On Prime-Gamma-Near-Rings with Generalized Derivations

Abstract

Let 𝑁 be a 2-torsion free prime Γ-near-ring with center 𝑍(𝑁). Let (𝑓,𝑑) and (𝑔,ℎ) be two generalized derivations on 𝑁. We prove the following results: (i) if 𝑓([𝑥,𝑦]_𝛼)=0 or 𝑓([𝑥,𝑦]_𝛼)=±[𝑥,𝑦]_𝛼 or 𝑓²(𝑥)∈𝑍(𝑁) for all 𝑥,𝑦∈𝑁, 𝛼∈Γ, then 𝑁 is a commutative Γ-ring. (ii) If 𝑎∈𝑁 and [𝑓(𝑥),𝑎]_𝛼=0 for all 𝑥∈𝑁, 𝛼∈Γ, then 𝑑(𝑎)∈𝑍(𝑁). (iii) If (𝑓𝑔,𝑑ℎ) acts as a generalized derivation on 𝑁, then 𝑓=0 or 𝑔=0.

1. Introduction

The derivations in Γ-near-rings have been introduced by Bell and Mason [1]. They studied basic properties of derivations in Γ-near-rings. Then Aşci [2] obtained commutativity conditions for a Γ-near-ring with derivations. Some characterizations of Γ-near-rings and regularity conditions were obtained by Cho [3]. Kazaz and Alkan [4] introduced the notion of two-sided Γ-𝛼-derivation of a Γ-near-ring and investigated the commutativity of a prime and semiprime Γ-near-rings. Uçkun et al. [5] worked on prime Γ-near-rings with derivations and they found conditions for a Γ-near-ring to be commutative. In [6] Dey et al. studied commutativity of prime Γ-near-ring with generalized derivations.

In this paper, we obtain the conditions of a prime Γ-near-ring to be a commutative Γ-ring. If 𝑎∈𝑁, and [𝑓(𝑥),𝑎]_𝛼=0 for all 𝑥∈𝑁,𝛼∈Γ, then 𝑑 is central. Also we prove that if (𝑓𝑔,𝑑ℎ) is the generalized derivation on 𝑁, then 𝑓 and 𝑔 are trivial.

2. Preliminaries

A Γ-near-ring is a triple (𝑁,+,Γ), where(i)(𝑁,+) is a group (not necessarily abelian);(ii)Γ is a nonempty set of binary operations on 𝑁 such that for each 𝛼∈Γ,(𝑁,+,𝛼) is a left near-ring;(iii)𝑥𝛼(𝑦𝛽𝑧)=(𝑥𝛼𝑦)𝛽𝑧, for all 𝑥,𝑦,𝑧∈𝑁 and 𝛼,𝛽∈Γ.

We will use the word Γ-near-ring to mean left Γ-near-ring. For a near-ring 𝑁, the set 𝑁₀={𝑥∈𝑁∶0𝛼𝑥=0,𝛼∈Γ} is called the zero-symmetric part of 𝑁. A Γ-near-ring 𝑁 is said to be zero-symmetric if 𝑁=𝑁₀. Throughout this paper, 𝑁 will denote a zero symmetric left Γ-near-ring with multiplicative centre 𝑍(𝑁). Recall that a Γ-near-ring 𝑁 is prime if 𝑥Γ𝑁Γ𝑦=0 implies 𝑥=0 or 𝑦=0. An additive mapping 𝑑∶𝑁→𝑁 is said to be a derivation on 𝑁 if 𝑑(𝑥𝛼𝑦)=𝑥𝛼𝑑(𝑦)+𝑑(𝑥)𝛼𝑦 for all 𝑥,𝑦∈𝑁,𝛼∈Γ, or equivalently, as noted in [1], that 𝑑(𝑥𝛼𝑦)=𝑑(𝑥)𝛼𝑦+𝑥𝛼𝑑(𝑦) for all 𝑥,𝑦∈𝑁,𝛼∈Γ. Further, an element 𝑥∈𝑁 for which 𝑑(𝑥)=0 is called a constant. For 𝑥,𝑦∈𝑁,𝛼∈Γ, the symbol [𝑥,𝑦]_𝛼 will denote the commutator 𝑥𝛼𝑦−𝑦𝛼𝑥, while the symbol (𝑥,𝑦) will denote the additive-group commutator 𝑥+𝑦−𝑥−𝑦. An additive mapping 𝑓∶𝑁→𝑁 is called a generalized derivation if there exits a derivation 𝑑 of 𝑁 such that 𝑓(𝑥𝛼𝑦)=𝑓(𝑥)𝛼𝑦+𝑥𝛼𝑑(𝑦) for all 𝑥,𝑦∈𝑁,𝛼∈Γ. The concept of generalized derivation covers also the concept of a derivation.

