On commutativity of quotient semirings through generalized derivations

1.   Introduction and preliminaries

In general commutators are undefined in semirings. However, this notion is well defined in a special class of semirings called MA-semirings, therefore Lie theory and other related notions can be studied in semirings. Javed et al. ^[1] defined MA-semiring as an additive inverse semiring $S$ with absorbing zero $'0'$ satisfying $A_2$ condition of Bandlet and Petrich ^[2] that is $u+u^{'}\in Z(S)\, \text{for all} \, u\in S$, where $u^{'}$ is the pseudo inverse of $u\in S$, and $Z(S)$ is the center of $S$. The class of MA-semirings properly contains the class of rings, meaning every ring is an MA-semiring while converse may not be true (for more detail, and examples of MA-semirings, we refer reader to ^[3,4,5,6,7]).

Theory of ideals has become a notion of special worth in certain algebras and ring theory. Several ideals, such as prime ideals, $k$-ideals, $Q$-ideals, Quasi-ideals, Jordan ideals, Lie ideals have been defined and studied for different types of algebraic structures (for ready reference one can see ^{[8,9,10,11,12,13]}). An ideal $I$ of a semiring $S$ is called prime if for $a, b\in S$, $aSb\subseteq I$ implies either $a\in I$ or $b\in I$. An ideal $I$ of a semiring $S$ is said to be a $Q$-ideal if there exists a partitioning subset $Q$ of $S$ such that $S = \bigcup\{q+I:q\in Q\}$ and if $q_1, q_2\in Q$, then $(q_1+I)\bigcap(q_2+I)\neq \phi$ if and only if $q_1 = q_2$ (see ^[9]). An ideal $I$ of a semiring $S$ is said to be $k$-ideal if $a+b\in I$ and $b\in I$, then $a\in I$. In fact, every $Q$-ideal is $k$-ideal but converse may not be true in general (see ^[8]). Throughout the sequel by $I_Q$, we mean prime $Q$-ideal unless mentioned otherwise.

Derivations and its generalized formats defined for any algebraic structure have become a very rich area of investigation and researchers have contributed a remarkable work on it (for reference, one can see ^{[1,3,14,15,16,17,18]}). A derivation is an additive mapping $d:S\longrightarrow S$ such that $d(ut) = d(u)t+ud(t)$. An additive mapping $G_d:S\longrightarrow S$ fulfilling $G_d(tu) = G_d(t)u+td(u)$ is a generalized derivation associated with a derivation $d$. The commutator and Jordan product of $t, u\in S$ are respectively defined as $[t, u] = tu+u^{'}t$ and $t\circ u = tu+ut$. For the sequel, we state some useful identities of MA-semirings: For all $t, u, v\in S$, we have $[tu, v] = t[u, v]+[t, v]u$, $[t, tu] = t[t, u]$, $[t, u]+[u, t] = u(t+t^{'}) = t(u+u^{'})$, $[t, uv] = [t, u]v+u[t, v]$, $[t, u]^{'} = [t, u^{'}] = [t^{'}, u]$, $(tu)^{'} = t^{'}u = tu^{'}$, $t\circ (u+v) = t\circ u+t\circ v$. For more on MA-semirings, we refer ^[1,3].

A ring $R$ is commutative if and only if $[u, v] = 0,$ for all $u, v\in R.$ However, in the case of an MA-semiring, the sufficiency of the above said statement is valid while the necessity part may not be true in general. One can find many reasonable examples in support of this statement. To make it more easy to understand: in rings $[u, u] = 0$ but in MA-semirings $[u, u] = 0$ may not hold in general. Nadeem and others ^[19] proved some results on commutativity of Quotient rings enforced by generalized derivations satisfying different identities on prime ideals. In this paper we prove these results for MA-semirings.

2.   Main results

Unless otherwise specified, in the sequel, we refer to $G_d$ as a generalized derivation associated with a derivation $d$.

Theorem 2.1. Let $(S, +, .)$ be an MA-semiring and $I_Q$ be a prime $Q$-ideal of $S$. Then the set $S/I_Q = \{u+I_Q:u\in S\}$ forms an MA-semiring with respect to the addition $\oplus$ and multiplication $\odot$ defined by

(1) $(u+I_Q)\oplus (t+I_Q) = u+t+I_Q$,

(2) $(u+I_Q)\odot (t+I_Q) = u.t+I_Q$,

for all $u, t \in S$.

