#  $(u,v)$-absorbing (prime) hyperideals in commutative multiplicative
hyperrings

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

In this paper, we will introduce the notion of $(u,v)$-absorbing hyperideals
in multiplicative hyperrings and we will show some properties of them. Then we
extend this concept to the notion of $(u,v)$-absorbing prime hyperideals and
then we will give some results about them.

###### Key words and phrases:

$(u,v)$-absorbing hyperideal, $AB$-$(u,v)$-absorbing hyperideal,
$(u,v)$-absorbing prime hyperideal.

###### 2010 Mathematics Subject Classification:

Primary 20N20; Secondary 16Y99

##  1\. Introduction

In 1934 [25], the French mathematician F. Marty presented the notion of
hyperstructures or multioperations in which an operation yields a set of
values rather than a single value. Many authors have contributed to the
advancement of this novel area of modern algebra [12, 13, 14, 15, 16, 26, 30,
36]. An importan class of the algebraic hyperstructures called the
multiplicative hyperring was defined by Rota in 1982 [32]. In this
hyperstructure, the multiplication is a hyperoperation however the addition is
an operation. Multiplicative hyperrings are deeply demonstrated and
characterized in [3, 4, 5, 22, 28, 29, 31, 35]. Dasgupta studied the prime and
primary hyperideals in multiplicative hyperrings in [17]. The notions of
1-absorbing and $v$-absorbing hyperideals as generalizations of the prime
hyperideals were studied in [29] and [5]. The idea of $(u,v)$-closed
hyperideals in a multiplicative hyperring was proposed in [6]. Our purpose of
this research work is to introduce and study the notion of $(u,v)$-absorbing
hyperideals as an extension of the $n$-absorbing hyperideals in a commutative
multiplicative hyperring. Several specific properties are given to illustrate
the structures of the new notion. Furthermore, we extend this concept to
$(u,v)$-absorbing prime hyperideals.

##  2\.  multiplicative hyperrings

In this section, we recall some background materials. A hyperoperation
$``\circ"$ on non-empty set $X$ is a mapping from $X\times X$ into $P^{*}(X)$
such that $P^{*}(X)$ is the family of all non-empty subsets of $X$. In this
case, $(X,\circ)$ is called hypergroupoid. Let $X_{1},X_{2}$ be two subsets of
$X$ and $x\in X$, then $X_{1}\circ X_{2}=\cup_{x_{1}\in X_{1},x_{2}\in
X_{2}}x_{1}\circ x_{2},$ and $X_{1}\circ x=X_{1}\circ\\{x\\}.$ This means that
the hyperoperation $``\circ"$ on $X$ can be extended to subsets of $X$. A
hypergroupoid $(X,\circ)$ is called a semihypergroup if $\cup_{a\in y\circ
z}x\circ a=\cup_{b\in x\circ y}b\circ z$ for all $x,y,z\in X$ which means
$\circ$ is associative. A semihypergroup $X$ is called a hypergroup if $x\circ
X=X=X\circ x$ for each $x\in X$. A non-empty subset $Y$ of a semihypergroup
$(X,\circ)$ refers to a subhypergroup if $x\circ Y=Y=Y\circ x$ for each $x\in
Y$ [19].

######  Definition 2.1.

[19] An algebraic structure $(A,+,\circ)$ refers to a commutative
multiplicative hyperring if

  * (1)

$(A,+)$ is a commutative group;

  * (2)

$(A,\circ)$ is a semihypergroup;

  * (3)

$x\circ(y+z)\subseteq x\circ y+x\circ z$ and $(y+z)\circ x\subseteq y\circ
x+z\circ x$ for every $x,y,z\in A;$

  * (4)

$x\circ(-y)=(-x)\circ y=-(x\circ y)$ for every $x,y\in A$;

  * (5)

$x\circ y=y\circ x$ for every $x,y\in A$.

If in (3), the equality holds then the multiplicative hyperring $A$ is called
strongly distributive.

Conseder the ring of integers $(\mathbb{Z},+,\cdot)$. For each subset $T\in
P^{\star}(\mathbb{Z})$ with $|T|\geq 2$, there exists a multiplicative
hyperring $(\mathbb{Z}_{T},+,\circ)$ where $\mathbb{Z}_{T}=\mathbb{Z}$ and
$a\circ b=\\{a.x.b\ |\ x\in T\\}$ for all $a,b\in\mathbb{Z}_{T}$ [17].

######  Definition 2.2.

[1] An element $e\in A$ is considered as an identity element if $a\in a\circ
e$ for all $a\in A$.

Throughout this paper, $A$ denotes a commutative multiplicative hyperring with
identity 1.

######  Definition 2.3.

[1] An element $x\in A$ is called unit, if there exists $y\in A$ such that
$1\in x\circ y$. Denote the set of all unit elements in $A$ by $U(A)$.

Moreover, a hyperring $A$ refers to a hyperfield if each non-zero element of
$A$ is unit.

######  Definition 2.4.

[19] A non-empty subset $B$ of $A$ is a hyperideal if

  * (i)

$x-y\in B$ for all $x,y\in B$;

  * (ii)

$r\circ x\subseteq B$ for all $x\in B$ and $r\in A$.

######  Definition 2.5.

[17] A proper hyperideal $B$ in $A$ refers to a prime hyperideal if $x\circ
y\subseteq B$ for $x,y\in A$, then $x\in B$ or $y\in B$.

The intersection of all prime hyperideals of $A$ containing a hyperideal $B$
is said to be the prime radical of $B$, denoted by $rad(B)$. If the
multiplicative hyperring $A$ has no prime hyperideal containing $B$, we define
$rad(B)=A$. Assume that $\mathcal{C}$ is the class of all finite products of
elements of $A$ that is $\mathcal{C}=\\{a_{1}\circ a_{2}\circ\cdots\circ
a_{n}\ |\ a_{i}\in A,n\in\mathbb{N}\\}\subseteq P^{\ast}(A)$ and $B$ is a
hyperideal of $A$. $B$ refers to a $\mathcal{C}$-hyperideal of $A$ if for each
$C\in\mathcal{C}$ and $B\cap C\neq\varnothing$ imply $C\subseteq B$. Notice
that $\\{a\in A\ |\ a^{n}\subseteq B\ \text{for some}\
n\in\mathbb{N}\\}\subseteq rad(A)$. The equality holds if $B$ is a
$\mathcal{C}$-hyperideal of $A$ (see Proposition 3.2 in [17]). Moreover, a
hyperideal $B$ of $A$ refers to a strong $\mathcal{C}$-hyperideal if for each
$D\in\mathfrak{U}$ and $D\cap B\neq\varnothing$ imply $D\subseteq B$ such that
$\mathfrak{U}=\\{\sum_{i=1}^{n}C_{i}\ |\
C_{i}\in\mathcal{C},n\in\mathbb{N}\\}$ and $\mathcal{C}=\\{a_{1}\circ
a_{2}\circ...\circ a_{n}\ |\ a_{i}\in A,n\in\mathbb{N}\\}$ (for more details
see [18]).

######  Definition 2.6.

[1] A proper hyperideal $B$ of $A$ is maximal in $A$ if for each hyperideal
$M$ of $A$ with $B\subset M\subseteq A$, then $M=A$.

The intersection of all maximal hyperideals of $A$ is denoted by $J(A)$. Also,
$A$ refers to a local multiplicative hyperring if it has just one maximal
hyperideal.

######  Definition 2.7.

[1] Assume that $B_{1}$ and $B_{2}$ are hyperideals of $H$. We define
$(B_{2}:B_{1})=\\{x\in A\ |\ x\circ B_{1}\subseteq B_{2}\\}$.

##  3\.  $(u,v)$-absorbing hyperideals

A proper hyperideal $Q$ in $A$ refers to a $v$-absorbing hyperideal for
$v\in\mathbb{N}$ if whenever $x_{1}\circ\cdots\circ x_{v+1}\subseteq Q$ for
$x_{1},\cdots,x_{v+1}\in A$, then there exist $v$ of the $x_{i}^{,}$s whose
product is a subset of $Q$. In this section, we aim to introduce the class of
the $(u,v)$-absorbing hyperideals as an extension of the $v$-absorbing
hyperideals and present some of their properties.

######  Definition 3.1.

Let $u,v\in\mathbb{N}$ with $u>v$. A proper hyperideal $Q$ of $A$ is called a
$(u,v)$-absorbing hyperideal if whenever $x_{1}\circ\cdots\circ x_{u}\subseteq
Q$ for $x_{1},\cdots,x_{u}\in A\backslash U(A)$, then there are $v$ of the
$x_{i}^{,}$s whose product is a subset of $Q$.

######  Example 3.2.

Consider the multiplicative hyperring $(\mathbb{Z}_{T},+,\circ)$ where
$T=\\{2,4\\}$. Then $Q=\langle 150\rangle$ is a $(4,3)$-absorbing hyperideal
of the multiplicative hyperring $(\mathbb{Z}_{T},+,\circ)$. However, it is not
$(3,2)$-absorbing as the fact that $3\circ 5^{2}=\\{300,600,1200\\}\subseteq
Q$ but $3\circ 5=\\{30,60\\}\nsubseteq Q$ and $5^{2}=\\{50,100\\}\nsubseteq
Q$.

######  Remark 3.3.

Assume that $Q$ is a proper hyperideal of $A$ and $k,r,u,v\in\mathbb{N}$.

  * (i)

Let $Q$ be a $(u,v)$-absorbing hyperideal of $A$. Then:

    * (1)

$Q$ is a $(k,r)$-absorbing hyperideal for every $k\geq u$ and $r\geq v$.

    * (2)

$Q$ is a $(u-1)$-absorbing hyperideal of $A$.

  * (ii)

$Q$ is a $(u,v)$-absorbing hyperideal of $A$ if and only if
$x_{1}\circ\cdots\circ x_{k}\subseteq Q$ for $x_{1},\cdots,x_{k}\in
A\backslash U(A)$ and $k\geq u$ imply that there are $v$ of the $x_{i}^{,}$s
whose product is a subset of $Q$.