3. Derivations on Γ-Near-Rings

In this section we prove that a few subsidiary results (Lemmas 3.1, 3.2, 3.4, 3.8, 3.9, 3.10 and 3.11) to use them for proving of our main results (Theorems 3.3, 3.5, 3.6, 3.12 and 3.13).

Lemma 3.1. Let 𝑑 be an arbitrary derivation on a Γ-near-ring 𝑁. Then 𝑁 satisfies the following partial distributive law: (𝑎𝛼𝑑(𝑏)+𝑑(𝑎)𝛼𝑏)𝛽𝑐=𝑎𝛼𝑑(𝑏)𝛽𝑐+𝑑(𝑎)𝛼𝑏𝛽𝑐 and (𝑑(𝑎)𝛼𝑏+𝑎𝛼𝑑(𝑏))𝛽𝑐=𝑑(𝑎)𝛼𝑏𝛽𝑐+𝑎𝛼𝑑(𝑏)𝛽𝑐 for all 𝑎,𝑏,𝑐∈𝑁,𝛼,𝛽∈Γ.

Proof. For all 𝑎,𝑏,𝑐∈𝑁,𝛼,𝛽∈Γ, we get 𝑑((𝑎𝛼𝑏)𝛽𝑐)=𝑎𝛼𝑏𝛽𝑑(𝑐)+(𝑎𝛼𝑑(𝑏)+𝑑(𝑎)𝛼𝑏)𝛽𝑐 and 𝑑(𝑎𝛼(𝑏𝛽𝑐))=𝑎𝛼𝑑(𝑏𝛽𝑐)+𝑑(𝑎)𝛼𝑏𝛽𝑐=𝑎𝛼(𝑏𝛽𝑑(𝑐)+𝑑(𝑏)𝛽𝑐)+𝑑(𝑎)𝛼𝑏𝛽𝑐=𝑎𝛼𝑏𝛽𝑑(𝑐)+𝑎𝛼𝑑(𝑏)𝛽𝑐+𝑑(𝑎)𝛼𝑏𝛽𝑐. Equating these two relations for 𝑑(𝑎𝛼𝑏𝛽𝑐) now yields the required partial distributive law.

Lemma 3.2. Let 𝑑 be a derivation on a Γ-near-ring 𝑁 and suppose 𝑢∈𝑁 is not a left zero divisor. If [𝑢,𝑑(𝑢)]_𝛼=0,𝛼∈Γ, then (𝑥,𝑢) is a constant for every 𝑥∈𝑁.

Proof. From 𝑢𝛼(𝑢+𝑥)=𝑢𝛼𝑢+𝑢𝛼𝑥, for all 𝑥∈𝑁, 𝛼∈Γ, we obtain 𝑢𝛼𝑑(𝑢+𝑥)+𝑑(𝑢)𝛼(𝑢+𝑥)=𝑢𝛼𝑑(𝑢)+𝑑(𝑢)𝛼𝑢+𝑢𝛼𝑑(𝑥)+𝑑(𝑢)𝛼𝑥, which reduces 𝑢𝛼𝑑(𝑥)+𝑑(𝑢)𝛼𝑢=𝑑(𝑢)𝛼𝑢+𝑢𝛼𝑑(𝑥), for all 𝛼∈Γ.
Since 𝑑(𝑢)𝛼𝑢=𝑢𝛼𝑑(𝑢), 𝛼∈Γ, this equation is expressible as 𝑢𝛼(𝑑(𝑥)+𝑑(𝑢)−𝑑(𝑥)−𝑑(𝑢))=0=𝑢𝛼𝑑((𝑥,𝑢)). Thus 𝑑((𝑥,𝑢))=0.

Theorem 3.3. Let 𝑁 be a Γ-near-ring having no nonzero divisors of zero. If 𝑁 admits a nontrivial commuting derivation 𝑑, then (𝑁,+) is abelian.

Proof. Let 𝑐 be any additive commutator. Then 𝑐 is a constant by Lemma 3.2. Moreover, for any 𝑤∈𝑁, 𝛼∈Γ, 𝑤𝛼𝑐 is an additive commutator, hence also a constant. Thus, 0=𝑑(𝑤𝛼𝑐)=𝑤𝛼𝑑(𝑐)+𝑑(𝑤)𝛼𝑐 and 𝑑(𝑤)𝛼𝑐=0, for all 𝛼∈Γ. Since 𝑑(𝑤)≠0 for all 𝑤∈𝑁, we conclude that 𝑐=0.