Proof. We only verify $A_2$ axiom of Bandlet and Petrich ^[2], the other axioms of the definition of MA-semiring are straightforward. For this let $u+I_Q, t+I_Q\in S/I_Q$. Then

Thus, $(S/I_Q, \oplus, \odot)$ is an MA-semiring. □

The following theorem is an extension of Lemma 1.2 of ^[19].

Theorem 2.2. Let $I_Q$ be a prime $Q$-ideal of an MA-semiring $S$. If

(1) $[t, u]\in I_Q$ or

(2) $t \circ u \in I_Q$,

for all $t, u \in S$, then $S/I_Q$ is a commutative MA-semiring.

Proof. (1). Based on the hypothesis, we have $[t, u]\in I_Q, \text{for all} \ t, u\in S$. As $0\in I_Q\subseteq S$, therefore $[t, u]+I_Q = I_Q$ and therefore by the addition and multiplication in $S/I_Q$, we can write $I_Q = (t+I_Q)(u+I_Q)+(u+I_Q)(t^{'}+I_Q)$. Hence $(t+I_Q)(u+I_Q) = (u+I_Q)(t+I_Q)$. This shows that $S/I_Q$ is commutative. (2). Using the same reasoning as above, we have

which further implies

In (2.1) substituting $ur$ for $u$, we get $[(t+I_Q)(u+I_Q)](r+I_Q)+(u+I_Q)[(r+I_Q)(t+I_Q)] = I_Q$ and using (2.2), we obtain $[(t+I_Q)(u+I_Q)](r+I_Q)+[(u+I_Q)(t^{'}+I_Q)](r+I_Q) = I_Q$. Therefore $[(t+I_Q), (u+I_Q)](r+I_Q) = I_Q$. As $I_Q$ is prime and $I_Q\neq S$, therefore $[t, u]+I_Q = [(t+I_Q), (u+I_Q)] = I_Q$. By the first part, we conclude that $S/I_Q$ is a commutative MA-semiring. □

Corollary 2.1. Let $S$ be a prime MA-semiring $S$. If

(1) $[t, u] = 0$ or

(2) $t \circ u = 0$,

for all $t, u \in S$, then $S$ is a commutative MA-semiring.

Remark 2.1. If $S/I_Q$ is commutative, then it has no nonzero zero divisors. Indeed, suppose that $I_Q = (t+I_Q)(u+I_Q)$. Then $I_Q = tu+I_Q$ which implies that $tu\in I_Q$. As $I_Q$ is prime, either $t\in I_Q$ or $u\in I_Q$ and therefore either $t+I_Q = I_Q$ or $u+I_Q = I_Q$.

Following result is an extension of Proposition 1.3 of ^[19].

Theorem 2.3. Let $I_Q$ be a prime $Q$-ideal of an MA-semiring $S$. If $G_d$ is a generalized derivation satisfying

for all $t \in S$, then either $S/I_Q$ is a commutative or $d(S)\subseteq I_Q$.

Proof. Linearizing (2.3), we get $[t, G_d(u)]+[t, G_d(t)]+[u, G_d(t)]+[u, G_d(u)]\in I_Q$. As $I_Q$ is prime Q-ideal, again using (2.3) and the fact that $I_Q$ is a $k$-ideal, we obtain

for all $t, u\in S$. Substituting $ut$ for $u$ in (2.4), we get $[t, G_d(ut)]+[ut, G_d(t)]\in I_Q$ and using MA-semiring identities, we can write

which further implies

Substituting $vu$ for $u$ in (2.5) and again using (2.5), we obtain $[t, v]ud(t)\in I_Q$ (i.e $[t, v]Sd(t)\subseteq I_Q$). As $I_Q$ is prime, we have either $[t, v]\in I_Q$ for all $t, v\in S$ or $d(t)\in I_Q$ for all $t\in S$. By Theorem 2.2, we have either $S/I_Q$ is commutative or $d(S)\subseteq I_Q.$ □

Theorems 2.4–2.8 are the extended version of the corresponding results proved in ^[19].

Theorem 2.4. Let $I_Q$ be a prime $Q$-ideal of an MA-semiring $S$. If $G_d$ is a generalized derivation of $S$ such that

for all $t, u \in S.$ Then either $S/I_Q$ is commutative or $d(S) \subseteq I_Q$.

