nonempty set $R$ with two hyperoperations $"+"$ and $"\circ"$ is called a
hyperring if $(R,+)$ is a canonical hypergroup , $(R,\circ)$ is a
semihypergroup with $r\circ 0=0\circ r=0$ for all $r\in R$ and the
hyperoperation $"\circ"$ is distributive with respect to $+$, i.e.,
$x\circ(y+z)=x\circ y+x\circ z$ and $(x+y)\circ z=x\circ z+y\circ z$ for all
$x,y,z\in R$.

######  Definition 2.1.

[20] A multiplicative hyperring is an abelian group $(R,+)$ in which a
hyperoperation $\circ$ is defined satisfying the following:

  * (i)

for all $a,b,c\in R$, we have $a\circ(b\circ c)=(a\circ b)\circ c$;

  * (ii)

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

  * (iii)

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

If in (ii) the equality holds then we say that the multiplicative hyperring is
strongly distributive. Recall that $R$ has a zero absorbing property if for
all $r\in R$, $\\{0\\}=0\circ r=r\circ 0$.
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 $rox\subseteq I$.

Let $(\mathbb{Z},+,\cdot)$ be the ring of integers. Corresponding to every
subset $A\in P^{\star}(\mathbb{Z})$ such that $|A|\geq 2$, there exists a
multiplicative hyperring $(\mathbb{Z}_{A},+,\circ)$ with
$\mathbb{Z}_{A}=\mathbb{Z}$ and for any $a,b\in\mathbb{Z}_{A}$, $a\circ
b=\\{a.r.b\ |\ r\in A\\}$.

######  Definition 2.2.

[17] 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.

[1] 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.

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 $R$([17],
proposition 3.2). In this paper, we assume that all hyperideals are
C-hyperideal.

######  Definition 2.4.

[17] A nonzero proper hyperideal $Q$ of $R$ is called a primary hyperideal if
$x\circ y\subseteq Q$ for $x,y\in R$ implies that $x\in Q$ or $y\in\sqrt{Q}$.
Since $\sqrt{Q}=P$ is a prime hyperideal of $R$ by Propodition 3.6 in [17],
$Q$ is referred to as a P-primary hyperideal of $R$.

######  Definition 2.5.

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

######  Definition 2.6.

[1] Let $R$ be a multiplicative hyperring. The element $x\in R$ is said to be
nilpotent if $0\in x^{n}$ for some integer $n>0$.

######  Definition 2.7.

A hyperring $R$ is called an integral hyperdomain, if for all $x,y\in R$,
$0\in x.y$ implies that $x=0$ or $y=0$.

######  Definition 2.8.

[2] An element $a\in R$ is said to be zero divizor if there exists $0\neq b\in
R$ such that $0\in a\circ b.$

######  Definition 2.9.

[20] Let $(R_{1},+_{1},\circ_{1})$ and $(R_{2},+_{2},\circ_{2})$ be two
multiplicative hyperrings. A mapping $f$ from $R_{1}$ into $R_{2}$ is said to
be a good homomorphism if for all $x,y\in R_{1}$, $f(x+_{1}y)=f(x)+_{2}f(y)$
and $f(x\circ_{1}y)=f(x)\circ_{2}f(y)$.
Moreover, the kernel of $f$ is defined by $Kerf=f^{-1}(\langle
0\rangle)=\\{x\in R_{1}\ |\ f(x)\in\langle 0\rangle\\}.$

##  3\.  $\alpha$-prime hyperideals

######  Definition 3.1.

Let $R$ be a multiplicative hyperring and let $\alpha:R\rightarrow R$ be a
fixed good endomorphism. We say that hyperideal $I$ of $R$ is $\alpha$-prime
if for all $x,y\in R$, $x\circ y\subseteq I$ implies $x\in I$ or $\alpha(y)\in
I$.

######  Example 3.2.

Assume that $(\mathbb{Z},+,\cdot)$ is the ring of integers. Consider the
multiplicative hyperring $(\mathbb{Z},+,\circ)$ in which $a\circ
b=\\{2ab,3ab\\}$, for all $a,b\in\mathbb{Z}$. Let $\alpha$ is an identity
mapping on $(\mathbb{Z},+,\circ)$. Then $\langle 2\rangle$ and $\langle
3\rangle$ are $\alpha$-prime hyperideals of $(\mathbb{Z},+,\circ)$.

######  Example 3.3.

Let $(\mathbb{Z},+,\cdot)$ is the ring of integers. Consider the
multiplicative hyperring $(\mathbb{Z},+,\langle 2\rangle)$. Define the mapping
$\alpha:\mathbb{Z}\longrightarrow\mathbb{Z}$ by $\alpha(x)=3x$. By Theorem
4.2.3 in [20], $\alpha$ is a homomorphism of the multiplicative hyperring
$(\mathbb{Z},+,\langle 2\rangle)$. In the multiplicative hyperring, the
hyperideal $\langle 2\rangle$ is $\alpha$-prime.

######  Theorem 3.4.

If $I$ is an $\alpha$-prime hyperideal of $R$, then $\alpha(I)\subseteq I$.

