A Study on Special Kinds of Derivations in Ordered Hyperrings

Abstract

In this study, we concentrate on an important class of ordered hyperstructures with symmetrical hyperoperations, which are called ordered Krasner hyperrings, and discuss strong derivations and homo-derivations. Additionally, we apply our results on nonzero proper hyperideals to the study of derivations of prime ordered hyperrings. This work is a pioneer in studies on structures such as hyperideals and homomorphisms of an ordered hyperring with the help of derivation notation. Finally, we prove some results on 2-torsion-free prime ordered hyperrings by using derivations. We show that if d is a derivation of 2-torsion-free prime hyperring R and the commutator set $[l,d(q\left)\right]$ is equal to zero for all q in R, then $l\in Z\left(R\right)$. Moreover, we prove that if the commutator set $\left(d\right(l),q)$ is equal to zero for all l in R, then $\left(d\right(R),q)=0$.

1. Introduction

The notion of Krasner hyperring goes back to Kasner’s paper [1] in 1983. Krasner hyperrings are generalizations of rings in which the addition is multivalued. Several books on hyperstructure theory have been published [2,3]. A multiring [4] is a ring with multivalued addition. The study of hypergroups was first undertaken by Marty (1934) in [5].

The study of ordered semihypergroups was started in [6,7,8,9] when the authors investigated constructions of this kind of hyperstructures using pseudo-orders and regular relations. The systematic study and comprehensive presentation of ordered semi-hypergroups began with the work of Heidari and Davvaz in [6]. Moreover, in [7], Davvaz et al. investigated pseudo-orders and strongly regular relations. Later, regular relations of ordered semi-hyper-groups appeared in the work of Gu and Tang [8]. Recently, Tang et al. [9] worked on weak pseudo-orders. Later on, Qiang et al. [10] defined the concept of the weak pseudo-orders in ordered semi-hyperrings and investigated quotient ordered semi-hyperrings.

The theory of derivation is one of the most fascinating research areas in rings and hyperrings. The first step in this direction was taken by Posner [11] in 1957 and Asokkumar [12] in 2013. Despite the short time since 2013, this topic has attracted considerable attention from many authors. Differential hyperrings introduced by Kamali Ardekani and Davvaz (2015) in [13] became one of the most interesting notions in the study of hyperrings. Rao et al. [14] generalized the notion of derivations on ordered semihyperrings. The concept of m-k-hyper-ideals in ordered semi-hyperrings was defined in [14]. Kou et al. continued the work on derivations on ordered semi-hyperrings [15] and various constructions of ordered hyper-structures are found in [15,16]. The study of semi-derivations in hyperrings began with the work of Yilmaz and Yazarli in [17]. The concept of derivations is used in coding theory [18,19].

Derivations on hyperrings can tell us about the structure of the hyperrings. We formally define the derivation of hyperrings, similar to what we define in number theory. This feature is based on the characteristic of the derivative of the Leibniz product rule of functions, which is defined based on the hyper-operation of hyper-addition and multiplication; that is, two hyper-operations are needed. Based on the derivation of the hyperrings, we can define some features of hyperrings.

Derivations can be useful in determining whether a ring is commutative [20,21]. Clearly, a ring R is commutative if and only if every inner derivation $d_a\left(x\right)=[a,x]=a·x-x·a$ on it is zero. In [22], Chvalina and Chvalinov$\acute{a}$ gave a construction of hyper-structures determined by quasi-orders defined by means of derivation operators on differential rings. They introduced a hyper-operation ∗ on a differential ring R such that $(R,\ast )$ is a hyper-group. In [14], Rao et al. proved that if d is a homo-derivation of an ordered semihyperring R, then $Ker\left(d\right)$ is a k-hyper-ideal of R. Hyperrings with derivations has been studied by Kamali Ardekani and Davvaz in [13], especially the relationships between derivations and the structure of hyperrings. Motivated by the above works on derivations, we study hyper-ideals of an ordered hyperring with the help of derivations.

Definition 1

([2]). Let ${\mathcal{P}}^{*}\left(R\right)$ be the family of all non-empty subsets of $R\ne \varnothing$. A mapping $\oplus :R\times R\to {\mathcal{P}}^{*}\left(R\right)$ is called a (binary) hyper-operation on R. A hyper-operation ⊕ is called symmetric if $x\oplus y=y\oplus x$ for all $x,y\in R$. If $\varnothing \ne J,K\subseteq R$ and $l\in R$, then

$$J\oplus K={\displaystyle \bigcup _{\begin{array}{c}\genfrac{}{}{0pt}{}{j\in J}{k\in K}\end{array}}}j\oplus k, l\oplus J=\left\{l\right\}\oplus J and K\oplus l=K\oplus \left\{l\right\}.$$

Now, $(R,\oplus )$ is called a semihyper-group if for every $j,k,l$ in R,

$$j\oplus (k\oplus l)=(j\oplus k)\oplus l.$$

A semihyper-group $(R,\oplus )$ is called a hyper-group if for every $l\in R$, $l\oplus R=R\oplus l=R$.

