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\begin{document}
\title{On regular multiplicative hyperrings}
% Please mark \corrauth after the name of the corresponding author.
\author{R. Ameri, A. Kordi}
\address{School of Mathematics, Statistics and Computer
Science, College of Sciences,\\ Department of Mathematics, University of Tehran, Tehran, Iran.}
\emails{{\tt rameri@ut.ac.ir}\ (R. Ameri), {\tt ali.kordi@gmail.com}\ (A. Kordi)}
\begin{abstract}
We introduce and study regular multiplicative hyperrings, as a
generalization of classical rings. Also, we use the fundamental
relation $\gamma ^{\star}$ on a given regular multiplicative hyperring $R$
and prove that the fundamental ring $R/\gamma ^{\star}$ of $R$
is a regular ring. Finally, we investigate the algebraic properties
of $M(R)$, the regular hyperideal of $R$, generated by all elements
of $R$ such that its generated hyperideal is regular.
\end{abstract}
\keywords{multiplicative hyperring, regular hyperring, fundamental relation, hyperideal, regular element, idempotent.}
\ams{20N20.}
\maketitle
\section{Introduction and Primary}
The theory of hyperstructures has been introduced by Marty in 1934
during the $8^{th}$ Congress of the Scandinavian Mathematicians
\cite{MA}. Marty introduced hypergroups as a generalization of
groups. He published some notes on hypergroups, using them in
different contexts as algebraic functions, rational fractions, non
commutative groups and then many researchers have been worked on
this new field of modern algebra and developed it. It was later
observed that the theory of hyperstructures has many applications in
both pure and applied sciences; for example, semi-hypergroups are
the simplest algebraic hyperstructures that possess the properties
of closure and
associativity. The theory of hyperstructures has been widely reviewed \cite{MA, CO1, CO2, DA3, VO1}.\\
In \cite{CO2} Corsini and Leoreanu-Fotea have collected numerous
applications of algebraic hyperstructures, especially those from the
last fifteen years to the following subjects: geometry, hypergraphs,
binary relations, lattices, fuzzy sets and rough sets, automata,
cryptography, codes, median algebras, relation algebras, artificial
intelligence, and probabilities. The hyperrings were introduced and
studied by Krasner \cite{KR}, Nakasis \cite{NA}, Massourouce
\cite{MA} and especially studied by Davvaz and Leoreanu-Fotea
\cite{DA4}, Zahedi and Ameri \cite{ZA}, Ameri and Norouzi \cite{A3,
A4} . The study on hyperrings in \cite{VO1} ends with an outline of
applications in chemistry and physics, analyzing several special
kinds of hyperstructures: $e$-hyperstructures and transposition
hypergroups. The theory of suitable modified hyperstructures can
serve as a mathematical background in the field of quantum
communication systems. A well-known type of a hyperring, called the
Krasner hyperring \cite{KR}. Krasner hyperrings are essentially
hyperrings, with approximately modified axioms in which addition is
a hyperoperation, while the multiplication is an operation. Then,
this concept has been studied by a variety of authors. Some
principal notions of hyperring theory can be found in \cite{DA3,
DA4, MI, SP, VO2}. The another type of hyperrings was introduced by
Rota in 1982 which the multiplication is a hyperoperation, while
the addition is an operation, and it is called it a multiplicative
hyperring( for more details see \cite{RO1, RO2, RO3, RO4}) which was
subsequently investigated by Olson and Ward \cite{OL} and many
others. De Salvo \cite{DE} introduced hyperrings in which the
additions and the multiplications are hyperoperations. Moreover,
there exists another types of hyperrings that both the addition and
multiplication are hyperoperations and instead associativity,
commutativity and distributivity satisfy in weak associativity, weak
commutativity and weak distributivity, which is called
$H_v$-hyperrings, this type of hyperrings can be seen in \cite{VO1,
VO2}. Also, there are other types of hyperrings which were
completely studied in \cite{DA3}. These hyperrings are studied by
Rahnamai Barghi \cite{RA1}. Procesi and Rota in \cite{PR1} have
studied ring of fractions in Krasner hyperrings and also they
conceptualized in \cite{PR2} the notion of primeness of hyperideal
in a multiplicative hyperring, and in \cite{DA}, Dasgupta extended
the prime and primary hyperideals in multiplicative hyperrings.
Asokkumar and Velrajan \cite{AS1, AS2} have studied Von Neumann
regularity in Krasner hyperrings.\\ A special equivalence relations
which is 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. The fundamental relation
${\beta}^{\ast}$ on hypergroups was defined by Koskas \cite{KO},
mainly studied by Corsini \cite{MA}, Freni \cite{FR1, FR2},
Vougiouklis \cite{VO2}( for more details about hyperrings and
fundamental relations on hyperrings see \cite{A3, A4, DA1, DA3, SP,
VO2}). In this paper we consider the classes of multiplicative
hyperring as a hyperstructures $(R, +, .)$, where $(R, +)$ is an
abelian group, $(R, +)$ is a semihypergroup and the hyperoperation
$"."$ is distributive with respect to the operation $"+"$, i.e.
$a.(b+c)\subseteq a.b +a.c$. The purpose of this paper is the study
study regular multiplicative hyperrings. In this regards we study
the properties of regular multiplicative hyperring $R$ and obtain
some results. We will proceed to use the fundamental relation
$\gamma^{\ast}$ on $R$ and prove that the fundamental ring
$R/\gamma^{\ast}$ of $R$ is regula whenever $R$ is regular. Also, we
show that this process make a functor from the category of regular
multiplicative hyperrings to the category of regular rings. Finally,
the notion of regular hyperideal $M(R)$, consisting of the elements
of $R$ such that the generated hyperideal by these elements are
regular hyperideals, are introduced and its basic properties are
investigated.
