#  $r$-Hyperideals and Generalizations of $r$-Hyperideals in Krasner
Hyperrings

Melis Bolat  Department of Mathematics, Yildiz Technical University, Istanbul,
Turkey. Orcid: 0000-0002-0667-524X<EMAIL_ADDRESS>,  Elif Kaya
Department of Mathematics and Science Education, Istanbul Sabahattin Zaim
University, Istanbul, Turkey. Orcid: 0000-0002-8733-2734<EMAIL_ADDRESS>,  Serkan Onar  Department of Mathematical Engineering, Yildiz Technical
University, Istanbul, Turkey. Orcid: 0000-0003-3084-7694<EMAIL_ADDRESS>,  Bayram Ali Ersoy  Department of Mathematics, Yildiz Technical University,
Istanbul, Turkey. Orcid: 0000-0002-8307-9644<EMAIL_ADDRESS>and  Kostaq
Hila  Department of Mathematical Engineering, Polytechnic University of
Tirana, Tirana, Albania. Orcid: 0000-0001-6425-2619<EMAIL_ADDRESS>
###### Abstract.

In this study, we examine some properties of $r$-hyperideals in the
commutative Krasner hyperrings. Some properties of $pr$-hyperideals are also
studied. The relation between prime hyperideals and $r$-hyperideals is
investigated. We show that the image and the inverse image of an
$r$-hyperideal is also an $r$-hyperideal. We also introduce a generalization
of r-hyperideals and we prove some properties of them.

###### Key words and phrases:

$r$-hyperideal, $pr$-hyperideal, $\phi-r$-hyperideal, $\phi-$prime hyperideal

###### 2000 Mathematics Subject Classification:

13A15, 13C05.

##  1\. Introduction

Prime ideals and primary ideals play a significant role in commutative ring
theory. Numerous generalizations of prime ideals have been studied by
different researchers. The concepts of $\phi$-prime, $\phi$-primary ideals are
examined in [2, 10].

French mathematician Marty [24] initated the study of hyperstructures, as an
expansion of classical algebraic structures, in special, of hypergroups, at
the 8th Congress of Scandinavian Mathematicians in 1934. Subsequently,
numerous articles and various books have been published about this issue. This
theory has been studied by Corsini [9], Vougiouklis [33], Mittas [27, 26],
Stratigopoulos [32], Davvaz [14, 12, 25], Krasner [21], Ameri [1], De Salvo
[15], Barghi [8] and Asokkumar and Velrajan [5, 6], etc.

Hyperrings $(\Re,+,\cdot)$ are of different types. For example, when the
addition + is a hyperoperation and the multiplication $\cdot$ is a binary
operation, we name them Krasner (additive) hyperrings [21]. If + is a binary
operation and the multiplication $\cdot$ is a hyperoperation, the hyperring is
called multiplicative hyperring [31].

Mohamadian investigated the $r$-ideals in [28]. In [3], Ugurlu studied
generalizations of $r$-ideals. The notion of $r$-hyperideals was briefly
mentioned in [30], as a generalization of $r$-ideals in commutative rings. In
this paper, our aim is to extend the notion of $r$-ideals to $r$-hyperideals
in Krasner hyperrings and generalize them. Let $\Re$ be a commutative Krasner
hyperring, $N$ be a proper hyperideal of $\Re$ and $\phi:Id(\Re)\rightarrow
Id(\Re)\cup\\{\emptyset\\}$ be a function. $Id(\Re)$ denotes the hyperideals
of $\Re$. $N$ is called $\phi-r$-hyperideal (resp., $\phi-pr$-hyperideal) if
$a\cdot b\in N-\phi(N)$ with $ann(a)=0$ implies that $b\in N$ (resp.,
$b^{n}\in N$), for $a,b\in\Re$. Some properties of them are provided. As a
result, we obtain that the union of directed collection of ascending chain
$\phi-r$-hyperideals of $R$ is also $\phi-r$-hyperideal of $\Re$, when $\phi$
preserves the order.

##  2\. Preliminaries

First of all, we will give the definitions of $\phi-$prime ideal,
$\phi-$primary ideal and $r$-ideal.

######  Definition 1.

[19] Let $\Re$ be a commutative ring. A reduction of ideals is a function
$\phi$ that leads any ideal $N$ of $\Re$ to other ideal $\phi(N)$ of $\Re$
such that the following statements hold:

$i)$ For all ideals $N$ of $\Re,$ $\phi(N)\subseteq N$,

$ii)$ If $N\subseteq M$ where $N$ and $M$ are ideals of $\Re$, then
$\phi(N)\subseteq\phi(M).$

######  Definition 2.

Let $\Re$ be a commutative ring. Let $N$ be an ideal of $\Re$. For
$a,b\in\Re$:

$i)$ $N$ is said to be a $\phi-$prime ideal, if $ab\in N-\phi(N)$, then $a\in
N$ or $b\in N$ [2],

$ii)$ $N$ is said to be a $\phi-$primary ideal, if $ab\in N-\phi(N)$, then
$a\in N$ or $b\in\sqrt{N}$ [10].

######  Definition 3.

[28] Let $\Re$ be a commutative ring. A proper ideal $N$ of $\Re$ is called an
$r$-ideal (resp., $pr$-ideal), if $ab\in N$ with $ann(a)=(0)$ implies that
$b\in N$ (resp., $b^{n}\in N,$ for any $n\in\mathbb{N}$), for each
$a,b\in\Re.$

######  Definition 4.

[3] Let $\Re$ be a commutative ring. A proper ideal $N$ of $\Re$ is called a
$\phi-r$-ideal, if $ab\in N-\phi(N)$ with $ann(a)=(0)$ implies that $b\in N$
for $a,b\in\Re.$ Furthermore, $N$ is called a $\phi-pr$-ideal, if $ab\in
N-\phi(N)$ with $ann(a)=(0)$ implies that $b^{n}\in N$ $(n\in\mathbb{N})$, for
$a,b\in\Re.$

In the following, we recall some notions regarding hyperstructures.

Let $G$ be a nonempty set and $\circ:G\times G\rightarrow\mathcal{P}^{*}(G)$
be a mapping, where $\mathcal{P}^{*}(G)$ denotes the set of all nonempty
subsets of $G$. Then, $\circ$ is called a binary (algebraic) hyperoperation on
$G$. An algebraic system $(G,\circ)$, where $\circ$ is a binary hyperoperation
defined on $G$, is called a hypergroupoid [24]. Let $(G,\circ)$ be a
hypergroupoid and $\emptyset\neq$ $A,B$ be subsets of $G.$

$A\circ B=\bigcup\limits_{\begin{subarray}{c}a\in A\\\ b\in
B\end{subarray}}a\circ b$ and for $x\in G$, $x\circ A=\left\\{x\right\\}\circ
A,$ $A\circ x=A\circ\left\\{x\right\\}.$ For each $x,y,z\in G,$ if
$x\circ(y\circ z)=(x\circ y)\circ z,$ which means, $\bigcup\limits_{u\in
x\circ y}u\circ z=\bigcup\limits_{v\in y\circ z}x\circ v$, then $G$ is called
a semihypergroup. When $(G,\circ)$ is a hypergroupoid, if there is $e\in G$
such that $x\in(e\circ x)\cap(x\circ e)$, for every $x\in G,$ then $e$ is
called identity element.

