A Study on k-Hyperideals in Ordered Semihyperrings

Abstract

In this study, we propose the concept of left extension of a hyperideal by generalizing the concept of k-hyperideals in ordered semihyperrings with symmetrical hyper-operation ⊕. By using the notion of extension of a k-hyperideal, we prove some results in ordered semihyperrings. The results of this paper can be viewed as a generalization for k-ideals of semirings.

1. Introduction

The notion of ordered semihypergroup was pioneered by Heidari and Davvaz [1] in 2011. In Ref. [2], Shi et al. attempted to study factorizable ordered hypergroupoids. In Ref. [3], Davvaz et al. initiated the study of pseudoorders in ordered semihypergroups. Gu and Tang in Ref. [4] and Tang et al. in Ref. [5] constructed the ordered semihypergroup from an ordered semihypergroup by using ordered regular relations.

The concept of hyperstructure was introduced by Marty [6] in 1934. In 1990, Vougiouklis [7] defined the notion of semihyperrings and discussed some of its properties. The theory of hyperideals in LA-hyperrings was studied by Rehman et al. in Ref. [8]. Many notions of algebraic geometry were extended to hyperrings in Ref. [9].

Some recent studies on ordered semihyperrings are on left k-bi-quasi hyperideals and right pure (bi-quasi-)hyperideals done by Rao et al. in Ref. [10] and Shao et al. in Ref. [11]. A study on w-pseudo-orders in ordered (semi)hyperrings was done in Ref. [12]. In Ref. [13], Kou et al. discussed the relationship between ordered semihyperrings by using homomorphisms and homo-derivations. Moreover, the connection between the ordered semihyperrings is explained by Omidi and Davvaz in Ref. [14].

In Ref. [15], Hedayati investigated some results in semihyperrings using k-hyperideals. In 2007, Ameri and Hedayati [16] introduced the notion of k-hyperideals in ordered semihyperrings. In this paper, we first define the left extension of a left hyperideal in an ordered semihyperring. The concept of extension of a k-ideal on a semiring R was introduced and studied by Chaudhari et al. in Refs. [17,18]. In the results of Chaudhari et al. [18], we replace the condition of extension of a k-ideal in semirings by extension of a k-hyperideal in ordered semihyperrings. By using the notion of extension of a k-hyperideal instead of k-hyperideal, we prove some results in ordered semihyperrings. Left extension of hyperideals are discovered to be a generalization of k-hyperideals. Let $Q,W$ be hyperideals of an ordered semihyperring $(R,\oplus ,\odot ,\le )$ such that $Q\subseteq W$. Then

$$\overline{W_Q}=\{r\in R\mid r\oplus P\subseteq W,\exists P\subseteq Q,0\in P\}$$

is the smallest left extension of Q containing W. Moreover, we proved that $\overline{W_Q}=W$ if and only if W is a left extension of Q. Some conclusions on extension of a k-hyperideal are gathered in the last section of the study.

2. Preliminaries

A mapping $\odot :R\times R\to {\mathcal{P}}^{*}\left(R\right)$ is called a hyperoperation on R. If $\varnothing \ne L,L^{\prime }\subseteq R$ and $x\in R$, then

$$L\odot L^{\prime }=\bigcup _{\begin{array}{c}\genfrac{}{}{0pt}{}{l\in L}{l^{\prime }\in L^{\prime }}\end{array}}l\odot l^{\prime },x\odot L=\left\{x\right\}\odot Land{L}^{\prime }\odot x=L^{\prime }\odot \left\{x\right\}.$$

$(R,\odot )$ is called a semihypergroup if for every $l,l^{\prime },x$ in R,

$$l\odot (l^{\prime }\odot x)=(l\odot l^{\prime })\odot x.$$

Definition 1. 

[7] A semihyperring is a triple $(R,\oplus ,\odot )$ such that for each $x,y,z\in R$,

(1) 

$(R,\oplus )$ is a commutative semihypergroup;

(2) 

$(R,\odot )$ is a semihypergroup;

(3) 

$x\odot (y\oplus z)=x\odot y\oplus x\odot z$ and $(x\oplus y)\odot z=x\odot z\oplus y\odot z$;

(4) 

There exists an element $0\in R$ such that $x\oplus 0=0\oplus x=\left\{x\right\}$ and $x\odot 0=0\odot x=\left\{0\right\}$ for all x in R.

Definition 2. 