Lemma 3.4. Let 𝑁 be a prime Γ-near-ring.(i)If 𝑧∈𝑍(𝑁)−{0}, then 𝑧 is not a zero divisor in 𝑁.(ii)If 𝑍(𝑁)−{0} contains an element 𝑧 for which 𝑧+𝑧∈𝑍(𝑁), then (𝑁,+) is abelian.(iii)Let 𝑑 be a nonzero derivation on 𝑁. Then 𝑥Γ𝑑(𝑁)={0} implies 𝑥=0, and 𝑑(𝑁)Γ𝑥={0} implies 𝑥=0.(iv)If 𝑁 is 2-torsion free and 𝑑 is a derivation on 𝑁 such that 𝑑²=0, then 𝑑=0.

Proof. (i) If 𝑧∈𝑍(𝑁)−{0} and 𝑧𝛼𝑥=0, 𝑥∈𝑁, 𝛼∈Γ, then 𝑧𝛼𝑟𝛽𝑥=0, 𝑥, 𝑟∈𝑁, 𝛼∈Γ. Thus we get 𝑧Γ𝑁Γ𝑥=0, by primeness of 𝑁, 𝑥=0.
(ii) Let 𝑧∈𝑍(𝑁)−{0} be an element such that 𝑧+𝑧∈𝑍(𝑁), and let 𝑥,𝑦∈𝑁,𝛼∈Γ. Since 𝑧+𝑧 is distributive, we get (𝑥+𝑦)𝛼(𝑧+𝑧)=𝑥𝛼(𝑧+𝑧)+𝑦𝛼(𝑧+𝑧)=𝑥𝛼𝑧+𝑥𝛼𝑧+𝑦𝛼𝑧+𝑦𝛼𝑧=𝑧𝛼(𝑥+𝑥+𝑦+𝑦).
On the other hand, (𝑥+𝑦)𝛼(𝑧+𝑧)=(𝑥+𝑦)𝛼𝑧+(𝑥+𝑦)𝛼𝑧=𝑧𝛼(𝑥+𝑦+𝑥+𝑦). Thus, 𝑥+𝑥+𝑦+𝑦=𝑥+𝑦+𝑥+𝑦 and therefore 𝑥+𝑦=𝑦+𝑥. Hence (𝑁,+) is abelian.
(iii) Let 𝑥Γ𝑑(𝑁)=0, and let 𝑟,𝑠 be arbitrary elements of 𝑁 and 𝛼,𝛽∈Γ. Then 0=𝑥𝛼𝑑(𝑟𝛽𝑠)=𝑥𝛼𝑟𝛽𝑑(𝑠)+𝑥𝛼𝑑(𝑟)𝛽𝑠=𝑥𝛼𝑟𝛽𝑑(𝑠). Thus 𝑥Γ𝑁Γ𝑑(𝑁)={0}, and since 𝑑(𝑁)≠{0},𝑥=0.
A similar argument works if 𝑑(𝑁)Γ𝑥={0}, since Lemma 3.1 provides enough distributivity to carry it through.
(iv) For arbitrary 𝑥,𝑦∈𝑁,𝛼∈Γ, we have 0=𝑑²(𝑥𝛼𝑦)=𝑑(𝑥𝛼𝑑(𝑦)+𝑑(𝑥)𝛼𝑦)=𝑥𝛼𝑑²(𝑦)+𝑑(𝑥)𝛼𝑑(𝑦)+𝑑(𝑥)𝛼𝑑(𝑦)+𝑑²(𝑥)𝛼𝑦=2𝑑(𝑥)𝛼𝑑(𝑦). Since 𝑁 is 2-torsion free, 𝑑(𝑥)𝛼𝑑(𝑦)=0,𝑥,𝑦∈𝑁,𝛼∈Γ. Thus 𝑑(𝑥)Γ𝑑(𝑁)={0} for each 𝑥∈𝑁, and (iii) yields 𝑑(𝑎)=0. Thus 𝑑=0.

Theorem 3.5. If a prime Γ-near-ring 𝑁 admits a nontrivial derivation 𝑑 for which 𝑑(𝑁)∈𝑍(𝑁), then (𝑁,+) is abelian. Moreover, if 𝑁 is 2-torsion free, then 𝑁 is a commutative Γ-ring.

