On Convex Ordered Hyperrings

Abstract

The concept of convex ordered hyperrings associated with a strongly regular relation was investigated in this study. In this paper, we first studied hyperatom elements of ordered hyperrings and then investigated characterizations of quotient ordered rings. Is there a strongly regular relation $\theta$ on a convex ordered hyperring R for which $R/\theta$ is a convex ordered ring? This leads to an ordered ring obtained from an ordered hyperring.

1. Introduction

The notion of a hypergroup was first developed by Marty in [1]. Later on, Krasner studied hyperrings in [2]. Krasner $(m,n)$-hyperrings were studied in [3].

The theory of ordered hyperstructures was first developed by Heidari and Davvaz [4] in 2011. There are some remarkable papers [5,6,7] on (weak) pseudo-orders of ordered hyperstructures.

The notion of $(m,n)$-hyperfilters in ordered hyperstructures was investigated by Rao et al. in [8], while Al-Tahan et al. [9] proposed the concept of $(m,n)$-quasi-filters in ordered structures. The convex ordered $\Gamma$-semi-hypergroups have been studied in connection with their strong regular relation [10]. Recently, pure hyperideals of an ordered semihyperring have been extensively investigated by Shao et al. in [11].

Posner [12] was the first to investigate derivations in rings. The strong derivation has roles in hyperring theory and semihyperring theory as we see in Asokkumar [13], Kamali and Davvaz [14], Rao et al. [15,16], etc.

The hyperatom elements in the content of an ordered semihyperring were investigated by Rao et al. [16]. Recently, Kou et al. [17] published an interesting article in an $L_R$-graph of an ordered semihyperring. Hyperatom elements are used in the $L_R$-graph of an ordered semihyperring, so it is very useful to study these elements. The concept of a convex ordered hyperring associated with a strongly regular relation was investigated in this study. Additionally, we studied hyperatom elements of ordered hyperrings and then investigated characterizations of quotient ordered rings. A construction of an ordered ring via a convex ordered hyperring was given.

Definition 1

([2,18]). A triple $(R,\oplus ,\odot )$ is a Krasner hyperring if:

(1) 

$(R,\oplus )$ is a canonical hypergroup;

(2) 

$(R,\odot )$ is a semigroup and $x\odot 0=0=0\odot x$ for all $x\in R$;

(3) 

The operation ⊙ is distributive with respect to the hyperoperation ⊕.

Throughout this paper we consider a Krasner hyperring $(R,\oplus ,\odot )$. For $\varnothing \ne E,F\subseteq R$, we set

$$E\overline{\overline{\theta }}F\iff \forall e\in E,\forall f\in F,e\theta f.$$

Moreover, $\theta$ is called a strongly regular relation on R [18] if $\forall p,q,s\in R$,

(i)

$p\theta q\Rightarrow (p\oplus s)\overline{\overline{\theta }}(q\oplus s)\mathrm{and}(s\oplus p)\overline{\overline{\theta }}(s\oplus q)$;

(ii)

$p\theta q\Rightarrow (p\odot s)\overline{\overline{\theta }}(q\odot s)\mathrm{and}(s\odot p)\overline{\overline{\theta }}(s\odot q)$.

Definition 2.

Let $(R,\oplus ,\odot )$ be a hyperring. If R admits a partial order relation ≤ such that for any $p,q,s\in R$,

$$p\le q\Rightarrow \left\{\begin{array}{c}p\oplus s⪯q\oplus s,\hfill \\ \\ s\oplus p⪯s\oplus q,\hfill \\ \\ p\odot s⪯q\odot s,\hfill \\ \\ s\odot p⪯s\odot q.\hfill \end{array}\right.$$

Then, $(R,\oplus ,\odot ,\le )$ is called an ordered hyperring. Here, for every $\varnothing \ne S_1,S_2\subseteq R$, we put

$$S_1⪯S_2\iff \forall s_1\in S_1,\exists s_2\in S_2;s_1\le s_2,$$

and

$$\left(S_1\right]:=\{q\in R|q\le s_1\mathrm{for}\mathrm{some}{s}_1\in S_1\}.$$

Theorem 1.