######  Proposition 3.4.

Let $Q_{i}$ be a proper hyperideal of $A$ and $u_{i},v_{i}\in\mathbb{N}$ for
every $i\in\\{1,\cdots,k\\}$. If $Q_{i}$ is a $(u_{i},v_{i})$-absorbing
hyperideal of $A$ for every $i\in\\{1,\cdots,k\\}$, then
$Q=\cap_{i=1}^{k}Q_{i}$ is a $(u,v)$-absorbing hyperideal of $A$ where
$v=v_{1}+\cdots+v_{k}$ and $u=\max\\{u_{1},\cdots,u_{k},v+1\\}$.

###### Proof.

Assume that $Q_{i}$ is a $(u_{i},v_{i})$-absorbing hyperideal of $A$ for every
$i\in\\{1,\cdots,k\\}$. By Remark 3.3 (i), we conclude that $Q_{i}$ is a
$(u,v_{i})$-absorbing hyperideal of $A$ where $v=v_{1}+\cdots+v_{k}$ and
$u=\max\\{u_{1},\cdots,u_{k},v+1\\}$. Let $x_{1}\circ\cdots\circ
x_{u}\subseteq Q$ for $x_{1},\cdots,x_{u}\in A\backslash U(A)$. Since $Q_{i}$
is a $(u,v_{i})$-absorbing hyperideal of $A$ for every $i\in\\{1,\cdots,k\\}$
and $x_{1}\circ\cdots\circ x_{u}\subseteq Q_{i}$, we get the result that
$\Pi_{\alpha\in I_{i}}x_{\alpha}\subseteq Q_{i}$ for each
$i\in\\{1,\cdots,k\\}$ such that $I_{i}\subseteq\\{1,\cdots,u\\}$ and
$|I_{i}|=v_{i}$. Put $I=\cup_{i=1}^{k}I_{i}$. Since $\Pi_{\alpha\in
I}x_{\alpha}\subseteq Q$ and $|I|\leq v$, we conclude that
$Q=\cap_{i=1}^{k}Q_{i}$ is a $(u,v)$-absorbing hyperideal of $A$. ∎

Let $u,v\in\mathbb{N}$. Then the smallest integer not less than $\frac{u}{v}$
is denoted by $\lceil\frac{u}{v}\rceil$.

######  Proposition 3.5.

Let $Q$ be a $(u,v)$-absorbing $\mathcal{C}$-hyperideal of $A$ for
$u,v\in\mathbb{N}$ with $u>v$. Then:

  * (i)

$rad(Q)=\\{x\in A\ |\ x^{v}\subseteq Q\\}$.

  * (ii)

$rad(Q)$ is a $(k,v)$-absorbing hyperideal of $A$ where
$k=\max\\{\lceil\frac{u}{v}\rceil,v+1\\}$.

###### Proof.

(i) Assume that $Q$ is a $(u,v)$-absorbing hyperideal of $A$. Take any $x\in
rad(Q)$. Then there exists a smallest element $n\in\mathbb{N}$ such that
$x^{n}\subseteq Q$. Let $n>v$. Remark 3.3(i)(2) shows that $Q$ is a
$(u-1)$-absorbing hyperideal of $A$. Then we conclude that $n\leq u-1$ by
Theorem 3.6 in [5]. This means $x^{u}\subseteq Q$ which means $x^{v}\subseteq
Q$ as $Q$ is a $(u,v)$-absorbing hyperideal of $A$. This is contradicts the
fact that $n$ is a smallest positive integer such that $x^{n}\subseteq Q$.
Therefore $n\leq v$ and $x^{v}\subseteq Q$.

(ii) Suppose that $x_{1}\circ\cdots\circ x_{k}\subseteq rad(Q)$ such that
$k=\max\\{\lceil\frac{u}{v}\rceil,v+1\\}$ and $x_{1},\cdots,x_{k}\in
A\backslash U(A)$. Then for any $a\in x_{1}\circ\cdots\circ x_{k}$,
$a^{v}\subseteq Q$. Again, $a^{v}\subseteq(x_{1}\circ\cdots\circ
x_{k})^{v}\subseteq x_{1}^{v}\circ\cdots\circ x_{k}^{v}\subseteq Q$. So,
$(x_{1}^{v}\circ\cdots\circ x_{k}^{v})\cap Q\neq\varnothing$ and then
$x_{1}^{v}\circ\cdots\circ x_{k}^{v}\subseteq Q$ as $Q$ is a
$\mathcal{C}$-hyperideal. The hyperideal $Q$ is a $(kv,v)$-absorbing
hyperideal of $A$ because $k\geq\lceil\frac{u}{v}\rceil$. Then we conclude
that $x_{i_{1}}^{r_{1}}\circ\cdots\circ x_{i_{v}}^{r_{v}}\subseteq Q$ for some
$\\{i_{1},\cdots,i_{v}\\}\subseteq\\{1,\cdots,k\\}$ such that $0\leq
i_{1},\cdots,i_{v}\leq v$ and $r_{1}+\cdots+r_{v}=v$. This implies that
$x_{i_{1}}^{v}\circ\cdots\circ x_{i_{v}}^{v}\subseteq Q$ which means
$x_{i_{1}}\circ\cdots\circ x_{i_{v}}\subseteq rad(Q)$. Consequently, $rad(Q)$
is a $(k,v)$-absorbing hyperideal of $A$. ∎

######  Theorem 3.6.

Assume that $Q$ is a hyperideal of $A$ and $u,v\in\mathbb{N}$ with $u>v$. Then
there exists a $(u,v)$-absorbing hyperideal of $A$ which is minimal in the set
of all $(u,v)$-absorbing hyperideals of $A$ containing $Q$.

###### Proof.

Suppose that $\Gamma=\\{P\ |\ P\ \text{is a $(u,v)$-absorbing hyperideal of
$A$ and $Q\subseteq P$}\\}$. Let $M$ be a maximal hyperideal of $A$ containing
$Q$. Then $M$ is a $(u,v)$-absorbing hyperideal for all $u,v\in\mathbb{N}$
with $u>v$ and so $M\in\Gamma$ which means $\Gamma\neq\varnothing$. Order
$\Gamma$ by $\leq$ such that $P_{1}\leq P_{2}$ if and only if $P_{1}\supseteq
P_{2}$ for every $P_{1},P_{2}\in\Gamma$. Let $\\{P_{i}\\}_{i\in\Delta}$ be a
non-empty chain of hyperideals in $\Gamma$. Put $I=\cap_{i\in\Delta}P_{i}$.
Assume that $x_{1}\circ\cdots\circ x_{u}\subseteq I$ for
$x_{1},\cdots,x_{u}\in A\backslash U(A)$ such that none of the product of the
terms $v$ of $x_{i}^{,}$s is a subset of $I$. Then that there exists $P_{j}$
in the chain such that none of the product of the terms $v$ of $x_{i}^{,}$s is
a subset of $P_{j}$ which is impossible. Hence there are $v$ of the
$x_{i}^{,}$s whose product is a subset of $I$. Now, by Zorn,s lemma, we
conclude that $(\Gamma,\leq)$ has a maximal element which means $\Gamma$ has a
minimal $(u,v)$-absorbing hyperideal of $A$ containing $Q$. ∎

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

######  Corollary 3.7.

Suppose that $Q$ is a hyperideal of $A$ and $v\in\mathbb{N}$. Then there
exists a minimal $v$-absorbing hyperideal $P$ of $A$ such that $Q\subseteq P$.

######  Theorem 3.8.

Assume that $Q$ is a $(u,v)$-absorbing hyperideal of $A$ and
$k,r\in\mathbb{N}$ with $2\leq k\leq r\leq u-1$. If $Q$ has $r$ minimal prime
hyperideals, then there are at least $\frac{r!}{(r-k)!k!}$ $k$-absorbing
hyperideals of $A$ that are minimal over $Q$.

###### Proof.

Suppose that $Q$ has $r$ minimal prime hyperideals and $k=2$. Then there exist
the distinct prime hyperideals $Q_{1},\cdots,Q_{r}$ of $A$ that are minimal
over $Q$. Hence $Q_{i}\cap Q_{j}$ is a 2-absorbing hyperideal of $A$ for each
$i,j\in\\{1,\cdots,r\\}$ and $i\neq j$ by Theorem 3.4 in [5]. Hence we have
$Q\subseteq Q_{ij}\subseteq Q_{i}\cap Q_{j}$ for some minmal 2-absorbing
hyperideal $Q_{ij}$ of $A$ by Corollary 3.7. Therefore we conclude that
$Q_{i}$ and $Q_{j}$ are only minimal prime hyperideals over $Q_{ij}$ by
Theorem 3.8 in [5]. Assume that ${Q_{ij}}^{,}$s are not distinct such that
$Q_{ij}=Q_{st}$ for some $i,j,s,t\in\\{1,\cdots,r\\}$ and
$\\{i,j\\}\neq\\{s,t\\}$. Then we conclude that $Q_{i}$, $Q_{j}$, $Q_{s}$ and
$Q_{t}$ are distinct minimal prime hyperideals over $Q_{ij}$ which contradicts
the fact that there are are at most two prime hyperideals of $A$ that are
minimal over the 2-absorbing hyperideal $Q_{ij}$. Hence the ${Q_{ij}}^{,}$s
are distinct. Thus we get the result that $Q$ has at least
$\frac{r!}{(r-k)!k!}$ minimal $2$-absorbing hyperideals. If $k>2$, one can
easily complete the proof by using a similar argument. ∎

######  Lemma 3.9.

Let $Q$ be a $(u,v)$-absorbing hyperideals of $A$, $Q_{1},\cdots,Q_{v}$ be
incomparable prime $\mathcal{C}$-hyperideals of $A$ such that
$Q\subseteq\cap_{i=1}^{v}Q_{i}$ and $k_{1},\cdots,k_{v}\in\mathbb{N}$. If
$x_{1}^{k_{1}}\circ\cdots\circ x_{v}^{k_{v}}\subseteq Q$ for $x_{i}\in
Q_{i}\backslash(\cup_{j\neq i}Q_{j})$, then $x_{1}\circ\cdots\circ
x_{v}\subseteq Q$.