\section{Regular multiplicative hyperring}
Recall that a {\it hyperoperation} $"."$ on nonempty set $H$ is a
mapping of $H\times H$ into the family of all nonempty subsets of
$H$. Let $"."$ be a hyperoperation on $H$. Then, $(H,.)$ is called a
{\it hypergroupoid}. we can extend the hyperoperation on $H$ to
subsets of $H$ as follows. For $A, B\subseteq H$ and $h\in H$, then
$AB=\cup_{a\in A,b\in B} ab,\\ Ah=A\{h\}, hB=\{h\}B$. A {\it
semihypergroup} is a hypergroupoid $(H,.)$, which is associative,
that is $(a.b).c=a.(b.c)$ or fall $a, b, c\in H$. A {\it
hypergroup} is a semihypergroup $(H,.)$, that satisfies the {\it
reproduction axioms}, that is $a.H=H=H.a$ for all $a\in H$
A non-empty set $R$ with two hyperoperations $+$ and $.$ is said to be a {\it hyperring} if $(R, +)$
is a {\it canonical hypergroup,} $(R,.)$ is a semihypergroup with $r.0=0.r=0$ for all $r\in R$ ($0$ as
a bilaterally absorbing element) and the hyperoperation $.$ is distributive
with respect to $+$, i.e., for every $a,b,c\in R; a(b+c)=ab+ac$ and $(a+b)c=ac+bc$.\\
A {\it multiplicative hyperring} is an additive commutative group
$(R,+)$ endowed with
a hyperoperation $.$ which satisfies the following conditions:\\
$(1.)$ $\forall a,b,c\in R: a(bc)=(ab)c$;\\
$(2.)$ $\forall a,b,c\in R: (a+b)c\subseteq ac+bc, a(b+c)\subseteq ab+ac$;\\
$(3.)$ $\forall a,b\in R: (-a)b=a(-b)=-(ab)$.\\
If in $(2)$ we have equalities instead of inclusions, then we say
that the multiplicative hyperring is {\it strongly distributive}.
\begin{definition}\label{some label}
Let $R$ be a multiplicative hyperring. Then
$(i)$ an element $e\in R$ is said to be a left(resp. right)identity if $a\in e.a$(resp.
$a\in a.e)$ for $a\in R$. An element $e$ is called an {\it identity
element} if it is both left and right identity element. $(ii)$ an
element $e\in R$ is said to be an scalar left(resp. right) scalar
identity if $a= e.a(resp.,~ a=a.e)$ for $a\in R$ $R$. An element $e$
is called an {\it scalar identity element} if it is both left and
right identity element.
$(iii)$ Let $R$ be a multiplicative hyperring with an identity $e$.
An element $A$ is called a left (right) invertible (with respect to
$e$), if there exists $x\in R$, such that $e\in xa (e\in ax)$ and
$a$ is called invertible if it is both a left and a right
invertible.
A multiplicative hyperring $R$ is called a {\it left (right)
invertible} if every element of $R$ has a left (right) invertible
and $R$ is called {\it invertible} if it is both a left and a right
invertible. Denote the set of all invertible elements in $R$ by
$U(R)$ (with respect to the identity $e$ by $U_e(R)$.
\end{definition}
\begin{definition}
Let $R$ be a multiplicative hyperring. We call $a\in R$ is {\it
regular} if there exists $x\in R$ such that $a\in axa$. So, we can
define that $R$ is regular multiplicative hyperring, if all of
elements in $R$ are regular elements. The set of all regular
elements in $R$ is denoted by $V(R)$.
\end{definition}
\begin{example} Let $(R,+,.)$ be the regular commutative ring with an unitary element.
For every subset $A\in P^{\ast}(R)=P(R)-\{\emptyset\}, |A|\geq 2$,
and $1\in A$, define a multiplicative hyperring $(R_A,+,\circ)$,
where $R_A=R$ and for all $x,y\in R_A, x\circ y=\{xay|a\in A\}$.
Then $(R_A,+,\circ)$ is a regular multiplicative hyperring. Since, for all $a\in R$,
there
exists $r\in R$ such that $a=ara$. Now, by setting $x=r$ we have,\\
$a\circ x\circ a=\{asx|s\in A\}\circ a=\{asxta|s,t\in
A\}=\{axast|s,t\in
A\}=\{ast|s,t\in A\}, $\\
since $1\in A$, we have $a\in a\circ x\circ a$. Hence
$(R_A,+,\circ)$ is regular.
\end{example}
\begin{example} Let $(R,+,.)$ be a non-zero regular ring and for all $a,b\in R$. Define
a hyperoperation $a\circ b=\{a.b, 2a.b, 3a.b,\dots\}$. Then
$(R,+,\circ)$ is a regular multiplicative hyperring, which is not
strongly distributive. Because, for all $a\in R$, there exists $r\in
R$ such that $a=ara$. Now by setting $x=r$ we have
\[a\circ x\circ a=\{ar,2ar,\cdots, nar,\cdots\}\circ a=\{ara,2ara,\cdots, nara,\cdots\}.
\]
Then $a\in a\circ x\circ a$.
\end{example}
\subsection{Applications of the $\gamma^{\ast}$-relation in regular multiplicative hyperrings}
Let $(R,+, .)$ be a hyperring. We define the relation $\gamma$ as follows:\\
$a\gamma b$ if and only if $\{a,b\}\subseteq U$ where $U$ is a
finite sum of finite products of elements of $R$, i.e.,
\[
a\gamma b\Leftrightarrow \exists z_1,\cdots,z_n\in
R~such~that~\{a,b\}\subseteq \sum_{j\in J}\prod_{i\in I_j}z_i;
~~I_j,J\subseteq \{1,\cdots,n\}.
\]
We denote the transitive closure of $\gamma$ by $\gamma^{\ast}$. The
relation $\gamma^{\ast}$ as the smallest equivalence relation on a
multiplicative hyperring $(R,+,.)$ such that the quotient
$R/\gamma^{\ast}$, the set of all equivalence classes, is a
fundamental ring. Let $\mathcal{U}$ be the set of all finite sums of
products of elements of $R$ we can rewrite the definition of
$\gamma^{\ast}$ on $R$ as follows:\\
$a\gamma^{\ast} b\Leftrightarrow \exists z_1,\cdots,z_{n+1}\in R$
with $z_1=a,z_{n+1}=b$ and $u_1,\cdots,u_n\in \mathcal{U}$ such that
$\{z_i,z_{i+1}\}\subseteq u_i$ for $i\in \{1,\cdots,n\}$.\\
Suppose that $\gamma^{\ast}(a)$ is the equivalence class containing
$a\in R$. Then, both the sum $\oplus$ and the product $\odot$ in
$R/\gamma^{\ast}$ are defined as
follows:$\gamma^{\ast}(a)\oplus\gamma^{\ast}(b)=\gamma^{\ast}(c)$
for all $c\in \gamma^{\ast}(a)+\gamma^{\ast}(b)$ and
$\gamma^{\ast}(a)\odot\gamma^{\ast}(b)=\gamma^{\ast}(d)$ for all
$d\in \gamma^{\ast}(a).\gamma^{\ast}(b)$. Then $R/\gamma^{\ast}$ is
a ring, which is called fundamental ring of $R$ (see also
\cite{VO1}).
\begin{theorem}\label{1}
Let $R$ be a regular multiplicative hyperring. Then
$R/\gamma^{\star}$ is regular ring.
\end{theorem}
\begin{proof} Assume that $x\in R/\gamma^{\ast}$.