If $(R,\oplus ,\odot ,\le )$ is a convex ordered hyperring associated with a strongly regular relation θ, then $(R/\theta ,\uplus ,\otimes ,⪯)$ is an ordered ring.

Proof. 

See Theorem 3.6 in [19]. □

Theorem 2.

If $(R,\oplus ,\odot ,\le )$ is a convex ordered hyperring associated with a d-strongly regular relation θ where d is an injective strong derivation of R, then there exists an injective strong derivation on $(R/\theta ,\uplus ,\otimes ,⪯)$.

Proof. 

See Theorem 3.7 in [19]. □

2. Results and Discussion

Definition 3.

We say that an element q of an ordered hyperring $(R,\oplus ,\odot ,\le )$ is a hyperatom element if

(1) 

for any $r\in R$, $r\le q$ implies $r=0$ or $r=q$;

(2) 

$q\odot r\ne 0$ implies $(q\odot r)\odot p=r$ for some $p\in R$.

Example 1.

Let $R=\{0,p,q,r\}$ and

$$\begin{array}{|ccccc|}\hline \oplus & 0& p& q& r\\ 0& 0& p& q& r\\ p& p& \{0,q\}& \{p,r\}& q\\ q& q& \{p,r\}& \{0,q\}& p\\ r& r& q& p& 0\\ \hline\end{array}$$

$$\begin{array}{|ccccc|}\hline \odot & 0& p& q& r\\ 0& 0& 0& 0& 0\\ p& 0& p& q& r\\ q& 0& q& q& 0\\ r& 0& r& 0& r\\ \hline\end{array}$$

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

Then $(R,\oplus ,\odot ,\le )$ is an ordered hyperring. Obviously, 0 is the only hyperatom element in R.

Example 2.

Let $R=\{0,p,q,r\}$ and

$$\begin{array}{|ccccc|}\hline \oplus & 0& p& q& r\\ 0& 0& p& q& r\\ p& p& \{0,p\}& r& \{q,r\}\\ q& q& r& 0& p\\ r& r& \{q,r\}& p& \{0,p\}\\ \hline\end{array}$$

$$\begin{array}{|ccccc|}\hline \odot & 0& p& q& r\\ 0& 0& 0& 0& 0\\ p& 0& 0& 0& 0\\ q& 0& 0& q& q\\ r& 0& 0& q& q\\ \hline\end{array}$$

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

Then $(R,\oplus ,\odot ,\le )$ is an ordered hyperring. Obviously, 0 and p are hyperatom elements in R.

Lemma 1.

Let

$$A\left(R\right):=\{q\in R|q\mathrm{is}\mathrm{a}\mathrm{hyperatom}\mathrm{element}\mathrm{of}(R,\oplus ,\odot ,\le )\}.$$

If $p\ominus q⪯p$ and $p\odot q\le p$ for all $p,q\in A\left(R\right)$, then $A\left(R\right)$ is a subhyperring of R.

Proof. 

Clearly, $0\in A\left(R\right)$. Let $p,q\in A\left(R\right)$. By hypothesis, $u\le p$ for every $u\in p\ominus q$. Since $p\in A\left(R\right)$, we get $u=0$ or $u=p$. Thus,

$$p\ominus q\subseteq \{0,p\}.$$

Hence,

$$p\ominus q\subseteq A\left(R\right).$$

Similarly,

$$p\odot q\in A\left(R\right).$$

If $t\in R$ and $t\le p\in A\left(R\right)$, then $t=0$ or $t=p$. Thus, $t\in A\left(R\right)$ and hence

$$\left(A\right(R\left)\right]\subseteq A\left(R\right).$$

Therefore, $A\left(R\right)$ is a subhyperring of R. □

Definition 4.

$(R,\le )$ is said to be a convex ordered hyperring associated with a strongly regular relation θ if

$$(p,q)\in \theta and p\le c\le q\Rightarrow (p,c)\in \theta .$$

Definition 5.

A mapping d of an ordered hyperring $(R,\oplus ,\odot ,\le )$ into itself is said to be a derivation if $\forall s_1,s_2\in R$,

(1) 

$d(s_1\oplus s_2)\subseteq d\left(s_1\right)\oplus d\left(s_2\right)$;

(2) 

$d(s_1\odot s_2)\in d\left(s_1\right)\odot s_2\oplus s_1\odot d\left(s_2\right)$;

(3) 

$s_1\le s_2\Rightarrow d\left(s_1\right)\le d\left(s_2\right)$.

