Generalizations of Prime Hyperideals via Hypersystems in Krasner Hyperrings

Abstract

The aim of this study is to investigate generalized prime hyperideals in the framework of Krasner hyperrings. To this end, new classes of hyperideals are introduced and analyzed based on multiplicatively closed properties. In particular, the concepts of s-m-hypersystems, f-hypersystems, and their associated s-prime and f-prime hyperideals are defined and examined. A subset $S\subseteq R$ of a Krasner hyperring is called an s-m-hypersystem if, for every $s\in S$, there exists a multiplicatively closed subset $S^{\ast }\subseteq S$ such that $⟨s⟩\cap S^{\ast }\ne \varnothing$. This concept extends the classical ideal of multiplicative compatibility to the setting of hyperrings. Furthermore, for each element $a\in R$, we define a hyperideal $f\left(a\right)$ satisfying the following conditions: (i) $a\in f\left(a\right)$, (ii) For any hyperideal K, if $x\in f\left(a\right)+K$, then $f\left(x\right)\subseteq f\left(a\right)+K$. Using this notion, a subset $S\subseteq R$ is defined to be an f-hypersystem if there exists a multiplicatively closed subset $S^{\ast }\subseteq S$ such that $f\left(a\right)\cap S^{\ast }\ne \varnothing$ for every $a\in S$. We provide characterizations and original examples of these hypersystems and their corresponding prime hyperideals. The relationships and distinctions between the s-m-hypersystems and f-hypersystems are also explored. Our findings offer a refined perspective on hyperideal theory and open new pathways for the algebraic analysis of hyperstructures.

1. Introduction

Prime ideals are used in a wide range of fields, from theoretical mathematics to applied areas such as cryptography, coding theory, and error correction codes [1]. In ring theory, we encounter many generalizations of prime ideals [2,3]. The concepts of m-systems and s-m-systems play an important role in the characterization of prime and s-prime ideals, respectively. Let R be a ring and $S\subseteq R$. If for every $a,b\in S,a·b\in S$, then the set S is called a multiplicatively closed set. In [4], McCoy generalized the definition of a multiplicatively closed set to give the definition of an m-system. Let R be a ring and $M\subseteq R$. If there exists an $r\in R$ such that for every $c,d\in M$, $c.r.d\in M$, then the set M is called an m-system. Based on this definition, “The primality of an ideal P is equivalent to the set $C\left(P\right)=R-P$ being an m-system”. Later, Van Der Walt defined the s-m-system and used this definition to characterize s-prime ideals [5].

Hypercompositional structures, a new generalization of algebraic structures, first appeared in the work of French mathematician F. Marty in 1934. In his article, the definition of a hyperoperation is given as follows. Let H be a nonempty set, and let $P^{\ast }\left(H\right)$ denote the set of all nonempty subsets of H. A mapping

$$\circ :H\times H⟶P^{\ast }\left(H\right)$$

is called a hyperoperation on H, and the pair $(H,\circ )$ is called a hypergroupoid. Moreover, in [6] for nonempty subsets K and L of H, and for any $h\in H$, the hyperoperation is extended as follows:

$$K\circ L=\bigcup _{k\in K,l\in L}k\circ lK\circ h=K\circ \left\{h\right\}\mathrm{and}\mathrm{h}\circ \mathrm{L}=\left\{\mathrm{h}\right\}\circ \mathrm{L}$$

Additionally, $(H,\circ )$ is a hypergroupoid and If conditions hold

(i)

$x\circ (y\circ z)=(x\circ y)\circ z$, for all $x,y,z\in H$

(ii)

$x\circ H=H\circ x=H$, for all $x\in H$

are satisfied then the set H with the hyperoperation “∘” is called a hypergroup in the sense of Marty.

Generalizations of the classical ring axioms have led to several definitions of hyperrings. General hyperrings, in which both “+” and “·” are hyperoperations, as well as additive hyperrings, in which only “+” is a hyperoperation, have been introduced, along with many other new hypercompositional structures. A special case of additive hyperrings was defined by Krasner [7].

Definition 1. 

Let R be a nonempty set with a hyperoperation “+” and a binary operation “·” such that the following conditions hold:

(1)

$(R,+)$ is a canonical hypergroup, i.e., for all $x,y,z\in R$,

(i)

$(x+y)+z=x+(y+z)$,

(ii)

$x+y=y+x$,

(iii)

If there exists an element $0\in R$ such that $x+0=0+x=\left\{x\right\}$,

(iv)

For every $x\in R$, there exists $x^{\prime }\in R$ such that $0\in x+x^{\prime }$,

(v)

If $z\in x+y$, then $y\in -x+z$ and $x\in z-y$

(2)

$(R,·)$ is a semigroup and for every $x\in R$, $0·x=x·0=0$,

(3)

The multiplication distributes over the hyperaddition from both sides; that is, for all $x,y,z\in R$,

(i)

$x·(y+z)=x·y+x·z$,

(ii)

$(x+y)·z=x·z+y·z$.

then R with hyperoperation “+” and operation “·” is called a Krasner hyperring. The triple $(R,+,·)$ denotes this structure.

Additionally, in a Krasner hyperring $(R,+,·)$, if there exists an element $1_R\in R$ such that for every $x\in R$, $x·1_R=1_R·x=x$ then $(R,+,·)$ is called a unital Krasner hyperring. The development of these structures gained momentum, particularly in the 1970s, with increasing interest in the algebraic properties of hypergroups and hyperrings [8,9,10]. The concept of hyperideals emerged as a generalization of classical ideals within the context of hyperrings and has become a fundamental tool in structural analysis. A foundational element of this progress was laid in 1972 with J. Mittas’s seminal work, “Hypergroupes canoniques” [11], which provided a detailed exploration of the canonical hypergroup, a cornerstone of algebraic hyperstructures. Mittas continued to solidify the theoretical basis of the subject in 1973 with “Sur certaines classes de structures hypercompositionnelles” [12], extending the research into broader classes of these structures. As the focus shifted towards the realization and application of hyperfields, C. Massouros’s 1985 paper, “Methods of constructing hyperfields” [13], contributed a crucial practical dimension by focusing on explicit methods for constructing such structures. This period of early, intensive research was then synthesized by A. Nakassis in 1988 with the publication of “Expository and survey article of recent in hyperring and hyperfield theory” [14]. This survey effectively summarized the key results of the preceding two decades, thereby establishing a robust foundation for subsequent advancements in the field of hypercompositional algebra. From the 1990s onward, various types of hyperideals such as prime, primary, and maximal were defined, facilitating a deeper investigation into the algebraic nature of hyperrings [15,16,17]. In recent years, the study of generalized hyperideals in Krasner hyperrings has led to the introduction of new classes of hyperideals, significantly enriching the field of hypercompositional algebra [2,3,14,18].

In this study, in Section 2, a brief literature review will be provided, including definitions and examples of Krasner hyperrings, hyperideal, and prime hyperideal over a Krasner hyperring. Section 3 will present the concepts of s-m-hypersystem and f-hypersystem, along with the definitions and examples of s-prime hyperideal and f-prime hyperideal given by these concepts. Furthermore, the f-hyperradical of a hyperideal will be defined, and some of its fundamental properties will be examined. Finally, examples and definitions of f-hyperrelated elements with a hyperideal will be provided.

2. Preliminaries

2.1. Krasner Hyperrings and Hyperideals of Krasner Hyperrings

The process of ascertaining whether a particular set of defined operations constitutes a Krasner hyperring can be rather time-consuming. In this subsection, we will introduce structures and examples that have been proven to be hyperrings. These structures and examples will be used throughout our work. For a more thorough exposition of the rudimentary concepts delineated herein, we kindly direct you to refer to [7,16,19].

Theorem 1 

([7]). Let $(R,+,·)$ be a unital ring, and let $G\subseteq R$ be such that G is a multiplicative group. For every $x,y\in R$, define the relation $x\sim y\iff x·G=y·G,$ where “∼” is an equivalence relation. The equivalence classes of this relation are given by $\overline{x}=x·G=\{y\in R:x\sim y\}=\{x·g:g\in G\}.$ The set of all equivalence classes is denoted by $R/G$. For each $\overline{x},\overline{y}\in R/G$, define the operations

$$\begin{array}{cc}\hfill \overline{x}\oplus \overline{y}& =\{\overline{z}:z\in x.g_1+y.g_2,g_1,g_2\in G\},\hfill \\ \hfill \overline{x}\odot \overline{y}& =\overline{x·y}.\hfill \end{array}$$

With these operations, the set $R/G$ forms a hyperring.

Example 1. 

Let $({\mathbb{Z}}_6,+,·)$ be a commutative ring and let $G=\{\overline{1},\overline{5}\}$ be a multiplicative group. Let $\tilde{0}=0·G=\left\{\overline{0}\right\}$, $\tilde{1}=1·G=\{\overline{1},\overline{5}\}$, $\tilde{2}=2·G=\{\overline{2},\overline{4}\}$, and $\tilde{3}=3·G=\left\{\overline{3}\right\}$. Then $H={\mathbb{Z}}_6/G=\{\tilde{0},\tilde{1},\tilde{2},\tilde{3}\}$ is a hyperring with the operations “⊕” and “·” given by the following tables.

⊕ $\tilde{0}$ $\tilde{1}$ $\tilde{2}$ $\tilde{3}$ $\tilde{0}$ $\left\{\tilde{0}\right\}$ $\left\{\tilde{1}\right\}$ $\left\{\tilde{2}\right\}$ $\left\{\tilde{3}\right\}$ $\tilde{1}$ $\left\{\tilde{1}\right\}$ $\{\tilde{0},\tilde{2}\}$ $\{\tilde{1},\tilde{3}\}$ $\left\{\tilde{2}\right\}$ $\tilde{2}$ $\left\{\tilde{2}\right\}$ $\{\tilde{1},\tilde{3}\}$ $\{\tilde{0},\tilde{2}\}$ $\left\{\tilde{1}\right\}$ $\tilde{3}$ $\left\{\tilde{3}\right\}$ $\left\{\tilde{2}\right\}$ $\left\{\tilde{1}\right\}$ $\left\{\tilde{0}\right\}$

· $\tilde{0}$ $\tilde{1}$ $\tilde{2}$ $\tilde{3}$ $\tilde{0}$ $\tilde{0}$ $\tilde{0}$ $\tilde{0}$ $\tilde{0}$ $\tilde{1}$ $\tilde{0}$ $\tilde{1}$ $\tilde{2}$ $\tilde{3}$ $\tilde{2}$ $\tilde{0}$ $\tilde{2}$ $\tilde{2}$ $\tilde{0}$ $\tilde{3}$ $\tilde{0}$ $\tilde{3}$ $\tilde{0}$ $\tilde{3}$

A new Krasner hyperring can be constructed using orthogonal idempotent elements in a ring. Let R be a ring. An element $a\in R$ is called an idempotent element if $a^2=a$. Furthermore, if a and b are nonzero idempotent elements in R such that $a·b=0_R$, then a and b are called orthogonal idempotent elements.