Proof. Let 𝑐 be an arbitrary constant, and let 𝑥 be anon-constant. Then 𝑑(𝑥𝛼𝑐)=𝑥𝛼𝑑(𝑐)+𝑑(𝑥)𝛼𝑐=𝑑(𝑥)𝛼𝑐∈𝑍(𝑁), 𝛼∈Γ. Since 𝑑(𝑥)∈𝑍(𝑁)−{0}, it follows easily that 𝑐∈𝑍(𝑁). Since 𝑐+𝑐 is a constant for all constants 𝑐, it follows from Lemma 3.4(ii) that (𝑁,+) is abelian, provided that there exists a nonzero constant.
Assume, then, that 0 is the only constant. Since 𝑑 is obviously commuting, it follows from Lemma 3.2 that all 𝑢 which are not zero divisors belong to the center of (𝑁,+), denoted by 𝑍(𝑁). In particular, if 𝑥≠0, 𝑑(𝑥)∈𝑍(𝑁). But then for all 𝑦∈𝑁, 𝑑(𝑦)+𝑑(𝑥)−𝑑(𝑦)−𝑑(𝑥)=𝑑((𝑦,𝑥))=0, hence (𝑦,𝑥)=0.
Now we assume that 𝑁 is 2-torsion free. By Lemma 3.1, (𝑎𝛼𝑑(𝑏)+𝑑(𝑎)𝛼𝑏)𝛽𝑐=𝑎𝛼𝑑(𝑏)𝛽𝑐+𝑑(𝑎)𝛼𝑏𝛽𝑐 for all 𝑎,𝑏,𝑐∈𝑁,𝛼,𝛽∈Γ, and using the fact that 𝑑(𝑎𝛼𝑏)∈𝑍(𝑁), 𝛼∈Γ, we get 𝑐𝛼𝑎𝛽𝑑(𝑏)+𝑐𝛼𝑑(𝑎)𝛽𝑏=𝑎𝛼𝑑(𝑏)𝛽𝑐+𝑎𝛼𝑑(𝑏)𝛽𝑐,𝛼,𝛽∈Γ. Since (𝑁,+) is abelian and 𝑑(𝑁)⊆𝑍(𝑁), this equation can be rearranged to yield 𝑑(𝑏)𝛼[𝑐,𝑎]_𝛽=𝑑(𝑎)𝛼[𝑏,𝑐]_𝛽 for all 𝑎,𝑏,𝑐∈𝑁,𝛼,𝛽∈Γ.
Suppose now that 𝑁 is not commutative. Choosing 𝑏,𝑐∈𝑁, with [𝑏,𝑐]_𝛽≠0, 𝛽∈Γ, and letting 𝑎=𝑑(𝑥), we get 𝑑²(𝑥)𝛼[𝑏,𝑐]_𝛽=0, for all 𝑥∈𝑁,𝛼,𝛽∈Γ, and since the central elements 𝑑²(𝑥) cannot be a nonzero divisor of zero, we conclude that 𝑑²(𝑥)=0 for all 𝑥∈𝑁. But by Lemma 3.4(iv), this cannot happen for nontrivial 𝑑.

Theorem 3.6. Let 𝑁 be a prime Γ-near-ring admitting a nontrivial derivation 𝑑 such that [𝑑(𝑥),𝑑(𝑦)]_𝛼=0 for all 𝑥,𝑦∈𝑁,𝛼∈Γ. Then (𝑁,+) is abelian. Moreover, if 𝑁 is 2-torsion free, then 𝑁 is a commutative Γ-ring.

