On the commutativity of quotient near-rings under specific identities

In the present paper, $N$ will denote a right near-ring with the multiplicative center $Z(N)$⁠. A near-ring $N$ is called zero-symmetric if x.0 = 0 for all $x∈N$ (recall that right distributive gives 0.x = 0). Recall that $N$ is 3-prime, that is, For all $x,y∈N,xNy={0}$ implies x = 0 or y = 0. $N$ is said to be 2-torsion free if whenever 2x = 0, with $x∈N$⁠, then x = 0. A non empty subset $I$ of $N$ is said to be a semigroup left ideal (resp. semigroup right ideal) ideal of $N$ if $NI⊆I$ (resp. $IN⊆I$⁠); and if I is both a semigroup left ideal and a semigroup right ideal, it is called a semigroup ideal of $N$⁠. An additive mapping $H:N→N$ is a multiplier, if H(xy) = xH(y) = H(x)y for all $x,y∈N$⁠. A derivation on $N$ is an additive endomorphism δ of $N$ satisfying the Leibnitz identity δ(xy) = xδ(y) + δ(x)y for all $x,y∈N$⁠; equivalently, δ(xy) = xδ(y) + δ(x)y, as noted by Wang (1994). An additive mapping $d:N→N$ is said to be a left derivation (resp. Jordan left derivation) if d(xy) = xd(y) + yd(x) (resp. d(r²) = 2rd(r)) holds for all $x,y∈N$⁠. It is important to note that, in general, the notion of a derivation is not equivalent to that of a left derivation (see, for instance, [[1], Example 3.3]), also a Jordan left derivation need not be a left derivation (see [[2], Example 1.1]).

The notions of left derivation and Jordan left derivation were introduced by Bres̆ar et al. in Ref. [3], where it was established that if a prime ring R of characteristic different from 2 and 3 admits a nonzero Jordan left derivation, then R is necessarily commutative. Later, Ashraf et al. [4] proved that the converse holds when the ring is prime and 2-torsion-free.

Further developments were made by Zaidi et al. in Ref. [2], where they considered a Jordan ideal J of a 2-torsion-free prime ring R admitting both a nonzero Jordan left derivation d and an automorphism T, satisfying d(r²) = 2T(r)d(r) for all r ∈ J. Under these assumptions, they showed that either J ⊆ Z(R) or d(J) = {0}. Given that every ring is, in particular, a near-ring, Lemma 3 of Boua et al. [5] shows that the first condition ensures that R is commutative.

A normal subgroup $P$ of $(N,+)$ called a left ideal (resp. a right ideal) if $PN⊆P$ (resp. $y(x+p)−yx∈P$ for all $x,y∈N,p∈P$⁠), and if $P$ is both a left ideal and a right ideal, then $P$ is said to be an ideal of $N$⁠. As defined by Groenewald [6], an ideal $P$ is said to be 3-prime if, for all $x,y∈N$⁠, the inclusion $xNy⊆P$ implies that $x∈P$ or $y∈P$⁠.

We denote by $N/P$ the quotient near-ring and $Z(N/P)$ its multiplicative center. Clearly, if $P$ is 3-prime, then $N/P$ is a 3-prime near-ring. In Ref. [7], the authors defined a special derivation $d̃$ on $N/P$ by $d̃(x¯)=d(x)¯$ for all $x∈N/P$⁠, where d is a derivation on $N$⁠. Motivated by this concept, we define a multiplier $H̃$ and a left derivation $d̃$ on $N/P$ as follows: $H̃(x¯)=H(x)¯$ and $d̃(x¯)=d(x)¯$ for all $x∈N$⁠, respectively.

In the literature, numerous studies have examined the commutativity of 3-prime and semiprime near-rings admitting derivations, generalized derivations, or multiplicative derivations that satisfy appropriate algebraic constraints on certain subsets. Further related results can be found in Refs. [8–18]. This work continues this line of investigation by considering quotient near-rings equipped with a left derivation $d̃$ and a multiplier $H̃$ that satisfy specific differential identities ensuring the commutativity of the structure.