Example 3.

In Example 1, obviously, $d:R\to R$, defined by

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

is a derivation on R.

Now, we are able to make the connections of ordered hyperrings with ordered rings.

Theorem 3.

Let θ be a strongly regular relation on an ordered hyperring $(R,\oplus ,\odot ,\le )$. If for every $(p,q)\in \theta$, q is a hyperatom element, then $(R/\theta ,\uplus ,\otimes ,⪯)$ is an ordered ring, where for all ${\left[t\right]}_{\theta },{\left[s\right]}_{\theta }\in R/\theta =\left\{{\left[t\right]}_{\theta }\right|t\in R\}$,

$${\left[t\right]}_{\theta }\uplus {\left[s\right]}_{\theta }=\left\{{\left[p\right]}_{\theta }\right|p\in t\oplus s\},$$

$${\left[t\right]}_{\theta }\otimes {\left[s\right]}_{\theta }={[t\odot s]}_{\theta },$$

$$\begin{array}{c}{\left[t\right]}_{\theta }⪯{\left[s\right]}_{\theta }\iff \forall t^{\prime }\in {\left[t\right]}_{\theta },\exists s^{\prime }\in {\left[s\right]}_{\theta }\mathrm{such}\mathrm{that}{t}^{\prime }\le s^{\prime }.\hfill \end{array}$$

Proof. 

Let $(p,q)\in \theta$ and $p\le c\le q$. We assert that

$$(p,c)\in \theta .$$

As q is a hyperatom element, we have

$$c=0 \mathrm{or} c=q.$$

Now,

Case 1.$c=0$.

Since R is positive, we have $p=0$. Thus, $(p,c)=(0,0)\in \theta$ in this case.

Case 2.$c=q$.

By this hypothesis, we obtain $(p,c)=(p,q)\in \theta$.

Therefore, R is a convex ordered hyperring associated with $\theta$. Now, by Theorem 3.6 in [19], $(R/\theta ,\uplus ,\otimes ,⪯)$ is an ordered ring. □

Theorem 4.

Let d be an injective strong derivation and θ a strongly regular relation on $(R,\oplus ,\odot ,\le )$ such that

(i) 

for every $(p,q)\in \theta$, q is a hyperatom element;

(ii) 

$(p,q)\in \theta \iff \left(d\right(p),d(q\left)\right)\in \theta$.

Then,

$$\psi :R/\theta \to R/\theta$$

defined by

$$\psi \left({\left[t\right]}_{\theta }\right)={\left[d\left(t\right)\right]}_{\theta }, \forall {\left[t\right]}_{\theta }\in R/\theta ,$$

is an injective strong derivation on $(R/\theta ,\uplus ,\otimes ,⪯)$.

Proof. 

By Theorem 3, $(R/\theta ,\uplus ,\otimes ,⪯)$ is an ordered ring. Now, let $(p,q)\in \theta$. Then, by condition (ii), we get

$$\left(d\right(p),d(q\left)\right)\in \theta .$$

Hence, $\psi \left({\left[p\right]}_{\theta }\right)=\psi \left({\left[q\right]}_{\theta }\right)$ and so $\psi$ is well-defined.

Let ${\left[p\right]}_{\theta },{\left[q\right]}_{\theta }\in R/\theta$ and $\psi \left({\left[p\right]}_{\theta }\right)=\psi \left({\left[q\right]}_{\theta }\right)$. Then ${\left[d\left(p\right)\right]}_{\theta }={\left[d\left(q\right)\right]}_{\theta }$ and hence $\left(d\right(p),d(q\left)\right)\in \theta$. As d is injective, we get $(p,q)\in \theta$. Thus ${\left[p\right]}_{\theta }={\left[q\right]}_{\theta }$ and so $\psi$ is injective. Now, by the proof of Theorem 3.7 in [19], $\psi$ is a strong derivation on $(R/\theta ,\uplus ,\otimes ,⪯)$. □

Theorem 5.