Indeed, hyper-groups characterize the symmetry of hyper-operations.

Definition 2.

A tuple $(R,\oplus ,\odot )$ is a Krasner hyperring [1,3] if

(1)

$(R,\oplus )$ is a canonical hyper-group [23], i.e., for all $x,y,z$ in R, (i) $x\oplus (y\oplus z)=(x\oplus y)\oplus z$, (ii) $x\oplus y=y\oplus x$, (iii) there exists $0\in R$ such that $0\oplus x=\left\{x\right\}$, (iv) for all x in R, there exists a unique element $x^{\prime }$ in R, such that $0\in x\oplus x^{\prime }$ (we shall write $\ominus x\in R$ for $x^{\prime }$, and we call it the inverse of x), (v) $(R,\oplus )$ is reversible, i.e., $z\in x\oplus y$ implies that $y\in \ominus x\oplus z$ and $x\in z\ominus y$;

(2)

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

(3)

The multiplication ⊙ is distributive with respect to the hyper-addition ⊕.

For all $a,b\in R$, we have

$$\ominus (\ominus a)=a,$$

$$\ominus (a\oplus b)=\ominus a\ominus b,$$

$$a\oplus R=R.$$

In addition, we have $\ominus A=\{\ominus a\mid a\in A\}$.

For all $a,b$ in a hyperring R, we will denote the commutator sets

(i)

$[a,b]=a\odot b\ominus b\odot a$;

(ii)

$(a,b)=a\odot b\oplus b\odot a$.

This study tries to consider the derivations based on homomorphisms and investigates some results based on $[a,b]$ and $(a,b)$. In this connection, several properties of (i) and (ii) are investigated. In this research, we succeeded in studying the concepts of strong derivations and homo-derivations of ordered hyperrings. In addition, the concepts of $\left[d\right(p),q]$ and $\left(d\right(p),q)$ are defined, and some results for these concepts with their complete proofs are investigated.

2. Special Kinds of Derivations in Ordered Hyperrings

Let us review some definitions and concepts on (ordered) hyperrings from [3,15,24], which we apply in what follows.

Definition 3.

We let R and ≤ be the Krasner hyperring and the (partial) order (reflexive, anti-symmetric, and transitive) relation, respectively. An ordered hyperring $(R,\oplus ,\odot ,\le )$ is compatible with ⊕ and ⊙, i.e., for all $l,l^{\prime },q$ in R,

$$l\le l^{\prime }\Rightarrow \left\{\begin{array}{c}l\oplus q⪯l^{\prime }\oplus q,\hfill \\ \\ q\oplus l⪯q\oplus l^{\prime },\hfill \\ \\ l\odot q\le l^{\prime }\odot q,\hfill & 0\le q\hfill \\ \\ q\odot l\le q\odot l^{\prime },\hfill & 0\le q\hfill \end{array}\right.$$

Here, $L⪯L^{\prime }$ means for any $l\in L$ there exists $l^{\prime }\in L^{\prime }$ such that $l\le l^{\prime }$.

If $\varnothing \ne Q\subseteq R$, then

$$\left(Q\right]:=\{l\in R|l\le q,\mathrm{for}\mathrm{some}q\in Q\}.$$

We denote the center of R by

$$Z\left(R\right)=\{q\in R|q\odot l=l\odot q,\mathrm{for}\mathrm{all}l\in R\}.$$

Definition 4.

$\varnothing \ne Q\subseteq R$ is a hyper-ideal of R if

(1)

$(Q,\oplus )$ is a canonical subhyper-group;

(2)

For every $q\in Q$ and $l\in R$, $l\odot q,q\odot l\in Q$;

(3)

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

We continue this section by mentioning that R is a prime ordered hyperring if

$$0=u\odot q\odot v, \mathrm{for} \mathrm{all}q \mathrm{in} R\Rightarrow u=0 \mathrm{or} v=0.$$

In addition, R is a 2-torsion-free ordered hyperring if

$$0\in q\oplus q, \mathrm{for} \mathrm{all}q \mathrm{in} R\Rightarrow q=0.$$

In the following, we clarify the concept of strong derivations as an expansion of derivations by some examples.

Definition 5.

A derivation of an ordered hyperring $(R,\oplus ,\odot ,\le )$ is a function $d:R\to R$ such that for all $l,l^{\prime }$ in R,

(1)

$d(l\oplus l^{\prime })\subseteq d\left(l\right)\oplus d\left(l^{\prime }\right)$;

(2)

$d(l\odot l^{\prime })\in d\left(l\right)\odot l^{\prime }\oplus l\odot d\left(l^{\prime }\right)$;

(3)

$l\le l^{\prime }\Rightarrow d\left(l\right)\le d\left(l^{\prime }\right)$.