Theorem 2 

([14]). Let $(H,·)$ be a group, let u and v be two orthogonal idempotent elements. Define $R=H\cup \{0,u,v\}$. Then R is a Krasner hyperring w.r.t following operations, for $a,b\in R$,

$$\begin{array}{cc}\hfill a\oplus b=& \left\{\begin{array}{cc}R-\{a,b,0\},\hfill & a\ne b\hfill \\ \left\{a\right\},\hfill & b=0\mathit{and}a\ne 0\hfill \\ \{a,0\},\hfill & b=a\hfill \end{array}\right.\hfill \end{array}\begin{array}{cc}\hfill a·b=& \left\{\begin{array}{cc}u,\hfill & \mathit{if}a=u\mathit{or}b=u\hfill \\ v,\hfill & \mathit{if}a=v\mathit{or}b=v\hfill \\ 0,\hfill & \mathit{if}a=u\mathit{and}b=v\hfill \\ 0,\hfill & \mathit{if}a=v\mathit{and}b=u\hfill \\ 0,\hfill & \mathrm{if}a=0\mathit{or}b=0\hfill \end{array}\right.\hfill \end{array}$$

Example 2. 

Let $({\mathbb{Z}}_{12},+,·)$ be a ring, with $H=U\left({\mathbb{Z}}_{12}\right)=\{\overline{1},\overline{5},\overline{7},\overline{11}\}$ the multiplicative group, and let $\overline{4}$ and $\overline{9}$ be orthogonal idempotent elements. Define $R=H\cup \{\overline{0},\overline{4},\overline{9}\}$. Then the set R, equipped with the operations “⊕” and “·” whose operation tables are given below, forms a Krasner hyperring.

⊕	$\overline{0}$	$\overline{1}$	$\overline{5}$	$\overline{7}$	$\overline{11}$	$\overline{4}$	$\overline{9}$
$\overline{0}$	$\left\{\overline{0}\right\}$	$\left\{\overline{1}\right\}$	$\left\{\overline{5}\right\}$	$\left\{\overline{7}\right\}$	$\left\{\overline{11}\right\}$	$\left\{\overline{4}\right\}$	$\left\{\overline{9}\right\}$
$\overline{1}$	$\left\{\overline{1}\right\}$	$\{\overline{0},\overline{1}\}$	$\{\overline{7},\overline{11},\overline{4},\overline{9}\}$	$\{\overline{5},\overline{11},\overline{4},\overline{9}\}$	$\{\overline{5},\overline{7},\overline{4},\overline{9}\}$	$\{\overline{5},\overline{7},\overline{11},\overline{9}\}$	$\{\overline{5},\overline{7},\overline{11},\overline{4}\}$
$\overline{5}$	$\left\{\overline{5}\right\}$	$\{\overline{7},\overline{11},\overline{4},\overline{9}\}$	$\{\overline{0},\overline{5}\}$	$\{\overline{1},\overline{11},\overline{4},\overline{9}\}$	$\{\overline{1},\overline{7},\overline{4},\overline{9}\}$	$\{\overline{1},\overline{7},\overline{11},\overline{9}\}$	$\{\overline{1},\overline{7},\overline{11},\overline{4}\}$
$\overline{7}$	$\left\{\overline{7}\right\}$	$\{\overline{5},\overline{11},\overline{4},\overline{9}\}$	$\{\overline{1},\overline{5},\overline{4},\overline{9}\}$	$\{\overline{0},\overline{7}\}$	$\{\overline{1},\overline{5},\overline{4},\overline{9}\}$	$\{\overline{1},\overline{5},\overline{11},\overline{9}\}$	$\{\overline{1},\overline{5},\overline{11},\overline{4}\}$
$\overline{11}$	$\left\{\overline{11}\right\}$	$\{\overline{5},\overline{7},\overline{4},\overline{9}\}$	$\{\overline{1},\overline{7},\overline{4},\overline{9}\}$	$\{\overline{1},\overline{5},\overline{4},\overline{9}\}$	$\{\overline{0},\overline{11}\}$	$\{\overline{1},\overline{5},\overline{7},\overline{9}\}$	$\{\overline{1},\overline{5},\overline{7},\overline{4}\}$
$\overline{4}$	$\left\{\overline{4}\right\}$	$\{\overline{5},\overline{7},\overline{11},\overline{9}\}$	$\{\overline{1},\overline{7},\overline{11},\overline{9}\}$	$\{\overline{1},\overline{5},\overline{11},\overline{9}\}$	$\{\overline{1},\overline{5},\overline{7},\overline{9}\}$	$\{\overline{0},\overline{4}\}$	$\{\overline{1},\overline{5},\overline{7},\overline{11}\}$
$\overline{9}$	$\left\{\overline{9}\right\}$	$\{\overline{5},\overline{7},\overline{11},\overline{4}\}$	$\{\overline{1},\overline{7},\overline{11},\overline{4}\}$	$\{\overline{1},\overline{5},\overline{11},\overline{4}\}$	$\{\overline{1},\overline{5},\overline{7},\overline{4}\}$	$\{\overline{1},\overline{5},\overline{7},\overline{11}\}$	$\{\overline{0},\overline{9}\}$

·	$\overline{0}$	$\overline{1}$	$\overline{5}$	$\overline{7}$	$\overline{11}$	$\overline{4}$	$\overline{9}$
$\overline{0}$	$\overline{0}$	$\overline{0}$	$\overline{0}$	$\overline{0}$	$\overline{0}$	$\overline{0}$	$\overline{0}$
$\overline{1}$	$\overline{0}$	$\overline{1}$	$\overline{5}$	$\overline{7}$	$\overline{11}$	$\overline{4}$	$\overline{9}$
$\overline{5}$	$\overline{0}$	$\overline{5}$	$\overline{1}$	$\overline{11}$	$\overline{7}$	$\overline{4}$	$\overline{9}$
$\overline{7}$	$\overline{0}$	$\overline{7}$	$\overline{11}$	$\overline{1}$	$\overline{5}$	$\overline{4}$	$\overline{9}$
$\overline{11}$	$\overline{0}$	$\overline{11}$	$\overline{7}$	$\overline{5}$	$\overline{1}$	$\overline{5}$	$\overline{6}$
$\overline{4}$	$\overline{0}$	$\overline{4}$	$\overline{4}$	$\overline{4}$	$\overline{4}$	$\overline{4}$	$\overline{0}$
$\overline{9}$	$\overline{0}$	$\overline{9}$	$\overline{9}$	$\overline{9}$	$\overline{9}$	$\overline{0}$	$\overline{9}$

Definition 2 

([19]). Let $(R,+,·)$ be a hyperring and let $I\subseteq R$. If

(1)

For all $a,b\in I$, we have $a-b\subseteq I$,

(2)

For all $a\in I$ and $r\in R$, we have $r·a\in I$ ($a·r\in I$),

then I is called a left(right) hyperideal. If I is both a left and a right hyperideal, then it is called a hyperideal.

In [19], let R be a hyperring and let X be a subset of R. The intersection of all hyperideals of R that contain X is called the hyperideal generated by X. This hyperideal is denoted by $\langle X\rangle$. Specifically, let R be a hyperring and $X=\left\{a\right\}\subseteq R$ then the hyperideal

$${\displaystyle <a>=\left\{t:t\in r·a+a·s+n·a+k·(a-a)+\sum _{i=1}^m{r}_i·a·s_i,r,s,r_i,s_i\in R,m\in {\mathbb{Z}}^{+},n,k\in \mathbb{Z}\right\}}$$

is the form. In addition to this point,

(1)

If R has a unit element then

${\displaystyle <a>=\left\{t:t\in k·(a-a)+\sum _{i=1}^m{r}_i·a·s_i,r_i,s_i\in R,m,k\in {\mathbb{Z}}^{+}\right\}}$.

(2)

If R has a unit element and commutative hyperring, then $<a>=R·a=a·R$

$(R,+,·)$ is a hyperring, A and B are hyperideals of R, then the sum and product of the hyperideals A and B defined as follows respectively,

$$A+B=\{x:x\in a+b,forsomea\in A,b\in B\}$$

$$A·B=\{x:x\in \sum _{i=1}^n{a}_i·b_i, a_i\in A,b_i\in B,n\in {\mathbb{Z}}^{+}\}$$

Furthermore, $A+B$ and $A·B$ are hyperideals over hyperring R.

2.2. Prime Hyperideals

Prime hyperideals, which are a generalization of prime ideals, an important building block of ring theory, also play an important role in Krasner hyperring theory. For this reason, prime hyperideals have been widely studied [3,7,18]. In ring theory, P is called a prime ideal of commutative ring R when P is proper ideal and whenever $a,b\in R$ with $a·b\in P$ then either $a\in P$ or $b\in P$ [20]. Ramaruban demonstrated that every prime ideal is a prime hyperideal of a Krasner hyperring [21]. It is, therefore, necessary to commence by recalling the definition of a prime hyperideal of a Krasner hyperring.

Definition 3 

([19]). Let $(R,+,·)$ be a commutative hyperring and B a hyperideal. For every $a,b\in R$, if $a·b\in B$ implies either $a\in B$ or $b\in B$, then B is called a prime hyperideal.

Example 3. 

Let $({\mathbb{Z}}_{10},+,·)$ be a ring, with $H=U\left({\mathbb{Z}}_{10}\right)=\{\overline{1},\overline{3},\overline{7},\overline{9}\}$ the multiplicative group, and let $\overline{5}$ and $\overline{6}$ be orthogonal idempotent elements. Define $R=H\cup \{\overline{0},\overline{5},\overline{6}\}$. Then the set R, equipped with the operations “⊕” and “·” whose operation tables are given below, forms a Krasner hyperring.