[10] Take a semihyperring $(R,\oplus ,\odot )$ and a partial order relation ≤. Then $(R,\oplus ,\odot ,\le )$ is called an ordered semihyperring if for any $q,q^{\prime },x\in R$,

(1) 

$q\le q^{\prime }\Rightarrow q\oplus x⪯q^{\prime }\oplus x$;

(2) 

$$q\le q^{\prime }\Rightarrow \left\{\begin{array}{c}q\odot x⪯q^{\prime }\odot x,\hfill \\ \\ x\odot q⪯x\odot q^{\prime }.\hfill \end{array}\right.$$

For every $\varnothing \ne L,L^{\prime }\subseteq R$, $L⪯L^{\prime }$ is defined by $\forall l\in L,\exists l^{\prime }\in L^{\prime }$ such that $l\le l^{\prime }$. Clearly, $L\subseteq L^{\prime }$ implies $L⪯L^{\prime }$, but the converse is not valid in general. In this definition, two types of relation are defined, one is between elements of R, which is denoted by ≤, and second one between subsets of R, which is ⪯.

Example 1. 

Let $\mathbb{N}$ be the set of natural numbers and ${\mathbb{N}}_0=\mathbb{N}\cup \left\{0\right\}$. Consider the semiring $({\mathbb{N}}_0,+,·)$ where + and · are usual addition and multiplication. Define

$$l\oplus l^{\prime }=\{l,l^{\prime }\}\mathrm{and}l\odot l^{\prime }=\{l{l}^{\prime },cl{l}^{\prime }\},\mathrm{where}c\in {\mathbb{N}}_0.$$

If ≤ is the natural ordering on ${\mathbb{N}}_0$, then $({\mathbb{N}}_0,\oplus ,\odot ,\le )$ is an ordered semihyperring.

Definition 3. 

We will say that $\varnothing \ne K\subseteq R$ is a left (resp. right) hyperideal of R if

(1) 

for all $a,b\in K$, $a\oplus b\subseteq K$;

(2) 

$R\odot K\subseteq K$ (resp. $K\odot R\subseteq K$);

(3) 

$\left(K\right]\subseteq K$.

The set $\left(K\right]$ is given by

$$\left(K\right]:=\{r\in R|r\le x\mathrm{for}\mathrm{some}x\in K\}.$$

Definition 4. 

We will say that a left hyperideal $\varnothing \ne W\subseteq R$ is a left k-hyperideal of R, if

$$\forall w\in W,\forall q\in R,(w\oplus q)\cap W\ne \varnothing \Rightarrow q\in W.$$

Remark 1. 

Clearly, every left k-hyperideal of R is a left hyperideal of R. The converse is not true, in general, that is, a left hyperideal may not be a left k-hyperideal of R (see Example 2).

3. Main Results

Now, we study the extension of a k-hyperideal in an ordered semihyperring.

Definition 5. 

Assume that $K,L$ are left hyperideals of an ordered semihyperring $(R,\oplus ,\odot ,\le )$ and $L\subseteq K$. Then K is said to be a left extension of L if

$$\forall l\in L,\forall q\in R,l\oplus q\subseteq K\Rightarrow q\in K,$$

or

$$\forall l\in L,\forall q\in R,(l\oplus q)\cap K\ne \varnothing \Rightarrow q\in K.$$

Remark 2. 

Every k-hyperideal K of $(R,\oplus ,\odot ,\le )$ with $K\supseteq L$ is a left extension of L, where L is a hyperideal of R.

Example 2. 

Let $R=\{0,p,q\}$ and define the symmetrical hyper-operations ⊕ and ⊙ as follows:

$$\begin{array}{cccc}\oplus & 0& p& q\\ 0& \left\{0\right\}& \left\{p\right\}& \left\{q\right\}\\ p& \left\{p\right\}& \{0,p\}& \{0,p,q\}\\ q& \left\{q\right\}& \{0,p,q\}& \{0,p\}\\ \end{array}$$

$$\begin{array}{cccc}\odot & 0& p& q\\ 0& \left\{0\right\}& \left\{0\right\}& \left\{0\right\}\\ p& \left\{0\right\}& \left\{0\right\}& \left\{0\right\}\\ q& \left\{0\right\}& \left\{0\right\}& \{0,p\}\\ \end{array}$$

$$\begin{array}{c}\le :=\left\{\right(0,0),(p,p),(q,q),(0,p),(0,q),(p,q\left)\right\}.\hfill \end{array}$$

Then, $(R,\oplus ,\odot ,\le )$ is an ordered semihyperring. Clearly, $L=\{0,p\}$ is a hyperideal of R, but it is not a k-hyperideal. Indeed:

$$R=(p\oplus q)\cap L\ne \varnothing \mathrm{and}p\in L\mathrm{but}q\notin L.$$

Obviously, L is a k-extension of $L^{\prime }=\left\{0\right\}$,

Example 3. 