Thus there is a $r\in R$ such that $x=\gamma^{\ast}(r)$. Since $R$
is a regular hyperring, then there exists a $r'\in R$ such that $r\in
rr'r$. So $\gamma^{\ast}(r)=\gamma^{\ast}(rr'r)=\gamma^{\ast}(r)
\odot\gamma^{\ast}(r')\odot\gamma^{\ast}(r)$. Therefore
$R/\gamma^{\ast}$ is a regular ring.
\end{proof}
\begin{remark}\label{2} The converse of Theorem \ref{1} is not valid.
For example let $(R,+,.)$ be a non regular ring. Consider $(R,+,.)$
as a hyperring under operators "$+$" and "$.$". Clearly
$R/\gamma^{\ast}\cong R$.
\end{remark}
\begin{definition} A multiplicative hyperring $R$ is said multiplicatively $n$-complete ($n$-MC) if for all
$x_1,\cdots, x_n\in R$ we have
\[\gamma(\prod_{i=1}^nx_i)=\prod_{i=1}^nx_i
\]
Also [$11$], we called that a hyperring $R$ is $n$-complete if for
all $(k_1,\cdots,k_n)\in \Bbb{N}^n$ and for all
$(x_{1j},\cdots,x_{ik_i})\in R^{k_i}$ we have
\[\gamma(\sum_{i=1}^n(\prod_{j=1}^{k_i}x_{ij}))=\sum_{i=1}^n(\prod_{j=1}^{k_i}x_{ij}).
\]
\end{definition}
\begin{theorem}\label{3} Let $R$ be a $3$-MC multiplicative hyperring.
If $R/\gamma^{\ast}$ is a regular ring then $R$ is a regular
multiplicative hyperring.
\end{theorem}
\begin{proof} Assume that $a\in R$. Then there exists $r\in R$ such that
\[\gamma^{\ast}(a)\odot\gamma^{\ast}(r)\odot\gamma^{\ast}(a)=\gamma^{\ast}(a).
\]
Then
\[\gamma^{\ast}(a)=\gamma^{\ast}(ara).\]
Thus $a\in \gamma^{\ast} (ara)=ara$, since $R$ is $3$-MC and hence
$a\in ara$.
\end{proof}
\begin{proposition}\label{4}(\cite{RA1}) Let $R$ be a commutative ring and $a\in V(R)$.
Then there is a unique $x\in R$ with $axa=a$ and $xax=x$.
\end{proposition}
Obviously, if $R$ is a regular commutative multiplicative hyperring, then
for each $a\in R$ there exists a unique element $\gamma^{\ast}(x)$
in $R/\gamma^{\ast}$ such that $\gamma^{\ast}(axa)=\gamma^{\ast}(a)$
and $\gamma^{\ast}(xax)=\gamma^{\ast}(x)$.
\begin{theorem}\label{5} If $R$ is a commutative $3$-MC multiplicative hyperring and $a\in R$ be
a regular element of $R$. Then there exists $x\in R$ such that
$a\in axa$ and $x\in xax$.
\end{theorem}
\begin{proof} Since $R$ is a commutative multiplicative hyperring then it is easy to check that
$R/\gamma^{\ast}$ is commutative ring. As $a\in R$ is a regular
element then $\gamma^{\ast}(a)$ is a regular element of
$R/\gamma^{\ast}$, by Theorem \ref{1}, then, by Proposition \ref{4},
there is a unique $\gamma^{\ast}(x)\in R/\gamma^{\ast}$ for $x\in
R$ such that\\
$\gamma^{\ast}(a)\odot\gamma^{\ast}(x)\odot\gamma^{\ast}(a)=\gamma^{\ast}(a)$
and
$\gamma^{\ast}(x)\odot\gamma^{\ast}(a)\odot\gamma^{\ast}(x)=\gamma^{\ast}(x)
$.\\
Then $\gamma^{\ast}(axa)=\gamma^{\ast}(a)$ and
$\gamma^{\ast}(xax)=\gamma^{\ast}(x).$ Therefore, $a\in axa$ and
$x\in xax$, since $R$ is $3$-MC.
\end{proof}
\begin{definition} Let $R$ be a multiplicative hyperring. Then we called
that $M_n(R)$, as the set of all hypermatixes of $R$. Also we called
that for all $\mathcal{A}=(A_{ij})_{n\times
n},\mathcal{B}=(B_{ij})_{n\times n}\in P^{\ast}(M_n(R))$,
$\mathcal{A}\subseteq \mathcal{B}$ if and only if $A_{ij}\subseteq
B_{ij}$.
\end{definition}
\begin{remark}\label{6} Let $R$ be a multiplicative hyperring with a
scalar identity $1$ and $M_n(R)$ denotes the set of all $n\times n$
matrices with entries in $R$. It is easy to verify that $M_n(R)$ is
a noncommutative multiplicative hyperring with unitary element under
usual matrix operations. Let $A=(a_{ij})_{n\times n}$ be a matrix,
where $a_{rs}=ab$ for some $1\leq r,s\leq n$ and in other positions
$a_{ij}=0$. Then $A=BC$, where $B=(b_{ij})_{n\times n}$ such that
$b_{rs}=a$ and $C=(c_{ij})$ such that $c_{ss}=b$ and in other
entries of $B,C$ we have $b_{ij}$ and $c_{ij}=0$.
\end{remark}
Recall that $R$ has a {\it zero absorbing property } if for all
$a\in R, 0\circ a=a\circ 0=\{0\}$.
\begin{theorem}\label{7} Let $(R,+,.)$ be a multiplicative hyperring such that it has zero
absorbing property. Then
\[M_n(R)/\gamma^{\ast}\cong M_n(R/\gamma^{\ast}).
\]
\end{theorem}
\begin{proof} Consider the projection homomorphism $\phi:M_n(R)\rightarrow M_n(R/\gamma^{\ast})$
defined by $\phi((a_{ij})_{n\times n})=(\gamma^{\ast}(a_{ij}))_{n\times n}$ for all $a_{ij}\in R$
and $1\leq i,j\leq n$. We denote the equivalence relation associated with $\phi$ by $\rho$. That is,
\begin{eqnarray*}
(a_{ij})_{n\times n}\rho (b_{ij})_{n\times n} \ \
&\Longleftrightarrow & \ \ ({\gamma}^{\ast}(a_{ij}))_{n\times n}\\
&=&({\gamma}^{\ast}(b_{ij}))_{n\times n},\\ \ \ \ \ \forall
a_{ij}, b_{ij}\in R, \ \ \ 1\leq i, j\leq n.
\end{eqnarray*}
In fact, $\rho=ker(\phi)$. Since $\phi$ is an epimorphism, we have
\[M_n(R)/\rho=M_n(R)/ker(\phi)\cong M_n(R/\gamma^{\ast}).