⊕	$\overline{0}$	$\overline{1}$	$\overline{3}$	$\overline{7}$	$\overline{9}$	$\overline{5}$	$\overline{6}$
$\overline{0}$	$\left\{\overline{0}\right\}$	$\left\{\overline{1}\right\}$	$\left\{\overline{3}\right\}$	$\left\{\overline{7}\right\}$	$\left\{\overline{9}\right\}$	$\left\{\overline{5}\right\}$	$\left\{\overline{6}\right\}$
$\overline{1}$	$\left\{\overline{1}\right\}$	$\{\overline{1},\overline{0}\}$	$\{\overline{5},\overline{6},\overline{7},\overline{9}\}$	$\{\overline{3},\overline{5},\overline{6},\overline{9}\}$	$\{\overline{3},\overline{5},\overline{6},\overline{7}\}$	$\{\overline{3},\overline{6},\overline{7},\overline{9}\}$	$\{\overline{3},\overline{5},\overline{7},\overline{9}\}$
$\overline{3}$	$\left\{\overline{3}\right\}$	$\{\overline{5},\overline{6},\overline{7},\overline{9}\}$	$\{\overline{0},\overline{3}\}$	$\{\overline{1},\overline{5},\overline{6},\overline{9}\}$	$\{\overline{1},\overline{5},\overline{6},\overline{7}\}$	$\{\overline{1},\overline{6},\overline{7},\overline{9}\}$	$\{\overline{1},\overline{5},\overline{7},\overline{9}\}$
$\overline{7}$	$\left\{\overline{7}\right\}$	$\{\overline{3},\overline{5},\overline{6},\overline{9}\}$	$\{\overline{1},\overline{5},\overline{6},\overline{9}\}$	$\{\overline{0},\overline{7}\}$	$\{\overline{1},\overline{3},\overline{5},\overline{6}\}$	$\{\overline{1},\overline{3},\overline{6},\overline{9}\}$	$\{\overline{1},\overline{3},\overline{5},\overline{9}\}$
$\overline{9}$	$\left\{\overline{9}\right\}$	$\{\overline{3},\overline{5},\overline{6},\overline{7}\}$	$\{\overline{1},\overline{5},\overline{6},\overline{7}\}$	$\{\overline{1},\overline{3},\overline{5},\overline{6}\}$	$\{\overline{0},\overline{9}\}$	$\{\overline{1},\overline{3},\overline{6},\overline{7}\}$	$\{\overline{1},\overline{3},\overline{5},\overline{7}\}$
$\overline{5}$	$\left\{\overline{5}\right\}$	$\{\overline{3},\overline{6},\overline{7},\overline{9}\}$	$\{\overline{1},\overline{6},\overline{7},\overline{9}\}$	$\{\overline{1},\overline{3},\overline{6},\overline{9}\}$	$\{\overline{1},\overline{3},\overline{6},\overline{7}\}$	$\{\overline{0},\overline{5}\}$	$\{\overline{1},\overline{3},\overline{7},\overline{9}\}$
$\overline{6}$	$\left\{\overline{6}\right\}$	$\{\overline{3},\overline{5},\overline{7},\overline{9}\}$	$\{\overline{1},\overline{5},\overline{7},\overline{9}\}$	$\{\overline{1},\overline{3},\overline{5},\overline{9}\}$	$\{\overline{1},\overline{3},\overline{5},\overline{7}\}$	$\{\overline{1},\overline{3},\overline{7},\overline{9}\}$	$\{\overline{0},\overline{6}\}$

·	$\overline{0}$	$\overline{1}$	$\overline{3}$	$\overline{7}$	$\overline{9}$	$\overline{5}$	$\overline{6}$
$\overline{0}$	$\overline{0}$	$\overline{0}$	$\overline{0}$	$\overline{0}$	$\overline{0}$	$\overline{0}$	$\overline{0}$
$\overline{1}$	$\overline{0}$	$\overline{1}$	$\overline{3}$	$\overline{7}$	$\overline{9}$	$\overline{5}$	$\overline{6}$
$\overline{3}$	$\overline{0}$	$\overline{3}$	$\overline{9}$	$\overline{1}$	$\overline{7}$	$\overline{5}$	$\overline{6}$
$\overline{7}$	$\overline{0}$	$\overline{7}$	$\overline{1}$	$\overline{9}$	$\overline{3}$	$\overline{5}$	$\overline{6}$
$\overline{9}$	$\overline{0}$	$\overline{9}$	$\overline{7}$	$\overline{3}$	$\overline{1}$	$\overline{5}$	$\overline{6}$
$\overline{5}$	$\overline{0}$	$\overline{5}$	$\overline{5}$	$\overline{5}$	$\overline{5}$	$\overline{5}$	$\overline{0}$
$\overline{6}$	$\overline{0}$	$\overline{6}$	$\overline{6}$	$\overline{6}$	$\overline{6}$	$\overline{0}$	$\overline{6}$

For the Krasner hyperring R, $\mathtt{I}=\left\{\overline{0}\right\}$, ${\mathtt{I}}_1=\{\overline{0},\overline{5}\}$, ${\mathtt{I}}_2=\{\overline{0},\overline{6}\}$ and R is a hyperideal. ${\mathtt{I}}_1$ and ${\mathtt{I}}_2$ are also prime hyperideals. Because for every $a,b\in R$, $a·b\in {\mathtt{I}}_1$, while $a\in {\mathtt{I}}_1$ or $b\in {\mathtt{I}}_1$, as can be clearly seen from the operation table. Similarly, when the hyperideal ${\mathtt{I}}_2$ is examined, ${\mathtt{I}}_1$ and ${\mathtt{I}}_2$ are prime hyperideals.

Example 4. 

We’ll consider the Krasner hyperring $(H=\{\tilde{0},\tilde{1},\tilde{2},\tilde{3}\},\oplus ,·)$ provided in Example 1 and hyperideal ${\mathtt{A}}_2=\{\overline{0},\overline{2}\}$ of H. For all $a,b\in H$, it can be shown that if $a·b\in {\mathtt{A}}_2$ then either $a\in {\mathtt{A}}_2$ or $b\in {\mathtt{A}}_2$. Thus, it is a prime hyperideal. Similarly, the hyperideal ${\mathtt{A}}_3=\{\overline{0},\overline{3}\}$ is also prime. However, in ${\mathtt{A}}_1=\left\{\overline{0}\right\}$, since $\overline{2}·\overline{3}=\left\{\overline{0}\right\}$ but $\overline{2}\notin {\mathtt{A}}_1$ and $\overline{3}\notin {\mathtt{A}}_1$, ${\mathtt{A}}_1$ is not a prime hyperideal.

Example 5. 

Let $R={\mathbb{Z}}_8$ be a ring and define $\overline{x}=\{x,-x\}$. Then, the set $\overline{R}$ is given by

$$\overline{R}=\left\{\overline{x}=\{x,-x\}\mid x\in R\right\}=\left\{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4}\right\}.$$

⊕	$\overline{0}$	$\overline{1}$	$\overline{2}$	$\overline{3}$	$\overline{4}$
$\overline{0}$	$\left\{\overline{0}\right\}$	$\left\{\overline{1}\right\}$	$\left\{\overline{2}\right\}$	$\left\{\overline{3}\right\}$	$\left\{\overline{4}\right\}$
$\overline{1}$	$\left\{\overline{1}\right\}$	$\{\overline{2},\overline{0}\}$	$\{\overline{3},\overline{1}\}$	$\{\overline{4},\overline{2}\}$	$\left\{\overline{3}\right\}$
$\overline{2}$	$\left\{\overline{2}\right\}$	$\{\overline{3},\overline{1}\}$	$\{\overline{4},\overline{0}\}$	$\{\overline{3},\overline{1}\}$	$\left\{\overline{2}\right\}$
$\overline{3}$	$\left\{\overline{3}\right\}$	$\{\overline{4},\overline{2}\}$	$\{\overline{3},\overline{1}\}$	$\{\overline{2},\overline{0}\}$	$\left\{\overline{1}\right\}$
$\overline{4}$	$\left\{\overline{4}\right\}$	$\left\{\overline{3}\right\}$	$\left\{\overline{2}\right\}$	$\left\{\overline{1}\right\}$	$\left\{\overline{0}\right\}$

·	$\overline{0}$	$\overline{1}$	$\overline{2}$	$\overline{3}$	$\overline{4}$
$\overline{0}$	$\overline{0}$	$\overline{0}$	$\overline{0}$	$\overline{0}$	$\overline{0}$
$\overline{1}$	$\overline{0}$	$\overline{1}$	$\overline{2}$	$\overline{3}$	$\overline{4}$
$\overline{2}$	$\overline{0}$	$\overline{2}$	$\overline{4}$	$\overline{2}$	$\overline{0}$
$\overline{3}$	$\overline{0}$	$\overline{3}$	$\overline{2}$	$\overline{1}$	$\overline{4}$
$\overline{4}$	$\overline{0}$	$\overline{4}$	$\overline{0}$	$\overline{4}$	$\overline{0}$

tables given by $\overline{x}\oplus \overline{y}=\{\overline{x+y},\overline{x-y}\}$ and $\overline{x}·\overline{y}=\left\{\overline{x·y}\right\}$ operations, $(\overline{R},\oplus ,·)$ is a Krasner hyperring. $I=\left\{\overline{0}\right\}$, ${\mathtt{I}}_1=\{\overline{0},\overline{4}\}$, ${\mathtt{I}}_2=\{\overline{0},\overline{2},\overline{4}\}$ and $\overline{R}$ are hyperideals. For every $a,b\in R$, if $a·b\in {\mathtt{I}}_2$, then either $a\in {\mathtt{I}}_2$ or $b\in {\mathtt{I}}_2$, so ${\mathtt{I}}_2$ is a prime hyperideal.

3. Results

In ring theory, m-systems and s-m-systems are important for characterizing prime and s-prime ideals, respectively. Let R be a ring and $S\subseteq R$. Then S is called a multiplicatively closed set if $a·b\in S$ for all $a,b\in S$. In [4], McCoy generalized this definition to introduce the concept of an m-system. Let M be a subset of the ring R, and for every $c,d\in M$, there exists an r in R such that $c·r·d\in M$, then M is called an m-system. From this definition, the primality of an ideal P is equivalent to $C\left(P\right)=R-P$ being an m-system. Later, Van der Walt (in [5]) defined an s-m-system and used the definition to characterize s-prime ideals.