Semihypergroup $(G,\circ)$ is said to be a hypergroup if $x\circ G=G\circ
x=G,$ $\forall x\in G$ [24]. If $(G,\circ)$ is a hypergroup, $\emptyset\neq H$
is a subset of $G\ $and $x\circ H=H\circ x=H,$ $\forall x\in H,$ then
$(H,\circ)$ is called a subhypergroup of $(G,\circ)$.

Let $(G,\circ)$ be a hypergroup. If $x\circ y=y\circ x,$ $\forall x,y\in G$,
then $(G,\circ)$ is called a commutative hypergroup.

######  Definition 5.

A non-empty set $\Re$ along with the hyperoperation $+$ is called a canonical
hypergroup if the following axioms hold:

i) $x+(y+z)=(x+y)+z,$ for $x,y,z\in\Re;$

ii) $x+y=y+x,$ for $x,y\in\Re;$

iii) there exists $0\in\Re$ such that $x+0=\\{x\\},$ for any $x\in\Re;$

iv) for any $x\in\Re,$ there exists a unique element $x^{\prime}\in\Re,$ such
that $0\in x+x^{\prime}$ ($x^{\prime}$ is called the inverse of $x$)

v) $z\in x+y$ implies that $y\in-x+z$ and $x\in z-y,$ that is, $(\Re,+)$ is
reversible [27].

######  Definition 6.

$(\Re,+,\cdot)$ is called a Krasner hyperring if it satisfies the following
conditions:

1) $(\Re,+)$ is a canonical hypergroup;

2) $(\Re,\cdot)$ is a semigroup having $0$ as a bilaterally absorbing element,
i.e., $x\cdot 0=0\cdot x=0,$ for all $x\in\Re;$

3) $(y+z)\cdot x=(y\cdot x)+(z\cdot x)$ and $x\cdot(y+z)=(x\cdot y)+(x\cdot
z),$ for all $x,y,z\in\Re$ [21].

######  Lemma 1.

[13] A non-empty subset $A$ of a Krasner hyperring $\Re$ is a left (resp.
right) hyperideal if and only if

i) $x-y\subseteq A$ $(x-y\subseteq A),$ for $x,y\in A$,

ii) $r\cdot x\in A$ $(resp.$ $x\cdot r\in A),$ for $x\in A,r\in\Re.$

######  Definition 7.

[20] Let $N$ be a hyperideal of a hyperring $\Re.$ Then
$\sqrt{N}=\\{r\in\Re:r^{n}\in N,$ for some $n\in\mathbb{N}\\}.$

Let $\Re$ be a hyperring. For $a\in\Re,$ we define
$ann(a)=\\{r\in\Re:ra=0\\}$. If $ann(a)=0$ (resp., $ann(a)\neq 0$), $a$ is
said to be regular (resp., zero divisor). We use the notion $r(\Re)$ (resp.,
$zd(\Re)$ to denote the set of all regular elements (resp., zero divisors)
[18].

######  Definition 8.

[30] A proper hyperideal $N$ of a commutative Krasner hyperring $\Re$ is
called an r-hyperideal (pr-hyperideal), if $a\cdot b\in N$ and $ann(a)=0$
implies that $b\in N$ ($b^{n}\in N$, for some $n\in\mathbb{N}$), for any
$a,b\in\Re$.

######  Example 1.

Clearly, $\mathbb{Z}_{6}$ is a Krasner hyperring with the usual addition and
multiplication and $\\{\hat{0},\hat{2},\hat{4}\\}=N$ is a proper hyperideal.

$\qquad ann(\hat{5})=0$ and $\hat{0}.\hat{5}=\hat{0}\in$ $N$ implies that
$\hat{0}\in N.$

$\qquad ann(\hat{5})=0$ and $\hat{2}.\hat{5}=\hat{4}\in$ $N$ implies that
$\hat{2}\in N$ .

$ann(\hat{5})=0$ and $\hat{4}.\hat{5}=\hat{2}\in$ $N$ implies that $\hat{4}\in
N$ .

Then $N$ is an $r$-hyperideal of $\mathbb{Z}_{6}.$

######  Example 2.

[17] Let $H=\\{0,1,a\\}$ and $B=\\{0,a\\}$. Then, $(H,+,\cdot)$ is a Krasner
hyperring with the hyperaddition and multiplication defined by

$+$ | 0 | 1 | $a$
---|---|---|---
0 | 0 | 1 | $a$
1 | 1 | $H$ | 1
$a$ | $a$ | 1 | $B$
$\cdot$ | 0 | 1 | $a$
---|---|---|---
0 | 0 | 0 | 0
1 | 0 | 1 | $a$
$a$ | 0 | $a$ | 0

Clearly, $B$ is an $r$-hyperideal of $H$.

######  Example 3.

[22] Let $H=[0,1]$. Then, $(H,\oplus,\cdot)$ is a Krasner hyperring where
$\cdot$ is the usual multiplication and the hyperaddition $\oplus$ is defined
by

$x\oplus y=\begin{cases}\\{\max\\{x,y\\}\\},&\text{if}\ x\neq y\\\
[0,x],&\text{if}\ x=y\end{cases}$

Clearly, $I=[0,a]$, where $H\ni a<1$, is an $r$-hyperideal of $H$.

$Min(N)$ is the set of all minimal prime hyperideals which contain $N$.
$Min((0))$ is stated by $Min(\Re).$

######  Definition 9.

[23] A ring $\Re$ satisfies:

  * •

Property A: if any finitely generated ideal $N\subseteq zd(\Re)$ has non-zero
annihilator;

  * •

Annihilator condition: if for any finitely generated ideal $N$ of $\Re$, there
exists an element $a\in\Re$ such that $ann(N)=ann(a)$;

  * •

Strong annihilator condition (briefly s.a.c.): if for any finitely generated
ideal $N$ of $\Re$, there exists an element $a\in N$ such that $ann(N)=ann(a)$
.

We use the same definition for a hyperideal $N$ of hyperring $\Re.$

$N$ is a $z^{0}$-hyperideal if $a\in N,$ $b\in\Re$ and $ann(a)=ann(b)$ imply
that $b\in N.$

$soc(\Re)$ denotes the sum of all minimal hyperideals of $\Re$. The socle of a
reduced hyperring $\Re$ is called $soc(\Re)=\oplus_{i\in A}e_{i}\Re,$ where
$\\{e_{i}:i\in A\\}$ is the set of idempotents of $\Re$ [11].

Let $A$ be a subset of $\Re$ and $a\in A$ such that $a\in a\Re a.$ Then $a$ is
called von Neumann regular element. Therefore, if all of the elements are von
Neumann regular, $A$ or $\Re$ is called von Neumann regular [7].