In Definition 5 (1), if the equality holds, then d is called a strong derivation. In addition, a strong derivation d is a homo-derivation if

$$d(l\odot l^{\prime })=d\left(l\right)\odot d\left(l^{\prime }\right).$$

Example 1.

Let $R=\{0,1,2\}$ and

$$\begin{array}{|cccc|}\hline \oplus & 0& 1& 2\\ 0& 0& 1& 2\\ 1& 1& R& 1\\ 2& 2& 1& \{0,2\}\\ \hline\end{array}$$

$$\begin{array}{|cccc|}\hline \odot & 0& 1& 2\\ 0& 0& 0& 0\\ 1& 0& 1& 2\\ 2& 0& 2& 0\\ \hline\end{array}$$

$$\begin{array}{c}\le :=\left\{\right(0,0),(0,1),(0,2),(1,1),(2,1),(2,2\left)\right\}.\hfill \end{array}$$

Then, $(R,\oplus ,\odot ,\le )$ is an ordered hyperring with two symmetrical hyper-operations ⊕ and ⊙. Obviously, $d:R\to R$ defined by

$$d\left(l\right)=\left\{\begin{array}{cc}0,\hfill & l=0\hfill \\ \\ 2,\hfill & l=1,2\hfill \end{array}\right.$$

is a derivation. Indeed:

$$d(1\oplus 1)=d\left(R\right)=\{0,2\}=2\oplus 2=d\left(1\right)\oplus d\left(1\right),$$

and

$$d(1\odot 1)=d\left(1\right)=2\in \{0,2\}=2\oplus 2=2\odot 1\oplus 1\odot 2=d\left(1\right)\odot 1\oplus 1\odot d\left(1\right).$$

In addition,

$$d(1\oplus 2)=d\left(1\right)=2\in \{0,2\}=2\oplus 2=d\left(1\right)\oplus d\left(2\right),$$

$$d(1\odot 2)=d\left(2\right)=2=0\oplus 2=2\odot 2+1\odot 2=d\left(1\right)\odot 2\oplus 1\odot d\left(2\right),$$

$$d(2\oplus 2)=d\left(\right\{0,2\left\}\right)=\{0,2\}=2\oplus 2=d\left(2\right)\oplus d\left(2\right),$$

$$d(2\odot 2)=d\left(0\right)=0=0\oplus 0=2\odot 2\oplus 2\odot 2=d\left(2\right)\odot 2\oplus 2\odot d\left(2\right),$$

and for $l,l^{\prime }\in R$,

$$l\le l^{\prime }\Rightarrow d\left(l\right)\le d\left(l^{\prime }\right).$$

Therefore, our claim is true.

Example 2.

Assume that $R=\{0,1,p,q\}$ and define

$$\begin{array}{|ccccc|}\hline \oplus & 0& 1& p& q\\ 0& 0& 1& p& q\\ 1& 1& R& \{1,p,q\}& \{1,p,q\}\\ p& p& \{1,p,q\}& R& \{1,p,q\}\\ q& q& \{1,p,q\}& \{1,p,q\}& R\\ \hline\end{array}$$

$$\begin{array}{|ccccc|}\hline \odot & 0& 1& p& q\\ 0& 0& 0& 0& 0\\ 1& 0& 1& p& q\\ p& 0& p& q& 1\\ q& 0& q& 1& p\\ \hline\end{array}$$

$$\begin{array}{cc}\le \hfill & =\left\{\right(0,0),(0,1),(0,p),(0,q),(1,1),(1,p),(p,p),\hfill \\ & (p,q),(q,q),(q,1)\}.\hfill \end{array}$$

Then, $(R,\oplus ,\odot ,\le )$ is an ordered hyperring with two symmetrical hyper-operations ⊕ and ⊙. Obviously, $d:R\to R$ defined by

$$d\left(l\right)=\left\{\begin{array}{cc}0,\hfill & l=0\hfill \\ \\ p,\hfill & l=1\hfill \\ \\ q,\hfill & l=p\hfill \\ \\ 1,\hfill & l=q\hfill \end{array}\right.$$

is a strong derivation. Indeed:

$$d(1\oplus 1)=d\left(R\right)=R=p\oplus p=d\left(1\right)\oplus d\left(1\right),$$

$$d(1\odot 1)=d\left(1\right)=p\in p\oplus p=p\odot 1\oplus 1\odot p=d\left(1\right)\odot 1\oplus 1\odot d\left(1\right),$$

$$d(1\oplus p)=d\left(\right\{1,p,q\left\}\right)=\{1,p,q\}=p\oplus q=d\left(1\right)\oplus d\left(p\right),$$

$$d(1\odot p)=d\left(p\right)=q\in q\oplus q=(p\odot p)\oplus (1\odot q)=\left(d\right(1)\odot p)\oplus (1\odot d(p\left)\right),$$