Let $x\in I$. Since $I$ is a ${\bf C}$-hyperideal of $R$ then $x\in 1\circ
x\subseteq I$. Since $I$ is an $\alpha$-prime hyperideal of $R$ and $1\notin
I$ then $\alpha(x)\in I$ for all $x\in I$. Thus we have $\alpha(I)\subseteq
I$.

######  Lemma 3.5.

If $I$ is an $\alpha$-prime hyperideal of $R$, then $\sqrt{I}$ is an
$\alpha$-prime hyperideal of $R$.

###### Proof.

Assume that $a\circ b\subseteq\sqrt{I}$ for $a,b\in R$. This means $(a\circ
b)^{n}=a^{n}\circ b^{n}\subseteq I$. Since $I$ is an $\alpha$-prime hyperideal
of $R$, then we get $a^{n}\subseteq I$ or
$\alpha(b^{n})=(\alpha(b))^{n}\subseteq I$. This implies that $a\in\sqrt{I}$
or $\alpha(b)\subseteq\sqrt{I}$. Thus $\sqrt{I}$ is an $\alpha$-prime
hyperideal of $R$. ∎

######  Theorem 3.6.

Let $I$ is an $\alpha$-prime hyperideal of $R$. Then $E=\\{s\in R\ |\
\alpha(s)\in I\\}$ is an $\alpha$-prime hyperideal of $R$ containing $I$.

###### Proof.

It is clear that $E$ is a hyperideal of $R$ containing $I$. Let $a\circ
b\subseteq E$ for some $a,b\in R$. This means $\alpha(a\circ
b)=\alpha(a)\circ\alpha(b)\subseteq I$. Since $I$ is an $\alpha$-prime
hyperideal of $R$ then $\alpha(a)\in I$ or $\alpha(\alpha(b))\in I$. This
implies that $a\in E$ or $\alpha(b)\in E$. Thus $E=\\{s\in R\ |\ \alpha(s)\in
I\\}$ is an $\alpha$-prime hyperideal of $R$. ∎

######  Lemma 3.7.

Let $I$ is an $\alpha$-prime hyperideal of $R$ and let $I$ be maximal with
respect to the fact that $s\in I$ implies $\alpha(s)\in I$. Then $I$ is prime.

###### Proof.

Assume that $I$ is not prime hyperideal of $R$. Then we get $x,y\in R$ with
$x\circ y\subseteq I$ such that neither $x\in I$ nor $y\in I$. Let $a=p+t\in
I+\langle x\rangle$ for $p\in I$ and $t\in\langle x\rangle$. Then there exists
$r\in R$ with $t\in r\circ x$ such that $a\in p+r\circ x$. Therefore we have
$a\circ y\subseteq(p+r\circ x)\circ y\subseteq p\circ y+r\circ x\circ
y\subseteq I$. Since $I$ is an $\alpha$-prime hyperideal of $R$ and $y\notin
I$ then $\alpha(a)\in I\subseteq I+\langle x\rangle$. By hypothesis, $I$ is
maximal. Hence $I=I+\langle x\rangle$ which implies $x\in I$. This a
contradiction. Thus $I$ is a prime hyperideal of $R$. ∎

######  Theorem 3.8.

The hyperideal $I$ of $R$ is $\alpha$-prime if and only if for any hyperideals
$I_{1},I_{2}$ of $R$, $\hskip 7.11317ptI_{1}\circ I_{2}\subseteq
I\Longrightarrow I_{1}\subseteq I\hskip 4.26773pt\text{or}\hskip
4.26773pt\alpha(I_{2})\subseteq I$.

###### Proof.

Let $I$ is an $\alpha$-prime hyperideal of $R$. Assume that $I_{1}\circ
I_{2}\subseteq I$ for some hyperideals $I_{1},I_{2}$ of $R$ such that
$I_{1}\nsubseteq I$. Then there exists $x\in I_{1}$ but $x\notin I$. Clearly,
$x\circ y\subseteq I_{1}\circ I_{2}\subseteq I$ for all $y\in I_{2}$. Since
the hyperideal $I$ is $\alpha$-prime and $x\notin I$ then $\alpha(y)\in I$ for
all $y\in I_{2}$ which means $\alpha(I_{2})\subseteq I$. Conversely, suppose
that $x\circ y\subseteq I$ for some $x,y\in R$. Hence $\langle
x\rangle\circ\langle y\rangle\subseteq\langle x\circ y\rangle\subseteq I$. By
hypothesis, we get $\langle x\rangle\subseteq I$ or $\alpha(\langle
y\rangle)\subseteq I$ which implies $x\in I$ or $\alpha(y)\in I$. Thus $I$ is
an $\alpha$-prime hyperideal of $R$. ∎

######  Theorem 3.9.

Let $I$ is an $\alpha$-prime hyperideal of $R$ and $S$ is a subset of $R$.
Then $(I:S)$ is an $\alpha$-prime hyperideal of $R$.

