On weakly semiprime segments of ordered semihypergroups

1.   Introduction and Preliminaries

The concept of prime segments of a right chain ring was introduced by Brungs and T$\ddot{o}$rner in ^[1]. Later on, the concept was extended to right chain semigroups ^[2] and right chain ordered semigroups ^[3] respectively. In 2006, Mazurek and T$\ddot{o}$rner generalized the concept to a ring and studied the semiprime segments of a ring ^[4]. Recently, the weakly semiprime segments of ordered semigroups was studied by Luangchaisri and Changphas ^[5]. The hyperideals of ordered semihypergroups were introduced by Changphas and Davvaz in ^[6]. In ^[7], Davvaz et al. introduced the concept of a pseudoorder on an ordered semihypergroup and used it to construct a strongly regular equivalence relation on an ordered semihypergroup for which the corresponding quotient structure is an ordered semigroup. Furthermore, Tang et al. ^[8] introduced the concept of a weak pseudoorder on an ordered semihypergroup and used it to construct an ordered regular equivalence relation on an ordered semihypergroup for which the corresponding quotient structure is an ordered semihypergroup. Recently, $(m, n)$-quasi-hyperideal and $(m, n)$-hyperideal were introduced and studied by A. Mahboob et al. in ^[9,10]. Motivated by the previous work on rings and (ordered) semigroups, we introduce the notions of weakly semiprime segments in ordered semihypergroups, and classify them into four subclasses.

We recall first some basic notions of ordered semihypergroups (see ^[11]).

A hypergroupoid $(S, \circ)$ is a nonempty set $S$ together with a hyperoperation or hypercomposition, that is a mapping $\circ: S\times S\rightarrow \mathcal{P}^{\ast}(S)$, where $\mathcal{P}^{\ast}(S)$ denotes the family of all nonempty subsets of $S$. If $x\in S$ and $A, B$ are nonempty subsets of $S$, then we denote $A\circ B = \bigcup_{a\in A, b\in B}a\circ b, x\circ A = \{x\}\circ A$ and $A\circ x = A\circ\{x\}$. A hypergroupoid $(S, \circ)$ is called a semihypergroup if $\circ$ is associative, that is $x\circ(y\circ z) = (x\circ y)\circ z$ for every $x, y, z\in S$.

An ordered semihypergroup $(S, \circ, \leq)$ is a semihypergroup $(S, \circ)$ with an order relation $\leq$ which is compatible with the hyperoperation $\circ$, meaning that for any $a, b, x\in S$, $a\leq b$ implies that $a\circ x\leq b\circ x$ and $x\circ a\leq x\circ b$. Here, let $A, B\in \mathcal{P}^{\ast}(S)$, then we say that $A\leq B$ if for every $a\in A$ there exists $b\in B$ such that $a\leq b$.

Let $S$ be an ordered semihypergroup and $I$ be a nonempty subset of $S$. We say that $I$ is a hyperideal of $S$ if (1) $S\circ I\subseteq I, I\circ S\subseteq I$ and (2) $a\in I, b\in S$ and $b\leq a$ imply that $b\in I$. For $\emptyset\neq H\subseteq S$, we use the notation

For convenience, we write $(a]$ instead of $(\{a\}]$. We denote by $I(a)$ the hyperideal of $S$ generated by $a$. One can easily verify that

2.   Main results

Let $(S, \circ, \leq)$ be an ordered semihypergroup. An element $e$ of $S$ is called an absolute identity if $e\circ a = a\circ e = \{a\}$ for every $a\in S$. It is easy to see that an ordered semihypergroup has at most an absolute identity. In this section, we introduce the concept of a weakly semiprime segment of an ordered semihypergroup, and classify weakly semiprime segments of an ordered semihypergroup into four cases.

Definition 1. Let $S$ be an ordered semihypergroup with an absolute identity. A pair $P_{2}\subset P_{1}$ of weakly semiprime hyperideals of $S$ is called a weakly semiprime segment if $\bigcap_{n\in\mathbb{N}}(I^{n}]\subseteq P_{2}$ for every hyperideal $I$ of $S$ with $P_{2}\subset I\subset P_{1}$.

Lemma 1. Let $(S, \circ, \leq)$ be an ordered semihypergroup with an absolute identity, and $P_{2}\subset P_{1}$ a weakly semiprime segment of $S$. Then exactly one of the following possibilities occurs.