$$d(1\oplus q)=d\left(\right\{1,p,q\left\}\right)=\{1,p,q\}=p\oplus 1=d\left(1\right)\oplus d\left(q\right),$$

$$d(1\odot q)=d\left(q\right)=1\in 1\oplus 1=R=(p\odot q)\oplus (1\odot 1)=\left(d\right(1)\odot q)\oplus (1\odot d(q\left)\right),$$

$$d(p\oplus p)=d\left(R\right)=R=q\oplus q=d\left(p\right)\oplus d\left(p\right),$$

$$d(p\odot p)=d\left(q\right)=1\in 1\oplus 1=(q\odot p)\oplus (p\odot q)=\left(d\right(p)\odot p)\oplus (p\odot d(p\left)\right),$$

$$d(p\oplus q)=d\left(\right\{1,p,q\left\}\right)=\{1,p,q\}=q\oplus 1=d\left(p\right)\oplus d\left(q\right),$$

$$d(p\odot q)=d\left(1\right)=p\in p\oplus p=(q\odot q)\oplus (p\odot 1)=\left(d\right(p)\odot q)\oplus (p\odot d(q\left)\right),$$

$$d(q\oplus q)=d\left(R\right)=R=1\oplus 1=d\left(q\right)\oplus d\left(q\right),$$

$$d(q\odot q)=d\left(p\right)=q\in q\oplus q=(1\odot q)\oplus (q\odot 1)=\left(d\right(q)\odot q)\oplus (q\odot d(q),$$

and for all $l,l^{\prime }\in R$,

$$l\le l^{\prime }\Rightarrow d\left(l\right)\le d\left(l^{\prime }\right).$$

Therefore, our claim is true.

Example 3.

In Example 2, $(R,\oplus ,\odot ,\le )$, where

$$\begin{array}{c}\le :=\left\{\right(0,0),(0,1),(0,p),(0,q),(1,1),(p,p),(q,q\left)\right\},\hfill \end{array}$$

is an ordered hyperring. Obviously, $d^{\prime }:R\to R$ defined by

$$d^{\prime }\left(l\right)=\left\{\begin{array}{cc}0,\hfill & l=0\hfill \\ \\ q,\hfill & l=1,p\hfill \\ \\ 1,\hfill & l=q\hfill \end{array}\right.$$

is a derivation of R, but it is not strong, since

$$d^{\prime }(1\oplus q)=\{1,q\}\ne \{1,p,q\}=q\oplus 1=d^{\prime }\left(1\right)\oplus d^{\prime }\left(q\right).$$

Example 4.

Let $(R,+,·,\le )$ be an ordered hyperring and

$$M\left(R\right)=\left\{\left(\begin{array}{cc}m& n\\ 0& 0\end{array}\right)\mid m,n\in R.\right\}$$

Define $\left(M\right(R),\oplus ,\odot )$ by the following hyper-operations:

$$\left(\begin{array}{cc}{a}_{11}& a_{12}\\ 0& 0\end{array}\right)\oplus \left(\begin{array}{cc}{b}_{11}& b_{12}\\ 0& 0\end{array}\right)=\left\{\left(\begin{array}{cc}u& v\\ 0& 0\end{array}\right)\mid u\in a_{11}+b_{11},v\in a_{12}+b_{12}\right\},$$

$$\left(\begin{array}{cc}{a}_{11}& a_{12}\\ 0& 0\end{array}\right)\odot \left(\begin{array}{cc}{b}_{11}& b_{12}\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}{a}_{11}{b}_{11}& a_{11}{b}_{12}\\ 0& 0\end{array}\right).$$

We define $U⪯V$ if and only if $u_{ij}\le v_{ij}$. Then, $\left(M\right(R),\oplus ,\odot ,⪯)$ is an ordered hyperring. Obviously, the function $d:M\left(R\right)\to M\left(R\right)$ defined by

$$d\left(\left(\begin{array}{cc}{a}_{11}& a_{12}\\ 0& 0\end{array}\right)\right)=\left(\begin{array}{cc}0& a_{12}\\ 0& 0\end{array}\right)$$

is not a homo-derivation of $M\left(R\right)$. Indeed:

$$\begin{array}{cc}d\left(\left(\begin{array}{cc}{a}_{11}& a_{12}\\ 0& 0\end{array}\right)\odot \left(\begin{array}{cc}{b}_{11}& b_{12}\\ 0& 0\end{array}\right)\right)\hfill & =d\left(\left(\begin{array}{cc}{a}_{11}·b_{11}& a_{11}·b_{12}\\ 0& 0\end{array}\right)\right)\hfill \\ \\ & =\left(\begin{array}{cc}0& a_{11}·b_{12}\\ 0& 0\end{array}\right).\hfill \end{array}$$

On the other hand,

$$\begin{array}{cc}d\left(\begin{array}{cc}{a}_{11}& a_{12}\\ 0& 0\end{array}\right)\odot d\left(\begin{array}{cc}{b}_{11}& b_{12}\\ 0& 0\end{array}\right)\hfill & =\left(\begin{array}{cc}0& a_{12}\\ 0& 0\end{array}\right)\odot \left(\begin{array}{cc}0& b_{12}\\ 0& 0\end{array}\right)\hfill \\ \\ & =\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right).\hfill \end{array}$$

So, $d(U\odot V)\ne d\left(U\right)\odot d\left(V\right)$, in general.