A two-sided hyperideal $N$ in a semihypergroup $\Re$ is called right pure
hyperideal if for each $a\in N$, there exists $b\in N$ such that $a\in ab$
[16]. Similarly, a hyperideal is called pure hyperideal in a Krasner
hyperring, if for each $a\in N$, there exists $b\in N$ such that $a=ab.$

A hyperring $\Re$ is called a reduced hyperring if there is no nilpotent
elements in $\Re$. If $x^{n}=0$ for $x\in\Re$,$\ n\in\mathbb{N}$, then $x=0$
[4].

Let $f:\Re\rightarrow S$ be a homomorphism, $N$ be a hyperideal of $\Re$ and
$M$ be a hyperideal of $S.$ Then, the hyperideal $\langle f(N)\rangle$ is said
to be the extension of $N$ and it is denoted by $N^{e}.$ The hyperideal
$f^{-1}(M)$ is called contraction of $M$ and denoted by $M^{c}$ [29] . The
mapping $\varphi:\Re\rightarrow S^{-1}\Re$ given by $a=a/1$ is a homomorphism.
If $N$ is a hyperideal of $\Re$, then $\varphi(N)=S^{-1}N=\\{\frac{i}{s}\in
S^{-1}\Re:i\in N,s\in S\\}$ is also a hyperideal of $S^{-1}\Re$ [14] . If
$S=r(\Re),$ then the hyperring $S^{-1}\Re$ is called quotient hyperring which
is denoted by $Q(\Re).$

##  3\. $r$-Hyperideals

In this section, we examine some properties of $r$-hyperideals in commutative
hyperrings. We compare $r$-hyperideals with prime and maximal hyperideals.
Throughout the section, $(\Re,+,\cdot)$ is a commutative Krasner hyperring.

Our main result regarding $r$-hyperideals is the following:

######  Theorem 1.

Let $\Re$ be a hyperring and $N$ be a hyperideal of $\Re$. Then the following
statements are equivalent:

a) $N$ is an $r$-hyperideal.

b) $(a\cdot\Re)\cap N=a\cdot N,$ for any $a\in r(\Re).$

c) $N=(N:a),$ for any $a\in r(\Re)\backslash N.$

d) $N=M^{c};$ where $M$ is a hyperideal of $Q(\Re).$

###### Proof.

($a\Rightarrow b)$ Let $N$ be an $r$-hyperideal and $a$ be a regular element.
Suppose that $x\in$ $(a\cdot\Re)\cap N.$ Then $x\in a\cdot\Re$ and $x\in N.$
Thus, $x=a\cdot a^{{}^{\prime}},$ for $a^{\prime}\in\Re.$ Since $x=a\cdot
a^{{}^{\prime}}\in N$, $ann(a)=0$ and $N$ is an $r$-hyperideal, $a^{\prime}\in
N.$ Hence, $a\cdot a^{\prime}\in a\cdot N$ and $(a\cdot\Re)\cap N\subseteq
a\cdot N.$

Therefore, for every $a\in\Re,$ $a\cdot N\subseteq(a\cdot\Re)\cap N,$
$(a\cdot\Re)\cap N=a\cdot N.$

$(b\Rightarrow c)$ We know that $N\subseteq(N:a),\ $for every $a\in\Re.$ Let
$a$ be regular, $a\notin N$ and $x\in(N:a).$ Hence $ann(a)=0$ and $x\cdot a\in
N.$ From (b), $a\cdot x\in(a\cdot\Re)\cap N$ and $a\cdot x\in a\cdot N.$ This
implies that $a\cdot x=a\cdot y,$ for $y\in N.$ Since $ann(a)=0,$ then $x=y\in
N.$ Hence, $x\in N.$ Thus, $(N:a)\subseteq N.$

$(c\Rightarrow d)$ Let $S$ be the set of regular elements and
$\varphi:\Re\rightarrow Q(\Re)$ be natural homomorphism. We know that
$N\subseteq\varphi^{-1}(M)=M^{c}$, for $M$ a hyperideal of $Q(\Re)$. Suppose
$x\in$ $\varphi^{-1}(M).$ Since $\varphi\left(x\right)=\frac{x}{s}\in
S^{-1}\Re,$ then $s\cdot x\in N,$ for $s\in S.$ From (c), $x\in(N:s)=N.$

$(d\Rightarrow a)$ Let $a\cdot x\in N$ and $ann(a)=0.$ We have $(a\cdot
x)/1=a/1\cdot x/1\in M$ and since $a$ is regular, then there exists $1/a$
which is the inverse of $a/1$ in $S.$ Thus $a/1\cdot x/1\cdot 1/a\in M$ and
$x/1\in M.$ Hence, $x\in N$ and so $N$ is an $r$-hyperideal. ∎

######  Corollary 1.

The following statements hold:

$a)$ The zero hyperideal is an $r$-hyperideal.

$b)$ The intersection of $r$-hyperideals is an $r$-hyperideal.

$c)$ When $N$ is an $r$-hyperideal, $N\subseteq zd(\Re).$

$d)$ Every $r$-hyperideal is a $pr$-hyperideal.

$e)$ A prime hyperideal is an $r$-hyperideal if and only if it consists all of
zerodivisors. As a result, every minimal prime hyperideal is an
$r$-hyperideal.

$f)$ Let $N$ is an $r$-hyperideal, $S\subseteq\Re$ and $S\nsubseteq N$. Then
$(N:S)$ is an $r$-hyperideal. Particularly, $ann(x)$ and $ann(S)$ are always
$r$-hyperideals.

$g)$ Every minimal hyperideal of a reduced hyperring is an $r$-hyperideal.

$h)$ Every pure hyperideal and every von Neumann regular hyperideal are
$r$-hyperideals.

$i)$ Suppose that $\Re$ satisfies the s.a.c. and $N$ is a hyperideal of $\Re.$
$N$ is an $r$-hyperideal if and only if for every hyperideal $J$ and $K$ of
$\Re$ such that $J$ is finitely generated, whenever $J\cdot K\subseteq N$ and
$ann(J)=0,$ then $K\subseteq N.$

$j)$ The sum of two $r$-hyperideals may not be an $r$-hyperideal.

######  Remark 1.

Let $K$ and $L$ be hyperideals of $Q(\Re).$ We know that $K^{c}\cdot
L^{c}\subseteq(K\cdot L)^{c}$ and $K^{c}+L^{c}\subseteq(K+L)^{c}.$ If $N$ and
$M$ are $r$-hyperideals of $\Re,$ then by Theorem 3.1(d), $N=K^{c}$ and
$M=L^{c}$, for some hyperideals $K$ and $L$ in $Q(\Re).$ It follows that:

$a)$ $N\cdot M$ is an $r$-hyperideal of $\Re$ if and only if $(K\cdot
L)^{c}\subseteq$ $K^{c}\cdot L^{c}.$(essentially $(K\cdot L)^{c}=K^{c}\cdot
L^{c}$)

$b)$ $N+M$ is an $r$-hyperideal of $\Re$ if and only if $(K+L)^{c}\subseteq$
$K^{c}+L^{c}$ (essentially $(K+L)^{c}=K^{c}+L^{c}$).