###### Proof.

Let $x\circ y\subseteq(I:S)$ for some $x,y\in R$. It is easy to see $(I:x\circ
S)=(I:S)\cup(I:x)$. Since $y\in(I:x\circ S)$ then we have $y\in(I:S)$ or
$y\circ x\subseteq I$. Hence we get $y\in(I:S)$ or $y\in I$ or $\alpha(x)\in
I$, since $I$ is an $\alpha$-prime hyperideal of $R$. This implies that
$y\in(I:S)$ or $\alpha(x)\in(I:S)$. Thus $(I:S)$ is an $\alpha$-prime
hyperideal of $R$. ∎

######  Lemma 3.10.

Let $I$ is an $\alpha$-prime hyperideal of $R$. If $x^{n}\subseteq I$ for some
$x\in R$, then $\alpha(x)\in I$.

###### Proof.

Let $x^{n}\subseteq I$ for some $x\in R$. Assume that $t\in x^{n-1}$. Then
$t\circ x\subseteq I$. Since $I$ is an $\alpha$-prime hyperideal of $R$ then
we have $t\in I$ or $\alpha(x)\in I$. Let $t\in I$. Since $t\in x^{n-1}$ and
$I$ is a C-hyperideal of $R$ then we get $x^{n-1}\subseteq I$. Continuing this
process, we obtain $x\in I$ or $\alpha(x)\in I$. If $x\in I$, then $x\in
1\circ x\subseteq I$, that is, $\alpha(x)\in I$. ∎

######  Theorem 3.11.

Let $I$ is an $\alpha$-prime hyperideal of $R$. If $(\alpha(y))^{n}\subseteq
I$ for some $y\in R$, then $\alpha^{2}(y)\in I$.

###### Proof.

Let $I$ is an $\alpha$-prime hyperideal of $R$ such that
$(\alpha(y))^{n}\subseteq I$ for some $y\in R$. Assume that $x=\alpha(y)$. Now
the claim follows by Lemma 3.10. ∎

######  Definition 3.12.

Let $R$ be a multiplicative hyperring. An element $x$ of $R$ is said to be
$\alpha$-nilpotent, if $0\in\alpha(x^{n})$, for some integer $n>0$. We denote
the set of $\alpha$-nilpotent elements of $R$ by $Nil_{\alpha}(R)$ and call it
the $\alpha$-nilradical of $R$.

######  Theorem 3.13.

The set $Nil_{\alpha}(R)$ of all $\alpha$-nilpotent elements of $R$ with
scalar identity 1, is a hyperideal.

###### Proof.

Suppose that $x\in Nil_{\alpha}(R)$. Then $0\in\alpha(x^{n})$ for some integer
$n>0$. Hence for all $r\in R$, we get
$0\in\alpha(r^{n})\circ\alpha(x^{n})=\alpha((r\circ x)^{n})$ which implies
$r\circ x\in Nil_{\alpha}(R)$. Now, Suppose that $x,y\in Nil_{\alpha}(R)$,
then there exist $n,m\in\mathbb{N}$ such that $0\in\alpha(x^{n})$ and
$0\in\alpha(y^{n})$. Thus we have $0\in\alpha((x-y)^{n+m})$. Therefore $x-y\in
Nil_{\alpha}(R)$. Consequently, $Nil_{\alpha}(R)$ is a hyperideal. ∎

######  Theorem 3.14.

Let $R_{1}$ and $R_{2}$ be two multiplicative hyperrings and
$f:R_{1}\rightarrow R_{2}$ a good homomorphism. If $\alpha\in End(R_{1})\cap
End(R_{2})$ such that $\alpha(f(r))=f(\alpha(r))$ for every $r\in R_{1}$, then
$f^{-1}(I_{2})$ is an $\alpha$-prime hyperideal of $R_{1}$ for some
$\alpha$-prime hyperideal $I_{2}$ of $R_{2}$.

###### Proof.

Assume that the hyperideal $I_{2}$ of $R_{2}$ is $\alpha$-prime. Let
$x\circ_{1}y\subseteq f^{-1}(I_{2})$ for some $x,y\in R_{1}$. Then
$f(x\circ_{1}y)=f(x)\circ_{2}f(y)\subseteq I_{2}$. Since $I_{2}$ is an
$\alpha$-prime hyperideal of $R_{2}$ then $f(x)\in I_{2}$ which implies $x\in
f^{-1}(I_{2})$ or $\alpha(f(y))=f(\alpha(y))\in I_{2}$ which implies
$\alpha(y)\in f^{-1}(I_{2})$. Thus $f^{-1}(I_{2})$ is an $\alpha$-prime
hyperideal of $R_{1}$. ∎

######  Lemma 3.15.

Let $\alpha\in End(R)$. Then $Ker\alpha\subseteq\bigcap_{I{\text{\ }is\
\alpha-prime\ of\ R}}I$.

###### Proof.

Let $r\in Ker\alpha$. Therefore $\alpha(r)\in\langle 0\rangle$. This means
$\alpha(r)$ is in every $\alpha$-prime hyperideal $I$ of $R$. Now by using
Theorem 3.14, the claim can be proved. ∎

######  Lemma 3.16.

Let $\langle 0\rangle$ be a prime hyperideal of $R$ and let $\alpha\in
End(R)$. Then $Ker\alpha$ is a prime hyprideal of $R$.