###### Proof.

Suppose that $k_{1},\cdots,k_{v}\in\mathbb{N}$. Put $k=k_{1}+\cdots+k_{v}$.
Let $k\geq u$. So $Q$ is a $(k,v)$-absorbing hyperideal of $A$ by Remark
3.3(i). Then there exist integers $r_{1},\cdots,r_{v}$ such that
$x_{1}^{r_{1}}\circ\cdots\circ x_{v}^{r_{v}}\subseteq Q$, $0\leq r_{i}\leq
k_{i}$ and $r_{1}+\cdots+r_{v}=v$. Assume that one of the $r_{i}^{,}$s is
zero. Then we have $x_{1}^{r_{1}}\circ\cdots x_{i-1}^{r_{i-1}}\circ
x_{i+1}^{r_{i+1}}\circ\cdots\circ x_{v}^{r_{v}}\subseteq Q\subseteq Q_{i}$.
Then there exists $j\in\\{1,\cdots,i-1,i+1,\cdots,v\\}$ such that $x_{j}\in
Q_{i}$ which is impossible. Hence $x_{1}\circ\cdots\circ x_{v}\subseteq Q$. ∎

######  Theorem 3.10.

Assume that $Q$ is a $(u,v)$-absorbing strong $\mathcal{C}$-hyperideal of $A$
for $u,v\in\mathbb{N}$ with $2\leq v<u$ and $Q_{1},\cdots,Q_{v}$ are the only
minimal prime $\mathcal{C}$-hyperideals over $Q$. If $z_{i}\in
Q_{i}\backslash(\cup_{j\neq i}Q_{j})$, then $Q_{j}\circ\Pi_{i\neq
j}z_{i}\subseteq Q$.

###### Proof.

Assume that $x\in P_{j}$. If $x\notin\cup_{j\neq i}Q_{j}$, then we have
$x\circ\Pi_{i\neq j}z_{i}\subseteq\cap_{i=1}^{v}Q_{i}=rad(Q)$ as the
$Q_{i}^{,}$s are the only minimal prime hyperideals over $Q$. Therefore we
conclude that $x\circ\Pi_{i\neq j}z_{i}\subseteq Q$ by Proposition 3.5 and
Lemma 3.9. Let $x\in Q_{j}\ \cap(\cup_{j\neq i}Q_{j})$. Then there exist $y\in
A$ and $z\in Q_{j}\backslash(\cup_{j\neq i}Q_{j})$ such that $w+x\in
Q_{j}\backslash(\cup_{j\neq i}Q_{j})$ for every $w\in y\circ z$ by Corollary
3.12 in [5]. It follows that $(w+x)\circ\Pi_{i\neq j}z_{i}\subseteq Q$ and so
$w\circ\Pi_{i\neq j}z_{i}+x\circ\Pi_{i\neq j}z_{i}\subseteq Q$ as $Q$ is a
strong $\mathcal{C}$-hyperideal of $A$. Since $w\circ\Pi_{i\neq
j}z_{i}\subseteq Q$, we obtain $x\circ\Pi_{i\neq j}z_{i}\subseteq Q$ which
means $Q_{j}\circ\Pi_{i\neq j}z_{i}\subseteq Q$. ∎

Let $Q$ be a $u$-absorbing hyperideal of $A$ for $u\in\mathbb{N}$. We define
$Abs(Q)=\min\\{v\ |\ Q\ \text{is a $v$-absorbing hyperideal of }A\\}$,
otherwise $Abs(Q)=\infty$ [5]. Moreover, we define $abs(Q)=\min\\{v\ |\ Q\
\text{is an $(Abs(Q)+1,v)$-absorbing hyperideal of }A\\}$. It is clear that
$abs(Q)\leq Abs(Q)$. Thus, in Proposition 3.4, we get $abs(\cap_{i=1}^{k})\leq
abs(Q_{1})+\cdots+abs(Q_{k})$ and
$Abs(\cap_{i=1}^{k}Q_{i})\leq\max\\{Abs(Q_{1}),\cdots,Abs(Q_{k}),abs(Q_{1})+\cdots+abs(Q_{k})\\}$.
Also, in Proposition 3.5, we have $abs(rad(Q))\leq abs(Q)$ and
$Abs(rad(Q))\leq\max\\{\lceil\frac{Abs(Q)+1}{abs(Q)}\rceil-1,abs(Q)\\}.$
Recall from [1] that the hyperideals $I$ and $J$ are coprime if $I+J=A$.

######  Theorem 3.11.

Assume that $Q$ is a $(u,v)$-absorbing strong $\mathcal{C}$-hyperideal of $A$
for $u,v\in\mathbb{N}$ with $2\leq v<u$ and $Q_{1},\cdots,Q_{v}$ are the only
minimal prime $\mathcal{C}$-hyperideals over $Q$. Then

  * (i)

$Q_{1}\circ\cdots\circ Q_{v}\subseteq Q$ and $abs(Q)=v$.

  * (ii)

If $Q_{1},\cdots,Q_{v}$ are pairwise coprime, then $Q_{1}\circ\cdots\circ
Q_{v}=Q$ and $abs(Q)=v$.

###### Proof.

It can be easily seen that the claim is true in a similar manner to the proofs
of Theorem 3.13 and Theorem 3.14 in [5]. ∎

######  Theorem 3.12.

Assume that $Q$ is a radical $\mathcal{C}$-hyperideal of $A$. Then $Q$ is a
$(u,v)$-absorbing hyperideal of $A$ if and only if $Q$ is a $v$-absorbing
hyperideal of $A$.

###### Proof.

($\Longrightarrow$) Suppose that $Q$ is a $(u,v)$-absorbing hyperideal of $A$
and $x_{1}\circ\cdots\circ x_{u-1}\subseteq Q$ for $x_{1},\cdots,x_{u-1}\in
A\backslash U(A)$. Since $x_{1}\circ\cdots\circ x_{k-1}\circ x_{k}^{2}\circ
x_{k+1}\circ\cdots\circ x_{u-1}\subseteq Q$, we conclude that either
$x_{i_{1}}\circ\cdots\circ x_{i_{v}}\subseteq Q$ for some
$i_{j}\in\\{1,\cdots,k-1,k,k+1,\cdots,u-1\\}$ which implies $Q$ is a
$(u-1,v)$-absorbing hyperideal of $A$ or $x_{k}^{2}\circ
x_{i_{1}}\circ\cdots\circ x_{i_{v-2}}\subseteq Q$ for some
$i_{j}\in\\{1,\cdots,k-1,k+1,\cdots,u-1\\}$ which means $x_{k}\circ
x_{i_{1}}\circ\cdots\circ x_{i_{v-2}}\subseteq rad(Q)$. Since $Q$ is a radical
hyperideal, we get $x_{k}\circ x_{i_{1}}\circ\cdots\circ x_{i_{v-2}}\subseteq
Q$ which implies $Q$ is a $(u-1,v)$-absorbing hyperideal of $A$. By continuing
this process, we conclude $Q$ is a $(v+1,v)$-absorbing hyperideal and so it is
a $v$-absorbing hyperideal of $A$ by Remark 3.3(i).

($\Longleftarrow$) Let $Q$ is a $v$-absorbing hyperideal of $A$ and
$x_{1}\circ\cdots\circ x_{u}\subseteq Q$ for $x_{1},\cdots,x_{u-1}\in
A\backslash U(A)$. Take any $c\in(x_{i_{1}}\circ\cdots\circ
x_{i_{u-v}})\backslash Q$ such that $i_{j}\in\\{1,\cdots,u\\}$. Then a product
of $v$ of the $x_{i}^{,}$s othere than the $x_{i_{j}}$ with $c$ is a subset of
$Q$ and so a product of $v$ of the $x_{i}^{,}$s othere than the $x_{i_{j}}$ is
a subset of $Q$ as $Q$ is a $v$-absorbing hyperideal of $A$. Thus $Q$ is a
$(u,v)$-absorbing hyperideal of $A$. ∎

Let us give the following theorem without proof as a result of Theorem 3.12.

######  Corollary 3.13.

Let $Q$ be a $(u,v)$-absorbing radical $\mathcal{C}$-hyperideal of $A$. Then
$abs(Q)=Abs(Q)$.

Recall from [17] that a proper hyperideal $Q$ of $A$ is primary if for any
$x,y\in A$, $x\circ y\subseteq Q$ implies that $x\in Q$ or $y^{t}\subseteq Q$
for some $t\in\mathbb{N}$. Since the radical of a primary
$\mathcal{C}$-hyperideal $Q$ is a prime hyperideal of $A$, say $P$, $Q$ is
referred to as a $P$-primary $\mathcal{C}$-hyperideal of $A$.

######  Theorem 3.14.

Assume that $u,v\in\mathbb{N}$ and $Q$ is a $P$-primary
$\mathcal{C}$-hyperideal of $A$ where $P$ is a prime hyperideal of $A$ with
$P^{v}\subseteq Q$. Then $Q$ is a $(u,v)$-absorbing hyperideal of $A$ for any
$u\geq v+1$. Furthermore, if $P^{v}$ is a $P$-primary $\mathcal{C}$-hyperideal
of $A$, then $P^{v}$ is a $(u,v)$-absorbing hyperideal of $A$ for any $u\geq
v+1$ and $abs(P^{v})=v$.

###### Proof.

The proof follows from Theorem 4.1 in [5]. ∎

##  4\. $AB$-$(u,v)$-absorbing hyperideals

This section is devoted to the study of a subclass of the $(u,v)$-absorbing
hyperideals called $AB$-$(u,v)$-absorbing hyperideals.