3.1. s-m-Hypersystems and s-Prime Hyperideals

In this section, we will investigate the correspondence of s-m-systems in a hyperring and provide examples.

Definition 4. 

Let R be a Krasner hyperring and S a subset of R. For every $s\in S$, if there exists a multiplicatively closed subset $S^{\ast }$ of S such that $<s>\cap S^{\ast }\ne \varnothing$, then the set S is called an s-m-hypersystem. Here, the set $S^{\ast }$ is called the kernel set of S.

Example 6. 

Let $({\mathbb{Z}}_6,+,·)$ be a ring, with $H=U\left({\mathbb{Z}}_6\right)=\{\overline{1},\overline{5}\}$ the multiplicative group, and let 3 and 4 be orthogonal idempotent elements. Define $R=H\cup \{\overline{0},\overline{3},\overline{4}\}$. Then the set R, equipped with the operations “⊕” and “·” whose operation tables are given below, forms a Krasner hyperring.

⊕ $\overline{0}$ $\overline{1}$ $\overline{3}$ $\overline{4}$ $\overline{5}$ $\overline{0}$ $\left\{\overline{0}\right\}$ $\left\{\overline{1}\right\}$ $\left\{\overline{3}\right\}$ $\left\{\overline{4}\right\}$ $\left\{\overline{5}\right\}$ $\overline{1}$ $\left\{\overline{1}\right\}$ $\{\overline{0},\overline{1}\}$ $\{\overline{4},\overline{5}\}$ $\{\overline{3},\overline{5}\}$ $\{\overline{3},\overline{4}\}$ $\overline{3}$ $\left\{\overline{3}\right\}$ $\{\overline{4},\overline{5}\}$ $\{\overline{0},\overline{3}\}$ $\{\overline{1},\overline{5}\}$ $\{\overline{1},\overline{4}\}$ $\overline{4}$ $\left\{\overline{4}\right\}$ $\{\overline{3},\overline{5}\}$ $\{\overline{1},\overline{5}\}$ $\{\overline{0},\overline{4}\}$ $\{\overline{1},\overline{3}\}$ $\overline{5}$ $\left\{\overline{5}\right\}$ $\{\overline{3},\overline{4}\}$ $\{\overline{1},\overline{4}\}$ $\{\overline{1},\overline{3}\}$ $\{\overline{0},\overline{5}\}$

· $\overline{0}$ $\overline{1}$ $\overline{3}$ $\overline{4}$ $\overline{5}$ $\overline{0}$ $\overline{0}$ $\overline{0}$ $\overline{0}$ $\overline{0}$ $\overline{0}$ $\overline{1}$ $\overline{0}$ $\overline{1}$ $\overline{3}$ $\overline{4}$ $\overline{5}$ $\overline{3}$ $\overline{0}$ $\overline{3}$ $\overline{3}$ $\overline{0}$ $\overline{3}$ $\overline{4}$ $\overline{0}$ $\overline{4}$ $\overline{0}$ $\overline{4}$ $\overline{4}$ $\overline{5}$ $\overline{0}$ $\overline{5}$ $\overline{3}$ $\overline{4}$ $\overline{1}$

• For the given R Krasner hyperring, let $S^{\ast }=\left\{\overline{3}\right\}$ be multiplicatively closed subset of $S=\{\overline{1},\overline{3},\overline{5}\}$.
• For $\overline{3}\in S$, $<\overline{3}>=\{\overline{t}\mid \overline{t}\in \overline{3·r},\overline{r}\in {\mathbb{Z}}_6\}=\{\overline{0},\overline{3}\}$ and $<\overline{3}>\cap S^{\ast }\ne \varnothing$
• For $s=\overline{1},\overline{5}\in S$, $<s>=R$ and $<s>\cap S^{\ast }\ne \varnothing$ Therefore, S is an s-m-hypersystem.

Example 7. 

Consider the hyperring $(R=\{\overline{0},\overline{1},\overline{5},\overline{7},\overline{11},\overline{4},\overline{9}\},\oplus ,·)$ given in Example 2 and $S=\{\overline{1},\overline{5},\overline{7},\overline{11},\overline{4}\}\subset R$. Let $S^{\ast }=\{\overline{1},\overline{5},\overline{4}\}$ be a multiplicatively closed subset of S. For $\overline{4}\in S$, $<\overline{4}>=\{\overline{0},\overline{4}\}$ and $<\overline{4}>\cap S^{\ast }\ne \varnothing$ and for all $x\in S-\left\{4\right\}$, $<x>=R$ and $<x>\cap S^{\ast }\ne \varnothing$, hence S is an s-m-hypersystem and the set $S^{\ast }$ is the core of S.

Example 8. 

Consider the hyperring $(R=\{\overline{0},\overline{1},\overline{3},\overline{4},\overline{5}\},\oplus ,·)$ as defined in Example 6, let us take the subset $S=\{\overline{1},\overline{3},\overline{4}\}$ of R. The multiplicatively closed subsets of the set S are: $S_1^{\ast }=\left\{\overline{1}\right\}$, $S_2^{\ast }=\left\{\overline{3}\right\}$, $S_3^{\ast }=\left\{\overline{4}\right\}$, $S_4^{\ast }=\{\overline{1},\overline{3}\}$ and $S_5^{\ast }=\{\overline{1},\overline{4}\}$.

• For $\overline{3}\in S$, $<\overline{3}>=\{\overline{0},\overline{3}\}$ then $S_1^{\ast }\cap <\overline{3}>=\varnothing$, $S_3^{\ast }\cap <\overline{3}>=\varnothing$ and $S_5^{\ast }\cap <\overline{3}>=\varnothing$.
• For $\overline{4}\in S$, $<\overline{4}>=\{\overline{0},\overline{4}\}$ then $S_2^{\ast }\cap <\overline{4}>=\varnothing$ and $S_4^{\ast }\cap <\overline{4}>=\varnothing$. Thus, since there is no multiplicatively closed subset $S^{\ast }$ such that $<s>\cap S^{\ast }\ne \varnothing$ for every $s\in S$, S is not an s-m-hypersystem.

Definition 5. 

Let R be a Krasner hyperring and P be a hyperideal of R. If the set $C\left(P\right)=R-P$ is an s-m-hypersystem, then P is called an s-prime hyperideal.

Example 9. 

Take the Krasner hyperring $R=\{\overline{0},\overline{1},\overline{3},\overline{4},\overline{5}\}$ in Example 6 and ${\mathtt{I}}_1=\{\overline{0},\overline{3}\}$ hyperideal of R, we have $C\left({\mathtt{I}}_1\right)=R-{\mathtt{I}}_1=\{\overline{1},\overline{4},\overline{5}\}$. Let $S^{\ast }=\{\overline{1},\overline{4}\}$ be a multiplicatively closed subset of $C\left({\mathtt{I}}_1\right)$. For $x=\overline{1},\overline{5}\in C\left({\mathtt{I}}_1\right)$, $<x>=R$, then $<x>\cap S^{\ast }\ne \varnothing$ and for $\overline{4}\in C\left({\mathtt{I}}_1\right)$, $<\overline{4}>=\{\overline{0},\overline{4}\}$ and $<\overline{4}>\cap S^{\ast }\ne \varnothing$. thus $C\left({\mathtt{I}}_1\right)$ is an s-m-hypersystem. By definition, ${\mathtt{I}}_1$ is an s-prime hyperideal of R.

Example 10. 

Let us consider the Krasner hyperring $(R,\oplus ,·)$ constructed with the set $R=\{\overline{0},\overline{1},\overline{3},\overline{4},\overline{5}\}$, as in Example 6. For the ideal $I=\left\{\overline{0}\right\}$, we have $C\left(I\right)=R\setminus I=\{\overline{1},\overline{3},\overline{4},\overline{5}\}.$ The multiplicatively closed subsets of $C\left(I\right)$ can be as follows $S_1^{\ast }=\left\{\overline{1}\right\},S_2^{\ast }=\{\overline{1},\overline{3}\},S_3^{\ast }=\{\overline{1},\overline{4}\},S_4^{\ast }=\{\overline{1},\overline{5}\},S_5^{\ast }=\{\overline{1},\overline{3},\overline{5}\},S_6^{\ast }=\{\overline{1},\overline{4},\overline{5}\},S_7^{\ast }=\left\{\overline{3}\right\},S_8^{\ast }=\left\{\overline{4}\right\}.$ Now consider the following:

• For $\overline{3}\in C\left(I\right),\langle \overline{3}\rangle =\{\overline{0},\overline{3}\}$, thus $\langle \overline{3}\rangle \cap S_1^{\ast }=\varnothing ,\langle \overline{3}\rangle \cap S_3^{\ast }=\varnothing ,\langle \overline{3}\rangle \cap S_4^{\ast }=\varnothing ,\langle \overline{3}\rangle \cap S_6^{\ast }=\varnothing$ and $\langle \overline{3}\rangle \cap S_8^{\ast }=\varnothing .$
• For $\overline{4}\in C\left(I\right),\langle \overline{4}\rangle =\{\overline{0},\overline{4}\},\langle \overline{4}\rangle \cap S_2^{\ast }=\varnothing ,\langle \overline{4}\rangle \cap S_5^{\ast }=\varnothing$ and $\langle \overline{4}\rangle \cap S_7^{\ast }=\varnothing$. Therefore, $C\left(I\right)$ is not an $s\text{-}m$-hypersystem. Since $C\left(I\right)$ is not an $s\text{-}m$-hypersystem, it follows that I is not an s-prime hyperideal.

Definition 6. 

Let R be a Krasner hyperring, and let $S\subseteq R$ be a multiplicatively closed subset. Suppose that P is a hyperideal of R such that $P\cap S=\varnothing$. If for all $a,b\in R$, the condition $a·b\in P$ implies that there exists an element $s\in S$ such that either $s·a\in P$ or $s·b\in P$, then P is called an S-prime hyperideal.

Example 11. 