(1) There are no further hyperideals of $S$ between $P_{2}$ and $P_{1}$, and $P_{2}$ is comparable with each hyperideal of $S$ contained in $P_{1}$;

(2) There exists a weakly semiprime hyperideal $Q$ of $S$ such that $P_{2}\subset Q\subset P_{1}$, and $Q$ is comparable with each hyperideal of $S$ contained in $P_{1}$;

(3) $((P_{1}\circ a\circ S)\cup (S\circ a\circ P_{1})]\subset(S\circ a\circ S]$ for all $a\in P_{1}\setminus P_{2}$;

(4) $((P_{1}\circ a\circ S)\cup (S\circ a\circ P_{1})] = (S\circ a\circ S]$ for some $a\in P_{1}\setminus P_{2}$ and $P_{1} = (S\circ a_{1}\circ S]\cup(S\circ a_{2}\circ S]\cup P_{2}$ for some $a_{1}, a_{2}\in P_{1}$ with $(S\circ a_{i}\circ S]\subset P_{1}$.

Proof. Let $a\in P_{1}\setminus P_{2}$. If $(P_{1}^{2}\cup P_{2}]\subset P_{1}$, then (3) occurs. Indeed: suppose that $a\in ((P_{1}\circ a\circ S)\cup (S\circ a\circ P_{1})]$. Then $a\in ((P_{1}^{n}\circ a\circ S)\cup(S\circ a\circ P_{1}^{n})]\subseteq(P_{1}^{n}]$ for any $n\in\mathbb{N}$. Thus $a\in\bigcap_{n\in\mathbb{N}}(P_{1}^{n}] = \bigcap_{n\in\mathbb{N}}(P_{1}^{2n}]\subseteq\bigcap_{n\in\mathbb{N}}((P_{1}^{2}\cup P_{2})^{n}]\subseteq P_{2}$, which contradicts that $a\notin P_{2}$. Hence, $a\notin((P_{1}\circ a\circ S)\cup (S\circ a\circ P_{1})]$ and so $((P_{1}\circ a\circ S)\cup (S\circ a\circ P_{1})]\subset(S\circ a\circ S]$. To finish the proof, we assume that $(P_{1}^{2}\cup P_{2}] = P_{1}$. Let $M$ be the union of all hyperideals $I$ of $S$ with $I\subset P_{1}$. Then $M$ is a hyperideal of $S$ and $P_{2}\subseteq M\subseteq P_{1}$. Next we consider three cases.

($i$) If $M = P_{2}$, then (1) occurs obviously.

($ii$) If $P_{2}\subset M\subset P_{1}$, then (2) occurs. In fact: Let $A$ be a hyperideal of $S$ with $A^{2}\subseteq M$. Then $A^{2}\subseteq P_{1}$. Since $P_{1}$ is weakly semiprime, we have $A\subseteq P_{1}$. Suppose that $A = P_{1}$. Then $P_{1} = (P_{1}^{2}\cup P_{2}] = (A^{2}\cup P_{2}]\subseteq M$ which contradicts that $M\subset P_{1}$. Thus $A\subset P_{1}$ and so $A\subseteq M$. Hence, $M$ is weakly semiprime.

($iii$) Let $M = P_{1}$. Then $(S\circ x\circ S]\subset P_{1}$ for any $x\in P_{1}$. Assume that (3) does not occur. Then there exists $a\in P_{1}\setminus P_{2}$ such that $((P_{1}\circ a\circ S)\cup (S\circ a\circ P_{1})] = (S\circ a\circ S]$. Thus $a\in((s\circ a\circ S)\cup(S\circ a\circ t)]$ for some $s, t\in P_{1}$. Set $I = (S\circ s\circ S]\cup(S\circ t\circ S]$. Then $I$ is a hyperideal of $S$ and $a\in ((s\circ a\circ S)\cup(S\circ a\circ t)]\subseteq((S\circ s\circ S\circ a\circ S)\cup(S\circ a\circ S\circ t\circ S)]\subseteq((I\circ a\circ S)\cup(S\circ a\circ I)]$. If $I\cup P_{2}\subset P_{1}$, then by the beginning of the proof, we have $a\in\bigcap_{n\in\mathbb{N}}(I^{n}]\subseteq\bigcap_{n\in\mathbb{N}}((I\cup P_{2})^{n}]\subseteq P_{2}$ which contradicts that $a\notin P_{2}$. Hence, $P_{1} = I\cup P_{2}$ and so (4) occurs.