######  Definition 4.1.

Let $u,v\in\mathbb{N}$ with $u>v$. A proper hyperideal $Q$ of $A$ is said to
be an $AB$-$(u,v)$-absorbing hyperideal if whenever $x_{1}\circ\cdots\circ
x_{u}\subseteq Q$ for $x_{1},\cdots,x_{u}\in A$, then there are $v$ of the
$x_{i}^{,}$s whose product is a subset of $Q$.

######  Theorem 4.2.

Let $Q$ be a proper strong $\mathcal{C}$-hyperideal of $A$ such that
$a+1\notin U(A)$ for some $a\in Q$ and $u,v\in\mathbb{N}$ with $u>v$. Then $Q$
is an $AB$-$(u,v)$-absorbing hyperideal if and only if $Q$ is a
$(u,v)$-absorbing hyperideal.

###### Proof.

$(\Longrightarrow)$ It is obvious.

$(\Longleftarrow)$ Assume that $x_{1}\circ\cdots\circ x_{u}\subseteq Q$ for
$x_{1},\cdots,x_{u}\in A$. Let $x_{1},\cdots,x_{s-1}$ be non-unit elememts in
$A$ and $x_{s},\cdots,x_{u}$ be unit elements in $A$ for some $s\geq v+2$.
Since $x_{1}\circ\cdots\circ x_{u}\subseteq Q$ and $x_{s},\cdots,x_{u}$ are
units in $A$, we get $x_{1}\circ\cdots,\circ x_{s}\subseteq Q$. Put
$y_{i}=1+a$ for each $i\in\\{1,\cdots,s\\}$. Then the $y_{i}^{,}$s are non-
unit. Since $Q$ is a $(u,v)$-absorbing hyperideal and $x_{1}\circ\cdots\circ
x_{s-1}\circ y_{s}\circ\cdots\circ y_{u}\subseteq Q$, we conclude that there
exist $v$ elements of the $x_{i}^{,}$ and the $y_{j}^{,}$s whose product is a
subset of $Q$. Take any $n\in\mathbb{N}$. Assume that $(a+1)^{n}\subseteq Q$.
This implies that $(\Sigma_{k=0}^{n}\dbinom{n}{k}a^{n-k}\circ 1^{k})\cap
Q\neq\varnothing$. Since $Q$ is a strong $\mathcal{C}$-hyperideal of $A$, we
have $(\Sigma_{k=0}^{n}\dbinom{n}{k}a^{n-k}\circ 1^{k})\subseteq Q$ and so
$1\in Q$, a contradiction. Then $(a+1)^{n}\nsubseteq Q$ for all $n\geq 1$.
Thus $x_{i_{1}}\circ\cdots\circ x_{i_{h}}\circ(a+1)^{v-h}\subseteq Q$ for some
$h\in\\{1,\cdots,v\\}$ and $i_{j}\in\\{1,\cdots,s-1\\}$. Clearly,
$\Sigma_{k=0}^{v-h}\dbinom{v-h}{k}a^{v-h-k}\circ 1^{k}=X+1$ where $X\subseteq
Q$. Since $x_{i_{1}}\circ\cdots\circ x_{i_{h}}\circ(X+1)\subseteq
Q\cap(x_{i_{1}}\circ\cdots\circ x_{i_{h}}\circ X)+(x_{i_{1}}\circ\cdots\circ
x_{i_{h}}\circ 1)$, we obtain $(x_{i_{1}}\circ\cdots\circ x_{i_{h}}\circ
X)+(x_{i_{1}}\circ\cdots\circ x_{i_{h}}\circ 1)\subseteq Q$. Hence we get the
result that $x_{i_{1}}\circ\cdots\circ x_{i_{h}}\subseteq Q$ which implies $Q$
is an $AB$-$(u,v)$-absorbing hyperideal of $A$. ∎

As a result of the previous theorems, we give the following result.

######  Corollary 4.3.

Let $Q$ be a proper strong $\mathcal{C}$-hyperideal of $A$ such that
$Q\nsubseteq J(A)$ and $u,v\in\mathbb{N}$ with $u>v$. Then $Q$ is an
$AB$-$(u,v)$-absorbing hyperideal if and only if $Q$ is a $(u,v)$-absorbing
hyperideal.

Let $X$ be a finite sum of finite products of elements of $A$. Consider the
relation $\gamma$ on a multiplicative hyperring $A$ defined as $x\gamma y$ if
and only if $\\{x,y\\}\subseteq X$, namely, $x\gamma y$ if and only if
$\\{x,y\\}\subseteq\sum_{j\in J}\prod_{i\in I_{j}}z_{i}$ for some
$z_{1},...,z_{n}\in A$ and $I_{j},J\subseteq\\{1,...,n\\}$. $\gamma^{\ast}$
denotes the transitive closure of $\gamma$. The relation $\gamma^{\ast}$ is
the smallest equivalence relation on $A$ such that the set of all equivalence
classes, i.e., the quotient $A/\gamma^{\ast}$, is a fundamental ring. Assume
that $\Sigma$ is the set of all finite sums of products of elements of $A$. We
can rewrite the definition of $\gamma^{\ast}$ on $A$, namely,
$x\gamma^{\ast}y$ if and only if there exist $z_{1},...,z_{n}\in A$ such that
$z_{1}=x,z_{n+1}=y$ and $u_{1},...,u_{n}\in\Sigma$ where
$\\{z_{i},z_{i+1}\\}\subseteq u_{i}$ for $1\leq i\leq n$. Suppose that
$\gamma^{\ast}(x)$ is the equivalence class containing $x\in A$. Define
$\gamma^{\ast}(x)\oplus\gamma^{\ast}(y)=\gamma^{\ast}(z)$ for every
$z\in\gamma^{\ast}(x)+\gamma^{\ast}(y)$ and
$\gamma^{\ast}(x)\odot\gamma^{\ast}(y)=\gamma^{\ast}(w)$ for every
$w\in\gamma^{\ast}(x)\circ\gamma^{\ast}(y)$. Then
$(A/\gamma^{\ast},\oplus,\odot)$ is a ring called a fundamental ring of $A$
[37].

######  Theorem 4.4.

Let $Q$ be a proper hyperideal of $A$. Then $Q$ is an $AB$-$(u,v)$-absorbing
hyperideal of $(A,+,\circ)$ if and only if $Q/\gamma^{\ast}$ is an
$AB$-$(u,v)$-absorbing hyperideal of $(A/\gamma^{\ast},\oplus,\odot)$.

###### Proof.

($\Longrightarrow$) Let $x_{1}\odot\cdots\odot x_{u}\in Q/\gamma^{\ast}$ for
some $x_{1},\cdots,x_{u}\in A/\gamma^{\ast}$. Then there exist $a_{i}\in A$
for every $i\in\\{1,\cdots,u\\}$ such that $x_{i}=\gamma^{\ast}(a_{i})$. This
implies that $x_{1}\odot\cdots\odot
x_{u}=\gamma^{\ast}(a_{1})\odot\cdots\odot\gamma^{\ast}(a_{u})=\gamma^{\ast}(a_{1}\circ\cdots\circ
a_{u})$. Since $\gamma^{\ast}(a_{1}\circ\cdots\circ a_{u})\in
Q/\gamma^{\ast}$, we get $a_{1}\circ\cdots\circ a_{u}\subseteq Q$. Since $Q$
is an $AB$-$(u,v)$-absorbing hyperideal of $A$, we get the result that there
is $v$ elements of the $a_{i}^{,}$s, say $a_{i_{1}},\cdots,a_{i_{v}}$, whose
product is a subset of $Q$. This implies that $x_{i_{1}}\odot\cdots\odot
x_{i_{v}}=\gamma^{\ast}(a_{i_{1}})\odot\cdots\odot\gamma^{\ast}(a_{i_{v}})=\gamma^{\ast}(a_{i_{1}}\circ\cdots\circ
a_{i_{v}})\in Q/\gamma^{\ast}$. Consequently, $Q/\gamma^{\ast}$ is an
$AB$-$(u,v)$-absorbing hyperideal of $A/\gamma^{\ast}$.

($\Longleftarrow$) Suppose that $a_{1}\circ\cdots\circ a_{u}\subseteq Q$ for
some $a_{1},\cdots,a_{u}\in A$. Then we have
$\gamma^{\ast}(a_{1}),\cdots,\gamma^{\ast}(a_{u})\in A/\gamma^{\ast}$ and so
$\gamma^{\ast}(a_{1})\odot\cdots\odot\gamma^{\ast}(a_{u})=\gamma^{\ast}(a_{1}\circ\cdots\circ
a_{u})\in Q/\gamma^{\ast}$. Since $Q/\gamma^{\ast}$ is an
$AB$-$(u,v)$-absorbing hyperideal of $A/\gamma^{\ast}$, there exist $v$
elements of the $\gamma^{\ast}(a_{i})^{,}$s, say
$\gamma^{\ast}(a_{i_{1}}),\cdots,\gamma^{\ast}(a_{i_{v}})$ whose product is in
$Q/\gamma^{\ast}$. This means $\gamma^{\ast}(a_{i_{1}}\circ\cdots\circ
a_{i_{v}})=\gamma^{\ast}(a_{i_{1}})\odot\circ\odot\gamma^{\ast}(a_{i_{v}})\in
Q/\gamma^{\ast}$ which implies $a_{i_{1}}\circ\cdots\circ a_{i_{v}}\subseteq
Q$. Thus $Q$ is an $AB$-$(u,v)$-absorbing hyperideal of $A$. ∎

Let $(A_{1},+_{1},\circ_{1})$ and $(A_{2},+_{2},\circ_{2})$ be two
multiplicative hyperrings with nonzero identity. The set $A_{1}\times A_{2}$
with the operation $+$ and the hyperoperation $\circ$ defined as

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

$(x_{1},x_{2})\circ(y_{1},y_{2})=\\{(x,y)\in A_{1}\times A_{2}\ |\ x\in
x_{1}\circ_{1}y_{1},y\in x_{2}\circ_{2}y_{2}\\}$
is a multiplicative hyperring [35]. Now, we present some characterizations of
$AB$-$(u,v)$-absorbing hyperideals on cartesian product of commutative
multiplicative hyperring.