Consider the ordered semihyperring $(R,\oplus ,\odot ,\le )$ with the symmetrical hyper-operation ⊕ and hyper-operation ⊙:

$$\begin{array}{ccccc}\oplus & 0& p& q& r\\ 0& \left\{0\right\}& \left\{p\right\}& \left\{q\right\}& \left\{r\right\}\\ p& \left\{p\right\}& \left\{p\right\}& \left\{p\right\}& \left\{p\right\}\\ q& \left\{q\right\}& \left\{p\right\}& \{0,q\}& \{0,q,r\}\\ r& \left\{r\right\}& \left\{p\right\}& \{0,q,r\}& \{0,r\}\\ \end{array}$$

$$\begin{array}{ccccc}\odot & 0& p& q& r\\ 0& \left\{0\right\}& \left\{0\right\}& \left\{0\right\}& \left\{0\right\}\\ p& \left\{0\right\}& \left\{p\right\}& \{0,q\}& \left\{0\right\}\\ q& \left\{0\right\}& \left\{0\right\}& \left\{0\right\}& \left\{0\right\}\\ r& \left\{0\right\}& \{0,r\}& \left\{0\right\}& \left\{0\right\}\\ \end{array}$$

$$\begin{array}{cc}\le \hfill & :=\left\{\right(0,0),(0,p),(0,q),(0,r),(p,p),(q,p),(q,q),(r,p),(r,r\left)\right\}.\hfill \end{array}$$

Clearly, $K=\{0,q,r\}$ is a left extension of $L=\{0,q\}$. In addition, L is a left extension of $\left\{0\right\}$, but it is not a k-hyperideal of R. Indeed:

$$(r\oplus q)\cap L\ne \varnothing \mathrm{and}q\in L\mathrm{but}r\notin L.$$

Example 4. 

Let $R=\{0,p,q,r\}$ be a set with the symmetrical hyper-addition ⊕ and the multiplication ⊙ defined as follows:

$$\begin{array}{ccccc}\oplus & 0& p& q& r\\ 0& \left\{0\right\}& \left\{p\right\}& \left\{q\right\}& \left\{r\right\}\\ p& \left\{p\right\}& \{p,q\}& \left\{q\right\}& \left\{r\right\}\\ q& \left\{q\right\}& \left\{q\right\}& \{0,q\}& \left\{r\right\}\\ r& \left\{r\right\}& \left\{r\right\}& \left\{r\right\}& \{0,r\}\\ \end{array}$$

$$\begin{array}{ccccc}\odot & 0& p& q& r\\ 0& \left\{0\right\}& \left\{0\right\}& \left\{0\right\}& \left\{0\right\}\\ p& \left\{0\right\}& \left\{p\right\}& \left\{p\right\}& \left\{p\right\}\\ q& \left\{0\right\}& \left\{q\right\}& \left\{q\right\}& \left\{q\right\}\\ r& \left\{0\right\}& \left\{r\right\}& \left\{r\right\}& \left\{r\right\}\\ \end{array}$$

$$\begin{array}{c}\le :=\left\{\right(x,x\left)\right|x\in R\}.\hfill \end{array}$$

Then, $(R,\oplus ,\odot ,\le )$ is an ordered semihyperring. Clearly, $K=\{0,r\}$ is a right hyperideal of R, but it is not a right k-hyperideal of R. Indeed:

$$r\oplus p=r\in K\mathrm{and}r\in K\mathrm{but}p\notin K.$$

Let $L=\left\{0\right\}$. Then, K is a right k-extension of L, but it is not a right k-hyperideal of R.

Remark 3. 

In the following, we consider the following condition:

$$\forall l\in L,\forall q\in R,l\oplus q\subseteq K\Rightarrow q\in K.$$

Definition 6. 

Assume that $Q,W$ are hyperideals of an ordered semihyperring $(R,\oplus ,\odot ,\le )$ such that $Q\subseteq W$. Then, we denote

$$\overline{Q}=\{r\in R\mid r\oplus P\subseteq Q,\exists P\subseteq Q,0\in P\},$$

$$\overline{W_Q}=\{r\in R\mid r\oplus P\subseteq W,\exists P\subseteq Q,0\in P\}.$$

$\overline{W_Q}$ will be called the k-closure of W with respect to Q.

Remark 4. 

We have

(1) 

$Q\subseteq \overline{Q}\subseteq \overline{W_Q}\subseteq \overline{W}$;

(2) 

$\overline{W_W}=\overline{W}$.

Lemma 1. 