Proof. By Lemma 3.4(ii), if both 𝑧 and 𝑧+𝑧 commute element-wise with 𝑑(𝑁), then 𝑧𝛼𝑑(𝑐)=0,𝛼∈Γ, for all additive commutators 𝑐. Thus, taking 𝑧=𝑑(𝑥), we get 𝑑(𝑥)𝛼𝑑(𝑐)=0 for all 𝑥∈𝑁,𝛼∈Γ, so 𝑑(𝑐)=0 by Lemma 3.4(iii). Since 𝑤𝛼𝑐 is also an additive commutator for any 𝑤∈𝑁, 𝛼∈Γ, we have 𝑑(𝑤𝛼𝑐)=0𝑑(𝑤)𝛼𝑐, and another application of Lemma 3.4(iii) gives 𝑐=0.
Now we assume that 𝑁 is 2-torsion free. By the partial distributive law, 𝑑(𝑑(𝑥)𝛼𝑦)𝛽𝑑(𝑧)=𝑑(𝑥)𝛼𝑑(𝑦)𝛽𝑑(𝑧)+𝑑²(𝑥)𝛼𝑦𝛽𝑑(𝑧) for all 𝑥,𝑦,𝑧∈𝑁,𝛼,𝛽∈Γ, hence, 𝑑²(𝑥)𝛼𝑦𝛽𝑑(𝑧)=𝑑(𝑑(𝑥)𝛼𝑦)𝛽𝑑(𝑧)−𝑑(𝑥)𝛼𝑑(𝑦)𝛽𝑑(𝑧)=𝑑(𝑥)𝛼𝑑(𝑦)𝛽𝑑(𝑧)=𝑑(𝑧)𝛼(𝑑(𝑑(𝑥)𝛽𝑦)−𝑑(𝑥)𝛽𝑑(𝑦))=𝑑(𝑧)𝛼𝑑²(𝑥)𝛽𝑦=𝑑²(𝑥)𝛼𝑑(𝑧)𝛽𝑦,𝛼,𝛽∈Γ. Thus 𝑑²(𝑥)𝛼(𝑦𝛽𝑑(𝑧)−𝑑(𝑧)𝛽𝑦)=0 for all 𝑥,𝑦𝑧∈𝑁, 𝛼,𝛽∈Γ.
Replacing 𝑦𝛿𝑡,𝛿∈Γ, we obtain 𝑑²(𝑥)𝛼𝑦𝛿𝑡𝛽𝑑(𝑧)=𝑑²(𝑥)𝛼𝑑(𝑧)𝛽𝑦𝛿𝑡=𝑑²(𝑥)𝛼𝑦𝛽𝑑(𝑧)𝛿𝑡, for all 𝑥,𝑦,𝑧,𝑡∈𝑁,𝛼,𝛽,𝛿∈Γ, so that 𝑑²(𝑥)𝛼𝑦𝛽[𝑡,𝑑(𝑧)]_𝛿=0 for all 𝑥,𝑦,𝑧,𝑡∈𝑁, 𝛼,𝛽,𝛿∈Γ. The primeness of 𝑁 shows that either 𝑑²=0 or 𝑑(𝑁)⊆𝑍(𝑁), and since the first of these conditions is impossible by Lemma 3.4(iv), the second must hold and 𝑁 is a commutative Γ-ring by Theorem 3.5.

Definition 3.7. Let 𝑁 be a Γ-near-ring and 𝑑 a derivation of 𝑁. An additive mapping 𝑓∶𝑁→𝑁 is said to be a right generalized derivation of 𝑁 associated with 𝑑 if 𝑓𝑥𝛼𝑦=𝑓(𝑥)𝛼𝑦+𝑥𝛼𝑑𝑦∀𝑥,𝑦∈𝑅,𝛼∈Γ, (3.1) and 𝑓 is said to be a left generalized derivation of 𝑁 associated with 𝑑 if 𝑓𝑥𝛼𝑦=𝑑(𝑥)𝛼𝑦+𝑥𝛼𝑓𝑦∀𝑥,𝑦∈𝑅,𝛼∈Γ. (3.2) Finally, 𝑓 is said to be a generalized derivation of 𝑁 associated with 𝑑 if it is both a left and right generalized derivation of 𝑁 associated with 𝑑.

Lemma 3.8. Let 𝑓 be a right generalized derivation of a Γ-near ring 𝑁 associated with 𝑑. Then(i)𝑓(𝑥𝛼𝑦)=𝑥𝛼𝑑(𝑦)+𝑓(𝑥)𝛼𝑦 for all 𝑥,𝑦∈𝑁,𝛼∈Γ;(ii)𝑓(𝑥𝛼𝑦)=𝑥𝛼𝑓(𝑦)+𝑑(𝑥)𝛼𝑦 for all 𝑥,𝑦∈𝑁,𝛼∈Γ.

Proof. (i) For any 𝑥,𝑦∈𝑁,𝛼∈Γ, we get 𝑓(𝑥)𝛼(𝑥)(𝑥)𝑦𝑦,𝑓(𝑥)𝑦(𝑥)𝑦.𝑥𝛼𝑦+𝑦=𝑓𝑦+𝑦+𝑥𝛼𝑑𝑦+𝑦=𝑓𝛼𝑦+𝑓𝛼𝑦+𝑥𝛼𝑑+𝑥𝛼𝑑𝑥𝛼𝑦+𝑥𝛼𝑦=𝑓𝛼𝑦+𝑥𝛼𝑑+𝑓𝛼𝑦+𝑥𝛼𝑑 (3.3) Comparing these two expressions, we obtain 𝑓(𝑥)𝛼𝑦+𝑥𝛼𝑑𝑦=𝑥𝛼𝑑𝑦+𝑓(𝑥)𝛼𝑦∀𝑥,𝑦∈𝑁,𝛼∈Γ, (3.4) and so, 𝑓𝑥𝛼𝑦=𝑥𝛼𝑑𝑦+𝑓(𝑥)𝛼𝑦∀𝑥,𝑦∈𝑁,𝛼∈Γ. (3.5)
(ii) In a similar way.