We begin by stating several key lemmas that are instrumental in proving our main results.

• If $x¯∈N/P$ and $x¯I¯={0¯}$ or $I¯x¯={0¯}$, then $x¯=0¯$⁠.

• If $x¯,y¯∈N/P$ and $x¯I¯y¯={0¯}$, then $x¯=0¯$ or $y¯=0¯$⁠.

Proof. Since $P$ is 3-prime, the quotient $N/P$ inherits the structure of a 3-prime near-ring. Consequently, the arguments for (i) and (ii) proceed analogously to those in [[19], Lemmas 1.3(i) & 1.4(i)], respectively. □

• [[7], Lemma 2(b)] If $Z(N/P)$ contains a nonzero element $z¯$ for which $z¯+z¯∈Z(N/P)$, then $(N/P,+)$ is abelian.

• [[7], Lemma 2(c)] If $z¯∈Z(N/P)\{0¯}$ and $x¯∈N/P$ such that $x¯z¯∈Z(N/P)$ or $z¯x¯∈Z(N/P)$, then $x¯∈Z(N/P)$⁠.

• If $Z(N/P)$ contains a nonzero semigroup left ideal or semigroup right ideal, then $N/P$ is a commutative ring.

• If there exists a nonzero element a such that d(a) = 0 then $a∈Z(N)$⁠.

• $(N,+)$ is abelian, if and only if $N$ is a commutative ring.

This section is devoted to investigating the structural behavior of a right near-ring $N$ under the action of certain algebraic identities, with the analysis restricted to two distinguished subsets of $N$⁠: a 3-prime ideal $P$ and a semigroup ideal $I$⁠, rather than the entire near-ring. The results obtained include, on the one hand, the commutativity of $N$⁠, and on the other, the nonexistence of a non-$P$-trivial left derivation d on $N$⁠, thereby highlighting the significance of the imposed assumptions. The main findings are summarized below.

Proof. Let $i∈I$⁠. By hypotheses, we have $d(i2)¯=(2i)¯d(i)¯∈Z(N/P)$⁠. In virtue of Lemma 2(ii), we find that

Suppose that there exists an element $k∈I$ such that $2k¯∉Z(N/P)$⁠. From Ref. [4] it follows that $d(k)¯=d̃(k¯)=0¯$⁠. Since $d̃$ is additive, we conclude that $d̃(2k¯)=d(2k)¯=0¯$⁠. By Lemma 3(i), we infer that $2k¯∈Z(N/P)$⁠; a contradiction. Therefore [4], reduces to $2i¯∈Z(N/P)$ for all $i∈I$⁠. Taking i = i² in the last relation and using Lemma 2(ii), we obtain $2i¯=0¯$ or $i¯∈Z(N/P)$ for all $i∈I$⁠. Assume that there exists $i0∈I$ such that $i0¯∉Z(N/P)$⁠, so that, $i0¯=−i0¯$⁠. Now, for all $n∈N$⁠, we have

and thus by Lemma 3(i), we deduce that $nഠi0¯∈Z(N/P)$ for all $n∈N$⁠. Putting n = ni₀ in the last relation and using Lemma 2(ii), we get $nഠi0¯=0¯$ for all $n∈N$ which means that $n¯i0¯=−i0¯n¯$ for all $n∈N$⁠. Replacing n by nt, where $t∈N$⁠, in the previous equation and using it again, we arrive at $[n¯,−i0¯]N/P={0¯}$ for all $n∈N$⁠. In view of Lemma 1(i), the above result yields $i0¯∈Z(N/P)$⁠, which contradicts our assumption. Therefore, we conclude that $i¯∈Z(N/P)$ for all $i∈I$⁠, which implies that $I¯⊆Z(N/P)$⁠. Hence, $N/P$ is a commutative ring by Lemma [20](iii). □

• $d(xy)+H([x,y]),tഠr¯∈Z(N/P)$ for all $x,y,t∈I,r∈N$,

• $(d(xy)+H([x,y]))ഠt,r¯∈Z(N/P)$ for all $x,y,r∈I,t∈N$,

• $(d(xy)+H([x,y]))ഠtഠr¯∈Z(N/P)$ for all $x,y∈I,t,r∈N$,

then $N/P$ is a commutative ring.