######  Theorem 4.5.

Assume that $A=A_{1}\times A_{2}$ such that $A_{1},A_{2}$ are commutative
multiplicative hyperrings, $Q=Q_{1}\times A_{2}$ such that $Q_{1}$ is a proper
hyperideal of $A_{1}$ and $u,v\in\mathbb{N}$ with $u>v$. Then the following
are equivalent.

  * (i)

$Q$ is a $(u,v)$-absorbing hyperideal of $A$.

  * (ii)

$Q_{1}$ is an $AB$-$(u,v)$-absorbing hyperideal of $A_{1}$.

  * (iii)

$Q$ is an $AB$-$(u,v)$-absorbing hyperideal of $A$.

###### Proof.

(i) $\Longrightarrow$ (ii) Suppose on the contrary, $Q_{1}$ is not an
$AB$-$(u,v)$-absorbing hyperideal of $A_{1}$. This means that we have
$x_{1}\circ_{1}\cdots\circ_{1}x_{u}\subseteq Q_{1}$ for some
$x_{1},\cdots,x_{u}\in A_{1}$ and none of product of the terms $v$ of the
$x_{i}^{,}$s is not a subset of $Q_{1}$. Since $Q$ is a $(u,v)$-absorbing
hyperideal of $A$, $(x_{1},0)\circ\cdots\circ(x_{u},0)\subseteq Q$ and none of
the $(x_{i},0)^{,}$s are unit, there exist $v$ of the $(x_{i},0)^{,}$s whose
is a subset of $Q$ which implies there exist $v$ of the $x_{i}^{,}$s whose is
a subset of $Q_{1}$ which is a contradiction. Hence $Q_{1}$ is an
$AB$-$(u,v)$-absorbing hyperideal of $A_{1}$.

(ii) $\Longrightarrow$ (iii) Let
$(x_{1},y_{1})\circ\cdots\circ(x_{u},y_{u})\subseteq Q$ for
$(x_{1},y_{1}),\cdots,(x_{u},y_{u})\in A$. This means that
$x_{1}\circ_{1}\cdots\circ_{1}x_{u}\subseteq Q_{1}$. Since $Q_{1}$ is an
$AB$-$(u,v)$-absorbing hyperideal of $A_{1}$, there exist $v$ of the
$x_{i}^{,}$s, say $x_{i_{1}},\cdots,x_{i_{v}}$, whose product is a subset of
$Q_{1}$. It follows that
$(x_{i_{1}},y_{i_{1}})\circ\cdots\circ(x_{i_{v}},y_{i_{v}})\subseteq Q$ and so
$Q$ is an $AB$-$(u,v)$-absorbing hyperideal of $A$.

(iii) $\Longrightarrow$ (i) The claim is true because every
$AB$-$(u,v)$-absorbing is a $(u,v)$-absorbing hyperideal. ∎

##  5\.  $(u,v)$-absorbing prime hyperideals

In this section, we aim to extend the class of the $(u,v)$-absorbing
hyperideals to $(u,v)$-absorbing prime hyperideals.

######  Definition 5.1.

Let $u,v\in\mathbb{N}$ with $u>v$. A proper hyperideal $Q$ of $A$ refers to a
$(u,v)$-absorbing prime hyperideal if $x_{1}\circ\cdots\circ x_{u}\subseteq Q$
for $x_{1},\cdots,x_{u}\in A\backslash U(A)$ implies either
$x_{1}\circ\cdots\circ x_{v}\subseteq Q$ or $x_{v+1}\circ\cdots\circ
x_{u}\subseteq Q$.

######  Example 5.2.

In Example 3.2, the hyperideal $\langle 15\rangle$ is a $(3,1)$-absorbing
prime hyperideal. However, since $3^{2}\circ
5=\\{180,360,720\\}\subseteq\langle 15\rangle$ but $3,5\notin\langle
15\rangle$, then $\langle 15\rangle$ is not a $(3,1)$-absorbing hyperideal

######  Remark 5.3.

Let $u,v\in\mathbb{N}$ with $u>v$ and $Q$ be a $\mathcal{C}$-hyperideal of
$A$.

  * (i)

$Q$ is a $(u,v)$-absorbing prime hyperideal of $A$ if and only if $Q$ is a
$(u,u-v)$-absorbing prime hyperideal.

  * (ii)

If $Q$ is a $(u,v)$-absorbing prime hyperideal of $A$, then $Q$ is a
$(u+1,v+1)$-absorbing prime hyperideal.

######  Proposition 5.4.

Assume that $Q$ is a strong $\mathcal{C}$-hyperideal of $A$ such that $A$ is
not local and $u,v\in\mathbb{N}$ with $u>v$. Then $Q$ is a $(u,v)$-absorbing
prime hyperideal of $A$ if and only if $Q$ is a $(u-1,v-1)$-absorbing prime
hyperideal.

###### Proof.

$(\Longrightarrow)$ Let $Q$ be a $(u,v)$-absorbing prime hyperideal of $A$.
Suppose on contrary that $Q$ is not a $(u-1,v-1)$-absorbing prime hyperideal
of $A$. Then there exist $x_{1},\cdots,x_{u-1}\in A\backslash U(A)$ such that
$x_{1}\circ\cdots\circ x_{u-1}\subseteq Q$ but neither $x_{1}\circ\cdots\circ
x_{v-1}\subseteq Q$ nor $x_{v}\circ\cdots\circ x_{u-1}\subseteq Q$. Take any
$y\in A\backslash U(A)$. Since $Q$ is a $(u,v)$-absorbing prime hyperideal of
$A$, $y\circ x_{1}\circ\cdots\circ x_{u-1}\subseteq Q$ and
$x_{v}\circ\cdots\circ x_{u-1}\nsubseteq Q$, we have $y\circ
x_{1}\circ\cdots\circ x_{v-1}\subseteq Q$. Take any $z\in U(A)$. If $y+z\notin
U(A)$, then we obtain $(y+z)\circ x_{1}\circ\cdots\circ x_{v-1}\subseteq Q$ as
$(y+z)\circ x_{1}\circ\cdots\circ x_{u-1}\subseteq Q$, $x_{v}\circ\cdots\circ
x_{u-1}\nsubseteq Q$ and $Q$ is a $(u,v)$-absorbing prime hyperideal of $A$.
Since $Q$ is a strong $\mathcal{C}$-hyperideal of $A$ and $(y+z)\circ
x_{1}\circ\cdots\circ x_{v-1}\subseteq y\circ x_{1}\circ\cdots\circ
x_{v-1}+z\circ x_{1}\circ\cdots\circ x_{v-1}$, we conclude that $y\circ
x_{1}\circ\cdots\circ x_{v-1}+z\circ x_{1}\circ\cdots\circ x_{v-1}\subseteq
Q$. This implies that $x_{1}\circ\cdots\circ x_{v-1}\subseteq Q$ because
$y\circ x_{1}\circ\cdots\circ x_{v-1}\subseteq Q$. This is a contradiction.
Then we have $y+z\in U(A)$. By Lemma 2.6 in [28], we get the result that $A$
is local which is a contradiction. Thus $Q$ is a $(u-1,v-1)$-absorbing prime
hyperideal of $A$.

$(\Longleftarrow)$ It is obvious by Remark 5.3 (ii). ∎

######  Proposition 5.5.

Assume that $u,v\in\mathbb{N}$ with $u>v$ and $Q$ is a strong
$\mathcal{C}$-hyperideal of $A$ such that $A$ is not local. Then $Q$ is a
$(u-v+1,1)$-absorbing prime hyperideal of $A$ if and only if $Q$ is a prime
hyperideal.

###### Proof.

($\Longrightarrow$)Let $Q$ be a $(u-v+1,1)$-absorbing prime hyperideal of $A$.
Then we get the result that $Q$ is a $(u-v+1,u-v)$-absorbing prime hyperideal
of $A$ by Remark 5.3 (i). Hence we conclude that $Q$ is a prime hyperideal by
Proposition 5.4.

($\Longleftarrow$) Straightforward. ∎

For a $\mathcal{C}$-hyperideal $Q$ of $A$ that the hyperring $A$ is not local,
it can be easily proved, by induction, that $Q$ is a $(u-1,v-1)$-absorbing
prime hyperideal for $u,v\in\mathbb{N}$ with $u>v$ if and only if $Q$ is a
$(u-v+1,1)$-absorbing prime hyperideal. Thus in view of Proposition 5.4 and
Proposition 5.5, we have the following result.

######  Corollary 5.6.

In a hyperring $A$ that is not local, every $(u,v)$-absorbing prime hyperideal
is prime and vice versa.

######  Theorem 5.7.

Assume that $Q_{1}$ and $Q_{2}$ are strong $\mathcal{C}$-hyperideals of the
hyperring $A_{1}$ and $A_{2}$, respectively. Let $u,v\in\mathbb{N}$ with
$u>v$. Then the following are equivalent:

  * (i)

$Q_{1}\times Q_{2}$ is a $(u,v)$-absorbing prime hyperideal of $A_{1}\times
A_{2}$.

  * (ii)

$Q_{1}\times Q_{2}$ is a prime hyperideal of $A_{1}\times A_{2}$.

  * (iii)

$Q_{1}$ is a prime hyperideal of $A_{1}$ and $Q_{2}=A_{2}$ or $Q_{2}$ is a
prime hyperideal of $A_{2}$ and $Q_{1}=A_{1}$.

###### Proof.

It can be easily seen that the idea is true by Corollary 5.6 as $A_{1}\times
A_{2}$ is not local. ∎

######  Proposition 5.8.

Let $u,v\in\mathbb{N}$ with $u>v$ and $Q$ be a strong $\mathcal{C}$-hyperideal
of $A$. Then $Q$ is a $(v+1,v)$-absorbing prime hyperideal of $A$ if and only
if $Q$ is a prime hyperideal or $A$ is local such that $M^{v}\subseteq Q$ for
the only maximal hyperideal $M$ of $A$.