Proposition 1.

Consider $d:R\to R$ as the homo-derivation. The function $\mathrm{\Theta }:M\left(R\right)\to M\left(R\right)$ defined by $\mathrm{\Theta }\left(A\right)=\left(\begin{array}{cc}d\left(a_{11}\right)& d\left(a_{12}\right)\\ 0& 0\end{array}\right)$ is a homo-derivation of $M\left(R\right)$.

Proof. 

Clearly, $\mathrm{\Theta }$ is a strong derivation of $M\left(R\right)$. Let $U=\left(\begin{array}{cc}{u}_{11}& u_{12}\\ 0& 0\end{array}\right)$ and $V=\left(\begin{array}{cc}{v}_{11}& v_{12}\\ 0& 0\end{array}\right)$. Then,

$$\begin{array}{cc}\mathrm{\Theta }(U\odot V)\hfill & =\mathrm{\Theta }\left(\left(\begin{array}{cc}{u}_{11}& u_{12}\\ 0& 0\end{array}\right)\odot \left(\begin{array}{cc}{v}_{11}& v_{12}\\ 0& 0\end{array}\right)\right)\hfill \\ \\ & =\mathrm{\Theta }\left(\left(\begin{array}{cc}{u}_{11}·v_{11}& u_{11}·v_{12}\\ 0& 0\end{array}\right)\right)\hfill \\ \\ & =\left(\begin{array}{cc}d(u_{11}·v_{11})& d(u_{11}·v_{12})\\ 0& 0\end{array}\right)\hfill \\ \\ & =\left(\begin{array}{cc}d\left(u_{11}\right)·d\left(v_{11}\right)& d\left(u_{11}\right)·d\left(v_{12}\right)\\ 0& 0\end{array}\right)\hfill \\ \\ & =\left(\begin{array}{cc}d\left(u_{11}\right)& d\left(u_{12}\right)\\ 0& 0\end{array}\right)\odot \left(\begin{array}{cc}d\left(v_{11}\right)& d\left(v_{12}\right)\\ 0& 0\end{array}\right)\hfill \\ \\ & =\mathrm{\Theta }\left(U\right)\odot \mathrm{\Theta }\left(V\right).\hfill \end{array}$$

It implies that $\mathrm{\Theta }(U\odot V)=\mathrm{\Theta }\left(U\right)\odot \mathrm{\Theta }\left(V\right)$. □

3. Main Results

Now, we prove that if Q is a nonzero proper hyper-ideal of a prime ordered hyperring $(R,\oplus ,\odot ,\le )$ and $d:R\to R$ is a derivation such that $d\left(q\right)=0,$ for all q in Q, then $d\left(a\right)=0,$ for all a in R. We denote by $\mathcal{T}$ the family of all 2-torsion-free prime ordered hyperrings. Let $R\in \mathcal{T},q\in R$ and, $d:R\to R$ be a nonzero derivation of R. We show that

(1)

If $[q,d(p\left)\right]=0,$ for all p in R, then $q\in Z\left(R\right)$.

(2)

If $\left(d\right(p),q)=0,$ for all p in R, then $d\left(\right(R,q\left)\right)=0$.

Lemma 1.

Let $d:R\to R$ be a derivation of a prime ordered hyperring $(R,\oplus ,\odot ,\le )$ and $l\in R$. Then

$$l\odot d\left(q\right)=0, for all q in R\Rightarrow l=0 or d=0.$$

Proof. 

Let $p,p^{\prime }\in R$ and $l\odot d\left(q\right)=0$ for all $q\in R$. Then

$$l\odot d(p\odot p^{\prime })=0.$$

Now, we obtain

$$\begin{array}{cc}0\hfill & =l\odot d(p\odot p^{\prime })\hfill \\ \\ & \in l\odot [d\left(p\right)\odot p^{\prime }\oplus p\odot d\left(p^{\prime }\right)]\hfill \\ \\ & \subseteq (l\odot d\left(p\right)\odot p^{\prime })\oplus (l\odot p\odot d\left(p^{\prime }\right))\hfill \\ \\ & =(0\odot p^{\prime })\oplus (l\odot p\odot d\left(p^{\prime }\right))\hfill \\ \\ & =0\oplus (l\odot p\odot d\left(p^{\prime }\right))\hfill \\ \\ & =l\odot p\odot d\left(p^{\prime }\right).\hfill \end{array}$$

So, $l\odot p\odot d\left(p^{\prime }\right)=0$. Since R is prime, we obtain $l=0$ or $d\left(p^{\prime }\right)=0,$ for all $p^{\prime }$ in R, i.e., $d=0$. □

Theorem 1.