###### Proof.

Let $a\circ b\subseteq Ker\alpha$ for some $a,b\in R$. This implies that
$\alpha(a\circ b)=\alpha(a)\circ\alpha(b)\subseteq\langle 0\rangle$. Since
$\langle 0\rangle$ is a a prime hyperideal of $R$, then we have
$\alpha(a)\in\langle 0\rangle$ or $\alpha(b)\in\langle 0\rangle$. This implies
that $a\in Ker\alpha$ or $a\in Ker\alpha$. Thus $Ker\alpha$ is a prime
hyprideal of $R$. ∎

######  Theorem 3.17.

Let $R$ be a multiplicative hyperring such that it has zero absorbing
property. If $\langle 0\rangle$ be a prime hyperideal of $R$, then

$Nil_{\alpha}(R)=\bigcap_{I{\text{\ }is\ \alpha-prime\ of\ R}}I.$

###### Proof.

Let $r\in Nil_{\alpha}(R)$. Then we have $0\in\alpha(r^{n})$ which means
$r^{n}\subseteq Ker\alpha$. Thus we get $r\in Ker\alpha$, by Lemma 3.16. Hence
we conclude that $r\in\bigcap_{I{\text{\ }is\ \alpha-prime\ of\ R}}I$. Then
$Nil_{\alpha}(R)$ is in the intersection of all $\alpha$-prime hyperideals of
$R$.
Now, assume that $r\in\bigcap_{I{\text{\ }is\ \alpha-prime\ of\ R}}I$ but
$r\notin Nil_{\alpha}(R)$. Consider the set

$\Sigma=\\{J\ |\ \text{J is a hyperideal of R and for all n>0,
$\alpha(r^{n})\nsubseteq$ J }\\}.$

Since $0\in\Sigma$ then $\Sigma\neq\varnothing$. Order $\Sigma$ by inclusion.
Assume that $\\{J_{i}\\}_{i\in\Delta}$ is a chain of hyperideals in $\Sigma$,
then for each pair of indices $t,s$ we have either $J_{t}\subseteq J_{s}$ or
$J_{s}\subseteq J_{t}$. Let $J=\bigcup_{i\in\Delta}J_{i}$. Clearly, $J$ is a
hyperideal and is an upper bound of the chain. Thus by Zorn’s lemma $\Sigma$
has a maximal element. Let $P$ is a maximal element of $\Sigma$. Assume that
$\alpha(x)\notin P$ and $\alpha(y)\notin P$ for some $x,y\in R$. Hence
$P+\langle\alpha(x)\rangle$ and $P+\langle\alpha(y)\rangle$ are not in
$\Sigma$. Then we have $\alpha(r^{m})\subseteq P+\langle\alpha(x)\rangle$ and
$\alpha(r^{n})\subseteq P+\langle\alpha(y)\rangle$ for some integers $m,n>0$.
Therefore $\alpha(r^{m+n})\subseteq P+\langle\alpha(x\circ y)\rangle$. This
means $P+\langle\alpha(x\circ y)\rangle\notin\Sigma$ which implies
$\alpha(x\circ y)=\alpha(x)\circ\alpha(y)\nsubseteq P$. Thus by Lemma 3.10
$x\circ y\nsubseteq P$. Since $P$ is an $\alpha$-prime hyperideal of $R$ and
$\alpha(r^{n})\nsubseteq P$ then $r\notin P$ which is a contradiction.
Therefore $r\in Nil_{\alpha}(R)$ and the proof is completed. ∎

######  Definition 3.18.

Let $J$ be a hyperideal of $R$ such that $R$ has zero absorbing property. The
$\alpha$-radical of $J$ is defined by

$\sqrt[\alpha]{J}=\\{r\in R\ |\ \alpha(r^{n})\subseteq J\ \text{for some}\
n\in\mathbb{N}\\}$

######  Theorem 3.19.

$Nil_{\alpha}(R)\subseteq\sqrt[\alpha]{\langle 0\rangle}$

###### Proof.

Let $r\in Nil_{\alpha}(R)$. Then there exists some $n\in\mathbb{N}$ such that
$0\in\alpha(r^{n})$. Since $\langle 0\rangle$ is a C-hyperideal and
$0\in\langle 0\rangle$ then $\alpha(r^{n})\subseteq\langle 0\rangle$.
Therefore $r\in\sqrt[\alpha]{\langle 0\rangle}$. Thus
$Nil_{\alpha}(R)\subseteq\sqrt[\alpha]{\langle 0\rangle}$. ∎

If $R$ is a multiplicative hyperring such that it has zero absorbing property,
then we have $Nil_{\alpha}(R)=\sqrt[\alpha]{\langle 0\rangle}$

######  Theorem 3.20.

Let $A,B$ be two hyperideals of $R$. Then we have the following statements:

  * (i)

If $A\subseteq B$, then $\sqrt[\alpha]{A}\subseteq\sqrt[\alpha]{B}$.

  * (ii)

$\sqrt[\alpha]{A+B}\subseteq\sqrt[\alpha]{\sqrt[\alpha]{A}+\sqrt[\alpha]{B}}$

  * (iii)

$\sqrt[\alpha]{A\circ B}=\sqrt[\alpha]{A\cap
B}=\sqrt[\alpha]{A}\cap\sqrt[\alpha]{B}$.