Let $(R,\oplus ,\odot ,\le )$ be a finite ordered hyperring and $\left|R\right|\ge 2$. Then,

$$\forall p\in R\backslash \left\{0\right\},\exists q_p\in A^{*}\left(R\right)=A\left(R\right)\backslash \left\{0\right\};q_p\le p.$$

Proof. 

Let $p\in R\backslash \left\{0\right\}$.

Case 1.p is a hyperatom element. Then take $q_p=p$.

Case 2.p is not a hyperatom element. Then

$$\exists p^{\prime }\in R\backslash \{0,p\};p^{\prime }\le p.$$

Subcase 1.$p^{\prime }$ is a hyperatom element. Then take $q_p=p^{\prime }$.

Subcase 2.$p^{\prime }$ is not a hyperatom element.

Then,

$$\exists p_1\in R\backslash \{0,p,p^{\prime }\};p_1\le p^{\prime }\le p.$$

As R is finite, we get

$$p_s\le p_{s-1}\le \cdots \le p_1\le p^{\prime }\le p,$$

where $p_s\in R\backslash \{0,p,p^{\prime },p_1,\cdots ,p_{s-1}\}$. Hence, $p_s=0$ if $0\in L_R\left(p\right)$ or $q_p=p_s\in A^{*}\left(R\right)$ and $q_p\le p$. □

Corollary 1.

Let $(R,\oplus ,\odot ,\le )$ be a finite ordered hyperring and $\left|R\right|\ge 2$. Then,

$$\forall p\in R\backslash \left\{0\right\},\exists q\in R\backslash \left\{0\right\};q\le p\iff A^{*}\left(R\right)=\left\{q\right\}.$$

Proof. 

(⇒): We have

$$\forall w\in A^{*}\left(R\right),\exists q\in R\backslash \left\{0\right\};q\le w.$$

As $q\ne 0$ and $w\in A^{*}\left(R\right)$, we obtain $q=w$. Thus, $A^{*}\left(R\right)=\left\{q\right\}$.

(⇐): Let $A^{*}\left(R\right)=\left\{q\right\}$ and $p\in R\backslash \left\{0\right\}$. By Theorem 5, $q\le p$. □

Remark 1.

Let

$${\theta }_{id}=\{(p,q)\in R\times R|p=q\}.$$

If R is an ordered ring, then R is a convex ordered ring associated with ${\theta }_{id}$. Clearly, ${\theta }_{id}$ on an ordered hyperring R is not a strongly regular relation (see Example 1). Thus, R is not a convex ordered hyperring associated with ${\theta }_{id}$.

Theorem 6.

Let $(R_{\lambda },{\oplus }_{\lambda },{\odot }_{\lambda },{\le }_{\lambda })$ be a convex ordered hyperring for all $\lambda \in \Lambda$. Then, ${\displaystyle {\displaystyle \prod }_{\begin{array}{c}\lambda \in \Lambda \end{array}}}{R}_{\lambda }$ is a convex ordered hyperring.

Proof. 

Clearly, ${\displaystyle \prod }_{\begin{array}{c}\lambda \in \Lambda \end{array}}{R}_{\lambda }=\left\{{\left(q_{\lambda }\right)}_{\lambda \in \Lambda }\right|q_{\lambda }\in R_{\lambda }\}$ is an ordered hyperring. Indeed: for any ${\left(p_{\lambda }\right)}_{\lambda \in \Lambda },{\left(q_{\lambda }\right)}_{\lambda \in \Lambda }\in {\displaystyle {\displaystyle \prod }_{\begin{array}{c}\lambda \in \Lambda \end{array}}}{R}_{\lambda }$, we set

(i)

${\left(p_{\lambda }\right)}_{\lambda \in \Lambda }\oplus {\left(q_{\lambda }\right)}_{\lambda \in \Lambda }=\left\{{\left(r_{\lambda }\right)}_{\lambda \in \Lambda }\right|r_{\lambda }\in p_{\lambda }{\oplus }_{\lambda }{q}_{\lambda }\}$;

(ii)

${\left(p_{\lambda }\right)}_{\lambda \in \Lambda }\odot {\left(q_{\lambda }\right)}_{\lambda \in \Lambda }={\left(p_{\lambda }{\odot }_{\lambda }{q}_{\lambda }\right)}_{\lambda \in \Lambda }$;

(iii)

${\left(p_{\lambda }\right)}_{\lambda \in \Lambda }\le {\left(q_{\lambda }\right)}_{\lambda \in \Lambda }\iff p_{\lambda }{\le }_{\lambda }{q}_{\lambda },\forall \lambda \in \Lambda$.