Let Q be a nonzero proper hyper-ideal of a prime ordered hyperring $(R,\oplus ,\odot ,\le )$ and $d:R\to R$ a derivation of R. Then,

$$d\left(x\right)=0,\forall x\in Q\Rightarrow d\left(l\right)=0,\forall l\in R.$$

Proof. 

If $0=x\in Q$, then clearly, $d\left(x\right)=d\left(0\right)=0$. Let $0\ne x\in Q$. By assumption, we obtain $d\left(x\right)=0$. Since Q is a hyper-ideal, we obtain

$$x\odot l\in Q, \forall l\in R.$$

Hence,

$$d(x\odot l)=0.$$

So,

$$\begin{array}{cc}0\hfill & =d(x\odot l)\hfill \\ \\ & \in d\left(x\right)\odot l\oplus x\odot d\left(l\right)\hfill \\ \\ & =0\odot l\oplus x\odot d\left(l\right)\hfill \\ \\ & =0\oplus x\odot d\left(l\right)\hfill \\ \\ & =x\odot d\left(l\right)\hfill \end{array}$$

Thus $x\odot d\left(l\right)=0,$ for all l in R. Now, by Lemma 1, we obtain $x=0$ or $d\left(l\right)=0,$ for all l in R. Since $0\ne x$, we obtain $d=0$. □

Example 5.

In Example 1, $Q=\{0,2\}$ is a proper hyper-ideal of a prime ordered hyperring R, and d is a nonzero derivation of Q.

Definition 6.

Let $d:R\to R$ be a derivation of an ordered hyperring $(R,\oplus ,\odot ,\le )$. We define

$$\left[d\right(p),q]=d\left(p\right)\odot q\ominus q\odot d\left(p\right).$$

Lemma 2.

Let $d:R\to R$ be a strong derivation of an ordered hyperring $(R,\oplus ,\odot ,\le )$. Then, for all $l,l^{\prime },q$ in R,

$$[d(l\oplus l^{\prime }),q]=[d\left(l\right),q]\oplus [d\left(l^{\prime }\right),q].$$

Proof. 

By hypothesis, we obtain

$$\begin{array}{cc}[d(l\oplus l^{\prime }),q]\hfill & =[d\left(l\right)\oplus d\left(l^{\prime }\right),q]\hfill \\ \\ & =(d\left(l\right)\oplus d\left(l^{\prime }\right))\odot q\ominus q\odot (d\left(l\right)\oplus d\left(l^{\prime }\right))\hfill \\ \\ & =d\left(l\right)\odot q\oplus d\left(l^{\prime }\right)\odot q\ominus q\odot d\left(l\right)\ominus q\odot d\left(l^{\prime }\right)\hfill \\ \\ & =(d\left(l\right)\odot q\ominus q\odot d\left(l\right))\oplus (d\left(l^{\prime }\right)\odot q\ominus q\odot d\left(l^{\prime }\right))\hfill \\ \\ & =[d\left(l^{\prime }\right),q]\oplus [d\left(l^{\prime }\right),q].\hfill \end{array}$$

□

Lemma 3.

Let $d:R\to R$ be a homo-derivation of an ordered hyperring $(R,\oplus ,\odot ,\le )$. Then, for all $l,l^{\prime },q$ in R,

$$[d(l\odot l^{\prime }),q]\subseteq [d\left(l\right),q]\odot d\left(l^{\prime }\right)\oplus d\left(l\right)\odot [d\left(l^{\prime }\right),q].$$

Proof. 

By hypothesis, we obtain

$$\begin{array}{cc}[d(l\odot l^{\prime }),q]\hfill & =[d\left(l\right)\odot d\left(l^{\prime }\right),q]\hfill \\ \\ & =(d\left(l\right)\odot d\left(l^{\prime }\right))\odot q\ominus q\odot (d\left(l\right)\odot d\left(l^{\prime }\right))\hfill \\ \\ & =d\left(l\right)\odot d\left(l^{\prime }\right)\odot q\ominus q\odot d\left(l\right)\odot d\left(l^{\prime }\right)\oplus \left\{0\right\}\hfill \\ \\ & \subseteq d\left(l\right)\odot d\left(l^{\prime }\right)\odot q\ominus q\odot d\left(l\right)\odot d\left(l^{\prime }\right)\oplus (d\left(l\right)\odot q\odot d\left(l^{\prime }\right)\ominus d\left(l\right)\odot q\odot d\left(l^{\prime }\right))\hfill \\ \\ & =(d\left(l\right)\odot q\odot d\left(l^{\prime }\right)\ominus q\odot d\left(l\right)\odot d\left(l^{\prime }\right))\oplus (d\left(l\right)\odot d\left(l^{\prime }\right)\odot q\ominus d\left(l\right)\odot q\odot d\left(l^{\prime }\right))\hfill \\ \\ & =(d\left(l\right)\odot q\ominus q\odot d\left(l\right))\odot d\left(l^{\prime }\right)\oplus d\left(l\right)\odot (d\left(l^{\prime }\right)\odot q\ominus q\odot d\left(l^{\prime }\right))\hfill \\ \\ & =[d\left(l\right),q]\odot d\left(l^{\prime }\right)\oplus d\left(l\right)\odot [d\left(l^{\prime }\right),q].\hfill \end{array}$$

□

Theorem 2.