It is easy to verify that the possibilities (1), (2), (3), (4) are mutually exclusive.

Theorem 1. Let $(S, \circ, \leq)$ be an ordered semihypergroup with an absolute identity, and $P_{2}\subset P_{1}$ a weakly semiprime segment of $S$. Then exactly one of the following possibilities occurs.

($a$) The semiprime segment $P_{2}\subset P_{1}$ is simple; that is, there are no further hyperideals of $S$ between $P_{2}$ and $P_{1}$, and $P_{2}$ is comparable with each hyperideal of $S$ contained in $P_{1}$;

($b$) The semiprime segment $P_{2}\subset P_{1}$ is exceptional; that is, there exists a weakly semiprime hyperideal $Q$ of $S$ such that $P_{2}\subset Q\subset P_{1}$, and $Q$ is comparable with each hyperideal of $S$ contained in $P_{1}$;

($c$) The semiprime segment $P_{2}\subset P_{1}$ is Archimedean; that is, for every $a\in P_{1}\setminus P_{2}$ there exists a hyperideal $I\subseteq P_{1}$ of $S$ such that $a\in I$ and $\bigcap_{n\in\mathbb{N}}(I^{n}]\subseteq P_{2}$;

($d$) The semiprime segment $P_{2}\subset P_{1}$ is decomposable; that is, the semiprime segment $P_{2}\subset P_{1}$ is not Archimedean and $P_{1} = A\cup B$ for some hyperideals $A, B$ of $S$ properly contained in $P_{1}$.

Proof. From Lemma 1, we know that there are four cases to consider. Clearly, in the case (1) the semiprime segment $P_{2}\subset P_{1}$ is simple; in the case (2) the semiprime segment $P_{2}\subset P_{1}$ is exceptional; and in the case (4) the segment $P_{2}\subset P_{1}$ is either Archimedean or decomposable.

Assume that the case (3) occurs and the segment $P_{2}\subset P_{1}$ is not Archimedean. Then there exists $a\in P_{1}\backslash P_{2}$ such that $\bigcap_{n\in\mathbb{N}}(I^{n}]\nsubseteq P_{2}$ for all hyperideals $I$ with $a\in I\subseteq P_{1}$. If $((S\circ a\circ S]^{2}]\cup P_{2}\subset P_{1}$, then $\bigcap_{n\in\mathbb{N}}((S\circ a\circ S]^{n}] = \bigcap_{n\in\mathbb{N}}((S\circ a\circ S]^{2n}]\subseteq\bigcap_{n\in\mathbb{N}}(((S\circ a\circ S]^{2}\cup P_{2})^{n}]\subseteq P_{2}$, which is a contradiction. Thus $((S\circ a\circ S]^{2}]\cup P_{2} = P_{1}$. Moreover, $((S\circ a\circ S]^{2}]\subset(P_{1}\circ a \circ S]\subset (S\circ a\circ S]\subseteq P_{1}$. Hence, the segment $P_{2}\subset P_{1}$ is decomposable.

It is easy to see that the possibilities ($a$), ($b$), ($c$), ($d$) are mutually exclusive.

3.   Conclusions

Let $S$ be an ordered semihypergroup with an absolute identity. A pair $P_{2}\subset P_{1}$ of weakly semiprime hyperideals of $S$ is called a weakly semiprime segment if $\bigcap_{n\in\mathbb{N}}(I^{n}]\subseteq P_{2}$ for every hyperideal $I$ of $S$ with $P_{2}\subset I\subset P_{1}$. In this paper, we classify weakly semiprime segments of an ordered semihypergroup into four cases which are simple, exceptional, Archimedean and decomposable.

Acknowledgments

I thank the referees whose comments led to significant improvements to this paper and the editor for his/her warm work.

This research was supported by the National Natural Science Foundation of China (No. 11701504, 11801081), the Characteristic Innovation Project of Department of Education of Guangdong Province (No. 2020KTSCX159), the innovative research team project of Zhaoqing University and the scientific research ability enhancement program for excellent young teachers of Zhaoqing University.

Conflict of interest

The authors declare no conflict of interest.