Lemma 3.9. Let 𝑓 be a right generalized derivation of a Γ-near ring 𝑁 associated with 𝑑. Then(i)(𝑓(𝑥)𝛼𝑦+𝑥𝛼𝑑(𝑦))𝛽𝑧=𝑓(𝑥)𝛼𝑦𝛽𝑧+𝑥𝛼𝑑(𝑦)𝛽𝑧, for all 𝑥,𝑦∈𝑁, 𝛼,𝛽∈Γ.(ii)(𝑑(𝑥)𝛼𝑦+𝑥𝛼𝑓(𝑦))𝛽𝑧=𝑑(𝑥)𝛼𝑦𝛽𝑧+𝑥𝛼𝑓(𝑦)𝛽𝑧, for all 𝑥,𝑦∈𝑁, 𝛼,𝛽∈Γ.

Proof. (i) For any 𝑥,𝑦,𝑧∈𝑁, 𝛼,𝛽∈Γ, we get 𝑓((𝑥𝛼𝑦)𝛽𝑧)=𝑓(𝑥𝛼𝑦)𝛽𝑧+𝑥𝛼𝑦𝛽𝑑(𝑧).
On the other hand, 𝑓𝑥𝛼𝑦𝛽𝑧=𝑓(𝑥)𝛼𝑦𝛽𝑧+𝑥𝛼𝑑𝑦𝛽𝑧=𝑓(𝑥)𝛼𝑦𝛽𝑧+𝑥𝛼𝑑𝑦𝛽𝑧+𝑥𝛼𝑦𝛽𝑑(𝑧). (3.6) From these two expressions of 𝑓(𝑥𝛼𝑦𝛽𝑧), we obtain that, for all 𝑥,𝑦,𝑧∈𝑁, 𝛼,𝛽∈Γ, 𝑓(𝑥)𝛼𝑦+𝑥𝛼𝑑𝑦𝛽𝑧=𝑓(𝑥)𝛼𝑦𝛽𝑧+𝑥𝛼𝑑𝑦𝛽𝑧. (3.7)
(ii) The proof is similar.

Lemma 3.10. Let 𝑁 be a prime Γ-near-ring, 𝑓 a nonzero generalized derivation of 𝑁 associated with the nonzero derivation 𝑑 and 𝑎∈𝑁. (i) If 𝑎Γ𝑓(𝑁)=0, then 𝑎=0. (ii) If 𝑓(𝑁)Γ𝑎=0, then 𝑎=0.

Proof. (i) For any 𝑥,𝑦∈𝑁,𝛼,𝛽∈Γ, we get 0=𝑎𝛽𝑓(𝑥𝛼𝑦)=𝑎𝛽𝑓(𝑥)𝛼𝑦+𝑎𝛽𝑥𝛼𝑑(𝑦)=𝑎𝛽𝑥𝛼𝑑(𝑦). Hence 𝑎Γ𝑁Γ𝑑(𝑁)=0. Since 𝑁 is a prime Γ-near-ring and 𝑑≠0, we obtain 𝑎=0.
(ii) A similar argument works if 𝑓(𝑁)Γ𝑎=0.

Lemma 3.11. Let 𝑁 be a prime Γ-near-ring. Let 𝑓 be a generalized derivation of 𝑁 associated with the nonzero derivation 𝑑. If 𝑁 is a 2-torsion free Γ-near-ring and 𝑓²=0, then 𝑓=0.

Proof. (i) For any 𝑥,𝑦∈𝑁, 𝛼∈Γ, we get 0=𝑓²𝑥𝛼𝑦=𝑓𝑓𝑥𝛼𝑦=𝑓𝑓(𝑥)𝛼𝑦+𝑥𝛼𝑑𝑦=𝑓²(𝑥)𝛼𝑦+2𝑓(𝑥)𝛼𝑑𝑦+𝑥𝛼𝑑²𝑦. (3.8) By the hypothesis, 2𝑓(𝑥)𝛼𝑑𝑦+𝑥𝛼𝑑²𝑦=0∀𝑥,𝑦∈𝑁,𝛼∈Γ. (3.9) Writing 𝑓(𝑥) by 𝑥 in (3.9), we get 𝑓(𝑥)𝛼𝑑²(𝑦)=0, for all 𝑥,𝑦∈𝑁,𝛼∈Γ.
By Lemma 3.9(ii), we obtain that 𝑑²(𝑁)=0 or 𝑓=0. If 𝑑²(𝑁)=0 then 𝑑=0 from Lemma 3.4(iv), a contradiction. So we find 𝑓=0.