Let $R\in \mathcal{T}$, $l\in R$ and $d:R\to R$ be a nonzero derivation of $(R,\oplus ,\odot ,\le )$. Then,

$$[l,d(q\left)\right]=0, for all q in R\Rightarrow l\in Z\left(R\right).$$

Proof. 

Assume that $l\notin Z\left(R\right)$. By Lemmas 2 and 3, we obtain

$$\begin{array}{cc}0\hfill & =[l,d(p\odot d\left(q\right)\left)\right]\hfill \\ \\ & \subseteq [l,d\left(p\right)\odot d\left(q\right)\oplus p\odot d^2\left(q\right)]\hfill \\ \\ & =[l,d\left(p\right)\odot d\left(q\right)]\oplus [l,p\odot d^2\left(q\right)]\hfill \\ \\ & \subseteq [l,d\left(p\right)]\odot d\left(q\right)\oplus d\left(p\right)\odot [l,d\left(q\right)]\oplus [l,p]\odot d^2\left(q\right)\oplus p\odot [l,d^2\left(q\right)]\hfill \\ \\ & =0\odot d\left(q\right)\oplus d\left(p\right)\odot 0\oplus [l,p]\odot d^2\left(q\right)\oplus p\odot [l,d^2\left(q\right)]\hfill \\ \\ & =0\oplus 0\oplus [l,p]\odot d^2\left(q\right)\oplus p\odot [l,d^2\left(q\right)]\hfill \\ \\ & =[l,p]\odot d^2\left(q\right)\oplus p\odot [l,d^2\left(q\right)].\hfill \end{array}$$

Now, let $d\left(q\right)=u$. Then,

$$p\odot [l,d^2\left(q\right)]=p\odot [l,d\left(u\right)]=p\odot 0=0.$$

Thus, $0\in [l,p]\odot d\left(u\right)$. So, $d\left(u\right)=0$ or $0\in [l,p]$. As $l\notin Z\left(R\right)$, we obtain $0\notin [l,p]$. Hence, $d\left(u\right)=0$ for all $u\in R$, which is a contradiction. Therefore, we obtain $l\in Z\left(R\right)$. □

Example 6.

Consider the 2-torsion-free prime ordered hyperring $R=\{0,u,v\}$:

$$\begin{array}{|cccc|}\hline \oplus & 0& u& v\\ 0& 0& u& v\\ u& u& \{u,v\}& R\\ v& v& R& \{u,v\}\\ \hline\end{array}$$

$$\begin{array}{|cccc|}\hline \odot & 0& u& v\\ 0& 0& 0& 0\\ u& 0& v& u\\ v& 0& u& v\\ \hline\end{array}$$

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

Obviously, $d:R\to R$ defined by

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

is a strong derivation of R. As $[u,d(q\left)\right]=0,$ for all q in R, we obtain $u\in Z\left(R\right)$.

Definition 7.

Let $d:R\to R$ be a derivation of an ordered hyperring $(R,\oplus ,\odot ,\le )$. We define

$$\left(d\right(p),q)=d\left(p\right)\odot q\oplus q\odot d\left(p\right).$$

Lemma 4.

Let $d:R\to R$ be a strong derivation of an ordered hyperring $(R,\oplus ,\odot ,\le )$. Then, for all $l,l^{\prime },q$ in R,

$$(d(l\oplus l^{\prime }),q)=(d\left(l\right),q)\oplus (d\left(l^{\prime }\right),q).$$

Proof. 

Straightforward. □

Lemma 5.

Let $(R,\oplus ,\odot ,\le )$ be an ordered hyperring. Then, for all $l,l^{\prime },q$ in R,

$$\begin{array}{cc}(l\odot l^{\prime },q)\hfill & \subseteq l\odot [l^{\prime },q]\oplus (l,q)\odot l^{\prime }\hfill \\ \\ & =l\odot (l^{\prime },q)\ominus [l,q]\odot l^{\prime }.\hfill \end{array}$$

Particularly, if $d:R\to R$ is a homo-derivation of R, then

$$\begin{array}{cc}(d(l\odot l^{\prime }),q)\hfill & \subseteq d\left(l\right)\odot [d\left(l^{\prime }\right),q]\oplus (d\left(l\right),q)\odot d\left(l^{\prime }\right)\hfill \\ \\ & =d\left(l\right)\odot (d\left(l^{\prime }\right),q)\ominus [d\left(l\right),q]\odot d\left(l^{\prime }\right).\hfill \end{array}$$

Proof. 