######  Lemma 2.

Let $\Re$ be a hyperring and $N$ be a hyperideal. Then the following
statements hold:

$a)$ $N$ is an $r$-hyperideal if and only if for any $J$, $K$ hyperideals of
$\Re$ with $J\cap r(\Re)\neq\emptyset$ and $J\cdot K\subseteq N$, then
$K\subseteq N.$

$b)$ Assume that $N\subseteq zd(\Re)$ is not an $r$-hyperideal. There exist
hyperideals $J$ and $K$ such that $J\cap r(\Re)\neq\emptyset,$ $N\subsetneqq
J,$ $K$ and $J\cdot K\subseteq N.$

###### Proof.

$a)$ Let $N$ be an $r$-hyperideal, $J$ and $K$ be hyperideals of $\Re$ such
that $J\cap r(\Re)\neq\emptyset$ and $J\cdot K\subseteq N.$ Let $a\in J\cap
r(\Re)$ and $b\in K$. Since $J$ and $K$ are hyperideals, then we can take
$a\cdot\Re\subseteq J$ and $b\cdot\Re\subseteq K.$ By our assumption, $a\cdot
b\cdot\Re\subseteq N.$ Since $J\cap r(\Re)\neq\emptyset,$ let we take
$ann(a)=0.$ Then $N$ is an $r$-hyperideal and so, $b\cdot\Re\subseteq N.$
Thus, $K\subseteq N.$

Conversely, assume that $J\cap r(\Re)\neq\emptyset$ and $J\cdot K\subseteq
N,K\subseteq N.$ Let $a\cdot\Re\subseteq J$ and $b\cdot\Re\subseteq K.$ Then
$a\cdot b\cdot\Re\subseteq N.$ Let $a\in J\cap r(\Re).$ So $ann(a)=0.$ At the
same time, since $K\subseteq N$, then $b\cdot\Re\subseteq N$.

$b)$ Assume that $N\subseteq zd(\Re)$ is not an $r$-hyperideal. Then there
exist $r\in r(\Re)$ and $x\notin N$ such that $r\cdot x\in N$. We have
$ann(r)=0.$ Let $J=(N:x)$ and $K=(N:J).$ It follows $r\in J.$ Since
$ann(r)=0$, then $J\cap r(\Re)\neq\emptyset$ and $r\notin N.$ Since
$N\subseteq zd(\Re),$ then $ann(x)\neq 0.$ Therefore, $x\in K$. Thus, there
exist $b,c\in\Re$ such that $b\cdot x\in N$ and $c\cdot J\subseteq N.$ Hence,
$b\cdot x\cdot c\cdot J\subseteq N.$ Then $x\cdot J\cdot K\subseteq N$ and
$J\cdot K\subseteq N.$ ∎

######  Proposition 1.

$a)$ Let $\Re$ be a hyperring and $N$ be a hyperideal of $\Re$ with $N\cap
r(\Re)\neq\emptyset.$ If $J$ and $K$ are $r$-hyperideals of $\Re$ such that
$N\cdot J=N\cdot K$ or $N\cap J=N\cap K,$ then $J=K.$

$b)$ Let $\Re$ be a hyperring and $N,M$ be hyperideals of $\Re$ with $M\cap
r(\Re)\neq\emptyset.$ If $N\cdot M$ is $r$-hyperideal of $\Re$, then $N=N\cdot
M.$ In addition, $N$ is an $r$-hyperideal.

###### Proof.

$a)$ Let $J$ and $K$ be $r$-hyperideals of $\Re$ and $N\cdot J=N\cdot K.$ From
Lemma 2, since $N\cdot J=N\cdot K\subseteq K,$ $J\subseteq K.$ Therefore
$K\subseteq J.$ Then, $J=K.$

$b)$ Let $N\cdot M$ be $r$-hyperideal of $\Re$ and $M\cap
r(\Re)\neq\emptyset.$ Since $N\cdot M\subseteq M\cdot N,$ from Lemma 2,
$N\subseteq N\cdot M.$ Therefore, obviously, $N\cdot M\subseteq N.$ Then
$N=N\cdot M.$ ∎

######  Theorem 2.

Let $I_{1},I_{2},...,I_{n}$ be prime hyperideals of $\Re,$ which are not
comparable. If $\ \bigcap\limits_{i=1}^{n}I_{i}$ is an $r$-hyperideal, then
$I_{i}$ is an $r$-hyperideal, for $i=1,...,n.$

###### Proof.

Let $r,x\in\Re$ such that $r\cdot x\in I_{i}$ and $ann(r)=0$. Let
$y\in(\prod\nolimits_{j\neq i}I_{j})\backslash I_{i}.$ Then, $r\cdot x\cdot
y\in\bigcap\limits_{i=1}^{n}I_{i}$ . We get that $x\cdot
y\in\bigcap\limits_{i=1}^{n}I_{i}$, since $\bigcap\limits_{i=1}^{n}I_{i}$ is
an $r$-hyperideal and $ann(r)=0.$ Thus, $x\cdot y\in I_{i}.$ We conclude that
$x\in I_{i},$ since $y\notin I_{i}$ and $I_{i}$’s are prime. Hence, $I_{i}$ is
an $r$-hyperideal. ∎

Let $\rho:\Re\rightarrow S$ be a homomorphism. We investigate whether the
image of an $r$-hyperideal and the inverse image of an $r$-hyperideal are
$r$-hyperideal.

######  Theorem 3.

Let $\rho:\Re\rightarrow S$ be a good epimorphism such that $Ker$
$\rho\subseteq N$. If $N$ is an $r$-hyperideal of $(\Re,+,\cdot),$ then
$\rho(N)$ is an $r$-hyperideal of the hyperring $(S,\oplus,\odot).$

###### Proof.

Obviously, $\rho(N)$ is hyperideal of $S.$ Let $b_{1}\odot b_{2}\in\rho(N)$
and $ann(b_{1})=0_{S},$ for $b_{1},b_{2}\in S.$ Since $\rho$ is onto, then
there exist $a_{1},a_{2}\in\Re$ such that $b_{1}=\rho(a_{1})$ and
$b_{2}=\rho(a_{2})$. Then $b_{1}\odot
b_{2}=\rho(a_{1})\odot\rho(a_{2})=\rho(a_{1}\cdot a_{2})=\rho(x)\in f(N),$ for
some $x\in N.$

$0\in$ $\rho(a_{1}\cdot a_{2})\ominus\rho(x)=\rho(a_{1}\cdot a_{2}-x).$ Then
there exists $t\in a_{1}\cdot a_{2}-x$ such that $\rho(t)=0.$ We have
$a_{1}\cdot a_{2}\in t+x\subseteq Ker\rho+N\subseteq N+N\subseteq N.$ Thus,
$a_{1}\cdot a_{2}\in N.$