Proof. (i) Assume that

Substituting $rd(xy)+H([x,y]),t$ for r in Ref. [20], and noting that mഠnm = (mഠn)m, we find that $d(xy)+H([x,y]),tഠr¯d(xy)+H([x,y]),t¯∈Z(N/P)$ for all $x,y,t∈I,r∈N$⁠. In view of Lemma 2(ii) together with [20], we deduce that

Let $x,y,t∈I$⁠, and suppose that $d(xy)+H([x,y]),tഠr¯=0¯$ for all $r∈N$⁠, which means that $−d(xy)+H([x,y]),t¯r¯=r¯d(xy)+H([x,y]),t¯$ for all $r∈N$⁠. Putting r = nr, where $n∈N$⁠, in the last expression, we arrive at $−d(xy)+H([x,y]),t¯n¯r¯=n¯−d(xy)+H([x,y]),t¯r¯$ which implies that $−d(xy)+H([x,y]),t¯,n¯N/P={0¯}$ for all $x,y,t∈I$⁠. By Lemma 1(i), it follows that $−d(xy)+H([x,y]),t¯∈Z(N/P)$⁠. Consequently [19], reduces to the following

Suppose that there are $x0,y0,t0∈N$ such that $d(x0y0)+H([x0,y0]),t0¯∉Z(N/P)$⁠. From Ref. [8], it necessarily follows that $−d(x0y0)+H([x0,y0]),t0¯∈Z(N/P)\{0¯}$⁠. Substituting r with $−d(x0y0)+H([x0,y0]),t0$ in Ref. [20] and applying Lemma 2(ii), we deduce that $2d(xy)+H([x,y]),t¯∈Z(N/P)$ for all $x,y,t∈I$⁠. Once again, choosing $r=d(xy)+H([x,y]),t$ in Ref. [20] and in the light of Lemma 2(ii), we conclude that

Let us now assume that there exist elements $x1,y1,t1∈N$ such that $2d(x1y1)+H([x1,y1]),t1¯=0¯$⁠. Taking x = x₁, y = y₁, t = t₁ and $r=rd(x1y1)+H([x1,y1]),t1$ in Ref. [20], we obtain

By combining Lemma 2(ii) with equation (2), the preceding result leads to $d(x1y1)+H([x1,y1]),t1ഠr¯=0¯$ or $d(x1y1)+H([x1,y1]),t1¯∈Z(N/P)$ for all $r∈N$⁠. Assume that $d(x1y1)+H([x1,y1]),t1ഠr¯=0¯$ for all $r∈N$⁠, it follows that

for all $r∈N$⁠, implying that $d(x1y1)+H([x1,y1]),t1¯∈Z(N/P)$⁠. From Ref. [9], we conclude that $d(xy)+H([x,y]),t¯∈Z(N/P)$ for all $x,y,t∈I$⁠. In particular, for x = x₀, y = y₀ and t = t₀, we have $d(x0y0)+H([x0,y0]),t0¯∈Z(N/P)$ thereby contradicting the assumption that $d(x0y0)+H([x0,y0]),t0¯∉Z(N/P)$ and therefore

Which leads to $d(xy)+H([x,y]),t,s¯=0¯$ for all $x,y,t∈I,s∈N$⁠. Replacing t by $td(xy)+H([x,y])$ in the latter relation and using [10], we obtain $d(xy)+H([x,y]),t¯d(xy)+H([x,y]),s¯=0¯$ for all $x,y,t,∈I,s∈N$⁠. On the other hand, by left-multiplying the above equation by $m∈Z(N/P)$ and using [10], we obtain $d(xy)+H([x,y]),t¯m¯d(xy)+H([x,y]),s¯={0¯}$ for all $x,y,t∈I,s,m∈N$ which mean that $d(xy)+H([x,y]),t¯N/Pd(xy)+H([x,y]),s¯={0¯}$ for all $x,y,t∈I,s∈N$⁠. By 3-primeness of $N/P$⁠, the previous relation forces