Assume that $Q,W,Y$ are hyperideals of an ordered semihyperring $(R,\oplus ,\odot ,\le )$ such that $Q\subseteq W\subseteq Y$. Then, $\overline{Y_Q}\subseteq \overline{Y_W}$.

Proof. 

Let W be a hyperideal of R such that $Q\subseteq W\subseteq Y$ and $x\in \overline{Y_Q}$. Then, there exists $P\subseteq Q\subseteq W$ such that $x\oplus P\subseteq Y$. So, $x\in \overline{Y_W}$. Therefore, $\overline{Y_Q}\subseteq \overline{Y_W}$. □

Proposition 1. 

$\overline{W_Q}$ is the smallest left extension of Q containing W.

Proof. 

Clearly, $\overline{W_Q}$ is a hyperideal of R.

Indeed: Let $q_1,q_2\in \overline{W_Q}$. By definition of $\overline{W_Q}$, there exist $P_1,P_2\subseteq Q$ such that $q_1\oplus P_1\subseteq W$ and $q_2\oplus P_2\subseteq W$. Now,

$$(q_1\oplus q_2)\oplus (P_1\oplus P_2)=q_1\oplus P_1\oplus q_2\oplus P_2\subseteq W\oplus W\subseteq W.$$

It means that $q_1\oplus q_2\subseteq \overline{W_Q}$.

Now, let $u\in \overline{W_Q}$ and $x\in R$. Then, there exists $P\subseteq Q$ such that $u\oplus P\subseteq W$. So,

$$x\odot u\oplus x\odot P=x\odot (u\oplus P)\subseteq R\odot W\subseteq W.$$

Since $x\odot P\subseteq Q$, we get $x\odot u\subseteq \overline{W_Q}$. Similarly, $u\odot x\subseteq \overline{W_Q}$.

Now, let $u\in \overline{W_Q}$ and $(v,u)\in \le$, where $v\in R$. By assumption, there exists $P\subseteq Q$ such that $u\oplus P\subseteq W$. Since R is an ordered semihyperring, we get $v\oplus p⪯u\oplus p$ for any $p\in P$. So, for any $x\in v\oplus p$, $x\le y$ for some $y\in u\oplus p\subseteq u\oplus P\subseteq W$. Since $\left(W\right]\subseteq W$, we obtain $x\in W$. So, $v\oplus p\subseteq W$ for each $p\in P$. Thus $v\oplus P\subseteq W$ and hence $v\in \overline{W_Q}$. Therefore, $\overline{W_Q}$ is a hyperideal of R.

Now, we prove that $\overline{W_Q}$ is a extension of Q. Let $q\in Q$ and $q\oplus r\subseteq \overline{W_Q}$, where $r\in R$. By assumption, $u\in \overline{W_Q}$ for all $u\in q\oplus r$. Hence, for any $u\in q\oplus r$, there exists $P_u\subseteq Q$ such that $u\oplus P_u\subseteq W$. Thus,

$$q\oplus r\oplus \bigcup _{\begin{array}{c}u\in q\oplus r\end{array}}{P}_u\subseteq \bigcup _{\begin{array}{c}u\in q\oplus r\end{array}}(u\oplus P_u)\oplus \bigcup _{\begin{array}{c}u\in q\oplus r\end{array}}{P}_u\subseteq W.$$

Since $q\oplus {\displaystyle \bigcup _{u\in q\oplus r}}{P}_u\subseteq Q$, it follows that $r\in \overline{W_Q}$. Therefore, $\overline{W_Q}$ is a left extension of Q.

Clearly, $W\subseteq \overline{W_Q}$. Now, let Y be a left extension of Q containing W and $q\in \overline{W_Q}$. Then, there exist $P\subseteq Q$ such that $q\oplus P\subseteq W\subseteq Y$. Since Y is a left extension of Q, we get $q\in Y$. Hence, $\overline{W_Q}\subseteq Y$. □

Theorem 1. 

Assume that $Q,W$ are hyperideals of an ordered semihyperring $(R,\oplus ,\odot ,\le )$ such that $Q\subseteq W$. Then, W is a left extension of Q if and only if $\overline{W_Q}=W$.

Proof. 

Necessity: Let W be a left extension of Q. By Proposition 1, $\overline{W_Q}$ is the smallest left extension of Q and $W\subseteq \overline{W_Q}$. Since W is a left extension of Q, we get $\overline{W_Q}\subseteq W$. So, $W\subseteq \overline{W_Q}\subseteq W$ and hence $\overline{W_Q}=W$.

Sufficiency: If $\overline{W_Q}=W$, then, since by Proposition 1, $\overline{W_Q}$ is a left extension of Q, it follows that W is a left extension of Q. □

Corollary 1. 