Straightforward. □

Theorem 3.

Let $R\in \mathcal{T}$, $q\in R$ and $d:R\to R$ be a nonzero derivation of $(R,\oplus ,\odot ,\le )$. If $\left(d\right(p),q)=0,$ for all p in R, then $d\left(\right(R,q\left)\right)=0$.

Proof. 

First of all, we prove that $d\left(q\right)=0$. If $q=0$, then $d\left(q\right)=d\left(0\right)=0$. Now, let $q\ne 0$. By hypothesis,

$$\left(d\right(p),q)=0,\forall p\in R.$$

So, by Lemmas 4 and 5, we obtain

$$\begin{array}{cc}0\hfill & =\left(d\right(p\odot q),q)\hfill \\ \\ & \subseteq \left(d\right(p)\odot q\oplus p\odot d(q),q)\hfill \\ \\ & =\left(d\right(p)\odot q,q)\oplus (p\odot d(q),q)\hfill \\ \\ & \subseteq d\left(p\right)\odot [q,q]\oplus \left(\right(d\left(p\right),q)\odot q)\oplus p\odot \left(d\right(q),q)\ominus [p,q]\odot d\left(q\right)\hfill \\ \\ & =0\oplus 0\oplus 0\ominus [p,q]\odot d\left(q\right).\hfill \end{array}$$

Thus, $0\in [p,q]\odot d\left(q\right)$ for all $p\in R$. Now, let $c\in R$. Then

$$\begin{array}{cc}0\hfill & \in [p\odot c,q]\odot d\left(q\right)\hfill \\ \\ & \subseteq \left(\right[p,q]\odot c\oplus p\odot [c,q\left]\right)\odot d\left(q\right)\hfill \\ \\ & =[p,q]\odot c\odot d\left(q\right)\oplus p\odot [c,q]\odot d\left(q\right)\hfill \\ \\ & \subseteq [p,q]\odot c\odot d\left(q\right).\hfill \end{array}$$

So, $0\in [p,q]\odot c\odot d\left(q\right)$. It implies that $u\odot R\odot d\left(q\right)=0$ for some $u\in [p,q]$. As R is prime, we obtain

Case 1.$u=0$.

If $u=0$, then $q\in Z\left(R\right)$. Thus $0=\left(d\right(q),q)=d\left(q\right)\odot q\oplus d\left(q\right)\odot q$ in this case. By hypothesis, we obtain $d\left(q\right)\odot q=0$. Since $q\ne 0$, we obtain $d\left(q\right)=0$.

Case 2.$d\left(q\right)=0$.

Now, for all $q^{\prime }\in R$, we have

$$\begin{array}{cc}d\left((q^{\prime },q)\right)\hfill & =d(q^{\prime }\odot q\oplus q\odot q^{\prime })\hfill \\ \\ & \subseteq d(q^{\prime }\odot q)\oplus d(q\odot q^{\prime })\hfill \\ \\ & \subseteq (d\left(q^{\prime }\right),q)\oplus (q^{\prime },d\left(q\right))\hfill \\ \\ & =0\oplus 0\hfill \\ \\ & =\left\{0\right\}.\hfill \end{array}$$

Therefore, $d\left(\right(R,q\left)\right)=0$. □

4. Conclusions

The current study has considered the concept of strong derivations of ordered hyperrings. This study tries to consider the derivations based on homomorphisms and investigates some results based on $\left[d\right(p),q]$ and $\left(d\right(p),q)$. Let $R\in \mathcal{T},q\in R$ and $d:R\to R$ be a nonzero derivation of R. We proved that

(1)

If $[q,d(p\left)\right]=0,$ for all p in R, then $q\in Z\left(R\right)$.

(2)

If $\left(d\right(p),q)=0,$ for all p in R, then $d\left(\right(R,q\left)\right)=0$.

$(R,\oplus ,\odot ,\le )$ is a negatively ordered hyperring if for all $l,l^{\prime }$ in R,

$$l\odot l^{\prime }\le l \mathrm{and}l\odot l^{\prime }\le l^{\prime }.$$

In Example 1, R is clearly a negatively ordered hyperring, but in Example 2, R is not a negatively ordered hyperring. Indeed:

$$(1\odot p,1)=(p,1)\notin \le .$$

Clearly, for any homo-derivation d of a negatively ordered hyperring R, we have

$$d\left(l\right)\odot d\left(l^{\prime }\right)⪯d\left(l\right)\oplus d\left(l^{\prime }\right),\forall l,l^{\prime }\in R.$$

In our future studies, we hope to obtain more results regarding derivation based on negatively ordered hyperrings and ordered multirings. In addition, in a future work, our aim is to study the relationship between the commutativity of an ordered hyperring and the existence of specific types of derivations.