\]
We know that $M_n(R/\gamma^{\ast})$ is a ring, and so $M_n(R)/\rho$
is a ring. Thus $\gamma^{\ast}\subseteq \rho$, since $\gamma^{\ast}$
is the smallest equivalence
relation on $M_n(R)$ such that $M_n(R)/\gamma^{\ast}$ is a ring. \\
Let $(a'_{ij})_{n\times n}\rho(a_{ij})_{n\times n}$ for all
$a_{ij}\in R$ and $1\leq i,j\leq n$. Hence
\[
(\gamma^{\ast}(a'_{ij}))_{n\times n}=
(\gamma^{\ast}(a_{ij}))_{n\times n}\Longleftrightarrow
\gamma^{\ast}(a'_{ij})= \gamma^{\ast}(a_{ij}), \ \ \ \forall
a_{ij}\in R
\]
Then we conclude that
\[
a_{ij}, a'_{ij} \in \sum_{s=1}^{m}\prod_{t=1}^{k_s}x_{st}
\]
for some $(x_{s1},...,x_{sk_s})\in R^{k_s}$ and $1\leq i,j\leq n$.
Then we have
\[
(a_{ij})_{n\times n},(a'_{ij})_{n\times n}\in (
\sum_{s=1}^{m}\prod_{t=1}^{k_s}x^{ij}_{st})_{n\times n}=
\sum_{s=1}^m(\prod_{t=1}^{k_s}x^{ij}_{st})_{n\times
n}=\sum_{s=1}^m\sum_{i=1,j=1}^nA^{ij} ,\]
where $A^{ij}=(b_{pq})_{n\times n}$,
such that
\[ b_{pq}=\left\{\!
\begin{array}{ll}
\prod_{t=1}^{k_s}x^{ij}_{st}&\text{if $p=i,q=j$},\\
0 &\text{; otherwise}
\end{array}\right.
\]
now by the Remark \ref{6}, we have $A^{ij}=B^{ij}(B^{jj})^{k_s-1}$
for $s=1,\cdots m$, where $B^{ij}=(c_{uv})_{n\times n}$, where
\[ c_{uv}=\left\{\!
\begin{array}{ll}
x^{ij}_{st}&\text{if $u=s,v=t$},\\
0 &\text{; otherwise}
\end{array}\right.
\]
so we have
\[\{(a_{ij})_{n\times n},(a'_{ij})_{n\times n}\}\in \sum_{s=1}^m\sum_{i=1,j=1}^nA^{ij}
=\sum_{s=1}^m\sum_{i=1,j=1}^nB^{ij}(B^{jj})^{k_s-1},\]
i.e., $(a_{ij})_{n\times n}\gamma (a'_{ij})_{n\times n}$. Hence
$(a_{ij})_{n\times n}\gamma^{\ast}(a'_{ij})_{n\times n}$. Consequently,
\[
(a'_{ij})_{n\times n}\in \gamma^{\ast}((a_{ij})_{n\times n})
\]
and therefore $\rho\subseteq \gamma^{\ast}$. Then
$\gamma^{\ast}=\rho$ and so $M_n(R)/\gamma^{\ast}\cong
M_n(R/\gamma^{\ast})$.
\end{proof}
\begin{theorem}\label{8} Let $R$ be a multiplicative hyperring. $R$ is $n$-complete ($n$-MC)
if and only if $M_n(R)$ is $n$-complete ($n$-MC).
\end{theorem}
\begin{proof} ($\Rightarrow$) Assume that $R$ is $n$-complete and
$T=(a_{ij})_{n\times n}\in \gamma(\sum_{s=1}^n(\prod_{t=1}^{k_s}
(x_{st})_{n\times n}))$, then
\[\{(a_{ij})_{n\times n}, \sum_{s=1}^n(\prod_{t=1}^{k_s} (x_{st})_{n\times n})\}\subseteq
\sum_{z=1}^n(\prod_{\ell=1}^{w_{z}} (y_{uv})^{z\ell}_{n\times n}).
\]
Now, for convenience, let $A=(A_{ij})_{n\times n}=\sum_{s=1}^n(\prod_{t=1}^{k_s} (x_{st})_{n\times n})$ and
$B=(B_{ij})_{n\times n}=\sum_{z=1}^n(\prod_{\ell=1}^{w_{z}} (y_{uv})^{z\ell}_{n\times n})$,
then $a_{ij}\in B_{ij}$, $A_{ij}\subseteq B_{ij}$, so $\gamma(a_{ij})=\gamma(B_{ij}),\gamma(A_{ij})=\gamma(B_{ij})$.
Since $R$ is $n$-complete, then $a_{ij}\in \gamma(B_{ij})=\gamma(A_{ij})=A_{ij}$, i.e., $a_{ij}\in A_{ij}$ for all $1\leq i,j\leq n$.
Hence $(a_{ij})_{n\times n}\in A$, and so $\gamma(\sum_{s=1}^n(\prod_{t=1}^{k_s} (x_{st})_{n\times n}))\subseteq A$.\\
($\Leftarrow$) Suppose that $M_n(R)$ is $n$-complete. Let $x\in
\gamma(\sum_{i=1}^{n}(\prod_{j=1}^{k_i}x_{ij}))$. Thus
\[\{x,\sum_{i=1}^{n}(\prod_{j=1}^{k_i}x_{ij})\} \subseteq \sum_{s=1}^{n}(\prod_{t=1}^{k_s}y_{st}).\]
Since $M_n(R)$ is $n$-complete then $x\in
\sum_{i=1}^{n}(\prod_{j=1}^{k_i}x_{ij})$, i.e., $R$ is $n$-complete.
\end{proof}
\begin{proposition}\label{10}(\cite{AS4}) Let $R$ be a strongly distributive multiplicative hyperring and $a\in R$.
If there exists $x\in R$ and $c\in axa-a$ such that $c$ is regular,
then $a$ is regular.
\end{proposition}
In 1950, Brown and McCoy $[6]$ proved that, $R$ is a regular ring if
and only if $M_n(R)$ is regular. Now we can extend it by the
following Theorem:
\begin{theorem}\label{11} Let $R$ be a strongly distributive multiplicative hyperring.
$R$ is regular if and only if $M_n(R)$ is regular multiplicative
hyperring.
\end{theorem}
\begin{proof} ($\Rightarrow$) First of all we'll proof it for $n=2$, and then we will extend it to arbitrary $n$.
Assume that
\[A=\left(\begin{array}{cc}
a & b\\
c & d
\end{array}\right)\in M_2(R),
\]
since $b$ is regular there exists $b'\in R$ such that $b\in bb'b$. If we set , $X=\left(\begin{array}{cc}
0 & 0\\
b' & 0
\end{array}\right)\in M_2(R)$
and denote $B=AXA-A$ then by an easy calculation we can see that
\[B'=\left(\begin{array}{cc}
bb'a-a & bb'b-b\\
db'a-c & db'b-d
\end{array}\right)\subseteq B.