###### Proof.

($\Longrightarrow$) Let $Q$ be a $(v+1,v)$-absorbing prime hyperideal of $A$
and the hyperring $A$ is not local. Then $Q$ is prime by Corollary 5.6.
Suppose that $A$ is local and $M$ is the only maximal hyperideal of $A$ such
that $M^{v}\nsubseteq Q$. If $Q$ is not a $(v,v-1)$-absorbing prime hyperideal
of $A$, then there exist $x_{1},\cdots,x_{v}\in A\backslash U(A)$ such that
$x_{1}\circ\cdots\circ x_{v}\subseteq Q$ but $x_{1}\circ\cdots\circ
x_{v-1}\nsubseteq Q$ and $x_{v}\notin Q$. Take any $a\in a_{1}\circ\cdots\circ
a_{v}$ for every $a_{1},\cdots,a_{v}\in M$. Since $Q$ is a $(v+1,v)$-absorbing
prime hyperideal of $A$, $a\circ x_{1}\circ\cdots\circ x_{v}\subseteq Q$ and
$x_{v}\notin Q$, we get $a\circ x_{1}\circ\cdots\circ x_{v-1}\subseteq Q$. Let
$b\in x_{1}\circ\cdots\circ x_{v-1}$. So $a\circ b\subseteq a\circ
x_{1}\circ\cdots\circ x_{v-1}\subseteq Q$. On the other hand, $a\circ
b\subseteq a_{1}\circ\cdots\circ a_{v}\circ b$. Then we get
$a_{1}\circ\cdots\circ a_{v}\circ b\subseteq Q$ as $Q$ is a strong
$\mathcal{C}$-hyperideal of $A$. If $b\in Q$, then $x_{1}\circ\cdots\circ
x_{v-1}\cap Q\neq\varnothing$ and so $x_{1}\circ\cdots\circ x_{v-1}\subseteq
Q$ which is impossible. Therefore $b\notin Q$. Since $Q$ is a
$(v+1,v)$-absorbing prime hyperideal of $A$, $a_{1}\circ\cdots\circ a_{v}\circ
b\subseteq Q$ and $b\notin Q$, we conclude that $a_{1}\circ\cdots\circ
a_{v}\subseteq Q$ which means $M^{v}\subseteq Q$, a contradiction. Then we get
the result that $Q$ is a $(v,v-1)$-absorbing prime hyperideal of $A$. By
induction, we conclude that $Q$ is prime as $M^{i}\nsubseteq Q$ for any $i\leq
v$.

($\Longleftarrow$) It is obvious. ∎

######  Corollary 5.9.

Assume that $u,v\in\mathbb{N}$ with $u>v$ and $Q$ is a $(v+1,v)$-absorbing
prime strong $\mathcal{C}$-hyperideal of $A$ that is not prime. If $P$ is a
hyperideal of $A$ containing $Q$, then $P$ is a $(u,v)$-absorbing prime
hyperideal of $A$.

###### Proof.

Suppose that $Q$ is a $(v+1,v)$-absorbing prime strong
$\mathcal{C}$-hyperideal of $A$ that is not prime. By Proposition 5.8, we get
the result that $A$ is local and $M^{v}\subseteq Q$ where $M$ is the only
maximal hyperideal of $A$. Then $M^{v}\subseteq P$ and so we are done by By
Proposition 5.8. ∎

######  Theorem 5.10.

Let $Q$ be a $(u,v)$-absorbing prime strong $\mathcal{C}$-hyperideal of $A$
where $u,v\in\mathbb{N}$ and $u>v$. Then the following statements hold:

  * (i)

$rad(Q)$ is a prime hyperideal of $A$.

  * (ii)

If $Q$ is not a prime hyperideal of $A$, then $A$ is local. In particular, if
$u=v+1$, then $rad(Q)=M$ where $M$ is the only maximal hyperideal of $A$.

###### Proof.

(i) Suppose that $Q$ is a $(u,v)$-absorbing prime strong
$\mathcal{C}$-hyperideal of $A$. Let $a\circ b\subseteq rad(Q)$ for $a,b\in
A$. We may assume $a,b\in A\backslash U(A)$, without loss of generality. Then
we have $a^{n}\circ b^{n}\subseteq Q$ for some $n\in\mathbb{N}$. Take any
$x\in a^{n}$ and $y\in b^{n}$. If $x$ or $y\in U(A)$, then we obtain
$b^{n}\subseteq e\circ b^{n}\subseteq x^{-1}\circ x\circ b^{n}\subseteq
x^{-1}\circ a^{n}\circ b^{n}\subseteq Q$ which means $b\in rad(Q)$ or
$a^{n}\subseteq a^{n}\circ e\subseteq a^{n}\circ y\circ y^{-1}\subseteq
a^{n}\circ b^{n}\circ y^{-1}\subseteq Q$ which means $a\in rad(Q)$. Now, we
assume that $x,y\notin U(A)$. Since $a^{v-1}\circ x\circ b^{u-v-1}\circ
y\subseteq Q$ and $Q$ is a $(u,v)$-absorbing prime hyperideal of $A$, we get
the result that $a^{v-1}\circ x\subseteq Q$ or $b^{u-v-1}\circ y\subseteq Q$.
In the first case, since $(a^{v-1}\circ a^{n})\cap Q\neq\varnothing$, we get
$a^{v+n-1}=a^{v-1}\circ a^{n}\subseteq Q$ which means $a\in rad(Q)$. In the
second case, since $(b^{u-v-1}\circ b^{n})\cap Q\neq\varnothing$, we have
$b^{u-v+n-1}=b^{u-v-1}\circ b^{n}\subseteq Q$ which implies $b\in rad(Q)$.
Thus $rad(Q)$ is a prime hyperideal of $A$.

(ii) Suppose that $Q$ is a $(u,v)$-absorbing prime strong
$\mathcal{C}$-hyperideal of $A$ that is not prime. Then we conclude that $A$
is local by Corollary 5.6. Let $u=v+1$ and $M$ be the only maximal hyperideal
of $A$. By Proposition 5.8, we conclude that $M^{v}\subseteq Q$ and so
$rad(Q)=M$. ∎

######  Lemma 5.11.

Let every hyperideal of $A$ be a prime $\mathcal{C}$-hyperideal. Then $A$ is a
hyperfield.

###### Proof.

Let $0\neq x\in A$. Since $\langle x^{2}\rangle$ is a prime hyperideal of $A$
and $x^{2}\subseteq\langle x^{2}\rangle$, we get $x\in\langle x^{2}\rangle$
which means $x\in x^{2}\circ a$ for some $a\in A$. It follows that $0\in
x^{2}\circ a-x$. Since $0\in x^{2}\circ a-x\cap\langle 0\rangle$, we get the
result that $(x\circ a-1)\circ x\subseteq x^{2}\circ a-x\subseteq\langle
0\rangle$. Let $t\in x\circ a$. Since $\langle 0\rangle$ is a prime
hyperideal, $(t-1)\circ x\subseteq\langle 0\rangle$ and $x\neq 0$, we obtain
$t-1=0$ and so $1\in x\circ a$ which means $x$ is unit. Thus we conclude that
$A$ is a hyperfield. ∎

######  Theorem 5.12.

Let every hyperideal of $A$ be a strong $\mathcal{C}$-hyperideal, $J(A)=M$ and
$v\in\mathbb{N}$. Then every proper hyperideal in $A$ is a $(v+1,v)$-absorbing
prime hyperideal if and only if $A$ is local and $M^{v}=0$.

###### Proof.

($\Longrightarrow$) Suppose on contrary that $A$ is not local. By Corollary
5.6, we conclude that every proper hyperideal in $A$ is prime. Therefore, by
Lemma 5.11, we get the result that $A$ is a hyperfield which is impossible.
Then $A$ is local. Let $M$ be the only maximal hyperideal of $A$. Since
$\langle 0\rangle$ is a $(v+1,v)$-absorbing prime hyperideal and $A$ is local,
we have $M^{v}=0$ by the hypothesis and Proposition 5.8.

($\Longleftarrow$) Let $Q$ be an arbitary hyperideal and
$x_{1}\circ\cdots\circ x_{v+1}\subseteq Q$ for $x_{1},\cdots,x_{v+1}\in
A\backslash U(A)$. Since $A$ is local, $M$ is the only maximal hyperideal of
$A$ and $x_{1},\cdots,x_{v}\in A\backslash U(A)$, we conclude that
$x_{1}\circ\cdots\circ x_{v}\subseteq M^{v}=0$ which means
$x_{1}\circ\cdots\circ x_{v}\subseteq Q$, as required. ∎

######  Theorem 5.13.

Let $Q$ be a proper hyperideal of $A$ and $u,v\in\mathbb{N}$ with $u>v$. Then
$Q$ is a $(u,v)$-absorbing prime hyperideal of $A$ if and only if for proper
hyperideals $Q_{1},\cdots,Q_{u}$ of $A$, $Q_{1}\circ\cdots\circ Q_{u}\subseteq
Q$ implies that $Q_{1}\circ\cdots\circ Q_{v}\subseteq Q$ or
$Q_{v+1}\circ\cdots\circ Q_{u}\subseteq Q$.

###### Proof.

($\Longrightarrow$) Let $Q$ be a $(u,v)$-absorbing prime hyperideal of $A$.
Assume that $Q_{1}\circ\cdots\circ Q_{u}\subseteq Q$ where
$Q_{1},\cdots,Q_{u}$ are proper hyperideals of $A$ such that
$Q_{v+1},\circ\cdots\circ Q_{u}\nsubseteq Q$. Then there exists $x_{i}\in
Q_{i}$ for each $i\in\\{v+1,\cdots,u\\}$ such that $x_{v+1}\circ\cdots\circ
x_{u}\nsubseteq Q$. Take any $y_{i}\in Q_{i}$ for each $i\in\\{1,\cdots,v\\}$.
Therefore we have $y_{1}\circ\cdots y_{v}\circ x_{v+1}\circ\cdots\circ
x_{u}\subseteq Q$. This implies that $y_{1}\circ\cdots y_{v}\subseteq Q$ as
$Q$ is a $(u,v)$-absorbing prime hyperideal of $A$ and
$x_{v+1}\circ\cdots\circ x_{u}\nsubseteq Q$. Hence $Q_{1}\circ\cdots\circ
Q_{v}\subseteq Q$, as needed.