###### Proof.

  * i.

Straightforward.

  * ii.

Since $A\subseteq\sqrt[\alpha]{A}$ and $B\subseteq\sqrt[\alpha]{B}$ then we
have $A+B\subseteq\sqrt[\alpha]{A}+\sqrt[\alpha]{B}$. Thus we get
$\sqrt[\alpha]{A+B}\subseteq\sqrt[\alpha]{\sqrt[\alpha]{A}+\sqrt[\alpha]{B}}$,
by (i).

  * iii.

Here $A\circ B\subseteq A\cap B$. Then $\sqrt[\alpha]{A\circ
B}\subseteq\sqrt[\alpha]{A\cap B}$. Now, let $r\in\sqrt[\alpha]{A\cap B}$. So
$\alpha(r^{n})\subseteq A\cap B$ for some $n\in\mathbb{N}$. Hence we have
$\alpha(r^{n})\circ\alpha(r^{n})=\alpha(r^{2n})\subseteq A\circ B$ which means
$r\in\sqrt[\alpha]{A\circ B}$. Finally, let
$r\in\sqrt[\alpha]{A}\cap\sqrt[\alpha]{B}$. This means $\alpha(r^{s})\subseteq
A$ and $\alpha(r^{t})\subseteq B$ for some $t,s\in\mathbb{N}$. Then we have
$\alpha(r^{m})\subseteq A\cap B$ for $m=max\\{t,s\\}$. This implies that
$r\in\sqrt[\alpha]{A\cap B}$. For the reverse inclusion, since $A\cap
B\subseteq A$ and $A\cap B\subseteq A$ then we get $\sqrt[\alpha]{A\cap
B}\subseteq\sqrt[\alpha]{A}\cap\sqrt[\alpha]{A}$.

∎

######  Theorem 3.21.

Let $A$ be a hyperideal of $R$.

  * (1)

If $\alpha(1)=1$, then $\sqrt[\alpha]{A}=R$ if and only if $I=R$.

  * (2)

If the hyperideal $A$ is $\alpha$-prime, then
$\sqrt[\alpha]{A^{n}}=\sqrt[\alpha]{A}$, for all $n\in\mathbb{N}$.

###### Proof.

  * (1)

Let $\sqrt[\alpha]{A}=R$. This means $1\in\sqrt[\alpha]{A}$. Hence
$\alpha(1^{n})\subseteq A$ which implis $\alpha(1)=1\in A$. Thus $A=R$

  * (2)

Let the hyperideal $A$ be $\alpha$-prime. Then
$\sqrt[\alpha]{A^{n}}=\sqrt[\alpha]{A}\cap...\cap\sqrt[\alpha]{A}$ for all
$n\in\mathbb{N}$, by Theorem 3.20 (iii). Thus
$\sqrt[\alpha]{A^{n}}=\sqrt[\alpha]{A}$.

∎

######  Theorem 3.22.

Let $R_{1}$ and $R_{2}$ be two multiplicative hyperrings and
$f:R_{1}\rightarrow R_{2}$ a good homomorphism such that $I_{1}$ and $I_{2}$
are hyperideals of $R_{1}$ and $R_{2}$, respectively. Assume that $\alpha\in
End(R_{1})\cap End(R_{2})$ such that $\alpha(f(r))=f(\alpha(r))$ for every
$r\in R_{1}$. Then

  * (1)

$f(\sqrt[\alpha]{I_{1}})\subseteq\sqrt[\alpha]{f(I_{1})}$.

  * (2)

$\sqrt[\alpha]{f^{-1}(I_{2})}\subseteq f^{-1}(\sqrt[\alpha]{I_{2}})$.

  * (3)

If $f$ is an isomorphism, then
$f(\sqrt[\alpha]{I_{1}})=\sqrt[\alpha]{f(I_{1})}$

###### Proof.

  * (1)

Let $y\in f(\sqrt[\alpha]{I_{1}})$. Then there exists some
$x\in\sqrt[\alpha]{I_{1}}$ such that $f(x)=y$. Hence we have
$\alpha(x^{n})\subseteq I_{1}$ for some $n\in\mathbb{N}$. Therefore
$\alpha(y^{n})=\alpha(f(x)^{n})=\alpha(f(x^{n}))$. Since $\alpha$ commutes
with $f$, then we get $\alpha(f(x^{n}))=f(\alpha(x^{n}))\subseteq f(I_{1})$.
Thus $y\in\sqrt[\alpha]{f(I_{1})}$ which means
$f(\sqrt[\alpha]{I_{1}})\subseteq\sqrt[\alpha]{f(I_{1})}$.

  * (2)

Let $x\in\sqrt[\alpha]{f^{-1}(I_{2})}$. Then we get $\alpha(x^{n})\subseteq
f^{-1}(I_{2})$ for some $n\in\mathbb{N}$. Therefore $f(\alpha(x^{n}))\subseteq
I_{2}$ which implies $\alpha(f(x)^{n})\subseteq I_{2}$. So $x\in
f^{-1}(\sqrt[\alpha]{I_{2}})$.