\]
Now, we can consider $B^{\ast}=\left(\begin{array}{cc}
s & 0\\
t & u
\end{array}\right)\in B'$, where $s\in bb'a-a, t\in db'a-c, u\in db'b-d$.
Since $s,u$ are regular there exist $s',u'\in R$ such that $s\in ss's,u\in uu'u$.
If $L=\left(\begin{array}{cc}
s' & 0\\
0 & u'
\end{array}\right)$,
then by simple calculation we can show that
\[C=B^{\ast}LB^{\ast}-B^{\ast}\supseteq \left(\begin{array}{cc}
ss's-s & 0\\
-t+ts's+uu't & uu'u-u
\end{array}\right)=C'.
\]
Let $C^{\ast}=\left(\begin{array}{cc}
0 & 0\\
m & 0
\end{array}\right)\in C'$ , where $m\in -t+ts's+uu't$.
Since $m$ is regular there exists $m'\in R$, such that $m\in mm'm$.
Finally, if $K= \left(\begin{array}{cc}
0 & m'\\
0 & 0
\end{array}\right)$
we can see that $0\in C^{\ast}KC^{\ast}-C^{\ast}$.
So, $C^{\ast}$ is regular, then by the Proposition \ref{10}, $B^{\ast}$
is regular and hence $A$ is regular. Therefore, for $n=2$, if $R$ is regular then $M_2(R)$ is regular.\\
Since $M_2(M_2(R)) \cong M_4(R)$, then $M_4(R)$ is regular.
Thus by continuing this process we can show that for any positive integer $k$,
$M_{2^k}(R)$ is regular. Now, assume that $n$ is an arbitrary positive integer,
choose $k$ such that $2^k \geq n$.
If $A\in M_n(R)$, let $A_1$ be the matrix of $M_{2^k}(R)$ with $A$ in
the upper left-hand corner and zeros elsewhere. Assume that $A_1\in M_{2^k}(R)$,
is regular, then there exists an element $T=\left(\begin{array}{cc}
B& C\\
D& E
\end{array}\right)$
of $M_{2^k}(R)$ such that $A_1\in A_1TA_1 $. However,
this implies that $A\in ABA$ and hence $A$ is regular.\\
($\Leftarrow$) Assume that $M_n(R)$ is regular and $a$ is an
arbitrary element of $R$, then
\[A=\begin{pmatrix}
a & 0& \ldots & 0\\
0& 0& \ldots & 0\\
\vdots &\vdots & \ddots & \vdots \\
0 & 0& \ldots & 0\\
\end{pmatrix}
\]
is in $M_n(R)$. Since $M_n(R)$ is regular, then there exists
\[B=\begin{pmatrix}
b_{11} & \ldots & b_{1n}\\
\vdots & \ddots & \vdots \\
b_{n1} & \ldots & b_{nn}\\
\end{pmatrix}
\in M_n(R)
\]
such that $A\in ABA$, this means that $a\in ab_{11}a$, i.e., $a$ is
regular.
\end{proof}
\begin{theorem}\label{12} Let $(R,+,.)$ be a commutative multiplicative hyperring with zero absorbing property. Then
\[R[x]/\gamma^{\ast}\cong (R/\gamma^{\ast})[x].
\]
\end{theorem}
\begin{proof} Consider the map $\phi:R[x]\rightarrow (R/\gamma^{\ast})[x]$ defined by
$\phi(\sum_{i=1}^na_ix^i)=\sum_{i=1}^n\gamma^{\ast}(a_i)x^i$. By
[\cite{DA3}, Theorem 5.6.5], $\phi$ is a projection homomorphism. We
denote the equivalence relation associated with $\phi$ by $\rho$.
That is,
\[(\sum_{i=1}^na_ix^i)\rho (\sum_{i=1}^nb_ix^i) \Longleftrightarrow
\sum_{i=1}^n\gamma^{\ast}(a_i)x^i=\sum_{i=1}^n\gamma^{\ast}(b_i)x^i,
\ \ \forall a_i,b_i\in R.
\]
Since $\phi$ is epimorphism we have
\[R[x]/\rho=R[x]/ker(\phi)\cong (R/\gamma^{\ast})[x].
\]
We know that $(R/\gamma^{\ast})[x]$ is a ring, then $R[x]/\rho$ is a
ring. Since $\gamma^{\ast}$ is the smallest equivalence relation on
$R[x]$ such that $R[x]/\gamma^{\ast}$ is a ring, then
$\gamma^{\ast}\subseteq \rho$. Now, let $(\sum_{i=1}^na_ix^i)\rho
(\sum_{i=1}^nb_ix^i)$ for all $b_i\in R$. Hence
\[\sum_{i=1}^n\gamma^{\ast}(a_i)x^i=\sum_{i=1}^n\gamma^{\ast}(b_i)x^i
\Longleftrightarrow \gamma^{\ast}(a_i)=\gamma^{\ast}(b_i), \ \
\forall a_i\in R.
\]
Thus, $\{a_i,b_i\}\subseteq \prod_{s=1}^mt_s$. Therefore,
\[\{\sum_{i=1}^na_ix^i,\sum_{i=1}^nb_ix^i\} \subseteq \sum_{i=1}^n(\prod_{s=1}^mt_s)_ix^i.
\]
This means that $(\sum_{i=1}^na_ix^i)\gamma(\sum_{i=1}^nb_ix^i)$.
Therefore $(\sum_{i=1}^na_ix^i)\gamma^{\ast}(\sum_{i=1}^nb_ix^i)$.
Hence $\rho\subseteq \gamma^{\ast}$, i.e., $\rho=\gamma^{\ast}$, and
so $R[x]/\gamma^{\ast}\cong (R/\gamma^{\ast})[x]$.
\end{proof}
\begin{remark}\label{13} If $R$ is a commutative multiplicative hyperring with zero absorbing property and $R[x]$ is regular, then $R$ is regular. But the converse is not true.
\end{remark}
\subsection{\textbf{ Some properties of regular multiplicative hyperrings}}
\begin{definition} Let $R$ be a multiplicative hyperring. A subset $A$ of $R$ is
idempotent if $A\subseteq A^2$. The set of all idempotent elements
of $R$ is denoted by $Idem(R)$.
\end{definition}
\begin{definition} We say that $I$ is a hyperideal of multiplicative hyperring $(R,+,.)$ if it satisfies the following conditions:\\
$(1)$ $I-I\subseteq I$,\\
$(2)$ $\forall x\in I$, $r\in R$, $xr\cup rx\subseteq I$.
\end{definition}
\begin{definition} Let $R$ be a multiplicative hyperring.