In the Krasner hyperring $R=\{\overline{0},\overline{1},\overline{5},\overline{7},\overline{11},\overline{4},\overline{9}\}$ given in Example 2, let $P=\left\{\overline{0}\right\}$ be a hyperideal of R. Consider the subset $S=\{\overline{1},\overline{5},\overline{7},\overline{11},\overline{9}\}\subseteq R$, which is multiplicatively closed. We observe that $P\cap S=\varnothing$. For all $a,b\in R$, if $a·b\in P$, then it must be that $a·b=\overline{0}$. If either $a=\overline{0}$ or $b=\overline{0}$, then for every $s\in S$, we have $s·a\in P$ or $s·b\in P$. In all other cases, according to the multiplication table, $a·b\in P$ holds only when either $(a=\overline{4}\mathrm{and}b=\overline{9})$ or $(a=\overline{9}\mathrm{and}b=\overline{4})$. In such cases, since $\overline{9}\in S$, we have either $\overline{9}·a\in P$ or $\overline{9}·b\in P$. Therefore, P is an S-prime hyperideal.

Remark 1. 

When we take the hyperideal $P=\left\{\overline{0}\right\}$ of the Krasner hyperring $R=\{\overline{0},\overline{1},\overline{5},\overline{7},\overline{11},\overline{4},\overline{9}\}$ given in Example 11, $\overline{4}·\overline{9}=\overline{0}\in P$ but $\overline{4}\notin P$ and $\overline{9}\notin P$. Therefore, P is not a prime hyperideal. This provides an example of an S-prime hyperideal which is not a prime hyperideal.

Example 12. 

In the Krasner hyperring $R=\{\overline{0},\overline{1},\overline{3},\overline{7},\overline{9},\overline{5},\overline{6}\}$ given in Example 3, let $P=\left\{\overline{0}\right\}$ be a hyperideal of R. Consider the subset $S=\{\overline{1},\overline{9}\}\subseteq R$, which is multiplicatively closed. For $\overline{5},\overline{6}\in R$, we have $\overline{5}·\overline{6}\in P$. However, the following hold $\overline{1}·\overline{5}=\overline{5}\notin P,\overline{1}·\overline{6}=\overline{6}\notin P,\overline{9}·\overline{5}=\overline{5}\notin P,\overline{9}·\overline{6}=\overline{6}\notin P.$ Therefore, P is not an S-prime hyperideal.

3.2. f-hypersystems and f-prime Hyperideals of Krasner Hyperrings

In this section, we present the generalization of the definitions of f-systems and f-prime ideals introduced to the setting of Krasner hyperrings by Murata et al. [22]. Throughout this section, let R be a commutative and unital Krasner hyperring. For an arbitrary element $a\in R$, the hyperideal determined singly by a and satisfying the following conditions will be denoted by $f\left(a\right)$.

(i)

$a\in f\left(a\right)$.

(ii)

For any hyperideal K, if $x\in f\left(a\right)+K$, then $f\left(x\right)\subseteq f\left(a\right)+K$.

Let us provide examples of the ideal $f\left(a\right)$ satisfying the above conditions. Let R be a Krasner hyperring and $\beta \in R$. For an arbitrary element $a\in R$, the hyperideal

$$f\left(a\right)=\langle a,\beta \rangle =\{t:t\in r_1·a+r_2·\beta ,r_1,r_2\in R\}$$

is an example of such a hyperideal because

(1)

it follows that $a\in f\left(a\right)$ since $1_R·a+0·\beta =\left\{a\right\}$.

(2)

Given any $x\in f\left(a\right)+K$ and since $f\left(a\right)+K={\bigcup }_i\{a_i+k_i\mid a_i\in f\left(a\right),k_i\in K\},$ then there exist $a_i\in f\left(a\right)$ and $k_i\in K$ such that $x=a_i+k_i$. Therefore,

$$f\left(x\right)=\langle x,\beta \rangle =\{t\mid t\in r_1·x+r_2·\beta \}\subseteq f\left(a\right)+K.$$

Consequently, the hyperideal defined by $f\left(a\right)=\langle a,\beta \rangle$ satisfies conditions (i) and (ii). Furthermore, it is clear that the conditions are satisfied when $f\left(a\right)=\langle a\rangle$.

Definition 7. 

Let R be a Krasner hyperring and let $S\subseteq R$. If there exists a multiplicatively closed subset $S^{\ast }\subseteq S$ such that for every $a\in S$, $f\left(a\right)\cap S^{\ast }\ne \varnothing ,$ then S is called an f-hypersystem.

Example 13. 

Consider the Krasner hyperring $(R,\oplus ,·)$ constructed with the set $R=\{\overline{0},\overline{1},\overline{5},\overline{7},\overline{11},\overline{4},\overline{9}\}$ as in Example 2. Let $S=\{\overline{1},\overline{7},\overline{9}\}$ be a subset of R. Consider the multiplicatively closed subset $S^{\ast }=\{\overline{1},\overline{9}\}\subseteq S$, and define for every $a\in S$ the hyperideal $f\left(a\right)=\langle a,\overline{9}\rangle =\{x:x\in a·r+\overline{9}·y,r,y\in R\}$. Then we have:

• If $a=\overline{9}\in S$, then $f\left(\overline{9}\right)\cap S^{\ast }=\{\overline{0},\overline{9}\}\cap \{\overline{1},\overline{9}\}=\left\{\overline{9}\right\}$.
• If $a=\overline{1},\overline{11}\in S$, then $f\left(a\right)\cap S^{\ast }=R\cap \{\overline{1},\overline{9}\}=\{\overline{1},\overline{9}\}$.
• Hence, since for every $a\in S$, $f\left(a\right)\cap S^{\ast }\ne \varnothing$, the set S is an f-hypersystem.

From the definition of the f-hypersystem, if $f\left(s\right)=\langle s\rangle$, then every s-m-hypersystem becomes an f-hypersystem, i.e., under this assumption, $f\left(s\right)$ coincides with the smallest hypersystem generated by s and therefore all closure and structural conditions required in the definition of an s-m-hypersystem naturally satisfy the conditions prescribed for an f-hypersystem. However, not every f-hypersystem is necessarily an s-m-hypersystem. The following Example 14 illustrates this.

Example 14. 

Consider the Krasner hyperring $(\overline{R},\oplus ,·)$ constructed with the set $\overline{R}=\{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4}\}$ as in Example 5. Let $S=\{\overline{1},\overline{2},\overline{4}\}$ be a subset of $\overline{R}$ and the multiplicatively closed subset $S^{\ast }=\left\{\overline{1}\right\}\subseteq S$. We have $\langle \overline{2}\rangle =\{\overline{0},\overline{2},\overline{4}\}$ and $\langle \overline{2}\rangle \cap S^{\ast }=\varnothing$. Therefore, S is not an s-m-hypersystem. However, if $f\left(s\right)=\langle s,\overline{3}\rangle$ then $\langle s,\overline{3}\rangle =\overline{R}$ for all $s\in S$, hence $\langle s,\overline{3}\rangle \cap S^{\ast }\ne \varnothing$, thus S is an f-hypersystem.

Example 15. 

Consider the Krasner hyperring $(R,\oplus ,·)$ constructed with the set $R=\{\overline{0},\overline{1},\overline{3},\overline{4},\overline{5}\}$ as in Example 6. Let $S=\{\overline{1},\overline{3},\overline{5}\}\subseteq R$ be a subset. Assume $f\left(a\right)=\langle a\rangle$ and consider the multiplicatively closed subset $S^{\ast }=\left\{\overline{3}\right\}$.

• For $a=\overline{1}$, we have $f\left(\overline{1}\right)=\langle \overline{1}\rangle =R$, so $f\left(\overline{1}\right)\cap S^{\ast }\ne \varnothing$.
• For $a=\overline{3}$, we have $f\left(\overline{3}\right)=\langle \overline{3}\rangle =\{\overline{0},\overline{3}\}$, so $f\left(\overline{3}\right)\cap S^{\ast }\ne \varnothing$.
• For $a=\overline{5}$, we have $f\left(\overline{5}\right)=\langle \overline{5}\rangle =R$, so $f\left(\overline{5}\right)\cap S^{\ast }\ne \varnothing$

Therefore, S is an f-hypersystem.

Example 16. 

Consider the Krasner hyperring $(\overline{R},\oplus ,·)$ constructed with the set $\overline{R}=\{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4}\}$ as in Example 5. Let $S=\{\overline{1},\overline{2},\overline{4}\}$ be a subset of $\overline{R}$. Assume $f\left(a\right)=\langle a\rangle$ and consider the multiplicatively closed subset $S^{\ast }=\left\{\overline{1}\right\}\subseteq S$. For $a=\overline{4}$, we have $\langle \overline{4}\rangle =\{\overline{0},\overline{4}\}$ and $\langle \overline{4}\rangle \cap S^{\ast }=\varnothing$. Therefore, S is not an f-hypersystem.

Definition 8. 

Let P be a hyperideal in the hyperring R. If $C\left(P\right)=R\setminus P$ is an f-hypersystem, then P is called an f-prime hyperideal.

Example 17. 

Consider the Krasner hyperring $(R,\oplus ,·)$ constructed with the set $R=\{\overline{0},\overline{1},\overline{3},\overline{4},\overline{5}\}$ as in Example 6. Let $P=\{\overline{0},\overline{3}\}$ be a prime hyperideal. Then, $C\left(P\right)=R\setminus P=\{\overline{1},\overline{4},\overline{5}\}$. Consider the multiplicatively closed subset $S^{\ast }=\left\{\overline{4}\right\}\subseteq C\left(P\right)$ and for each $a\in C\left(P\right)$ define $f\left(a\right)=\langle a+\overline{1}\rangle$.

Then for all $a\in C\left(P\right)$:

• For $a=\overline{1}$, we have $\overline{1}+\overline{1}=\{\overline{0},\overline{1}\}$, and

  $$f\left(\overline{1}\right)=\langle \{\overline{0},\overline{1}\}\rangle =\{t:t\in r_1·\overline{0}+r_2·\overline{1},r_1,r_2\in R\}=R,$$

  thus $f\left(\overline{1}\right)\cap S^{\ast }\ne \varnothing$.
• For $a=\overline{4}$, we have $\overline{1}+\overline{4}=\{\overline{3},\overline{5}\}$, and

  $$f\left(\overline{4}\right)=\langle \{\overline{3},\overline{5}\}\rangle =\{t:t\in r_1·\overline{3}+r_2·\overline{5},r_1,r_2\in R\}=R,$$

  thus $f\left(\overline{4}\right)\cap S^{\ast }\ne \varnothing$.
• For $a=\overline{5}$, we have $\overline{1}+\overline{5}=\{\overline{3},\overline{4}\}$, and

  $$f\left(\overline{5}\right)=\langle \{\overline{3},\overline{4}\}\rangle =\{t:t\in r_1·\overline{3}+r_2·\overline{4},r_1,r_2\in R\}=R,$$

  thus $f\left(\overline{5}\right)\cap S^{\ast }\ne \varnothing$.