  * (3)

Let $f$ is an isomorphism. The claim follows by (1).

∎

######  Theorem 3.23.

Let $\alpha\in End(R)$. Assume that $I$ is a hyperideal of $R$ such that for
all $a,b\in R$, $a\circ b\subseteq I$ implies $a\in I$ or
$\alpha(b^{n})\subseteq I$ for some $n\in\mathbb{N}$. Then $\sqrt[\alpha]{I}$
is an $\alpha$-prime hyperideal of $R$.

###### Proof.

Let $x\circ y\subseteq\sqrt[\alpha]{I}$ for some $x,y\in R$. This means
$\alpha((x\circ y)^{n})=\alpha(x^{n})\circ\alpha(y^{n})\subseteq I$ for some
$n\in\mathbb{N}$. Let $t\in x^{n}$ and $s\in y^{n}$ for some $t,s\in R$.
Therefore
$\alpha(t)\circ\alpha(s)\subseteq\alpha(x^{n})\circ\alpha(y^{n})\subseteq I$.
By assumption, we get $\alpha(t)\in I$ or $\alpha(\alpha(s)^{m})\subseteq I$
for some $m\in\mathbb{N}$. Since $I$ is a C-hyperideal of $R$ and
$\alpha(x^{n})\cap I\neq\varnothing$ or $\alpha(\alpha(y)^{nm})\cap
I\neq\varnothing$, then we have $\alpha(x^{n})\subseteq I$ or
$\alpha(\alpha(y)^{nm})\subseteq I$. This implies that $x\in\sqrt[\alpha]{I}$
or $\alpha(y)\in\sqrt[\alpha]{I}$ which means $\sqrt[\alpha]{I}$ is an
$\alpha$-prime hyperideal of $R$. ∎

######  Theorem 3.24.

Let $I$ be a hyperideal of $R$. Then $I$ is $\alpha$-prime if and only if
every zero divizor of $R/I$ is in $Ker\alpha$.

###### Proof.

Let $I$ be an $\alpha$-prime hyperideal of $R$. Let $0_{R/I}\neq y+I$ be a
zero divizor of $R/P$. Then there exists $0_{R/I}\neq x+I$ such that
$0_{R/I}\in(x+I)(y+I)=x\circ y+I$. This implies that $x\circ y\subseteq I$.
Since the hyperideal $I$ of $R$ is $\alpha$-prime, then we get $x\in I$ or
$\alpha(y)\in I$. Since $0_{R/I}\neq x+I$, then we have $\alpha(y)\in I$ and
so $\alpha(y+I)\subseteq I$which means $y+I$ is in $Ker\alpha$. Conversely,
Let $x\circ y\subseteq I$ such that $x,y\notin I$ for some $x,y\in R$. Then
$I\in x\circ y+I=(x+I)(y+I)$. Then $y+I$ is a zero divizor of $R/P$. By
hypothesis, $\alpha(y+I)\subseteq I$ which means $\alpha(y)\in I$, as claimed.
∎

The following lemma is needed in the proof of our next result.

######  Lemma 3.25.

Let $I$ be a hyperideal of $R$. Then $I$ is prime if and only if $R/I$ has no
zero divisors.

###### Proof.

Let $I$ be a prime hyperideal of $R$. Let $I\neq x+I$ is a zero divisor of
$R/I$. Then there exists $I\neq y+I$ such that $I\in(x+I)(y+I)=x\circ y+I$
which means $x\circ y\subseteq I$. Since $I$ is a prime hyperideal of $R$,
then we get $x\in I$ or $y\in I$, contradicion. Conversely, let for some
$x,y\in R$, $x\circ y\subseteq I$. Then $I\in x\circ y+I=(x+I)(y+I)$. Since
$R/I$ has no zero divisors, then we have $I=x+I$ or $I=y+I$ which means $x\in
I$ or $y\in I$. ∎

######  Theorem 3.26.

Let $I$ be a hyperideal of $R$. Then $I$ is $\alpha$-prime if and only if
$I/Ker\alpha$ is prime in $R/Ker\alpha$.

###### Proof.

By Lemma 3.15, we conclude that $R/I\cong\frac{R/Ker\alpha}{I/Ker\alpha}$.
Now, the claim follows by Lemma 3.25 and Theorem 3.24. ∎

######  Definition 3.27.

A hyperring $R$ is called an $\alpha$-integral hyperdomain, if for all $x,y\in
R$, $0\in x\circ y$ implies that $x=0$ or $\alpha(y)=0$.

######  Theorem 3.28.

Let $I$ be a hyperideal of $R$. Then $I$ is $\alpha$-prime if and only if
$R/I$ is an $\alpha$-integral hyperdomain.