Assume that $Q,W$ are hyperideals of an ordered semihyperring $(R,\oplus ,\odot ,\le )$ such that $Q\subseteq W$. Then, $\overline{{\left(\overline{W_Q}\right)}_Q}=\overline{W_Q}$.

Proof. 

The proof obtains from Proposition 1 and Theorem 1. □

Theorem 2. 

Assume that $Q,W,Y$ are hyperideals of an ordered semihyperring $(R,\oplus ,\odot ,\le )$ such that $Q\subseteq W,Y$. Then,

$$\overline{{(W\cap Y)}_Q}=\overline{W_Q}\cap \overline{Y_Q}.$$

Proof. 

Let $a\in \overline{{(W\cap Y)}_Q}$. Then, there exists $P\subseteq Q$ such that

$$a\oplus P\subseteq W\cap Y\subseteq W.$$

So, $a\in \overline{W_Q}$. Therefore, $\overline{{(W\cap Y)}_Q}\subseteq \overline{W_Q}$. Similarly,

$$\overline{{(W\cap Y)}_Q}\subseteq \overline{Y_Q}.$$

Hence,

$$\overline{{(W\cap Y)}_Q}\subseteq \overline{W_Q}\cap \overline{Y_Q}.$$

Now, let $x\in \overline{W_Q}\cap \overline{Y_Q}$. Then, there exist $P,P^{\prime }\subseteq Q$ such that $x\oplus P\subseteq W$ and $x\oplus P^{\prime }\subseteq Y$. Since $P^{\prime }\subseteq Q\subseteq W$ and W is a hyperideal of R, we have

$$x\oplus P\oplus P^{\prime }\subseteq W\oplus W\subseteq W.$$

Similarly, $x\oplus P\oplus P^{\prime }\subseteq Y$. So, $x\oplus P\oplus P^{\prime }\subseteq W\cap Y$. This implies that $x\in \overline{{(W\cap Y)}_Q}$. Therefore, $\overline{W_Q}\cap \overline{Y_Q}\subseteq \overline{{(W\cap Y)}_Q}$. □

Theorem 3. 

Assume that $Q,W,Y$ are hyperideals of an ordered semihyperring $(R,\oplus ,\odot ,\le )$ such that $Q\subseteq W,Y$. If $W,Y$ are left extensions of Q, then $W\cap Y$ is a left extension of Q.

Proof. 

By Theorem 2, we have

$$\overline{{(W\cap Y)}_Q}=\overline{W_Q}\cap \overline{Y_Q}.$$

Since $W,Y$ are left extensions of Q, then by Theorem 1, we get

$$\overline{W_Q}\cap \overline{Y_Q}=W\cap Y.$$

Hence,

$$\overline{{(W\cap Y)}_Q}=W\cap Y.$$

Now, by Theorem 1, $W\cap Y$ is a left extension of Q. □

Definition 7. 

Assume that $K,L$ are left hyperideals of an ordered semihyperring $(R,\oplus ,\odot ,\le )$ and $L\subseteq K$. Then K is said to be a left m-extension of L if

$$\forall l\in L,\forall q\in R,l\odot q\subseteq K\Rightarrow q\in K.$$

Theorem 4. 

Assume that $K,L$ are hyperideals of an ordered semihyperring $(R,\oplus ,\odot ,\le )$ and $L\subseteq K$ such that $L\oplus R\subseteq L$. If K is a m-extension of L, then K is an extension of L.

Proof. 

Let K be a m-extension of L. Consider $l\oplus q\subseteq K$, $l\in L$ and $q\in R$. Since K is a hyperideal of R, we get

$$(l\oplus q)\odot q\subseteq K\odot R\subseteq K.$$

So, for any $p\in l\oplus q$, $p\odot q\subseteq K$. Since K is a m-extension of L, we have $q\in K$. Thus, K is an extension of L. □

4. Conclusions

The concept of left extension of hyperideals in ordered semihyperrings is introduced in this study. Left extension of hyperideals are discovered to be a generalization of k-hyperideals. Let $Q,W$ be hyperideals of an ordered semihyperring $(R,\oplus ,\odot ,\le )$ such that $Q\subseteq W$. Then

$$\overline{W_Q}=\{r\in R\mid r\oplus P\subseteq W,\exists P\subseteq Q,0\in P\}$$

is the smallest left extension of Q containing W. In addition, we proved that $\overline{W_Q}=W$ if and only if W is a left extension of Q. By using the concept of extension of a k-hyperideal, we discussed some results in ordered semihyperrings. Some further works can be done on left extension of a fuzzy hyperideal in ordered semihyperrings.