($\Longleftarrow$) Let $x_{1}\circ\cdots\circ x_{u}\subseteq Q$ for some
$x_{1},\cdots,x_{u}\in A\backslash U(A)$ such that $x_{1}\circ\cdots\circ
x_{v}\nsubseteq Q$. Put $Q_{i}=\langle x_{i}\rangle$ for each
$i\in\\{1,\cdots,u\\}$. Hence we have $Q_{1}\circ\cdots\circ Q_{u}\subseteq Q$
and $Q_{1}\circ\cdots\circ Q_{v}\nsubseteq Q$. By the hypothesis, we conclude
that $Q_{v+1}\circ\cdots\circ Q_{u}\subseteq Q$ which implies
$x_{v+1}\circ\cdots\circ x_{u}\subseteq Q$. This means that $Q$ is a
$(u,v)$-absorbing prime hyperideal of $A$. ∎

######  Proposition 5.14.

Let $u,v\in\mathbb{N}$ with $u>v$, $Q$ a $(u,v)$-absorbing prime hyperideal of
$A$ and $x\in A\backslash(Q\cup U(A))$. Then $(Q:x)$ is a
$(u-1,v-1)$-absorbing prime hyperideal of $A$.

###### Proof.

Let $x_{1}\circ\cdots\circ x_{u-1}\subseteq(Q:x)$ for $x_{1},\cdots,x_{u-1}\in
A\backslash U(A)$ such that $x_{1}\circ\cdots\circ x_{v-1}\nsubseteq(Q:x)$. So
we have $x\circ x_{1}\circ\cdots\circ x_{u-1}\subseteq Q$. Since $Q$ is a
$(u,v)$-absorbing prime hyperideal of $A$ and $x\circ x_{1}\circ\cdots\circ
x_{v-1}\nsubseteq Q$, we get the result that $x_{v}\circ\cdots x_{u-1}\
\subseteq Q\subseteq(Q:x)$. This shows that $(Q:x)$ is a $(u-1,v-1)$-absorbing
prime hyperideal of $A$. ∎

Assume that $A$ is a multiplicative hyperring. Then the set of all
hypermatrices of $A$ is denoted by $M_{m}(A)$. Let $I=(I_{ij})_{m\times
m},J=(J_{ij})_{m\times m}\in P^{\star}(M_{m}(A))$. Then $I\subseteq J$ if and
only if $I_{ij}\subseteq J_{ij}$[1].

######  Theorem 5.15.

Let $u,v\in\mathbb{N}$ with $u>v$ and $Q$ be a hyperideal of $A$. If
$M_{m}(Q)$ is a $(u,v)$-absorbing prime hyperideal of $M_{m}(A)$, then $Q$ is
a $(u,v)$-absorbing prime hyperideal of $A$.

###### Proof.

Let $x_{1}\circ\cdots\circ x_{u}\subseteq Q$ for $x_{1},\cdots,x_{u}\in
A\backslash U(A)$. Therefore we have

$\begin{pmatrix}x_{1}\circ\cdots\circ x_{u}&0&\cdots&0\\\ 0&0&\cdots&0\\\
\vdots&\vdots&\ddots\vdots\\\ 0&0&\cdots&0\end{pmatrix}\subseteq M_{m}(Q).$

Since $M_{m}(Q)$ is a $(u,v)$-absorbing prime hyperideal of $M_{m}(A)$ and

$\begin{pmatrix}x_{1}\circ\cdots\circ x_{u}&0&\cdots&0\\\ 0&0&\cdots&0\\\
\vdots&\vdots&\ddots\vdots\\\
0&0&\cdots&0\end{pmatrix}=\begin{pmatrix}x_{1}&0&\cdots&0\\\ 0&0&\cdots&0\\\
\vdots&\vdots&\ddots\vdots\\\
0&0&\cdots&0\end{pmatrix}\circ\cdots\circ\begin{pmatrix}x_{u}&0&\cdots&0\\\
0&0&\cdots&0\\\ \vdots&\vdots&\ddots\vdots\\\ 0&0&\cdots&0\end{pmatrix}$

we conclude that

$\begin{pmatrix}x_{1}&0&\cdots&0\\\ 0&0&\cdots&0\\\
\vdots&\vdots&\ddots\vdots\\\
0&0&\cdots&0\end{pmatrix}\circ\cdots\circ\begin{pmatrix}x_{v}&0&\cdots&0\\\
0&0&\cdots&0\\\ \vdots&\vdots&\ddots\vdots\\\
0&0&\cdots&0\end{pmatrix}=\begin{pmatrix}x_{1}\circ\cdots\circ
x_{v}&0&\cdots&0\\\ 0&0&\cdots&0\\\ \vdots&\vdots&\ddots\vdots\\\
0&0&\cdots&0\end{pmatrix}\subseteq M_{m}(Q)$

or

$\begin{pmatrix}x_{v+1}&0&\cdots&0\\\ 0&0&\cdots&0\\\
\vdots&\vdots&\ddots\vdots\\\
0&0&\cdots&0\end{pmatrix}\circ\cdots\circ\begin{pmatrix}x_{u}&0&\cdots&0\\\
0&0&\cdots&0\\\ \vdots&\vdots&\ddots\vdots\\\
0&0&\cdots&0\end{pmatrix}=\begin{pmatrix}x_{v+1}\circ\cdots\circ
x_{u}&0&\cdots&0\\\ 0&0&\cdots&0\\\ \vdots&\vdots&\ddots\vdots\\\
0&0&\cdots&0\end{pmatrix}\subseteq M_{m}(Q).$

Thus we get the result that $x_{1}\circ\cdots\circ x_{v}\subseteq Q$ or
$x_{v+1}\circ\cdots\circ x_{u}\subseteq Q$. Hence $Q$ is a $(u,v)$-absorbing
prime hyperideal of $A$. ∎

Suppose that $(A,+,\circ)$ is a commutative multiplicative hyperring and $x$
is an indeterminate. Then $(A[x],+,\diamond)$ is a polynomail multiplicative
hyperring where $tx^{n}\diamond sx^{m}=(t\circ s)x^{n+m}$ [11].

######  Theorem 5.16.

Let $u,v\in\mathbb{N}$ with $u>v$. If $Q$ is a $(u,v)$-absorbing prime
hyperideal of $(A,+,\circ)$, then $Q[x]$ is a $(u,v)$-absorbing prime
hyperideal of $(A[x],+,\diamond)$.

###### Proof.

Assume that $t_{1}(x)\diamond\cdots\diamond t_{u}(x)\subseteq Q[x]$. Without
loss of generality, we may assume that $t_{i}(x)=a_{i}x^{n_{i}}$ for each
$i\in\\{1,\cdots,u\\}$ and $a_{i}\in A$. Therefore $a_{1}\circ\cdots\circ
a_{u}x^{n_{1}+\cdots+n_{u}}\subseteq Q[x]$. This implies that
$a_{1}\circ\cdots\circ a_{u}\subseteq Q$. Since $Q$ is a $(u,v)$-absorbing
prime hyperideal of $(A,+,\circ)$, we obtain $a_{1}\circ\cdots\circ
a_{v}\subseteq Q$ or $a_{v+1}\circ\cdots\circ a_{u}\subseteq Q$ which implies
$t_{1}(x)\diamond\cdots\diamond t_{v}(x)=a_{1}x^{n_{1}}\diamond\cdots\diamond
a_{v}x^{n_{v}}=a_{1}\circ\cdots\circ a_{v}x^{n_{1}+\cdots+n_{v}}\subseteq
Q[x]$ or $t_{v+1}(x)\diamond\cdots\diamond
t_{u}(x)=a_{v+1}x^{n_{1}}\diamond\cdots\diamond
a_{u}x^{n_{u}}=a_{v+1}\circ\cdots\circ a_{u}x^{n_{v+1}+\cdots+n_{u}}\subseteq
Q[x]$. Thus $Q[x]$ is a $(u,v)$-absorbing prime hyperideal of
$(A[x],+,\diamond)$. ∎

In view of Theorem 5.16, we get the following result.

######  Corollary 5.17.

Suppose that $Q$ is a $(u,v)$-absorbing prime hyperideal of $A$. Then $Q[x]$
is a $(u,v)$-absorbing prime hyperideal of $A[x]$.

Recall from [19] that a mapping $\theta$ from the multiplicative hyperring
$(A_{1},+_{1},\circ_{1})$ into the multiplicative hyperring
$(A_{2},+_{2},\circ_{2})$ is a hyperring good homomorphism if
$\theta(x+_{1}y)=\theta(x)+_{2}\theta(y)$ and
$\theta(x\circ_{1}y)=\theta(x)\circ_{2}\theta(y)$ for all $x,y\in A_{1}$.

######  Theorem 5.18.

Assume that $A_{1}$ and $A_{2}$ are two multiplicative hyperrins such that
$\theta:A_{1}\longrightarrow A_{2}$ is a hyperring good homomorphism and
$\theta(x)\notin U(A_{2})$ for every $x\in A_{1}\backslash U(A_{1})$. Let
$u,v\in\mathbb{N}$ with $u>v$. Then the following hold.

  * (i)

If $Q_{2}$ is a $(u,v)$-absorbing prime hyperideal of $A_{2}$, then
$\theta^{-1}(Q_{2})$ is a $(u,v)$-absorbing prime hyperideal of $A_{1}$.

  * (ii)

If $\theta$ is surjective and $Q_{1}$ is a is a $(u,v)$-absorbing prime
$\mathcal{C}$-hyperideal of $A_{1}$ with $Ker(\theta)\subseteq Q_{1}$, then
$\theta(Q_{1})$ is a $(u,v)$-absorbing prime hyperideal of $A_{2}$.