###### Proof.

Let the hyperideal $I$ of $R$ be $\alpha$-prime. Assume that
$I\in(x+I)(y+I)=x\circ y+I$ for some $x,y\in R$. Then $x\circ y\subseteq I$.
Therefore we get $x\in I$ or $\alpha(y)\in I$, since $I$ is a $\alpha$-prime
hyperideal of $R$. Hence we conclude that $x+I=I$ or $\alpha(y)+I=I$ which
implies $x+I=I$ or $\alpha(y+I)=I$. Consequently, $R/I$ is an
$\alpha$-integral hyperdomain. Conversely, Let $R/I$ be an $\alpha$-integral
hyperdomain. Suppose that $x\circ y\subseteq I$ for some $x,y\in R$. Then
$I\in x\circ y+I$. This means $I\in(x+I)(y+I)$. Thus we have $I=x+I$ or
$I=\alpha(y+I)$, since $R/I$ is an $\alpha$-integral hyperdomain. This means
$x\in I$ or $\alpha(y)\in I$. Thus the hyperideal $I$ of $R$ is
$\alpha$-prime. ∎

######  Theorem 3.29.

Let $R_{1}$ and $R_{2}$ be two multiplicative hyperrings and
$f:R_{1}\rightarrow R_{2}$ a good epimorphism and $\alpha\in End(R_{1})\cap
End(R_{2})$ such that $\alpha(f(r))=f(\alpha(r))$ for every $r\in R_{1}$. Let
$I_{1}$ be a hyperideal of $R_{1}$ with $Ker\alpha\subseteq I_{1}$. Then the
hyperideal $I_{1}$ is $\alpha$-prime if and only if the hyperideal $f(I_{1})$
of $R_{2}$ is $\alpha$-prime.

###### Proof.

Let $a_{2}\circ b_{2}\subseteq f(I_{1})$ for some $a_{2},b_{2}\in R_{2}$. Then
for some $a_{1},b_{1}\in R_{1}$ we have $f(a_{1})=a_{2}$ and $f(b_{1})=b_{2}$.
So $f(a_{1})\circ f(b_{1})=f(a_{1}\circ b_{1})\subseteq f(I_{1})$. Now, take
any $u\in a_{1}\circ b_{1}$. Then $f(u)\in f(a_{1}\circ b_{1})\subseteq
f(I_{1})$ and so there exists $w\in I_{1}$ such that $f(u)=f(w)$. This means
that $f(u-w)=0$, that is, $u-w\in Kerf\subseteq I_{1}$ and then $u\in I_{1}$.
Since $I_{1}$ is a C-hyperideal of $R_{1}$, then we get $a_{1}\circ
b_{1}\subseteq I_{1}$. Since $I_{1}$ is an $\alpha$-prime hyperideal of
$R_{1}$, then we obtain $a_{1}\in I_{1}$ or $\alpha(b_{1})\in I_{1}$. This
implies that $f(a_{1})=a_{2}\in f(I_{1})$ or
$\alpha(b_{2})=\alpha(f(b_{1}))=f(\alpha(b_{1}))\in f((I_{1})$. Thus
$f(I_{1})$ is an $\alpha$-prime hyperideal of $R_{2}$. The converse part is
follows by 3.14. ∎

In view of Theorem 3.29, we have the following result.

######  Corollary 3.30.

Let $I$ and $J$ be two hyperideals of $R$ with $J\subseteq I$. Assume that
$\alpha\in End(R)$ and $\alpha^{\star}$ is the induced mapping on $R/J$ from
$\alpha$. Then $I$ is an $\alpha$-prime hyperideal of $R$ if and only if $I/J$
is an $\alpha^{\star}$-prime hyperideal of $R/J$.

Let $(R_{1},+_{1},\circ_{1})$ and $(R_{2},+_{2},\circ_{2})$ be two
multiplicative hyperrings with non zero identity. [ul] 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

$(x_{1},x_{2})\circ(y_{1},y_{2})=\\{(x,y)\in R_{1}\times R_{2}\ |\ x\in
x_{1}\circ_{1}y_{1},y\in x_{2}\circ_{2}y_{2}\\}$.
Assume that $\alpha_{1}\in End(R_{1})$ and $\alpha_{2}\in End(R_{2})$. We
define the map $\bar{\alpha}:R_{1}\times R_{2}\longrightarrow R_{1}\times
R_{2}$ by $\bar{\alpha}(r_{1},r_{2})=(\alpha_{1}(r_{1}),\alpha_{2}(r_{1}))$.
It is easy to see that $\bar{\alpha}\in End(R_{1}\times R_{2})$.

######  Theorem 3.31.

Let $(R_{1},+_{1},\circ_{1})$ and $(R_{2},+_{2},\circ_{2})$ be two
multiplicative hyperrings with non zero identity such that $\alpha_{1}\in
End(R_{1})$ and $\alpha_{2}\in End(R_{2})$. Let $I_{1}$ be a hyperideal of
$R_{1}$. Then $I_{1}$ is an $\alpha_{1}$-prime hyperidea of $R_{1}$ if and
only if $I_{1}\times R_{2}$ is an $\bar{\alpha}$-prime hyperideal of
$R_{1}\times R_{2}$.