The element $a\in R$ is nilpotent, if there exists an $n$ such that
$a^n=\{0\}$. Denote the set of all nilpotent elements of $R$ by
$nil(R)$.
\end{definition}
\begin{definition} Let $R$ be a multiplicative hyperring and $x\in R$. Then a left(right)
annihilator of $x$ is $Ann(x)=\{r\in R|rx=0\}$ ($Ann(x)=\{r\in
R|xr=0\}$). For a non-empty subset $B$ of a multiplicative hyperring
$R$, the {\it annihilator} of $B$ is $Ann(B)=\cap \{Ann(x)|x\in
B\}$.
\end{definition}
\begin{theorem}\label{14} Let $R$ be a commutative multiplicative hyperring with a scalar identity $1$, and $a\in R$.
Then we have the following statements:\\
$(1)$ If $a\in ara$ for $r\in R$, then $ar\in Idem(R)$.\\
$(2)$ $V(R)\cap nil(R)=\{0\}$.\\
\end{theorem}
\begin{proof} $(1)$ This is clear.\\
$(2.)$ Suppose $a\in V(R)\cap nil(R)$. Then there is a $r\in R$ and
$n\in \mathbb{N}$ such that $a\in ara$ and $a^n=\{0\}$. Thus we
have,
\[a\in ara\subseteq (ara)r(ara)=a^4r^3=(ara)a^2r^2\subseteq ...\subseteq a^nr'=\{0\}r'=\{0\},
\]
because $0.r'=(1-1).r'\subseteq 1.r'-1.r'=\{r'\}-\{r'\}=\{0\}$.
Therefore, $a=0$.
\end{proof}
\begin{theorem}\label{15} Suppose that $R$ is a commutative multiplicative hyperring with a scalar identity $1$.
Then we have the following statements are satisfied:\\
$(1)$ If for some $u\in U(R)$ and $a\in R$, $a\in aua$, then $a\in ve$ for some $v\in U(R)$ and $e\in Idem(R)$.\\
$(2)$ If $\zeta=ue$ for some $u\in U(R)$ and $e\in Idem(R)$, then for some $v\in U(R)$, $\zeta\subseteq \zeta v\zeta$.\\
$(3)$ If for $a\in R$, there exists $b\in R$ such that $ab=0$, with $a+b\in U(R)$, then $a$ is regular.\\
$(4)$ If $\eta=ue$ for some $u\in U(R)$ and $e\in Idem(R)$ and
$|\eta^2|=1$, then there exists $\ell \in \eta$ such that $\ell b=0$ for $b\in R$ with $\ell+b\in U(R)$.
\end{theorem}
\begin{proof} $(1)$ Assume that $a\in aua$ for some $u\in U(R)$. Since $u$ is invertible
then there is a $v\in U(R)$ such that $1\in uv$. Let $e=au$, thus we
have
\[e^2=(au)^2=(au)(au)=(aua)u\supseteq au=e,
\]
then $e\in Idem(R)$. Hence $ve=v(au)=a(vu)\supseteq a.1\ni a$.\\
$(2)$ Since $u\in U(R)$, then there exists $v\in U(R)$ such that
$1\in vu$. Thus we have:
\[e= 1.e\subseteq (vu)e=v(ue)=v\zeta \Rightarrow e\subseteq e^2\subseteq ev\zeta\Rightarrow \zeta=ue\subseteq uev\zeta=\zeta v\zeta,
\]
i.e., $\zeta\subseteq \zeta v\zeta$ for some $v\in U(R)$.\\
$(3)$ Let $u=a+b$. Then $au=a(a+b)\subseteq a^2+ab=a^2$. Since $u$
is invertible, so there is a $v\in U(R)$ such that $1\in uv$.
Therefore, we have:
\[a\in a.1\subseteq a(uv)=(au)v\subseteq a^2v=ava.
\]
Hence $a$ is regular.\\
$(4)$ Assume that $\eta=ue$ for some $u\in U(R)$ and $e\in Idem(R)$.
Let $\tau=u(1-e)$. Then $\eta \tau=ue(u(1-e))\subseteq
ueu1-ueue=ueu-ueue\subseteq ueue-ueue=\eta^{2}-\eta^{2}=0$, i.e., for all $\ell\in \eta, b\in \tau$, $\ell b=0$. Now, we
need to show that $\eta+b\cap U(R)\neq \emptyset$. We have $1\in
u.1.u\subseteq u(e+(1-e))u\subseteq (ue+u(1-e))u=(\eta+b)u$, i.e.,
$1\in (\eta+b)u$, so there exist $b\in \tau, \ell\in \eta$ such that $1\in b+\ell$. Therefore $\ell+b\in U(R)$.
\end{proof}
\begin{definition} Let $R$ be a multiplicative hyperring. Then $R$ is said to be
$\pi$-regular if for all $a\in R$ there are $r\in R$ and an integer
$n\geq 1$, such that $a^n\subseteq a^nra^n$.
\end{definition}
Clearly, a regular multiplicative hyperring is $\pi$-regular multiplicative hyperring. Also, if $a\in R$ is $\pi$-regular
multiplicative hyperring then for an $n\geq 1$, $a^n$ is regular.
\begin{theorem}\label{16} Let $R$ be a commutative multiplicative hyperring with a scalar identity $1$.
Then for $a\in R$, the following hold:\\
$(1)$ $a$ is a $\pi$-regular if and only if for some $n\geq 1$, $a^n$ is regular.\\
$(2)$ If $a^n\subseteq a^nra^n$ for $r\in R$ and $n\geq 1$, then $a^nr\subseteq Idem(R)$.\\
$(3)$ If $a^n=ue$ for some $u\in U(R)$ and $e\in Idem(R)$ and
$|a^{2n}|=1$ then there exists $\ell\in a^n$ such that $\ell$ is $\pi$-regular.
\end{theorem}
\begin{proof} It's straightforward by Theorem \ref{15}.
\end{proof}
For each hyperideal $I$ of multiplicative hyperring $R$, letting $O_I=\{a\in I: a\in aI\}$. Then $O_I$ is called a {\it pure part}
of $I$. An hyperideal $I$ is called a {\it pure hyperideal} if $I=O_I$.
\begin{theorem}\label{17} Let $R$ be a commutative multiplicative hyperring with a scalar identity $1$ and
$a\in R$ and also for $r\in R$ there exists $n\in \mathbb{N}$ such
that $|ar^n|=1$. If hyperideal $$ is pure hyperideal,
then $R=+Ann(a)$.