Theorem 3.12. Let 𝑁 be a prime Γ-near-ring with a nonzero generalized derivation 𝑓 associated with 𝑑. If 𝑓(𝑁)⊆𝑍(𝑁), then (𝑁,+) is abelian. Moreover, if 𝑁 is 2-torsion free, then 𝑁 is commutative Γ-ring.

Proof. Suppose that 𝑎∈𝑁, such that 𝑓(𝑎)≠0. So, 𝑓(𝑎)∈𝑍(𝑁)−{0} and 𝑓(𝑎)+𝑓(𝑎)∈𝑍(𝑁)−{0}. For all 𝑥,𝑦∈𝑁, 𝛼∈Γ, we have (𝑥+𝑦)𝛼(𝑓(𝑎+𝑓(𝑎))=(𝑓(𝑎+𝑓(𝑎))𝛼(𝑥+𝑦).
That is, 𝑥𝛼𝑓(𝑎)+𝑥𝛼𝑓(𝑎)+𝑦𝛼𝑓(𝑎)+𝑦𝛼𝑓(𝑎)=𝑓(𝑎)𝛼𝑥+𝑓(𝑎)𝛼𝑥+𝑓(𝑎)𝛼𝑦+𝑓(𝑎)𝛼𝑦, for all 𝑥,𝑦∈𝑁, 𝛼∈Γ.
Since 𝑓(𝑎)∈𝑍(𝑁), we get 𝑓(𝑎)𝛼𝑥+𝑓(𝑎)𝛼𝑦=𝑓(𝑎)𝛼𝑦+𝑓(𝑎)𝛼𝑥, and so, 𝑓(𝑎)𝛼(𝑥,𝑦)=0 for all 𝑥,𝑦∈𝑁,𝛼∈Γ.
Since 𝑓(𝑎)∈𝑍(𝑁)−{0} and 𝑁 is a prime Γ-near-ring, it follows that (𝑥,𝑦)=0, for all 𝑥,𝑦∈𝑁. Thus (𝑁,+) is abelian.
Using the hypothesis, for any 𝑥,𝑦,𝑧∈𝑁,𝛼,𝛽∈Γ, 𝑧𝛼𝑓(𝑥𝛽𝑦)=𝑓(𝑥𝛽𝑦)𝛼𝑧. By Lemma 3.4(ii), we have 𝑧𝛼𝑑(𝑥)𝛽𝑦+𝑧𝛼𝑥𝑓(𝑦)=𝑑(𝑥)𝛼𝑦𝛽𝑧+𝑥𝛼𝑓(𝑦)𝛽𝑧. Using 𝑓(𝑁)⊂𝑍(𝑁) and (𝑁,+) being abelian, we obtain that 𝑧𝛼𝑑(𝑥)𝛽𝑦−𝑑(𝑥)𝛼𝑦𝛽𝑧=[𝑥,𝑧]_𝛼𝛽𝑓𝑦,∀𝑥,𝑦∈𝑁,𝛼,𝛽∈Γ. (3.10) Substituting 𝑓(𝑧) for 𝑧 in (3.10), we get 𝑓(𝑧)𝛽[𝑑(𝑥),𝑦]_𝛼=0 for all 𝑥,𝑦∈𝑁,𝛼,𝛽∈Γ.
Since 𝑓(𝑧)∈𝑍(𝑁) and f a nonzero generalized derivation with associated with 𝑑, we get 𝑑(𝑁)⊂𝑍(𝑁). So, 𝑁 is a commutative Γ-ring by Theorem 3.3.

Theorem 3.13. Let 𝑁 be a prime Γ-near-ring with a nonzero generalized derivation 𝑓 associated with 𝑑. If [𝑓(𝑁),𝑓(𝑁)]_𝛼=0, 𝛼∈Γ, then (𝑁,+) is abelian. Moreover, if 𝑁 is 2-torsion free, then 𝑁 is commutative Γ-ring.