Therefore, $C\left(P\right)$ is an f-hypersystem, and consequently, P is an f-prime hyperideal.

Example 18. 

Consider the Krasner hyperring $(R,\oplus ,·)$ constructed with the set $R=\{\overline{0},\overline{1},\overline{3},\overline{7},\overline{9},\overline{5},\overline{6}\}$ as in Example 3. Let $P=\{\overline{0},\overline{5}\}$ be a prime hyperideal. Then, $C\left(P\right)=R\setminus P=\{\overline{1},\overline{3},\overline{7},\overline{9},\overline{6}\}$. Consider the multiplicatively closed subset $S^{\ast }=\{\overline{1},\overline{6}\}\subseteq C\left(P\right)$, and for each $a\in C\left(P\right)$ define

$$f\left(a\right)=\langle a,\overline{6}\rangle =\{x:x\in a·r+\overline{6}·y,r,y\in R\}.$$

Then:

If $a=\overline{6}$, then $f\left(\overline{6}\right)=\langle \overline{6},\overline{6}\rangle =\{\overline{0},\overline{6}\}$. Hence,

$$f\left(\overline{6}\right)\cap S^{\ast }=\{\overline{0},\overline{6}\}\cap \{\overline{1},\overline{6}\}\ne \varnothing .$$

If $a\ne \overline{6}$, then $f\left(a\right)=\langle a,\overline{6}\rangle =R$. Therefore,

$$f\left(a\right)\cap S^{\ast }=R\cap \{\overline{1},\overline{6}\}\ne \varnothing .$$

Consequently, for every $a\in C\left(P\right)$, there exists a multiplicatively closed subset $S^{\ast }$ such that $f\left(a\right)\cap S^{\ast }\ne \varnothing$. Thus, $C\left(P\right)$ is an f-hypersystem, and hence P is an f-prime hyperideal.

Example 19. 

Consider the Krasner hyperring $(\overline{R},\oplus ,·)$ constructed from the set $\overline{R}=\{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4}\}$ as in Example 5. Let $I=\left\{\overline{0}\right\}$ be a hyperideal of $\overline{R}$. Then, $C\left(I\right)=\overline{R}\setminus I=\{\overline{1},\overline{2},\overline{3},\overline{4}\}.$ Define $f\left(a\right)=\langle a\rangle$ for each $a\in C\left(I\right)$, and consider the multiplicatively closed subsets $S_1^{\ast }=\left\{\overline{1}\right\}\mathrm{and}{S}_2^{\ast }=\{\overline{1},\overline{3}\}.$ Then: For $a=\overline{2}$, we have $\langle \overline{2}\rangle =\{\overline{0},\overline{2},\overline{4}\},\mathrm{and}\langle \overline{2}\rangle \cap S_1^{\ast }=\varnothing .$

• For $a=\overline{4}$, we have $\langle \overline{4}\rangle =\{\overline{0},\overline{4}\},\mathrm{and}\langle \overline{4}\rangle \cap S_2^{\ast }=\varnothing .$

Therefore $C\left(I\right)$ is not an f-hypersystem then it is not an f-prime hyperideal.

Remark 2. 

If on a Krasner hyperring R, for every $a\in R$, we define $f\left(a\right)=\langle a\rangle$, then prime hyperideals are also f-prime hyperideals. However, when $f\left(a\right)\ne \langle a\rangle$, f-prime hyperideals are not necessarily prime hyperideals.

Example 20. 

Consider the Krasner hyperring $(\overline{R},\oplus ,·)$ constructed from the set $\overline{R}=\{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4}\}$ as in Example 5. Take the hyperideal $I=\left\{\overline{0}\right\}$. Then, $C\left(I\right)=\{\overline{1},\overline{2},\overline{3},\overline{4}\}$. For $f\left(\overline{a}\right)=\langle \overline{a},\overline{3}\rangle$ and the multiplicatively closed subset $S^{\ast }=\{\overline{1},\overline{3}\}\subseteq C\left(I\right)$, we have:

• For $a=\overline{1}$, $\langle \overline{1},\overline{3}\rangle =\overline{R}$ and $\langle \overline{1},\overline{3}\rangle \cap S^{\ast }\ne \varnothing$.
• For $a=\overline{2}$, $\langle \overline{2},\overline{3}\rangle =\overline{R}$ and $\langle \overline{2},\overline{3}\rangle \cap S^{\ast }\ne \varnothing$.
• For $a=\overline{3}$, $\langle \overline{3},\overline{3}\rangle =\overline{R}$ and $\langle \overline{3},\overline{3}\rangle \cap S^{\ast }\ne \varnothing$.
• For $a=\overline{4}$, $\langle \overline{4},\overline{3}\rangle =\overline{R}$ and $\langle \overline{4},\overline{3}\rangle \cap S^{\ast }\ne \varnothing$.

Therefore, $C\left(I\right)$ is an f-hypersystem and consequently I is an f-prime hyperideal. However, since $\overline{2}·\overline{4}=\overline{0}\in I$ but $\overline{2}\notin I$ and $\overline{4}\notin I$, the hyperideal I is not a prime hyperideal. This is an example showing that f-prime hyperideals are not necessarily prime hyperideals.

Lemma 1. 

Let $S\left(S^{\ast }\right)$ be an f-hypersystem in R, and let K be a hyperideal of R that does not intersect S. Then there exists a maximal hyperideal J containing K among all hyperideals that are disjoint from S. Moreover, this hyperideal J is necessarily an f-prime hyperideal.

Proof. 

If $S=\varnothing$, the proof is immediate. Hence, we may assume that $S\ne \varnothing$. The existence of J follows from Zorn’s lemma. We now show that $C\left(J\right)$ forms an f-hypersystem with core $S^{\ast }+J$.

For any $k\in C\left(J\right)$, the maximality of J implies that $f\left(k\right)+J$ contains an element of S. Therefore, we can choose $s^{\ast }\in S^{\ast }$ such that $f\left(s\right)\subseteq f\left(k\right)+J$. Let $j^{\ast }\in J$ and $k^{\ast }\in f\left(k\right)$; since $s^{\ast }\in k^{\ast }+j^{\ast }$, it follows that $k^{\ast }\in s^{\ast }-j^{\ast }$. Consequently, $k^{\ast }\in S^{\ast }+J$, which completes the proof. □

Theorem 3. 

Let K be a hyperideal of R. Then K is contained in every maximal f-prime hyperideal belonging to it.

Proof. 

Let J be a maximal f-prime hyperideal belonging to K. It suffices to show that $K+J$ is f-hyperrelated. We define the set $K+J$ as:

$$K+J=\{x\mid x\in k+j,k\in K,j\in J\}$$

Since $k+j\in K+f\left(j\right)$ for every $k\in K$ and $j\in J$, we obtain the set inclusion $f(k+j)\subseteq K+f\left(j\right)$. Consequently, for every $x^{\prime }\in f(k+j)$, there exist $k^{″}\in K$ and $j^{″}\in f\left(j\right)$ such that $x^{\prime }\in k^{″}+j^{″}$. Since $k^{″}\in K$, and K is assumed to be a hyperideal, we have $k^{″}·c\in K$. As shown, $j^{″}·c\in K$. Thus, we obtain the containment:

$$x^{\prime }·c\in k^{″}·c+j^{″}·c\subseteq K+K=K$$

Therefore, for any $x^{\prime }\in f(k+j)$ and $c\notin K$, we have $x^{\prime }·c\in K$. This confirms that $K+J$ satisfies the condition for being f-hyperrelated. The proof is completed. □

Definition 9. 

In a Krasner hyperring, the f-hyperradical of a hyperideal A of R is defined as the set of all elements $a\in R$ such that every S, an f-hypersystem containing a, intersects A. In other words,

$$r_f\left(A\right)=\{a\in R\mid a\in S,S\cap A\ne \varnothing \}.$$

Example 21. 

Consider the Krasner hyperring $(R,\oplus ,·)$ constructed from the set $R=\{\overline{0},\overline{1},\overline{3},\overline{7},\overline{9},\overline{5},\overline{6}\}$ as in Example 3. Let $A=\{\overline{0},\overline{5}\}$ be a hyperideal and define $f\left(a\right)=\langle a,\overline{6}\rangle$. We aim to find the f-hyperradical of A.

For $S=\{\overline{0},\overline{5}\}$, take the multiplicatively closed subset $S^{\ast }=\left\{\overline{0}\right\}$. Then $f\left(\overline{0}\right)=\{\overline{0},\overline{6}\}$ and since $\{\overline{0},\overline{6}\}\cap S^{\ast }\ne \varnothing$, S is an f-hypersystem. Because $S\cap A\ne \varnothing$, for every f-hypersystem containing S and $\overline{0}$, we have $S\cap A\ne \varnothing$, thus $\overline{0}\in r_f\left(A\right)$.

Also, $f\left(\overline{5}\right)=R$ and since $R\cap S^{\ast }\ne \varnothing$, S is an f-hypersystem. Since $S\cap A\ne \varnothing$, all f-hypersystems containing S and $\overline{5}$ satisfy $S\cap A\ne \varnothing$, so $\overline{5}\in r_f\left(A\right)$.

Consider $S=\{\overline{1},\overline{3},\overline{7},\overline{9},\overline{6}\}$ with multiplicatively closed subset $S^{\ast }=\{\overline{1},\overline{6}\}$. For $a\ne \overline{6}$, $f\left(a\right)=R$ and $R\cap S^{\ast }\ne \varnothing$, and for $a=\overline{6}$, $f\left(a\right)=\{\overline{0},\overline{6}\}$ satisfies $\{\overline{0},\overline{6}\}\cap S^{\ast }\ne \varnothing$, hence S is an f-hypersystem. Since $S\cap A=\varnothing$, we have $\overline{1},\overline{3},\overline{7},\overline{9},\overline{6}\notin r_f\left(A\right)$.