Proof. In (2.6) substituting $uw$ for $u$, we get $G_d(tuw)+G_d(t)G_d(uw) = G_d(tu)w+tud(w)+G_d(t)G_d(u)w+G_d(t)ud(w)\in I_Q$ and using (2.6) again and the fact that $I_Q$ is a $k$-ideal, we have

In (2.7) substituting $ts$ for $t$, we obtain $tsud(w)+G_d(ts)ud(w)\in I_Q$ and therefore, which can be further written as

In (2.7) substituting $su$ for $u$, we obtain $tsud(w)+G_d(t)sud(w)\in I_Q$ and therefore

As $I_Q$ is Q-ideal, therefore using (2.8) and (2.9), we obtain $td(s)ud(w)\in I_Q$ and therefore $[t, r]d(s)Sd(w)\in I_Q$. As $I_Q$ is prime ideal, therefore either $[t, r]d(s)\in I_Q$ or $d(w)\in I_Q$. If $d(w)\in I_Q$, then $d(S)\subseteq I_Q$. On the other hand, if $[t, r]d(s)\in I_Q$, then using commutator identities, we find $[t, r]Sd(s)\in I_Q$. Using the same arguments as before, we find either $d(S)\subseteq I_Q$ or $[t, r]\in I_Q$ and employing Theorem 2.2, we obtain $S/I_Q$ is commutative. □

We can establish Theorem 2.5 by similar reasoning used in the proof of Theorem 2.4.

Theorem 2.5. Let $I_Q$ be a prime $Q$-ideal of an MA-semiring $S$. If $G_d$ is a generalized derivation of $S$ such that

for all $t, u \in S,$ then either $S/I_Q$ is commutative or $d(S) \subseteq I_Q$.

Theorem 2.6. Let $I_Q$ be a prime $Q$-ideal of an MA-semiring $S$. If $G_d$ is a generalized derivation of $S$ such that

for all $t, w \in S,$ then either $S/I_Q$ is commutative or $d(S) \subseteq I_Q$.

Proof. In (2.10) writing $tw$ in place of $t$, we obtain $G_d(tw)w+twd(w)+G_d(w)G_d(t)w^{'}+G_d(w)t^{'}d(w) \in I_Q$. As $I_Q$ is Q-ideal, therefore

In (2.11) substituting $st$ for $t$, we obtain $stwd(w)+G_d(w)st^{'}d(w) \in I_Q$ and therefore $stwd(w)+I_Q+G_d(w)st^{'}d(w)+I_Q = I_Q$, which further implies

Left multiplying (2.11) by $s$, we obtain

Using (2.12) in (2.13), we get $[G_d(w), s]td(w)+I_Q = I_Q$, therefore $[G_d(w), s]Sd(w)\subseteq I_Q$. As $I_Q$ is prime, we get either $[G_d(w), s]\in I_Q$ or $d(S)\subseteq I_Q$. As a result of Theorem 2.3 $S/I_Q$ is commutative or $d(S)\subseteq I_Q$. □

Using the proof of Theorem 2.6, we can establish the following result.

Theorem 2.7. Let $I_Q$ be a prime $Q$-ideal of an MA-semiring $S$. If $G_d$ is a generalized derivation of $S$ such that

for all $t, w \in S,$ then either $S/I_Q$ is commutative or $d(S) \subseteq I_Q$.

Following result provide a generalized form of Theorem 1.5 of ^[19].

Theorem 2.8. Let $I_Q$ be a prime $Q$-ideal of an MA-semiring $S$. If $G_d$ is a generalized derivation meeting one of the conditions given below:

(1) $G_d(tw)+tw\in I_Q,$

(2) $G_d(tw)+t^{'}w\in I_Q,$

(3) $G_d(tw)+wt\in I_Q,$

(4) $G_d(tw)+w^{'}t\in I_Q,$

(5) $G_d(t)G_d(w)+tw\in I_Q,$

(6) $G_d(t)G_d(w)+t^{'}w\in I_Q,$

(7) $G_d(t)G_d(w)+wt\in I_Q,$

(8) $G_d(t)G_d(w)+w^{'}t\in I_Q,$

for all $t, w\in S,$ then either $S/I_Q$ is commutative or $d(S)\subseteq I_Q$.

Proof. (1). Based on the hypothesis, we have

If $G_d = 0$, then from (2.14), we have $tw\in I_Q$ and interchanging $t$ and $w$, we get $wt\in I_Q$. From the last two expressions, we have $t\circ w \in I_Q$. Employing Theorem 2.2, we conclude that $S/I_Q$ is commutative. Next, we consider the case when $G_d\neq 0.$ In (2.14), writing $ws$ in place of $w$, we get $(G_d(tw)+tw)s+twd(s)\in I_Q$. As $I_Q$ is Q-ideal, therefore using (2.14), we obtain $twd(s)\in I_Q$ and therefore substituting $wr$ for $w$, we obtain

In (2.15) interchanging $t$ and $w$, we get

From (2.15) and (2.16), we can write $(t\circ w)Sd(s)\subseteq I_Q$ and by the primeness of $I_Q$, we obtain either $t\circ w\in I_Q$ or $d(S)\subseteq I_Q.$ Again by the Theorem 2.2, we conclude either $S/I_Q$ is commutative or $d(S)\subseteq I_Q.$

By the similar arguments, we can prove (2).