Now, let ${\left(p_{\lambda }\right)}_{\lambda \in \Lambda }\le {\left(q_{\lambda }\right)}_{\lambda \in \Lambda }$. If ${\left(u_{\lambda }\right)}_{\lambda \in \Lambda }\in {\left(r_{\lambda }\right)}_{\lambda \in \Lambda }\oplus {\left(p_{\lambda }\right)}_{\lambda \in \Lambda }$, then $u_{\lambda }\in r_{\lambda }{\oplus }_{\lambda }{p}_{\lambda }$. As ${\left(p_{\lambda }\right)}_{\lambda \in \Lambda }\le {\left(q_{\lambda }\right)}_{\lambda \in \Lambda }$, we obtain

$$p_{\lambda }{\le }_{\lambda }{q}_{\lambda },\forall \lambda \in \Lambda .$$

Hence,

$$u_{\lambda }\in r_{\lambda }{\oplus }_{\lambda }{p}_{\lambda }{\le }_{\lambda }{r}_{\lambda }{\oplus }_{\lambda }{q}_{\lambda }.$$

Thus, there exists $v_{\lambda }\in r_{\lambda }{\oplus }_{\lambda }{q}_{\lambda }$ such that $u_{\lambda }{\le }_{\lambda }{v}_{\lambda }$. It means that

$${\left(u_{\lambda }\right)}_{\lambda \in \Lambda }\le {\left(v_{\lambda }\right)}_{\lambda \in \Lambda }.$$

So,

$${\left(r_{\lambda }\right)}_{\lambda \in \Lambda }\oplus {\left(p_{\lambda }\right)}_{\lambda \in \Lambda }\le {\left(r_{\lambda }\right)}_{\lambda \in \Lambda }\oplus {\left(q_{\lambda }\right)}_{\lambda \in \Lambda }.$$

Similarly,

$${\left(r_{\lambda }\right)}_{\lambda \in \Lambda }\odot {\left(p_{\lambda }\right)}_{\lambda \in \Lambda }\le {\left(r_{\lambda }\right)}_{\lambda \in \Lambda }\odot {\left(q_{\lambda }\right)}_{\lambda \in \Lambda }.$$

Therefore, ${\displaystyle {\displaystyle \prod }_{\begin{array}{c}\lambda \in \Lambda \end{array}}}{R}_{\lambda }$ is an ordered hyperring.

Claim: If ${\theta }_{\lambda }$ is a strongly regular relation on $R_{\lambda }$ for all $\lambda \in \Lambda$, then $\theta$ is a strongly regular relation on ${\displaystyle {\displaystyle \prod }_{\begin{array}{c}\lambda \in \Lambda \end{array}}}{R}_{\lambda }$, where

$${\left(p_{\lambda }\right)}_{\lambda \in \Lambda }\theta {\left(q_{\lambda }\right)}_{\lambda \in \Lambda }\iff (p_{\lambda },q_{\lambda })\in {\theta }_{\lambda },\forall p_{\lambda },q_{\lambda }\in R_{\lambda },\forall \lambda \in \Lambda .$$

Let ${\left(p_{\lambda }\right)}_{\lambda \in \Lambda }\theta {\left(q_{\lambda }\right)}_{\lambda \in \Lambda }$. Then

$$(p_{\lambda },q_{\lambda })\in {\theta }_{\lambda },\forall p_{\lambda },q_{\lambda }\in R_{\lambda },\forall \lambda \in \Lambda .$$

Hence,

$$\left(p_{\lambda }{\oplus }_{\lambda }{r}_{\lambda }\right)\overline{\overline{{\theta }_{\lambda }}}\left(q_{\lambda }{\oplus }_{\lambda }{r}_{\lambda }\right), \forall r_{\lambda }\in R_{\lambda },\forall \lambda \in \Lambda .$$