Therefore,

$$r_f\left(A\right)=\{a\in R\mid a\in S\mathit{where}S\mathit{is}\mathit{an}f\text{-}\mathit{hypersystem}\mathit{with}S\cap A\ne \varnothing \}=\{\overline{0},\overline{5}\}.$$

Theorem 4. 

The f-hyperradical of a hyperideal A is the intersection of all f-prime hyperideals containing A.

$$r_f\left(A\right)=\bigcap _{\begin{array}{c}P\mathit{is}\mathit{an}f\text{-}\mathit{prime}\mathit{hyperideal}\\ A\subseteq P\end{array}}P$$

Proof. 

If P is an f-prime hyperideal containing A, then we show that A is contained in P. Suppose that A is not contained in P. Then there exists an element $x\in A$ such that $x\notin P$. Since P is an f-hypersystem, it follows that $C\left(P\right)\cap A\ne \varnothing$. However, this contradicts the fact that A is contained in P. Therefore, $r_f\left(A\right)$ lies in the intersection of all f-prime hyperideals containing A.

Conversely, suppose that a is an element of R but $a\notin r_f\left(A\right)$. Then there exists an f-hypersystem $S\left(S^{\ast }\right)$ containing a and disjoint from A. By Lemma 1, there exists an f-prime hyperideal P containing A but disjoint from S. Hence, P does not contain a and so a is not in the intersection of all f-prime hyperideals containing A. □

Example 22. 

For the f-hyperradical in Example 21, the only f-prime hyperideal containing the hyperideal A is A itself. Thus, $r_f\left(A\right)=A=\{\overline{0},\overline{5}\}$ is an f-prime hyperideal.

Theorem 5. 

The f-hyperradical of a hyperideal is also a hyperideal.

Proof. 

Let $S\left(S^{\ast }\right)$ be an f-hypersystem in R and let A be an ideal disjoint from S. By Zorn’s Lemma, there exists a maximal f-hypersystem $S_1^{\ast }$ containing $S^{\ast }$ and disjoint from A.

Consider the set $S_1=\{x\in R\mid f\left(x\right)\cap S_1^{\ast }\ne \varnothing \}\cap C\left(A\right).$ Then, $S_1$ is an f-hypersystem with core $S_1^{\ast }$ and is disjoint from A. By Lemma 1, there exists an f-prime hyperideal P containing A and disjoint from $S_1^{\ast }$. The set $C\left(P\right)$ is an f-hypersystem with core $S_1^{\ast }+P$, and the maximality of $S_1^{\ast }$ implies $S_1^{\ast }+P=S_1^{\ast }.$ Therefore, by definition, we have $C\left(P\right)=S_1.$ □

Theorem 6. 

Let R be a Krasner hyperring and let A and B be two distinct f-hyperideals of R. Then the following propositions hold.

(i)

$r_f\left(A\right)=r_f\left(r_f\left(A\right)\right)$

(ii)

$r_f\left(A+B\right)\subseteq r_f\left(A\right)+r_f\left(B\right)$

(iii)

$r_f\left(A\cap B\right)\subseteq r_f\left(A\right)+r_f\left(B\right)$

(iv)

$r_f\left(A·B\right)\subseteq r_f\left(A\right)+r_f\left(B\right)$

Proof. 

(i)

$$r_f\left(A\right)=\{x\in R\mid x\in S,S\cap A\ne \varnothing \}$$

$$r_f\left(A\right)=\bigcap _{A\subseteq P}P\Rightarrow A\subseteq P\Rightarrow A\subseteq r_f\left(A\right)\Rightarrow r_f\left(A\right)\subseteq r_f\left(r_f\left(A\right)\right)$$

If $x\in r_f\left(A\right)$ then there exists an S such that $x\in S$ and $S\cap A\ne \varnothing$.

Assume that $x\in r_f(r_f\left(A\right))$ and $x\notin r_f\left(A\right)$. Then $x\notin S\mathrm{or}\mathrm{S}\cap \mathrm{A}=\varnothing$. Since S is an f-hypersystem and $S\cap A\ne \varnothing$, it follows that $x\notin S$ is impossible. Hence $x\in r_f\left(A\right)$.

$$\Rightarrow r_f\left(r_f\left(A\right)\right)\subseteq r_f\left(A\right)$$

Therefore $r_f\left(A\right)=r_f\left(r_f\left(A\right)\right)$.

(ii)

Let $x\in r_f\left(A+B\right)$. By definition, there exists an f-hypersystem S such that $x\in S\mathrm{and}\mathrm{S}\cap (\mathrm{A}+\mathrm{B})\ne \varnothing$. Hence $S\cap (A+B)\ne \varnothing$ implies that for some $a\in A$ and $b\in B$ we have $a+b\in S$. Define $(S-b)=\{t\in R\mid t+b\in S\}$. By our assumption $(S-b)$ is also an f-hypersystem. Moreover, $a=(a+b)-b\in (S-b),$ so $(S-b)\cap A\ne \varnothing$. Since $x\in S$ we have $x-b\in (S-b),$ and because $(S-b)\cap A\ne \varnothing$ it follows that $x-b\in r_f\left(A\right)$. Similarly, by considering $(S-a)$ one obtains $b\in r_f\left(B\right)$. Therefore $x=(x-b)+b,$ with $x-b\in r_f\left(A\right)$ and $b\in r_f\left(B\right)$, so $x\in r_f\left(A\right)+r_f\left(B\right)$. Thus, for every $x\in r_f\left(A+B\right)$ we have $x\in r_f\left(A\right)+r_f\left(B\right)$; hence $r_f\left(A+B\right)\subseteq r_f\left(A\right)+r_f\left(B\right).$

(iii)

Let $x\in \left(rA\cap B\right)$. By definition, $x\in S\mathrm{and}\mathrm{S}\cap (\mathrm{A}\cap \mathrm{B})\ne \varnothing$. Since $S\cap A\ne \varnothing$ and $S\cap B\ne \varnothing$, $x\in r_f\left(A\right)$ and $x\in r_f\left(B\right)$. It can be demonstrated that, in view of the definition of $r_f\left(A\right)+r_f\left(B\right)$ as $r_f\left(A\right)+r_f\left(B\right)=\{y\mid y\in a+b,a\in r_f\left(A\right),b\in r_f\left(B\right)\}$ and $x\in r_f\left(A\right)$, it can be deduced that x is an element of $r_f\left(A\right)+r_f\left(B\right)$. Hence $r_f\left(A\cap B\right)\subseteq r_f\left(A\right)+r_f\left(B\right)$.

(iv)

Since $r_f\left(A·B\right)=\bigcap _{A·B\subseteq P}P$, $A·B\subseteq A$ and $A·B\subseteq B$, thus $x\in r_f\left(A·B\right)\Rightarrow x\in r_f\left(A\right)$ and $x\in r_f\left(B\right).$ By definition of $r_f\left(A\right)+r_f\left(B\right)$, $r_f\left(A·B\right)\subseteq r_f\left(A\right)+r_f\left(B\right).$

□

A notable point to emphasize in the forthcoming example is that, in Theorem 6, it cannot be guaranteed that the converses of conditions (ii), (iii), and (iv) are necessarily true.

Example 23. 

Consider hypering R in Example 3 and $f\left(a\right)=<a>$. Let $A=\{\overline{0},\overline{5}\}$ and $B=\{\overline{0},\overline{6}\}$ hyperideals of R. Then $r_f\left(A\right)+r_f\left(B\right)=\{\overline{0},\overline{5}\}+\{\overline{0},\overline{6}\}=R$.

• Clearly, $r_f\left(A·B\right)=\left\{\overline{0}\right\}$ then $r_f\left(A\right)+r_f\left(B\right)⊈r_f\left(A·B\right)$.
• Since $r_f\left(A\cap B\right)=\left\{\overline{0}\right\}$ then $r_f\left(A\right)+r_f\left(B\right)⊈r_f\left(A\cap B\right)$
• If $r_f\left(A+B\right)=\varnothing$ then $r_f\left(A\right)+r_f\left(B\right)⊈r_f\left(A+B\right)$

3.3. Left f-related Elements with a Hyperideal

In this section, the definition of f-hyperrelated elements, which establishes the relationship between an element of the hyperring and a hyperideal using f-hypersystems, will be given, and examples will be provided.

Definition 10. 

Let R be a Krasner hyperring, A a hyperideal of R, and for each $a\in R$, let $f\left(a\right)$ be a hyperideal in R. If for every $a^{\prime }\in f\left(a\right)$ there exists $c\in C\left(A\right)$ such that $a^{\prime }·c\in A$ (respectively, $c·a^{\prime }\in A$), then a is called a left (right) f-hyperrelated element to A.

Example 24. 

For the Krasner hyperring $(R,\oplus ,·)$ constructed from the set $R=\{\overline{0},\overline{1},\overline{5},\overline{7},\overline{11},\overline{4},\overline{9}\}$ as in Example 2, consider the prime hyperideal $P=\{\overline{0},\overline{9}\}$ and thus $C\left(P\right)=\{\overline{1},\overline{5},\overline{7},\overline{11},\overline{4}\}$. We define

$$f\left(a\right)=\langle a,\overline{9}\rangle =\{x:x\in a·r+\overline{9}·y,forallr,y\in R\}.$$

• For $a=\overline{0}$, $f\left(\overline{0}\right)=\{\overline{0},\overline{9}\}$. Since for $\overline{4}\in C\left(P\right)$, $\overline{0}·\overline{4}=\overline{0}\in P$ and $\overline{9}·\overline{4}=\overline{0}\in P$, it follows that $\overline{0}$ is left f-hyperrelated to P.
• For $a=\overline{9}$, $f\left(\overline{9}\right)=\{\overline{0},\overline{9}\}$. Since for $\overline{4}\in C\left(P\right)$, $\overline{0}·\overline{4}=\overline{0}\in P$ and $\overline{9}·\overline{4}=\overline{0}\in P$, it follows that $\overline{9}$ is left f-hyperrelated to P.

Example 25. 