(3). An appeal to the hypothesis, we have

For $G_d = 0$, we have $wt\in I_Q$. We can prove that $S/I_Q$ is commutative using the same reasoning as in the proof of (1). Next we consider the case when $G_d\neq 0$. In (2.17) substituting $wt$ for $t$, we obtain $(G_d(tw)+wt)t+twd(t)\in I_Q$ and using (2.17) again, we obtain $twd(t)\in I_Q$. Remaining part follows through similar arguments of the proof of (1).

Part (4) can be followed similarly as part (3).

(5). Based on the hypothesis

for all $t, w\in S.$ The case when $G_d = 0$ is straightforward. For the second case when $G_d\neq 0$, substituting $ws$ for $w$ in (2.18), we get $(G_d(t)G_d(w)+tw)s+G_d(t)wd(s)\in I_Q$. As $I_Q$ is Q-ideal, using (2.18) again, $G_d(t)wd(s)\in I_Q$. Since $I_Q$ is prime, therefore either $G_d(t)\in I_Q$ or $d(S)\subseteq I_Q$. If

In (2.19) substituting $tw$ for $t$, we get $G_d(t)w+td(w)\in I_Q$. As $I_Q$ is Q-ideal, using (2.19) we obtain $td(w)\in I_Q$ and writing $[r, s]t$ in place of $t$, we further get $[r, s]Sd(w)\subseteq I_Q$. As $I_Q$ is prime, we obtain $[r, s]\in I_Q$ or $d(w)\in I_Q$. In view of Theorem 2.2, we conclude that $S/I_Q$ is commutative or $d(S)\subseteq I_Q.$

Proof of (6) is not quite different from the proof of (5).

(7). Based on the hypothesis, we have

The case when $G_d = 0$ is straightforward. We consider the case when $G_d\neq 0$. In (2.20) replacing $w$ by $wt$, we obtain $(G_d(t)G_d(w)+wt)t+G_d(t)wd(t)\in I_Q$. As $I_Q$ is Q-ideal, using (2.20) again, we obtain $G_d(t)wd(t)\in I_Q$. Remaining part can be followed by the similar arguments of the proof of part (5).

On the similar lines of (7), we can establish (8). □

The following is an extension of Theorem 1.6 of ^[19].

Theorem 2.9. Let $I_Q$ be a prime $Q$-ideal of an MA-semiring $S$. If $G_d$ is a generalized derivation meeting one of the conditions below:

(1) $G_d(tw)+[t, w]^{'} \in I_Q,$

(2) $G_d(tw)+[t, w] \in I_Q,$

(3) $G_d(tw)+t^{'}\circ w \in I_Q,$

(4) $G_d(tw)+t\circ w \in I_Q,$

(5) $G_d(t)G_d(w)+[t, w]^{'} \in I_Q,$

(6) $G_d(t)G_d(w)+[t, w] \in I_Q,$

(7) $G_d(t)G_d(w)+t^{'}\circ w \in I_Q,$

(8) $G_d(t)G_d(w)+t\circ w \in I_Q,$

for all $t, w\in S,$ then $S/I_Q$ is a commutative.

Proof. (1). Based on the hypothesis, we have

If $G_d = 0$, then by Theorem 2.2, we obtain the required result. Suppose that $G_d\neq 0$. In (2.21) substituting $ws$ for $w$ and using MA-semiring identities, we obtain $(G_d(tw)+[t, w]^{'})s+twd(s)+w[t, s]^{'} \in I_Q$. As $I_Q$ is Q-ideal, therefore using (2.21) again, we get

In (2.22) writing $rw$ in place of $w$, we obtain

From (2.22) we can also write $trwd(s)+I_Q+rw[t, s]^{'}+I_Q = I_Q$, which further implies

Multiplying (2.22) by $r$ from the left, we obtain $rtwd(s)+rw[t, s]^{'} \in I_Q$ which can further written as

Using (2.24) into (2.25), we obtain $[r, t]wd(s)+I_Q = I_Q$ and therefore $[r, t]Sd(s)\subseteq I_Q$. As $I_Q$ is prime, therefore either $[r, t]\in I_Q$ or $d(S)\subseteq I_Q$. For the first possibility, employing Theorem 2.2, we conclude that $S/I_Q$ is commutative. On the other hand if $d(S)\subseteq I_Q$, then from (2.23), we have $rw[t, s] \in I_Q$ and hence $[t, s]S[t, s] \subseteq I_Q$. In view of the primeness of $I_Q$, using Theorem 2.2, we get the required result.