Let $ann(a_{1})\neq 0.$ Then there exists $0\neq c\in\Re$ such that
$a_{1}\cdot c=0_{\Re}.$ So, $\rho(a_{1}\cdot
c)=\rho(a_{1})\odot\rho(c)=\rho(0_{\Re})=0_{S}.$ Since $\rho(c)\neq 0_{S}$,
$ann(\rho(a_{1}))=ann(b_{1})=0_{S}$ and this is a contradiction. This implies
that $ann(a_{1})=0.$ Since $N$ is an $r$-hyperideal, then $a_{2}\in N.$
Therefore, $b_{2}=\rho(a_{2})\in\rho(N).$ ∎

######  Theorem 4.

Let $\rho:\Re\rightarrow S$ be a good monomorphism. If $M$ is an
$r$-hyperideal of $(S,\oplus,\odot),$ then $\rho^{-1}(M)=M^{c}$ is an
$r$-hyperideal of $(\Re,+,\cdot).$

###### Proof.

$\rho^{-1}(M)$ is a hyperideal [30]. Let $r\cdot x\in\rho^{-1}(M)$ and
$ann(r)=0.$ Then $\rho(r)\odot f(x)=\rho(r\cdot x)\in M.$ If $ann(\rho(r))\neq
0,$ then there exists $0\neq\rho(s)\in M$ such that
$\rho(r)\odot\rho(s)=\rho(r\cdot s)=0_{S}.$ This means that there exists a
$0\neq s\in\rho^{-1}(M)$ such that $r\cdot s=0_{\Re}.$ Then, $ann(r)\neq 0$.
This is a contradiction, so $ann(\rho(r))=0.$ Since $M$ is an $r$-hyperideal,
then $\rho(x)\in M$ and therefore $x\in\rho^{-1}(M).$ ∎

######  Theorem 5.

Let $\Re$ be a hyperring. The following statements are equivalent:

$a)$ $\Re$ is a hyperdomain.

$b)$ The only $r$-hyperideal of $\Re$ is the zero hyperideal.

$c)$ $ann(x\cdot y)=$ $ann(x)\cup ann(y),$ for each $x,y\in\Re.$

###### Proof.

($a)\Rightarrow(b)$ Let us suppose that $\Re$ is hyperdomain and $(0)\neq N$
is a proper hyperideal of $\Re.$ Then, there exists $0\neq\Re\in N$. Since
$\Re$ is a hyperdomain, then we have $ann(r)=0$. This contradicts the ideal
being proper. Then the zero ideal is the only $r$-hyperideal.

($b)\Rightarrow(c)$ From Corollary 1(f), $ann(x)$ is an $r$-hyperideal. Assume
that the zero hyperideal is the only $r$-hyperideal of $\Re.$ Hence,
$ann(x)=0.$ So $ann(x\cdot y)=$ $ann(x)\cup ann(y)=0,$ for every $x,y\in\Re.$

($c)\Rightarrow(a)$ Let $x\cdot y=0,$ $x,y\in\Re.$ Thus, $ann(x\cdot
y)=ann(0)=$ $ann(x)\cup ann(y)=\Re$. This means $1_{\Re}\in ann(x)\cup
ann(y).$ Then $x=\\{0\\}$ or $y=\\{0\\}.$ Hence, $\Re$ is a hyperdomain. ∎

######  Proposition 2.

Let $\Re$ be a hyperring and $\\{e_{i}:i\in\Lambda\\}$ be a set of idempotents
of $\Re$. If $N$ is proper and $N=\sum\limits_{i\in\Lambda}e_{i}\cdot\Re$,
then $N$ is an $r$-hyperideal.

###### Proof.

Let $r\cdot x\in N$ and $ann(r)=0.$ We get that $r\cdot
x\in\sum\limits_{k=1}^{n}e_{i_{k}}r_{i_{k}}$, since
$N=\sum\limits_{i\in\Lambda}e_{i}\cdot\Re,$ for $i_{1},...,i_{n}\in\Lambda$
and $r_{i_{1}},...,r_{i_{k}}\in\Re.$ Let
$y=\prod\limits_{k=1}^{n}(1-e_{i_{k}}).$ Since $r\cdot x\cdot y=0$ and
$ann(r)=0,$ then $x\cdot y=0.$ Thus there exists $s\in N$ such that $y\in
1-s.$ If $x\cdot y\in x\cdot(1-s)=0,$ then $x=x\cdot s\in N.$ ∎

######  Proposition 3.

$a)$ Let $\Re$ be a hyperring and $x,y\in\Re$ with $1_{\Re}\in x+y.$ Then
$N=ann(x)+ann(y)$ is an $r$-hyperideal.

$b)$ Let $\Re$ be a reduced hyperring, $I\in Min(\Re)$ and $e\in\Re$ be an
idempotent element. We infer that $N=I+ann(e)$ is an $r$-hyperideal.

###### Proof.

$a)$ Let $a\cdot b\in N$ and $ann(a)=0.$ Then, there exists $r\in ann(x)$ and
$s\in ann(y)$ such that $a\cdot b\in r+s.$ Obviously, $r\cdot x=0$ and $s\cdot
y=0.$ We get that $a\cdot b\cdot x\cdot y\in(r+s)\cdot(x\cdot y)=r\cdot(x\cdot
y)+s\cdot(x\cdot y)=y\cdot(x\cdot r)+x\cdot(y\cdot s)=b\cdot 0+x\cdot 0=0$ and
by our assumption, $b\cdot x\cdot y=0$. Thus, $b\cdot x\in ann(y)$ or $b\cdot
y\in ann(x).$ Hence, $b=b\cdot 1_{\Re}\in b\cdot(x+y)=(b\cdot x)+(b\cdot
y)\subseteq ann(x)+ann(y)=N$ and therefore $b\in N.$

$b)$ Let $r\cdot x\in N$ and $ann(r)=0.$ We have $r\cdot x\in a+b$ such that
$a\in I$ and $b\in ann(e).$ Since $I\in Min(\Re),$ then $I$ is an
$r$-hyperideal. There is $y\notin I$ such that $a\cdot y=0.$ $e\cdot y\cdot
r\cdot x\in e\cdot y\cdot(a+b)=(e\cdot y\cdot a)+(e\cdot y\cdot b)=0.$ Since
$\Re$ is a reduced hyperring, then every idempotent element is $0.$ By our
assumption, $e\cdot y\cdot x\in I.$ Since $y\notin I,$ then $e\cdot x\in I.$
We write $x\in x+(e\cdot x)-(e\cdot x),$ $x\in(e\cdot x)+x-(e\cdot x)=e\cdot
x+x\cdot(1_{\Re}-e)\subseteq I+ann(e)=N.$ Then $N$ is an $r$-hyperideal. ∎

######  Corollary 2.

a) If there is a hyperideal $K$ of $\Re$ such that $ann(N)+ann(M)=ann(K),$ for
$N,M$ hyperideals, then $ann(N)+ann(M)$ is $r$-hyperideal.

b) The direct sum of $r$-hyperideals is $r$-hyperideal. Therefore, if
$N=M\oplus K$ and $N$ is $r$-hyperideal, then $M$ and $K$ are $r$-hyperideals

c) Let $\Re$ be a reduced hyperring. Then $soc(\Re)$ is an $r$-hyperideal.
Therefore, there is a hyperideal $M$ of $\Re$ such that $soc(\Re)=M^{c}.$

######  Proposition 4.

Let $N$ be a hyperideal of $\Re.$ $N$ is a $pr$-hyperideal if and only if
$\sqrt{N}$ is $r$-hyperideal.