In both cases, we get

Taking y = x, where $x∈I$⁠, in Ref. [5], we find that $d(x2)¯∈Z(N/P)$ for all $x∈I$⁠. Putting y = x³ in Ref. [5], we obtain $d(x4)¯=(2x¯2)d(x2)¯∈Z(N/P)$ for all $x∈I$⁠. By Lemma 2(ii), we obtain $2x2¯∈Z(N/P)$ or $d(x2)¯=0¯$ for all $x∈I$⁠. Invoking Lemma 3(i), it follows that $2x¯2∈Z(N/P)$ or $x¯2∈Z(N/P)$ for all $x∈I$⁠.

1. Suppose there exists an element $x0∈I$ such that $x0¯2≠0¯$⁠. Then, by Lemma 2(i), the additive group $(N/P,+)$ is abelian. Moreover, using Lemma 3(ii), we conclude that $N/P$ is a commutative ring.

2. Assume that $x¯2=0¯$ for all $x∈I$⁠. Replacing x by xy, where $y∈I$⁠, in Ref. [5], we obtain $H(−y)¯x¯y¯∈Z(N/P)$ for all $x,y∈I$⁠. Putting x = tx and y = ty, where $t∈I$ in the last relation, we get $H(−t)¯y¯t¯x¯t¯y¯∈Z(N/P)$ for all $x,y,t∈I$⁠. In virtue of Lemma 2(ii), it follows that

Suppose that $H(−t)¯y¯t¯=0¯$ for $y,t∈I$⁠, which can be written as $H(−t)¯I¯t¯={0¯}$ for all $t∈I$⁠. Using Lemma 1(ii), we find that $H(N)⊆P$⁠, contradicting the assumption that H is not $P$-trivial. Then, from Ref. [3], there exist $y0,t0∈I$ such that $−H(t0y0)¯t0¯≠0¯$ and $x¯t0¯y0¯∈Z(N/P)$ for all $x∈I$⁠. Substituting zx for x, where $z∈N$⁠, in the last relation, we obtain $z¯x¯t0¯y0¯∈Z(N/P)$ for all $x∈I,z∈N$⁠. In view of Lemma 2(ii), the previous result shows that $x¯t0¯y0¯=0¯$ or $z¯∈Z(N/P)$ for all $x∈I,z∈N$⁠. If the first condition holds, in this case we can conclude that $I¯t0¯y0¯={0¯}$ which, in virtue of Lemma 1(i), implies that $t0¯y0¯=0¯$ and hence $−H(t0y0)¯t0¯=0¯$⁠; a contradiction. Accordingly, $z¯∈Z(N/P)$ for all $z∈N$ and thus $N/P$ is a commutative ring by Lemma 2(iii).

(ii) Suppose that

Replacing r by $r(d(xy)+H([x,y]))ഠt$ in Ref. [11] and invoking Lemma 2(ii), we find that for all $x,y,r∈I,t∈N$⁠, $(d(xy)+H([x,y]))ഠt,r¯=0¯$ or $(d(xy)+H([x,y]))ഠt¯∈Z(N/P)$⁠. By expanding the first condition, we reach the same conclusion as in the second one, thereby reinforcing the result

1. In the next step, our objective is to show that $Z(N/P)≠{0¯}$⁠. Indeed, suppose that $Z(N/P)={0¯}$⁠. From Ref. [1], it follows that $(d(xy)+H([x,y]))ഠt¯=0¯$ for all $x,y∈I,t∈N$ which leads to $−d(xy)+H([x,y])¯t¯=t¯d(xy)+H([x,y])¯$ for all $x,y∈I,t∈N$⁠. Taking t = ts, where $s∈N$⁠, in the previous equation and applying it again, we get for all $x,y∈I$ and $t,s∈N$⁠, $−d(xy)+H([x,y])¯t¯s¯=t¯−d(xy)+H([x,y])¯s¯$⁠. So that, $−d(xy)+H([x,y])¯,t¯N/P={0¯}$ for all $x,y∈I,t∈N$⁠. By applying Lemma 1(i) and under the assumption that $Z(N/P)={0¯}$⁠, we obtain $d(xy)+H([x,y])¯=0¯$ for all $x,y∈I$⁠. In particular, putting x = y in the last relation, we get $d(y2)¯=0¯$ for all $y∈I$ which, in view of Lemma 3(i), implies that $y¯2∈Z(N/P)$⁠. Using the same arguments as those used in the first part, we arrive at the conclusion that $N/P$ is a commutative ring. Hence, $N/P=Z(N/P)={0¯}$⁠, which contradicts the fact that d and H are not P-trivial.