Part (2) can be established on the similar lines of Part (1).

(3). Based on the hypothesis, we have

For $G_d = 0$, from Theorem 2.2, we obtain the required result. Assume that $G_d\neq 0$. In (2.26) substituting $wr$ for $w$, we obtain $G_d(tw)r+twd(r)+t^{'}wr+wrt^{'} \in I_Q$ and using the definition of MA-semiring, we can write

Therefore $G_d(tw)r+(t^{'}\circ w)r+twd(r)+w[t, r]\in I_Q$. As $I_Q$ is Q-ideal, using (2.26), we obtain

In (2.27) substituting $sw$ for $w$ and using MA-semiring identities, we obtain $tswd(r)+sw[t, r]\in I_Q$, which further gives $tswd(r)+I_Q+sw[t, r]+I_Q = I_Q$ and therefore

Multiplying (2.27) by $s$ from the left, we obtain $stwd(r)+sw[t, r]\in I_Q$ and therefore

Using (2.28) into (2.29), we get $stwd(r)+I_Q+t^{'}swd(r)+I_Q = I_Q$ and therefore

As $I_Q$ is prime, therefore either $[s, t]\in I_Q$ or $d(S)\subseteq I_Q$. Repeating the same arguments as above, we obtain the required result.

Part (4) can be proved on the similar lines of Part (3).

(5). Based on the hypothesis, we have

In view of Theorem 2.2, the case when $G_d = 0$, is straightforward. Assume that $G_d\neq 0$. In (2.31) substituting $wv$ for $w$ and using MA-semiring commutator identities, we get $G_d(t)G_d(w)v+G_d(t)wd(v)+w[t, v]^{'}+[t, w]v^{'}\in I_Q$. Using (2.31) again, we obtain

In (2.32), substituting $sw$ for $w$, we obtain $G_d(t)swd(v)+sw[t, v]^{'}\in I_Q$, which further implies $G_d(t)swd(v)+I_Q+sw[t, v]^{'}+I_Q = I_Q$ and therefore

Multiplying (2.32) by $s$ from the left, we obtain $sG_d(t)wd(v)+sw[t, v]^{'}\in I_Q$ which further gives

Using (2.33) in (2.34), we get $[s, G_d(t)]wd(v)+I_Q = I_Q$, therefore $[s, G_d(t)]Sd(v)\subseteq I_Q$. By the primeness of $I_Q$, we have either $[s, G_d(t)]\in I_Q$ or $d(S)\subseteq I_Q$. If $d(S)\subseteq I_Q$, then from (2.32), we have

In (2.35) substituting $[t, v]w$ for $w$, we obtain $[t, v]S[t, v]\subseteq I_Q$. As $I_Q$ is prime, $[t, v]\in I_Q$ and hence by Theorem 2.2, $S/I_Q$ is commutative. Second, suppose $[s, G_d(t)]\in I_Q$. By Theorem 2.3, we have either $S/I_Q$ is commutative or $d(S)\subseteq I_Q$. If $d(S)\subseteq I_Q$, then by the same arguments, we again find $S/I_Q$ is commutative.

Part (6) can be followed on the similar lines of Part (5).

(7). Based on the hypothesis, we have

If $G_d = 0$, then $t\circ w \in I_Q, \ \text {for all} \ t, w\in S$ and hence by Theorem {2.2} $S/I_Q$ is commutative. Next, we suppose that $G_d\neq 0$. In (2.36) writing $ws$ in place of $w$, we obtain

By the definition of MA-semiring, we have

Therefore (2.37) becomes

As $I_Q$ is a $Q$-ideal, therefore using (2.36) again, we

Remaining part can be followed through the similar arguments of Part (5).

Part (8) can be followed on the same lines of the proof of Part (7). □

3.   Conclusions

We discussed some notions about generalized derivations and prime $Q$-ideals of MA-semirings. Some interesting results about the commutativity of quotient MA-semirings under certain differential identities are proved. Taking $I_Q = \{0\}$, one can obtain important consequences of the results for prime MA-semirings.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

We are thankful to the refrees for their valuable comments for the improvement of this paper.

Conflict of Interest

The authors declare that they have no conflict of interest.