###### Proof.

Similarly with [28], let $N$ be $pr$-hyperideal, $x\cdot y\in\sqrt{N}$ and
$ann(x)=0.$ There is $n\in\mathbb{N}$ such that $(x\cdot y)^{n}\in N.$ Hence
$x^{n}\cdot y^{n}\in N$ and $ann(x^{n})=0.$ Since $N$ is $pr$-hyperideal, then
$y^{nm}\in N,$ for $m\in\mathbb{N}.$ Thus, $y\in\sqrt{N}.$

Conversely, let $x\cdot y\in N$ and $ann(x)=0.$ Since $N\subseteq\sqrt{N},$
then $x\cdot y\in\sqrt{N}.$ Therefore, since $\sqrt{N}$ is $r$-hyperideal,
then $b\in\sqrt{N}.$ Thus, $y^{n}\in N,$ for $n\in\mathbb{N}.$ ∎

######  Proposition 5.

Let $\Re$ be a hyperring and $nil(\Re)$ be a set of nilpotents of $\Re$. Then
$nil(\Re)$ is an $r$-hyperideal.

###### Proof.

$nil(\Re)=\\{a\in\Re:0=a^{n},n\in\mathbb{N}\\}=\sqrt{0}.$ Obviously,$\sqrt{0}$
is $r$-hyperideal. Then $nil(\Re)$ is also $r$-hyperideal. ∎

From Theorem 1 and Proposition 4, we obtain the next corollary.

######  Corollary 3.

Let $N$ be a hyperideal of $\Re.$ The following statements are equivalent:

a) $N$ is $pr$-hyperideal.

b) $(r\cdot\Re)\cap\sqrt{N}=r\cdot\sqrt{N},$ for any $r\in\Re(\Re).$

c) $\sqrt{N}=\sqrt{(N:r)},$ for any $r\in r(\Re)$ and $r\notin N.$

d) $N=M^{c},$ $M$ is a primary hyperideal of $Q(\Re).$

######  Theorem 6.

Every $z^{0}$-hyperideal is an $r$-hyperideal.

###### Proof.

Similar to [28], let $x\cdot y\in N$ and $ann(x)=0.$ Then, $ann(x\cdot
y)=ann(y).$ Since $N$ is $z^{0}$-hyperideal, then $y\in N$ and $N$ is an
$r$-hyperideal. ∎

######  Theorem 7.

Every hyperideal which contains zero divisors is contained by a prime
$r$-hyperideal.

###### Proof.

It is trivial. ∎

######  Theorem 8.

If $N$ is a hyperideal and $I\in Min(N)$ in hyperring $\Re,$ then $I$ is an
$r$-hyperideal.

###### Proof.

Let $x\cdot y\in I$ and $ann(x)=0.$ There exists $a\notin I$ and
$n\in\mathbb{N}$ such that $a\cdot(x\cdot y)^{n}=a\cdot x^{n}\cdot y^{n}\in
N.$ Since $N$ is $r$-hyperideal and $ann(x^{n})=0,$ then $a\cdot y^{n}\in
N\subseteq I.$ Since $a\notin I,$ then $y^{n}\in I.$ Thus $y\in I.$ ∎

######  Proposition 6.

Let $N$ be an $r$-hyperideal and $N\subseteq M.$ If $M/N$ is $r$-hyperideal of
$\Re/N,$ then $M$ is $r$-hyperideal of $\Re.$

###### Proof.

Let $M/N$ be $r$-hyperideal of $\Re/N,$ $(a+N)\cdot(b+N)\subseteq M/N$ and
$ann(a+N)=0$ for $a+N,b+N\in\Re/N.$

$(a+N)\cdot(b+N)\subseteq a\cdot(b+N)+N\cdot(b+N)\subseteq(a\cdot b)+(a\cdot
N)+(N\cdot b)+N\subseteq(a\cdot b)+N\subseteq M+N.$ So, $a\cdot b\in M.$ Since
$ann(a)=0$ and $b\in M$, $M$ is an $r$-hyperideal. ∎

In the following, we investigate the maximal hyperideal and $r$-hyperideal
relation.

######  Proposition 7.

Every maximal $r$-hyperideal is a prime hyperideal.

###### Proof.

Obvious from [28]. ∎

######  Proposition 8.

If every prime hyperideal of $\Re$ is an $r$-hyperideal, every maximal
hyperideal of $\Re$ is an $r$-hyperideal.

###### Proof.

Suppose that $M$ is a maximal hyperideal of $\Re$. We know that every maximal
hyperideal is a prime hyperideal. Let $a\cdot b\in M$ with $ann(a)=0.$ Then,
$b\in(M:a).$ Since $M$ is maximal, then $M=(M:a)$ and $b\in M.$ ∎

######  Proposition 9.

Let $\Re$ be a reduced hyperring satisfying the Property A. Then every maximal
$r$-hyperideal is an $z^{0}$-hyperideal.

###### Proof.

Let $I$ be a maximal $r$-hyperideal, $I\subseteq zd(\Re).$ There exists a
$z^{0}$-hyperideal $J$ such that $I\subseteq J.$ Then $J$ is an
$r$-hyperideal. Since $I$ is maximal, then $I=J.$ Thus, $I$ is an
$z^{0}$-hyperideal. ∎

##  4\. Generalizations of $r$-Hyperideals

In this section, we introduce a generalization of $r$-hyperideals in
commutative Krasner hyperrings. $Id(\Re)$ denotes the hyperideals of $\Re.$
Initially, we give the definition of $\phi-$prime and $\phi-$primary
hyperideal. Similarly, we give the definitions of $\phi-r$-hyperideal, $\phi-
pr$-hyperideal, $\phi-$pure hyperideal and $\phi-$von Neumann regular
hyperideal in $\Re.$ Afterwards, we investigate some properties of
$\phi-r$-hyperideal of $\Re$.

######  Definition 10.

Let $\Re$ be a Krasner hyperring, $\phi:Id(\Re)\rightarrow
Id(\Re)\cup\\{\emptyset\\}$ be a function and $\emptyset\neq N\in Id(\Re).$
Then $N$ is said to be a $\phi-$prime (resp. $\phi-$primary) hyperideal of
$\Re$ if whenever $r,s\in\Re$ and $\Re\cdot s\in N-$ $\phi(N),$ then $r\in N$
or $s\in N$ (resp. $r\in N$ or $s^{n}\in N$).