\end{theorem}
\begin{proof} Suppose that $$ is pure hyperideal, then there exists $x=sa+\sum_{i=1}^na\cdot x_i\subseteq $, such that $a\in ax$, where $s\in \mathbb{N}, r,x_i\in R$. So, there exists $\ell\in x$ such that $a\in a\ell$. Thus we have $a(1-\ell)\subseteq a-a\ell\subseteq
a-a\ell^2\subseteq...\subseteq a-a\ell^n=0$, i.e.,
$a(1-\ell)=0$, which implies that $1-\ell\in Ann(a)$. Therefore
$1= \ell+(1-\ell)\in +Ann(a)$. Hence $R=+Ann(a)$.
\end{proof}
\begin{theorem}\label{18} Let $R$ be a multiplicative hyperring with a scalar identity $1$ and $M$ be
a maximal hyperideal of $R$. Also, for $a\in M$ and $r\in R$ there
exists $s\in \mathbb{N}$, such that $|ar^s|=1$.
Then the following statements:\\
$(1)$ if $a\in V(R)$ then for $a\in M$, $a\in O_M$,\\
$(2)$ $a\in O_M$ for $a\in M$ if and only if $Ann(a)$ is not contained in $M$.
\end{theorem}
\begin{proof} $(1)$ Since $a\in V(R)$, then for some $r\in R, a\in ara$.
Now, for a maximal hyperideal $M$ such that $a\in M$, we have $a\in
ara=a(ra)\subseteq aM$. Hence $a\in O_M$.\\
$(2)$: $(\Rightarrow)$ Suppose that $a\in O_M$ and $Ann(a)\subseteq
M$. Then there is a $m\in M$ such that $a\in am$. Thus
$a(1-m)\subseteq a-am\subseteq a-am^2\subseteq... \subseteq
a-am^s=0$, i.e., $a(1-m)=0$. Therefore, $1-m\in Ann(a)\subseteq M$,
i.e., $1-m\in M$ and it's contradiction. Hence $Ann(a)\nsubseteq M$.\\
$(\Leftarrow)$ Assume that $Ann(a)$ is not contained in $M$.
Then $R=M+Ann(a)$. So, there exist $m\in M, x\in Ann(a)$ such that
$1=m+x$, then $a\in a.1=a(m+x)\subseteq am+ax=am$, i.e., $a\in am$.
Hence $a\in O_M$.
\end{proof}
\section{\textbf{Properties of $M(R)$}}
\begin{definition} Let $R$ be a multiplicative hyperring. Denote by $M(R)$ the set of all elements in $R$ such that the
generated hyperideals by each of these elements are regular.
Clearly, $M(R)$ is a regular hyperring.
\end{definition}
\begin{lemma}\label{19} If $a$ and $b$ are two regular elements in a commutative multiplicative
hyperring $R$. Then the following statements hold:\\
$(1)$ $\gamma^{\ast}(ab)$ is regular in
$R/\gamma^{\ast}$,\\
$(2)$ Moreover, if $|ab|=1$, then so is $ab$.
\end{lemma}
\begin{proof} $(1)$ Since $a,b$ are regular in $R$, then there
exist $r_1, r_2\in R$, such that $a\in ar_1a,b\in br_2b$. So
\[{\gamma}^{\ast}(ab)={\gamma}^{\ast}(ab)\odot{\gamma}^{\ast}(r_1r_2)\odot
{\gamma}^{\ast}(ab).\]
Hence ${\gamma}^{\ast}(ab)$ is regular in $R/{\gamma}^{\ast}.$\\
$(2)$ By definition of regular element, there exist $r_1, r_2\in R$,
such that $a\in ar_1a$ and $b\in br_2b$. Since $|ab|=1$, we have
$ab\in (ar_1a)(br_2b)=(ab)r_1r_2(ab)$.
\end{proof}
\begin{theorem}\label{20} Let $R$ be a strongly distributive multiplicative hyperring and $a\in R$.
Thus we have the following statements:\\
$(1.)$ If there is a regular element $c$ in $a^2-a$, then $a$ and $1-a$ are regular.\\
$(2.)$ If $a$ and $1-a$ are regular in $R$ such that $|a(1-a)|=1$,
then so is $a(1-a)$.
\end{theorem}
\begin{proof} $(1.)$. It is immediately follows by Proposition $2.15$.\\
$(2.)$ By Lemma \ref{19}$(2)$, it is clear.
\end{proof}
\begin{theorem}\label{21} If $R$ be a strongly distributive multiplicative hyperring,
then a right hyperideal $I$ in the hyperring $M(R)$ is a right hyperideal in $R$.
\end{theorem}
\begin{proof} Suppose $a\in I, r\in R$, then $ar\subseteq M(R)$, hence for some element
$r'\in R, ar\subseteq arr'ar$. But $rr'ar\subseteq M(R)$, so $ar\subseteq I$. Thus $I$ is a right hyperideal in $R$.
\end{proof}
\begin{theorem}\label{22} Let $R$ be a strongly distributive multiplicative hyperring.
Then $M(R)$ is a hyperideal of $R$.
\end{theorem}
\begin{proof} Let $z\in M(R)$ and $r\in R$. Since $\subseteq $ and $\subseteq $,
we have $zr\subseteq M(R)$ and $rz\subseteq M(R)$, thus $zr \cup
rz\subseteq M(R)$. Now, assume that $t_1,t_2\in M(R)$. We need to
prove that all of elements in $$ is regular. For achieving
to it, let $a\in $. Then for some $u\in $ and $v\in
$ we have $a=u-v$. As $$ is regular then there exists
$r\in R$ such that $u\in uru$. Thus by distributive property of $R$ we have
\begin{eqnarray*}
ara &=& (u-v)r(u-v)-u+v\\
&=& uru-urv-vru+vrv-u+v\\
&\subseteq & uru-uru+v-urv-vru+vrv\\
&=& u(r-r)u+v-urv-vru+vrv\\
&=& u0u+v-urv-vru+vrv\\
&=& u(v-v)u+v-urv-vru+vrv\\
&=& uvu-uvu+v-urv-vru+vrv,
\end{eqnarray*}
since the right side is in $$, then $ara-a\subseteq $,
i.e., $ara-a$ is regular and by Proposition \ref{6}, $a$ is regular.
Hence $$ is a regular hyperideal and $t_1-t_2\in M(R)$,
i.e., $M(R)-M(R)\subseteq M(R)$.
\end{proof}
Let $(R,+,.)$ be a multiplicative hyperring and $I$ be a hyperideal of it. We consider the usual addition of cosets and the
multiplication defined as follows:
\[(a+I)\ast(b+I)=\{c+I|c\in a.b\},
\]
On the set $R/I=\{r+I|r\in R\}$ of all cosets of $I$. Then
$(R/I,+,\ast)$ is a multiplicative hyperring.
\begin{proposition}\label{23}([19]) If $(R,+,.)$ is a strongly distributive multiplicative hyperring
and $I$ is a hyperideal of $R$, then $(R/I,+,\ast)$ is ring.