2. Now, letting $0¯≠z0¯∈Z(N/P)$ and replacing t by z₀ in Ref. [1], we obtain $2d(xy)+H([x,y])¯z0¯∈Z(N/P)$ for all $x,y∈I$⁠. In view of Lemma 2(ii) and the fact that $z0¯≠0¯$⁠, we deduce that $2d(xy)+H([x,y])¯∈Z(N/P)$ for all $x,y∈I$⁠. Once again, replacing t by d(xy) + H([x, y]) in Ref. [1] and invoking Lemma 2(ii), we get

Suppose that there exist $x1,y1∈I$ such that $2d(x1y1)+H([x1,y1])¯=0¯$⁠, it follows that $d(x1y1)+H([x1,y1])¯=−(d(x1y1)+H([x1,y1])¯)$⁠. Taking x = x₁, y = y₁ and t = t(d(x₁y₁) + H([x₁, y₁])) in Ref. [1], we get $(d(x1y1)+H([x1,y1])ഠt)¯(d(x1y1)+H([x1,y1]))¯∈Z(N/P)$ for all $t∈N$ which, in view of Lemma 2(ii), implies that $(d(x1y1)+H([x1,y1]))ഠt¯=0¯$ or $d(x1y1)+H[x1,y1]¯∈Z(N/P)$ for all $t∈N$⁠. Assume that $(d(x1y1)+H([x1,y1]))ഠt¯=0¯$ for all $t∈N$⁠, that is, $t¯(d(x1y1)+H([x1,y1])¯)=(−(d(x1y1)+H([x1,y1])¯))t¯=(d(x1y1)+H([x1,y1])¯)t¯$ for all $t∈N$ which leads to $d(x1y1)+H([x1,y1])¯∈Z(N/P)$ and therefore [21] reduces to $d(xy)+H([x,y])¯∈Z(N/P)$ for all $x,y∈I$⁠. The rest of the proof follows the same steps as those used after relation [5] in the proof of (i), we can establish that $N/P$ is a commutative ring.

To prove (iii), we assume that

Suppose that $Z(N/P)={0¯}$⁠. It follows that $r¯((d(xy)+H([x,y]))ഠt¯)=(−((d(xy)+H([x,y]))ഠt¯))r¯$ for all $x,y∈I,t,r∈N$⁠. Putting r = rs, where $s∈N$⁠, in the previous equation and using it again, we find that $r¯(−((d(xy)+H([x,y]))ഠt¯))s¯=(−((d(xy)+H([x,y]))ഠt¯))r¯s¯$ for all $x,y∈I,r,t,s∈N$⁠, which implies that $[r¯,−((d(xy)+H([x,y]))ഠt¯)]N/P={0¯}$ for all $x,y∈I,t,s∈N$⁠. In view of the Lemma 1(i), we obtain $−((d(xy)+H([x,y]))ഠt¯)∈Z(N/P)={0¯}$ which implies that $(d(xy)+H([x,y]))ഠt¯=0 for all x,y∈I,t∈N$⁠. Since this result coincides with relation [1], a contradiction follows, and hence $Z(N/P)≠{0¯}$⁠. Let $0¯≠z0¯∈Z(N/P)$⁠. Substituting r = z₀ into [6] and invoking Lemma 2(ii), we obtain $2(d(xy)+H([x,y]))ഠt¯∈Z(N/P)$ for all $x,y∈I,t∈N$⁠. Replacing r by $d(xy)+H([x,y])ഠt$ in Ref. [6], and applying Lemma 2(ii), we obtain