######  Definition 11.

Let $\Re$ be a Krasner hyperring, $N$ be a proper hyperideal of $\Re$ and
$\phi:Id(\Re)\rightarrow Id(\Re)\cup\\{\emptyset\\}$ be a function. $N$ is
called $\phi-r$-hyperideal (resp., $\phi-pr$-hyperideal) if $a\cdot b\in
N-\phi(N)$ with $ann(a)=0$ implies that $b\in N$ (resp., $b^{n}\in N$), for
$a,b\in\Re.$

Let $J$ be a $\phi-r$-hyperideal of $\Re.$ Then:

i) If $\phi(J)=\varnothing,$ for all $J\in Id(\Re),$ then $J$ is called
$\phi_{\emptyset}-r$-hyperideal of $\Re$ ($r$-hyperideal).

ii) If $\phi(J)=0,$ for all $J\in Id(\Re),$ then $J$ is called
$\phi_{0}-r$-hyperideal of $\Re$ (weakly $r$-hyperideal).

iii) If $\phi(J)=J,$ for all $J\in Id(\Re),$ then $J$ is called
$\phi_{1}-r$-hyperideal of $\Re$ (any hyperideal).

iv) If $\phi(J)=J^{n},$ for all $J\in Id(\Re)$ and $n\geq 2,$ then $N$ is
called $\phi_{n}-r$-hyperideal ($n$-almost $r$-hyperideal). If $n=2,$ then $J$
is called almost $r$-hyperideal of $\Re$.

v) If $\phi(J)=\cap_{i=1}^{\infty}J^{i},$ for all $J\in Id(\Re),$ then $J$ is
called $\phi_{\omega}-r$-hyperideal ($\omega-r$-hyperideal) of $\Re.$

######  Definition 12.

Let $\Re$ be a Krasner hyperring, $N$ be a proper ideal of $\Re$ and
$\phi:Id(\Re)\rightarrow Id(\Re)\cup\\{0\\}$ be a function.

(1) $N$ is called $\phi-$pure hyperideal if there exists $b\in N$ such that
$a=a\cdot b,$ for every $a\in N-\phi(N).$

(2) $N$ is called $\phi-$von Neumann regular hyperideal if there exists $b\in
N$ such that $a=a^{2}\cdot b,$ for every $a\in N-\phi(N).$

######  Theorem 9.

Let $N\in Id(\Re).$ Then the following statements hold:

(i) Let $\psi_{1}$ and $\psi_{2}$ be two functions.
$\psi_{1,2}:Id(\Re)\longrightarrow Id(\Re)\cup\\{\emptyset\\}$ such that
$\psi_{1}\leq\psi_{2}.$ If $N$ is a $\psi_{1}-r$-hyperideal, then $N$ is a
$\psi_{2}-r$-hyperideal.

(ii) Every $\phi-r$-hyperideal is a $\phi-pr$-hyperideal.

(iii) $N$ is an $r$-hyperideal $\Rightarrow N$ is a weakly $r$-hyperideal
$\Rightarrow$ $N$ is an $\omega-r$-hyperideal $\Rightarrow$ $N$ is an
$(n+1)$-almost $r$-hyperideal $\Rightarrow N$ is an $n$-almost $r$-hyperideal
$\Rightarrow N$ is an almost $r$-hyperideal.

(iv) Every $\phi-$pure hyperideal is a $\phi-r$-hyperideal.

(v) Every $\phi-$von Neumann regular hyperideal is a $\phi-r$-hyperideal.

###### Proof.

Straighforward. ∎

######  Theorem 10.

Let $\Re$ be a Krasner hyperring, $N$ be a proper ideal of $\Re.$ Then the
following statements are equivalent:

i) $N$ is $\phi-r$-hyperideal;

ii) $(N:r)=N\cup(\phi(N):r),$ $r\in r(\Re)\backslash N;$

iii) $(N:r)=N$ or $(N:r)=(\phi(N):r),$ $r\in r(\Re)\backslash N;$

###### Proof.

($i)\Rightarrow(ii)$ We know that $N\subseteq(N:r)$ and
$(\phi(N):r)\subseteq(N:r).$ Let $s\in(N:r).$ Then $r\cdot s\in N.$ If $r\cdot
s\in$ $\phi(N),$ then $s\in(\phi(N):r).$ On the other hand, if $r\cdot
s\notin$ $\phi(N),$ since $N$ is $\phi-r$-hyperideal, then $s\in N.$

($ii)\Rightarrow(iii)$ It is evident.

($iii)\Rightarrow(i)$ Let $r,s\in\Re$ such that $r\cdot s\in N-$ $\phi(N)$ and
$r$ be regular. Then, $(N:r)=N$ or $(N:r)=(\phi(N):r)$.

Case 1: Assume that $(N:r)=N.$ Since $r\cdot s\in N,$ then $s\in$ $(N:r)=N.$

Case 2: Assume that $(N:r)=(\phi(N):r)$. Since $r\cdot s\in N,$ then
$s\in(N:r).$ Thus $r\cdot s\in$ $\phi(N).$ This is a contradiction. ∎

######  Definition 13.

Let $N$ be a proper hyperideal of $\Re$, $J$ and $K$ be hyperideals of $\Re$
such that $J\cdot K\subseteq N,$ $J\cdot K\nsubseteq\phi(N)$ and $ann(J)=0$
implies that $K\subseteq N$. Then $N$ is called a strongly
$\phi-r$-hyperideal.

######  Theorem 11.

Every strongly $\phi-r$-hyperideal is an $\phi-r$-hyperideal.

###### Proof.

Similar to [3], let $a,b\in\Re$ such that $a\cdot b\in N-$ $\phi(N)$ and
$ann(a)=0.$ Then $<a>\cdot<b>\subseteq N-\phi(N).$ Obviously, $ann(<a>)=0.$
Since $N$ is strongly $\phi-r$-hyperideal, then $<b>\subseteq N.$ Therefore,
$b\in N.$ ∎

######  Theorem 12.

Let $\Re$ satisfy the s.a.c., $N$ be a proper hyperideal of $\Re$ and
$\phi(N)$ be an $r$-hyperideal of $\Re.$ $N$ is $\phi-r$-hyperideal if and
only if for every hyperideal $J$ and $K$ of $\Re$ such that $J$ is finitely
generated, whenever $J\cdot K\subseteq N$, $J\cdot K\nsubseteq\phi(N)$ and
$ann(J)=0,$ implies that $K\subseteq N.$

###### Proof.

$\Rightarrow:$ Let $N$ be a $\phi-r$-hyperideal, $\phi(N)$ be an
$r$-hyperideal, $J$ be finitely generated, $J$ and $K$ be hyperideals of $\Re$
such that $J\cdot K\subseteq N$ and $J\cdot K\nsubseteq\phi(N).$ Assume that
$K\nsubseteq N.$ Then there exists an element $0\neq b\in K-N.$ Since $\Re$
satisfy the s.a.c and $J$ is finitely generated, then there is an element
$a\in J$ with $ann(J)=ann(a)=0.$ Let $a\cdot b\in N.$ If $a\cdot
b\notin\phi(N)$, since $N$ is $\phi-r$-hyperideal, then $b\in N$ which is a
contradiction. If $a\cdot b\in\phi(N),$ since $\phi(N)$ is an $r$-hyperideal,
then $b\in\phi(N)$ which is a contradiction. Thus, $K\subseteq N.$

$\Leftarrow:$ It is obvious. ∎

Now, we investigate the relation between the hyperideal $N$ and the set of all
zero divisors, while $\phi(N)$ is an $r$-hyperideal of $\Re.$

######  Theorem 13.

Let $\phi(N)$ be an $r$-hyperideal of $\Re$ and $N$ be a proper hyperideal of
$\Re.$ If $N$ is $\phi-r$-hyperideal, then $N\subseteq zd(\Re)$.