\end{proposition}
\begin{corollary}\label{24} If $(R,+,.)$ is a strongly distributive multiplicative hyperring,
then $(R/M(R),+,\ast)$ is ring.
\end{corollary}
\begin{theorem}\label{25} If $(R,+,.)$ is a strongly distributive multiplicative hyperring,
then $M(R/M(R))=\{0\}$.
\end{theorem}
\begin{proof} Assume that $a+M(R)$ denote the residue class modulo $M(R)$
which contains the element $a$ of $R$. If $b+M(R)\in M(R/M(R))$ and
$a\in $, then $a+M(R)\in $. Since $$ is a regular
ideal in $R/M(R)$, then $a+M(R)$ is regular. Thus, for some
$x+M(R)\in R/M(R), a+M(R)=(a+M(R))\ast(x+M(R))\ast(a+M(R))$ or
$a+M(R)=axa+M(R)$, i.e., $axa-a\subseteq M(R)$. Therefore $axa-a$ is
regular element of strongly distributive multiplicative hyperring
$R$ and by Proposition \ref{10}, $a$ is regular. Thus $$ is
regular hyperideal and hence $b\in M(R)$. So, $b+M(R)=0_{R/M(R)}$.
\end{proof}
\begin{theorem}\label{26} Let $B$ be a hyperideal of a strongly distributive multiplicative hyperring $R$.
Then $M(B)=B\cap M(R)$.
\end{theorem}
\begin{proof} Assume that $B$ is a hyperideal of $R$ and let $b$ be a element of $B$ such
that generated a regular hyperideal $^*$ in the strongly
distributive multiplicative hyperring $B$.
Suppose $$ be the hyperideal in $R$ generated by the element $b$, and let\\
$\indent\indent\indent\indent\indent c=nb+rb+bs+\Sigma{r_ibs_i}; \ \ \ where\ \ n\in \mathbb{Z}, r,s,r_i,s_i\in R\\
$
\noindent be any element of $$. Since $b$ is regular in $B$ we have for some $x\in B$, $b\in bxb$. Hence\\
$ c\subseteq nb+r(bxb)+(bxb)s+\Sigma {r_i(bxb)s_i}\\
\indent\ \ =nb+r(bxb)+(bxb)s+\Sigma{r_i(bxbxb)s_i} \\
\indent\ \ =nb+r(bxb)+(bxb)s+\Sigma{(r_ibx)b(xbs_i)}, \\
$
\noindent and thus $c\subseteq ^{\ast}$. As $$ coincides with
$^*$, then $$ is regular. Hence, for $b\in M(B)$ we have $b\in
B\cap M(R)$, i.e.,
$M(B)\subseteq B\cap M(R)$.\\
Conversely, let $b\in B\cap M(R)$, then $b\in B,~b\in M(R)$. As $b$
is regular in $R$, then there exists $r\in R$ such that $b\in brb$
and since $B$ is a hyperideal and $b\in B$ we have $rb\subseteq B$
then $b\in brb\subseteq b(rbr)b$, where $rbr\subseteq B$. It shows
that $b$ is regular in $B$. Therefore $B\cap M(R)$ is regular hyper
ideal in the strongly distributive multiplicative hyperring $B$.
Thus $B\cap M(R)\subseteq M(B)$. It complete the proof.
\end{proof}
\begin{theorem}\label{27} If $R$ is a multiplicative hyperring and it has zero absorbing property,
then $M(R)\cap Ann(M(R))=\{0\}$.
\end{theorem}
\begin{proof} For all $a\in M(R)\cap Ann(M(R))$ we have $a\in M(R),~a\in Ann(M(R))$.
As $a\in M(R)$ thus there exists $r\in R$ such that $a\in ara$.
Since $ra\subseteq M(R)$, we have $a\in a(ra)=\{0\}$. Therefore
$a=0$.
\end{proof}
\begin{proposition}\label{28}([19]) Every strongly distributive hyperring $(R, +, .)$ with a scalar identity $1$ is a ring.
%
\end{proposition}
\begin{theorem}\label{29} Let $R$ be a multiplicative hyperring and $I$ be a strongly distributive hyperideal
of $R$ such that it has a scalar identity $e$ and for all $x\in R,
|xe|=1$, then $R=I+Ann(I)$.
\end{theorem}
\begin{proof}Let $x$ be an arbitrary element of $R$, then $xe\cup ex\subseteq I$.
So $(xe)e=e(xe)$, i.e., $xe=exe$ and similarly $ex=exe$. Thus
$xe=ex$ and $e$ is in center of $R$. As, by Proposition \ref{28},
$ex\in I$ and for all $y\in I$,
$y(x-ex)=ye(x-ex)=y(ex-e^2x)=y(ex-ex)=y0=0$, i.e., $x-ex\in Ann(I)$.
Thus we have,
\[x=ex+(x-ex).
\]
Hence for all $x\in R$ as a sum of elements $ex\in I$ and $x-ex\in
I$ of $Ann(I)$.
It's complete the proof.
\end{proof}
The following example show that under Theorem \ref{29}, $R$ is not a
ring:
\begin{example} Let $R=\Bbb{Z}\bigoplus\Bbb{Z}$ and define
\[(a,b)\circ(c,d)=\left\{
\begin{array}{ll}
(ac,\Bbb{Z});\ \ bd\neq 0\\
(ac,0) ; \ \ bd=0 \end{array}\right.
\]
Then $(R,+,\circ)$ is a multiplicative hyperring such that is not
strongly distributive, because by considering
$a=(1,1),b=(0,2),c=(0,-2)$, we have
$a\circ(b+c)=(1,1)\circ(0,0)=(0,0)$ but $a\circ b+a\circ
c=(0,\Bbb{Z})+(0,\Bbb{Z})=(0,\Bbb{Z}+\Bbb{Z})$. Now, let
$I=(\Bbb{Z},0)$ be a strongly distributive hyperideal of $R$ and
$e=(1,0)$ is scalar identity element of $I$ such that for all $r\in
R$ we have $|xe|=1$. It is clear that $Ann(I)=(0,\Bbb{Z})$ and
$R=I+Ann(I)$ but $(R,+,\circ)$ is not a ring.
\end{example}
\begin{corollary}\label{30}(see \cite{BR}) If an ideal $I$ in a ring $R$ has a unit element $e$, then $R=I+Ann(I)$.
\end{corollary}
\paragraph{{\bf}ACKNOWLEDGEMENTS} The first author partially has been supported by "Algebraic
Hyperstructure Excellence, Tarbiat Modares University, Tehran, Iran"
and "Research Center in Algebraic Hyperstructures and Fuzzy
Mathematics, University of Mazandaran, Babolsar, Iran".
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