So,

$${\left(p_{\lambda }\right)}_{\lambda \in \Lambda }\oplus {\left(r_{\lambda }\right)}_{\lambda \in \Lambda }\overline{\overline{\theta }}{\left(q_{\lambda }\right)}_{\lambda \in \Lambda }\oplus {\left(r_{\lambda }\right)}_{\lambda \in \Lambda }.$$

Similarly,

$${\left(p_{\lambda }\right)}_{\lambda \in \Lambda }\odot {\left(r_{\lambda }\right)}_{\lambda \in \Lambda }\overline{\overline{\theta }}{\left(q_{\lambda }\right)}_{\lambda \in \Lambda }\odot {\left(r_{\lambda }\right)}_{\lambda \in \Lambda }.$$

Hence, $\theta$ is a strongly regular relation on ${\displaystyle {\displaystyle \prod }_{\begin{array}{c}\lambda \in \Lambda \end{array}}}{R}_{\lambda }$.

Let $({\left(p_{\lambda }\right)}_{\lambda \in \Lambda },{\left(r_{\lambda }\right)}_{\lambda \in \Lambda })\in \theta$ and ${\left(p_{\lambda }\right)}_{\lambda \in \Lambda }\le {\left(q_{\lambda }\right)}_{\lambda \in \Lambda }\le {\left(r_{\lambda }\right)}_{\lambda \in \Lambda }$. Then, $(p_{\lambda },r_{\lambda })\in {\theta }_{\lambda }$ and $p_{\lambda }{\le }_{\lambda }{q}_{\lambda }{\le }_{\lambda }{r}_{\lambda }$. As $R_{\lambda }$ is a convex ordered hyperring, we get $(p_{\lambda },q_{\lambda })\in {\theta }_{\lambda }$. Therefore, $({\left(p_{\lambda }\right)}_{\lambda \in \Lambda },{\left(q_{\lambda }\right)}_{\lambda \in \Lambda })\in \theta$. Hence, ${\displaystyle {\displaystyle \prod }_{\begin{array}{c}\lambda \in \Lambda \end{array}}}{R}_{\lambda }$ is a convex ordered hyperring. □

Remark 2.

Suppose that θ is a pseudo-order in a convex ordered hyperring $(R,\oplus ,\odot ,\le )$. Then, $(R/{\theta }^{*},\uplus ,\otimes ,{⪯}_{{\theta }^{*}})$ is a convex ordered ring, where

$${\theta }^{*}=\{(p,q)\in R\times R|(p,q)\in \theta \mathrm{and}(q,p)\in \theta \},$$

and

$${⪯}_{{\theta }^{*}}:=\{({\theta }^{*}\left(x\right),{\theta }^{*}\left(y\right))\in R/{\theta }^{*}\times R/{\theta }^{*}|\exists p\in {\theta }^{*}\left(x\right),\exists q\in {\theta }^{*}\left(y\right)\mathrm{such}\mathrm{that}(p,q)\in \theta \}.$$

Indeed: Let $p\le q\le r$ and $(p,r)\in {\theta }^{*}$. Then ${\left[p\right]}_{{\theta }^{*}}{⪯}_{{\theta }^{*}}{\left[q\right]}_{{\theta }^{*}}$ and ${\left[q\right]}_{{\theta }^{*}}{⪯}_{{\theta }^{*}}{\left[r\right]}_{{\theta }^{*}}={\left[p\right]}_{{\theta }^{*}}$. So, ${\left[p\right]}_{{\theta }^{*}}={\left[q\right]}_{{\theta }^{*}}$. Thus $(p,q)\in {\theta }^{*}$ and hence $R/{\theta }^{*}$ is convex.

3. Conlusions

Strongly regular relations are interesting topics in hyperring theory. The concept of a convex ordered hyperring associated with a strongly regular relation was investigated in this study. Researchers in hyperstructure theory in recent years have seriously investigated the construction of ordered structures. Considering the convex ordered hyperring, a method was suggested to construct ordered rings using the strongly regular relations. In the future, we will study convex ordered superrings.