###### Proof.

(i) Let $x_{1}\circ_{1}\cdots\circ_{1}x_{u}\subseteq\theta^{-1}(Q_{2})$ for
some $x_{1},\cdots,x_{u}\in A_{1}$. Then we get
$\theta(x_{1}\circ_{1}\cdots\circ_{1}x_{u})=\theta(x_{1})\circ_{2}\cdots\circ_{2}\theta(x_{u})\subseteq
Q_{2}$ because $\theta$ is a good homomorphism. Since $Q_{2}$ is a
$(u,v)$-absorbing prime hyperideal of $A_{2}$, we obtain
$\theta(x_{1}\circ_{1}\cdots\circ_{1}x_{v})=\theta(x_{1})\circ_{2}\cdots\circ_{2}\theta(x_{v})\subseteq
Q_{2}$ which means
$x_{1}\circ_{1}\cdots\circ_{1}x_{v}\subseteq\theta^{-1}(Q_{2})$ or
$\theta(x_{v+1}\circ_{1}\cdots\circ_{1}x_{u})=\theta(x_{v+1})\circ_{2}\cdots\circ_{2}\theta(x_{u})\subseteq
Q_{2}$ which implies
$x_{v+1}\circ_{1}\cdots\circ_{1}x_{u}\subseteq\theta^{-1}(Q_{2})$. Thus
$\theta^{-1}(Q_{2})$ is a $(u,v)$-absorbing prime hyperideal of $A_{1}$.

(ii) Let $y_{1}\circ_{2}\cdots\circ_{2}y_{u}\subseteq\theta(Q_{1})$ for
$y_{1},\cdots,y_{u}\in A_{2}$. Then there exists $x_{i}\in A_{1}$ such that
$\theta(x_{i})=y_{i}$ for each $i\in\\{1,\cdots,u\\}$ because $\theta$ is
surjective. Therefore
$\theta(x_{1}\circ_{1}\cdots\circ_{1}x_{u})=\theta(x_{1})\circ_{2}\cdots\circ_{2}\theta(x_{u})\subseteq\theta(Q_{1})$.
Now, pick any $t\in x_{1}\circ_{1}\cdots\circ_{1}x_{u}$. Then
$\theta(t)\in\theta(x_{1}\circ_{1}\cdots\circ_{1}x_{u})\subseteq\theta(Q_{1})$
and so there exists $s\in Q_{1}$ such that $\theta(t)=\theta(s)$. Then we get
$\theta(t-s)=0$ which means $t-s\in Ker(\theta)\subseteq Q_{1}$ and so $t\in
Q_{1}$. Therefore $x_{1}\circ_{1}\cdots\circ_{1}x_{u}\subseteq Q_{1}$ as
$Q_{1}$ is a $\mathcal{C}$-hyperideal. Since $Q_{1}$ is a $(u,v)$-absorbing
prime hyperideal of $A_{1}$, we get the result that
$x_{1}\circ_{1}\cdots\circ_{1}x_{v}\subseteq Q_{1}$ or
$x_{v+1}\circ_{1}\cdots\circ_{1}x_{u}\subseteq Q_{1}$. This means
$y_{1}\circ_{2}\cdots\circ_{2}y_{v}=\theta(x_{1}\circ_{1}\cdots\circ_{1}x_{v})\subseteq\theta(Q_{1})$
or
$y_{v+1}\circ_{2}\cdots\circ_{2}y_{u}=\theta(x_{v+1}\circ_{1}\cdots\circ_{1}x_{u})\subseteq\theta(Q_{1})$.
Consequently, $\theta(Q_{1})$ is a $(u,v)$-absorbing prime hyperideal of
$A_{2}$. ∎

Now, we have the following result.

######  Corollary 5.19.

Let the hyperideal $Q_{1}$ of $A$ be subset the $\mathcal{C}$-hyperideal
$Q_{2}$ of $A$, $x+Q_{1}\notin U(A/Q_{1})$ for all $x\in A\backslash U(A)$ and
$u,v\in\mathbb{N}$ with $u>v$. Then $Q_{2}$ is a $(u,v)$-absorbing prime
hyperideal of $A$ if and only if $Q_{2}/Q_{1}$ is a $(u,v)$-absorbing prime
hyperideal of $A/Q_{1}$.

###### Proof.

Consider the homomorphism $\pi:A\longrightarrow A/Q_{1}$ defined by
$\pi(a)=a+Q_{1}$. Since $\pi$ is a good epimorphism, the claim follows from
Theorem 5.18 ∎

Let $S$ be a non-empty subset of $A$. If $S$ is closed under the
hypermultiplication and $S$ contains $1$, then $S$ refers to a multiplicative
closed subset (briefly, MCS)[1]. Consider the set $(A\times S/\sim)$ of
equivalence classes denoted by $S^{-1}A$ such that
$(x_{1},r_{1})\sim(x_{2},r_{2})$ if and only if there exists $r\in S$ with
$r\circ r_{1}\circ x_{2}=r\circ r_{2}\circ x_{1}$. The equivalence class of
$(x,r)\in A\times S$ is denoted by $\frac{x}{r}$. The triple
$(S^{-1}A,\oplus,\odot)$ is a multiplicative hyperring where

$\hskip
28.45274pt\frac{x_{1}}{r_{1}}\oplus\frac{x_{2}}{r_{2}}=\frac{r_{1}\circ
x_{2}+r_{2}\circ x_{1}}{r_{1}\circ r_{2}}=\\{\frac{a+b}{c}\ |\ a\in r_{1}\circ
x_{2},b\in r_{2}\circ x_{1},c\in r_{1}\circ r_{2}\\}$

$\hskip 28.45274pt\frac{x_{1}}{r_{1}}\odot\frac{x_{2}}{r_{2}}=\frac{x_{1}\circ
a_{2}}{r_{1}\circ r_{2}}=\\{\frac{a}{b}\ |\ a\in x_{1}\circ x_{2},b\in
r_{1}\circ r_{2}\\}$

The localization map $\pi:A\longrightarrow S^{-1}A$, defined by
$a\mapsto\frac{a}{1}$, is a homomorphism of hyperrings. Moreover, if $I$ is a
hyperideal of $A$, then $S^{-1}I$ is a hyperideal of $S^{-1}A$ [27].

######  Theorem 5.20.

Assume that $Q$ is a $\mathcal{C}$-hyperideal of $A$, $S$ a MCS such that
$Q\cap S=\varnothing$ and $u,v\in\mathbb{N}$ with $u>v$. If $Q$ is a
$(u,v)$-absorbing prime hyperideal of $A$, then $S^{-1}Q$ is a
$(u-1,v-1)$-absorbing prime hyperideal of $S^{-1}A$.

###### Proof.

Let
$\frac{x_{1}}{r_{1}}\odot\cdots\odot\frac{x_{u-1}}{r_{u-1}}=\frac{x_{1}\circ\cdots\circ
x_{u-1}}{r_{1}\circ\cdots\circ r_{u-1}}\subseteq S^{-1}Q$ for
$x_{1},\cdots,x_{u-1}\in A\backslash U(A)$ and $r_{1},\cdots,r_{u-1}\in S$
such that
$\frac{x_{1}}{r_{1}}\odot\cdots\odot\frac{x_{v-1}}{r_{v-1}}\nsubseteq
S^{-1}Q$. Take any $a\in x_{1}\circ\cdots\circ x_{u-1}$ and $r\in
r_{1}\circ\cdots\circ r_{u-1}$. Hence
$\frac{a}{r}\in\frac{x_{1}\circ\cdots\circ x_{u-1}}{r_{1}\circ\cdots\circ
r_{u-1}}$ and so $\frac{a}{r}=\frac{a^{\prime}}{r^{\prime}}$ for some
$a^{\prime}\in Q$ and $r^{\prime}\in S$. Then there exists $t\in S$ such that
$t\circ a\circ r^{\prime}=t\circ a^{\prime}\circ r$. This means $t\circ a\circ
r^{\prime}\subseteq Q$. Since $a\in x_{1}\circ\cdots\circ x_{u-1}$, we
conclude that $t\circ a\circ r^{\prime}\subseteq t\circ x_{1}\circ\cdots\circ
x_{u-1}\circ r^{\prime}$. Therefore we have $t\circ x_{1}\circ\cdots\circ
x_{u-1}\circ r^{\prime}\subseteq Q$ as $Q$ is a $\mathcal{C}$-hyperideal of
$A$ and $t\circ x_{1}\circ\cdots\circ x_{u-1}\circ r^{\prime}\cap
Q\neq\varnothing$. Take any $s\in t\circ r^{\prime}$. If $s\circ
x_{1}\circ\cdots\circ x_{v-1}\subseteq Q$, then we get
$\frac{x_{1}}{r_{1}}\odot\cdots\odot\frac{x_{v-1}}{r_{v-1}}=\frac{x_{1}\circ\cdots\circ
x_{v-1}}{r_{1}\circ\cdots\circ r_{v-1}}=\frac{s\circ x_{1}\circ\cdots\circ
x_{v-1}}{s\circ r_{1}\circ\cdots\circ r_{v-1}}\subseteq S^{-1}Q$ which is
impossible. Since $Q$ is a $(u,v)$-absorbing prime hyperideal of $A$, $s\circ
x_{1}\circ\cdots\circ x_{u-1}\subseteq Q$ and $s\circ x_{1}\circ\cdots\circ
x_{v-1}\nsubseteq Q$, we have $x_{v}\circ\cdots\circ x_{u-1}\subseteq Q$ which
implies
$\frac{x_{v}}{r_{v}}\odot\cdots\odot\frac{x_{u-1}}{r_{u-1}}=\frac{x_{v}\circ\cdots\circ
x_{u-1}}{r_{v}\circ\cdots\circ r_{u-1}}\subseteq S^{-1}Q$. This shows that
$S^{-1}Q$ is a $(u-1,v-1)$-absorbing prime hyperideal of $S^{-1}A$. ∎

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