For the Krasner hyperring $(R,\oplus ,·)$ constructed from the set $R=\{\overline{0},\overline{1},\overline{3},\overline{7},\overline{9},\overline{5},\overline{6}\}$ as in Example 3, consider the prime hyperideal $P=\{\overline{0},\overline{5}\}$. Then, $C\left(P\right)=\{\overline{1},\overline{3},\overline{7},\overline{9},\overline{6}\}$. Define

$$f\left(a\right)=\langle a,\overline{6}\rangle =\{x:x\in a·r+\overline{6}·y,r,y\in R\}.$$

• If $a\ne \overline{6}$ and $a\ne \overline{0}$, then $f\left(a\right)=R$; if $a=\overline{6}$ or $a=\overline{0}$, then $f\left(a\right)=\{\overline{0},\overline{6}\}$. Now, let us find the elements that are not left f-hyperrelated to the hyperideal P:
• For $a=\overline{1}$: $f\left(\overline{1}\right)=R$. Since $\overline{6}\in f\left(\overline{1}\right)$, and $\overline{6}·\overline{0}=\overline{0}\in P$ or $\overline{6}·\overline{5}=\overline{0}\in P$, but $\overline{0}\notin C\left(P\right)$ and $\overline{5}\notin C\left(P\right)$, $\overline{1}$ is not left f-hyperrelated to P.
• For $a=\overline{3}$: $f\left(\overline{3}\right)=R$. Since $\overline{6}\in f\left(\overline{3}\right)$, and $\overline{6}·\overline{0}=\overline{0}\in P$ or $\overline{6}·\overline{5}=\overline{0}\in P$, but $\overline{0}\notin C\left(P\right)$ and $\overline{5}\notin C\left(P\right)$, $\overline{3}$ is not left f-hyperrelated to P.
• For $a=\overline{6}$: $f\left(\overline{6}\right)=\{\overline{0},\overline{6}\}$. Since $\overline{6}\in f\left(\overline{6}\right)$, and $\overline{6}·\overline{0}=\overline{0}\in P$ or $\overline{6}·\overline{5}=\overline{0}\in P$, but $\overline{0}\notin C\left(P\right)$ and $\overline{5}\notin C\left(P\right)$, $\overline{6}$ is not left f-hyperrelated to P.
• For $a=\overline{7}$: $f\left(\overline{7}\right)=R$. Since $\overline{6}\in f\left(\overline{7}\right)$, and $\overline{6}·\overline{0}=\overline{0}\in P$ or $\overline{6}·\overline{5}=\overline{0}\in P$, but $\overline{0}\notin C\left(P\right)$ and $\overline{5}\notin C\left(P\right)$, $\overline{7}$ is not left f-hyperrelated to P.
• For $a=\overline{9}$: $f\left(\overline{9}\right)=R$. Since $\overline{6}\in f\left(\overline{9}\right)$, and $\overline{6}·\overline{0}=\overline{0}\in P$ or $\overline{6}·\overline{5}=\overline{0}\in P$, but $\overline{0}\notin C\left(P\right)$ and $\overline{5}\notin C\left(P\right)$, $\overline{9}$ is not left f-hyperrelated to P.

Therefore, the elements $\overline{1},\overline{3},\overline{6},\overline{7}$, and $\overline{9}$ are not left f-hyperrelated to the hyperideal P.

Definition 11. 

Let A and B be hyperideals of R. If for every $b\in B$, b is left f-hyperrelated to A, then B is called a left f-hyperrelated hyperideal to A.

Example 26. 

In Example 3, consider the Krasner hyperring $(R,\oplus ,·)$ constructed from the set $R=\{\overline{0},\overline{1},\overline{3},\overline{7},\overline{9},\overline{5},\overline{6}\}$. Let $P=\{\overline{0},\overline{6}\}$ be a prime hyperideal. Then $C\left(P\right)=\{\overline{1},\overline{3},\overline{7},\overline{9},\overline{5}\}$. Define

$$f\left(a\right)=\langle a,\overline{6}\rangle =\{x:x\in a·r+\overline{6}·y,r,y\in R\}.$$

• If $a\ne \overline{6}$ and $a\ne \overline{0}$, then $\langle a,\overline{6}\rangle =R$.
• If $a=\overline{0}$ or $a=\overline{6}$, then $\langle a,\overline{6}\rangle =\{\overline{0},\overline{6}\}$.
• When $f\left(a\right)$ is defined as above, let us find the left f-hyperrelated and non-left f-hyperrelated hyperideals of R with respect to P:
• For the hyperideal $I=\left\{\overline{0}\right\}$ and $\overline{0}\in I$, since $f\left(\overline{0}\right)=\{\overline{0},\overline{6}\}$, and $\overline{0}·\overline{5}=\overline{0}\in P$ and $\overline{6}·\overline{5}=\overline{0}\in P$, and $\overline{5}\in C\left(P\right)$, it follows that I is left f-hyperrelated to P.
• For the hyperideal $P=\{\overline{0},\overline{6}\}$, similarly, $\overline{0},\overline{6}\in P$ are left f-hyperrelated to P since $f\left(\overline{0}\right)=f\left(\overline{6}\right)=\{\overline{0},\overline{6}\}$ and for $\overline{5}\in C\left(P\right)$, $\overline{0}·\overline{5}=\overline{0}\in P$ and $\overline{6}·\overline{5}=\overline{0}\in P$. Thus, every $a\in P$ is left f-hyperrelated to P, and therefore P is left f-hyperrelated hyperideal to itself.
• For the hyperideal ${\mathtt{I}}_1=\{\overline{0},\overline{5}\}$ and $a=\overline{5}$, $f\left(\overline{5}\right)=R$. For $\overline{5}\in f\left(\overline{5}\right)$, $\overline{5}·\overline{0}=\overline{0}\in P$ or $\overline{5}·\overline{6}=\overline{0}\in P$, but $\overline{0}\notin C\left(P\right)$ and $\overline{6}\notin C\left(P\right)$, so $\overline{5}$ is not left f-hyperrelated to P. Hence, ${\mathtt{I}}_1$ is a hyperideal not left f-hyperrelated to P.
• For the whole hyperring R and $a=\overline{3}$, $f\left(\overline{3}\right)=R$. For $\overline{5}\in f\left(\overline{3}\right)$, $\overline{5}·\overline{0}=\overline{0}\in P$ or $\overline{5}·\overline{6}=\overline{0}\in P$, but again $\overline{0}\notin C\left(P\right)$ and $\overline{6}\notin C\left(P\right)$, so $\overline{3}$ is not left f-hyperrelated to P. Therefore, R is a hyperideal not left f-hyperrelated to P.

Proposition 1. 

Let R be a commutative Krasner hyperring. Then the following conditions are equivalent.

(i)

Every hyperideal K is f-hyperrelated to itself.

(ii)

The zero hyperideal is f-hyperrelated to every hyperideal of R

Proof. 

Assume that 0 is f-hyperrelated to hyperideal K of R. Since $k\in K+f\left(0\right)$, it follows that $f\left(k\right)\subseteq K+f\left(0\right)$. For any $k^{\prime }\in f\left(k\right)$, we may write $k^{\prime }\in k^{″}+m^{″}$ with $k^{″}\in K$ and $m^{″}\in f\left(0\right)$. Because 0 is f-hyperrelated to K, we have $k^{″}\in K$, and hence $k^{\prime }\in K$. Thus, $f\left(k\right)\subseteq K$. For an arbitrary $c\in R$, we have $k^{\prime }·c\in k^{″}·c+m^{″}·c$, which implies that K is f-hyperrelated to itself.

Conversely, every hyperideal K is f-hyperrelated to itself. Since K is a hyperideal, $0\in K$. As K is f-hyperrelated to itself, it follows by definition that 0 is f-hyperrelated to K. □

Definition 12. 

A maximal hyperideal in a class of hyperideals is said to be a maximal f-prime hyperideal if it is associated with a hyperideal J that is f-hyperrelated to it.

Proposition 2. 

Let K be a hyperideal of R. Then the set S of all elements of R that are not f-hyperrelated to K forms an f-hypersystem.

Proof. 

For each $k\in S$, consider $f\left(k\right)$. For any $c\notin K$, let $k^{\prime }·c\notin K$. The set $S^{\ast }$ consisting of all such $k^{\prime }$ is closed under multiplication. Therefore, S together with its kernel $S^{\ast }$ constitutes an f-hypersystem. □

Theorem 7. 

Let K be a hyperideal of R. Then every f-hyperrelated element to K and every f-hyperrelated hyperideal to K are contained in a maximal f-prime hyperideal belonging to K.

Proof. 

Clearly, an element k is f-hyperrelated to K if and only if $f\left(k\right)$ is f-hyperrelated to K. Therefore, it suffices to prove the statement for an f-hyperrelated hyperideal.

Let L be f-hyperrelated hyperideal to K, and let S be the f-hypersystem consisting of all elements of R that are not f-hyperrelated to K. Then $L\cap S=\varnothing$. Thus, by Lemma 1, L is contained in a maximal f-prime hyperideal belonging to K. □

Proposition 3. 

Let K be a hyperideal of R. Then the f-hyperradical $r_f\left(K\right)$ of K is f-hyperrelated to K.

Proof. 

Let S denote the set of all elements that are not f-hyperrelated to K. If $r_f\left(K\right)$ contains an element that is not f-hyperrelated to K, then by the definition of the f-hyperradical, we must have $S\cap K\ne \varnothing$. Hence, for any hyperideal K of R, the f-hyperradical $r_f\left(K\right)$ is f-hyperrelated to K. □

4. Conclusions

In this study, a range of structures associated with Krasner hyperrings and prime hyperideals were examined through the generalization of multiplicatively closed subsets in Krasner hyperrings. Consequently, this investigation yielded several original concepts and findings. The concept of an s–m-hypersystem was introduced, accompanied by illustrative examples. Secondly, the concept of an s-prime hyperideal was defined, and representative examples were provided. Thirdly, the notions of f-hypersystem and f-prime hyperideal were proposed, accompanied by novel examples. Notably, an example of an f-prime hyperideal that is not a prime hyperideal was constructed. Furthermore, the notion of an f-hyperradical was proposed, and it was demonstrated to be a hyperideal, accompanied by the presentation of an illustrative example. The connection between the briefly defined prime hyperideals was found to be as follows:

In conclusion, the notions of left f-hyperrelated element and left f-hyperrelated hyperideal were defined. Their interrelations were investigated, and original examples were provided to illustrate these new concepts. These findings contribute to the structural understanding of Krasner hyperrings and pave the way for further exploration in hyperideal theory.