###### Proof.

($\Longrightarrow$) Let $(x_{1},x_{2})\circ(y_{1},y_{2})\subseteq I_{1}\times
R_{2}$ for some $(x_{1},x_{2}),(y_{1},y_{2})\in R_{1}\times R_{2}$. This means
$x_{1}\circ_{1}y_{1}\subseteq I_{1}$. Since $I_{1}$ is a $\alpha_{1}$-prime
hyperideal of $R_{1}$, then we get $x_{1}\in I_{1}$ or $\alpha_{1}(y_{1})\in
I_{1}$. This implies that $(x_{1},x_{2})\in I_{1}\times R_{2}$ or
$\bar{\alpha}(y_{1},y_{2})=(\alpha_{1}(y_{1}),\alpha_{2}(y_{2})\in I_{1}\times
R_{2}$. Consequently, $I_{1}\times R_{2}$ is an $\bar{\alpha}$-prime
hyperideal of $R_{1}\times R_{2}$.

($\Longleftarrow$) Assume on the contrary that $I_{1}$ is not a
$\alpha_{1}$-prime hyperideal of $R_{1}$. So $x_{1}\circ_{1}y_{1}\subseteq
I_{1}$ with $x_{1},y_{1}\in R_{1}$ implies that $x_{1}\notin I_{1}$ and
$\alpha_{1}(y_{1})\notin I_{1}$. It is clear that
$(x_{1},1_{R_{2}})\circ(y_{1},1_{R_{2}})\subseteq I_{1}\times R_{2}$. Since
$I_{1}\times R_{2}$ is an $\bar{\alpha}$-prime hyperideal of $R_{1}\times
R_{2}$, then we have $(x_{1},1_{R_{2}})\in I_{1}\times R_{2}$ or
$\bar{\alpha}(y_{1},1_{R_{2}})\in I_{1}\times R_{2}$ which means
$(x_{1},1_{R_{2}})\in I_{1}\times R_{2}$ or
$(\alpha_{1}(y_{1}),\alpha_{2}(1_{R_{2}})\in I_{1}\times R_{2}$. Hence we get
$x_{1}\in I_{1}$ or $\alpha_{1}(y_{1})\in I_{1}$ which is a contradiction.
Thus, $I_{1}$ is an $\alpha_{1}$-prime hyperideal of $R_{1}$. ∎

######  Theorem 3.32.

Let $(R_{1},+_{1},\circ 1)$ and $(R_{2},+_{2},\circ_{2})$ be two
multiplicative hyperrings with non zero identity such that $\alpha_{1}\in
End(R_{1})$ and $\alpha_{2}\in End(R_{2})$. Let $I_{1}$ and $I_{2}$ be some
hyperideals of $R_{1}$ and $R_{2}$, respectively. Then the following
statements are equivalent:

  * (1)

$I_{1}\times I_{2}$ is an $\bar{\alpha}$-prime hyperideal of $R_{1}\times
R_{2}$.

  * (2)

$I_{1}=R_{1}$ and $I_{2}$ is an $\alpha_{2}$-prime hyperideal of $R_{2}$ or
$I_{2}=R_{2}$ and $I_{1}$ is an $\alpha_{1}$-prime hyperideal of $R_{1}$.

###### Proof.

(1) $\Longrightarrow$ (2) Assume that $I_{1}=R_{1}$. Then $I_{2}$ is a
$\alpha_{2}$-primary hyperideal of $R_{2}$, by Theorem 3.31.

(2) $\Longrightarrow$ (1) This can be proved by using Theorem 3.31. ∎

######  Example 3.33.

Suppose that $(\mathbb{Z},+,.)$ is the ring of integers. Then
$(\mathbb{Z},+,\circ_{1})$ is a multiplicative hyperring with a hyperoperation
$a\circ_{1}b=\\{ab,7ab\\}$. Also, $(\mathbb{Z},+,\circ_{2})$ is a
multiplicative hyperring with a hyperoperation $a\circ_{2}b=\\{ab,5ab\\}$.
Note that $(\mathbb{Z}\times\mathbb{Z},+,\circ)$ is a multiplicative hyperring
with a hyperoperation $(a,b)\circ(c,d)=\\{(x,y)\in\mathbb{Z}\times\mathbb{Z}\
|\ x\in a\circ_{1}c,y\in b\circ_{2}d\\}$. Let $\alpha_{1}$ and $\alpha_{2}$
are the identity maps on $(\mathbb{Z},+,\circ_{1})$ and
$(\mathbb{Z},+,\circ_{2})$, respectively. Clearly, $7\mathbb{Z}=\\{7t\ |\
t\in\mathbb{Z}\\}$ and $5\mathbb{Z}=\\{5t\ |\ t\in\mathbb{Z}\\}$ are
$\alpha_{1}$-prime and $\alpha_{2}$-prime of $(\mathbb{Z},+,\circ_{1})$ and
$(\mathbb{Z},+,\circ_{2})$, respectively. Since $(5,0)\circ(0,7)\subseteq
7\mathbb{Z}\times 5\mathbb{Z}$ but $(5,0),(0,7)\notin 7\mathbb{Z}\times
5\mathbb{Z}$ and $\bar{\alpha}(5,0),\bar{\alpha}(0,7)\notin 7\mathbb{Z}\times
5\mathbb{Z}$, then $7\mathbb{Z}\times 5\mathbb{Z}$ is not a
$\bar{\alpha}$-prime hyperideal of $\mathbb{Z}\times\mathbb{Z}$.

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