###### Proof.

Let $N\nsubseteq zd(\Re).$ Then there exists an element $a\in N$ $-zd(\Re).$
So, $ann(a)=0.$ We have $a=a\cdot 1_{\Re}\in N.$ If $a\notin\phi(N),$ then
$1_{\Re}\in N,$ which is a contradiction. If $a\in\phi(N),$ then
$1_{\Re}\in\phi(N)\subseteq N,$ since $\phi(N)$ is an $r$-hyperideal. ∎

######  Theorem 14.

Let $\phi(N)$ be an $r$-hyperideal of $\Re$ and $N$ be a prime hyperideal of
$\Re.$ $N$ is $\phi-r$-hyperideal if and only if $N\subseteq zd(\Re)$.

###### Proof.

Let $\phi(N)$ be an $r$-hyperideal of $\Re$ and $N$ be a prime hyperideal.

$\Rightarrow:$ It is obvious from previous theorem.

$\Leftarrow:$ Let $a\cdot b\in N-$ $\phi(N)$ and $ann(a)=0.$ Since $N$ is a
prime hyperideal, $a\in N$ or $b\in N.$ Since $N\subseteq zd(\Re)$, then
$a\notin N.$ Therefore $b\in N.$ ∎

In a hyperideal $N,$ nilpotent elements are also zero divisors. So we can
change the theorem as if $N$ is a hyperideal of which elements are nilpotent,
then $N$ is $\phi-r$-hyperideal if and only if $N$ is $\phi$-primary
hyperideal.

######  Theorem 15.

If $N$ is $\phi-r$-hyperideal, then $N-\phi(N)\subseteq zd(\Re).$

###### Proof.

Let us suppose that $N-\phi(N)\nsubseteq zd(\Re).$ Then there exists an
element $a$ in $N-\phi(N)$ such that $ann(a)=0.$ Since $a=a\cdot 1\in
N-\phi(N)$ and $N$ is $\phi-r$-hyperideal, then $1\in N.$ That is a
contradiction. ∎

######  Theorem 16.

Let $\phi(N)$ be an $r$-hyperideal. $N$ is $\phi-r$-hyperideal if and only if
$N$ is an $r$-hyperideal.

###### Proof.

($\Rightarrow:)$ Let $a\cdot b\in N$ and $ann(a)=0.$ If $a\cdot b\in\phi(N),$
since $\phi(N)$ is an $r$-hyperideal, then $b\in\phi(N)\subseteq N.$ If
$a\cdot b\notin\phi(N),$ since $N$ is $\phi-$r-hyperideal, then $b\in N.$

($\Leftarrow:)$ It is obvious. ∎

Similar to [3], if $N$ is a hyperideal of $\Re,$ we define
$\phi_{N}:Id(\Re/N)\longrightarrow Id(\Re/N)\cup\\{\emptyset\\}$ by
$\phi_{N}(M/N)=$ $(\phi(M)+N)/N$ for each hyperideal $N\subseteq M$. If
$\phi(N)=\emptyset,$ $\phi_{N}(M/N)=\emptyset$. We see that
$\phi_{N}(M/N)\subseteq M/N.$

######  Theorem 17.

Suppose that $N\subseteq r(\Re)$ and $M$ are hyperideals in $\Re$ such that
$N\subseteq M.$ If $M$ is $\phi-r$-hyperideal of $\Re$, then $M/N$ is
$\phi_{N}-r$-hyperideal of $\Re/N.$

###### Proof.

Suppose that $(r+N)\cdot(s+N)\subseteq$ $M/N-\phi_{M}(M/N)=M/N-(\phi(M)+N)/N$
and $ann_{\Re/N}(r+N)=0.$ Then $r\cdot s\in M-\phi(M)$ and as $N\subseteq
r(\Re),$ $ann_{\Re}(r)=0$. Since $M$ is $\phi-r$-hyperideal, then $s\in M.$
Therefore, $s+N\subseteq$ $M/N.$ ∎

######  Theorem 18.

Suppose that $N$ is an $r$-hyperideal and $M$ is a hyperideal in $\Re$ such
that $N\subseteq M.$ If $M/N$ is $r$-hyperideal of $\Re/N$, then $M$ is
$\phi-r$-hyperideal of $\Re.$

###### Proof.

Suppose that $r\cdot s\in$ $M-\phi(M)$ with $ann(r)=0.$ If $r\cdot s\in N$,
since $N$ is an $r$-hyperideal, then $s\in N\subseteq M.$ If $r\cdot s\in
M-N,$ then $r\cdot s+N=(r+N)\cdot(s+N)\subseteq M/N.$ In addition, since
$ann(r)=0,$ then $ann(r+N)=0.$ Therefore, since $M/N$ is $r$-hyperideal of
$\Re/N,$ then $s+N\subseteq M/N.$ Hence, $s\in M,$ as desired. ∎

Let $\phi:Id(\Re)\rightarrow Id(\Re)\cup\\{0\\}$ be a function. If $N$ and $M$
are hyperideals of $\Re$ such that $N\subseteq M$ implies that
$\phi(N)\subseteq\phi(M)$, then $\phi$ is said to preserve the order.

######  Theorem 19.

Let $A$ be a subset of $\Re,$ $A\neq\emptyset$ and $\phi$ preserves the order.
If $N$ is $\phi-r$-hyperideal such that $A\nsubseteq N,$ then $(N:A)$ is
$\phi-r$-hyperideal.

###### Proof.

Suppose that $a\cdot b\in(N:A)-\phi(N:A)$ and $ann(a)=0.$ Since $\phi$
preserves order, then $\phi(N)\subseteq\phi(N:A).$ We have $a\cdot b\cdot
A\subseteq N-\phi(N)$. Therefore, $b\cdot A\subseteq N$ and $b\in(N:A).$ ∎

######  Corollary 4.

Let $\\{N_{i}\\}_{i\in\Lambda}$ directed collection of ascending chain
$\phi-r$-hyperideals of $\Re$ and $\phi$ preserves the order. Then
$N=\underset{N\in\Lambda}{\cup}N_{i}$ is $\phi-r$-hyperideal.

###### Proof.

Let $a\cdot b\in N-\phi(N)$ and $ann(a)=0.$ Then $a\cdot b\in N_{k}$ and
$a\cdot b\notin\phi(N_{N}),$ for $k\in\Lambda.$ Since $N_{k}$ is
$\phi-r$-hyperideal, then $b\in N_{k}\subseteq N.$ ∎

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