Proof. By the same argument as in Theorem 3.12, it is shown that if both 𝑧 and 𝑧+𝑧 commute elementwise with 𝑓(𝑁), then we have 𝑧𝛼𝑓𝑥,𝑦=0∀𝑥,𝑦∈𝑁,𝛼∈Γ. (3.11) Substituting 𝑓(𝑡),𝑡∈𝑁 for 𝑧 in (3.11), we get 𝑓(𝑡)𝛼𝑓(𝑥,𝑦)=0, 𝛼∈Γ. By Lemma 3.9(i), we obtain that 𝑓(𝑥,𝑦)=0 for all 𝑥,𝑦∈𝑁, 𝛼∈Γ. For any 𝑤∈𝑁, 𝛽∈Γ, we have 0=𝑓(𝑤𝛽𝑥,𝑤𝛽𝑦)=𝑓(𝑤𝛽(𝑥,𝑦))=𝑑(𝑤)𝛽(𝑥,𝑦)+𝑤𝛽𝑓(𝑥,𝑦) and so, we obtain 𝑑(𝑤)𝛽(𝑥,𝑦)=0, for any 𝑤∈𝑁, 𝛽∈Γ. From Lemma 3.4(iii), we get (𝑥,𝑦)=0 for any 𝑥,𝑦∈𝑁.
Now we assume that 𝑁 is 2-torsion free. By the assumption [𝑓(𝑁),𝑓(𝑁)]_𝛼=0, 𝛼∈Γ, we have 𝑓(𝑧)𝛼𝑓𝑓(𝑥)𝛽𝑦=𝑓𝑓(𝑥)𝛽𝑦𝛼𝑓(𝑧)∀𝑥,𝑦,𝑧∈𝑁,𝛼,𝛽∈Γ. (3.12) Hence we get 𝑓(𝑧)𝑓(𝑥)(𝑧)(𝑥)𝑦𝑓(𝑥)(𝑧)(𝑥)𝑦(𝑧),𝑓(𝑧)𝑓(𝑥)(𝑥)(𝑧)𝑦𝑓(𝑥)(𝑧)(𝑥)(𝑧)𝑦,𝛼𝑑𝛽𝑦+𝑓𝛼𝑓𝛽𝑓=𝑑𝛼𝑦𝛽𝑓+𝑓𝛼𝑓𝛽𝑓𝛼𝑑𝛽𝑦+𝑓𝛼𝑓𝛽𝑓=𝑑𝛼𝑦𝛽𝑓+𝑓𝛼𝑓𝛽𝑓 (3.13) and so, 𝑓(𝑧)𝛼𝑑𝑓(𝑥)𝛽𝑦=𝑑𝑓(𝑥)𝛼𝑦𝛽𝑓(𝑧)∀𝑥,𝑦,𝑧∈𝑁,𝛼,𝛽∈Γ. (3.14) If we take 𝑦𝛿𝑤 instead of 𝑦 in (3.14), then 𝑑𝑓(𝑥)𝛼𝑦𝛿𝑤𝛽𝑓(𝑧)=𝑓(𝑧)𝛼𝑑𝑓(𝑥)𝛽𝑦𝛿𝑤=𝑑𝑓(𝑥)𝛼𝑦𝛿𝑓(𝑧)𝛽𝑤∀𝑥,𝑦,𝑧∈𝑁,𝛼,𝛽,𝛿∈Γ, (3.15) and so, 𝑑𝑓(𝑥)𝛼𝑦𝛿𝑤𝛽𝑓(𝑧)−𝑑𝑓(𝑥)𝛼𝑦𝛿𝑓(𝑧)𝛽𝑤=𝑑𝑓(𝑥)𝛼𝑦𝛿𝑓(𝑧),𝑤_𝛿=0∀𝑥,𝑦,𝑧∈𝑁,𝛼,𝛽,𝛿∈Γ. (3.16) Thus we get 𝑑(𝑓(𝑥))Γ𝑁Γ[𝑓(𝑧),𝑤]_𝛿=0, for all 𝑥,𝑦,𝑧∈𝑁, 𝛼,𝛽,𝛿∈Γ. Since 𝑁 is a prime Γ-near-ring, we have 𝑑(𝑓(𝑁))=0 or 𝑓(𝑁)⊂𝑍(𝑁). Let us assume that 𝑑(𝑓(𝑁))=0. Then 0=𝑑𝑓𝑥𝛼𝑦=𝑑𝑑(𝑥)𝛼𝑦+𝑥𝛼𝑓𝑦 (3.17) and so, 𝑑²(𝑥)𝛼𝑦+𝑑(𝑥)𝛼𝑑𝑦+𝑑(𝑥)𝛼𝑓𝑦=0,∀𝑥,𝑦∈𝑁,𝛼∈Γ. (3.18) Replacing 𝑦 by 𝑦𝛽𝑧, 𝛽∈Γ, in (3.18), we get 0=𝑑²(𝑥)𝛼𝑦𝛽𝑧+𝑑(𝑥)𝛼𝑑𝑦𝛽𝑧+𝑑(𝑥)𝛼𝑓𝑦𝛽𝑧=𝑑²(𝑥)(𝑥)𝑦(𝑥)(𝑧)(𝑥)𝑦(𝑥)(𝑧)=𝑑𝛼𝑦𝛽𝑧+𝑑𝛼𝑑𝛽𝑧+𝑑𝛼𝑦𝛽𝑑+𝑑𝛼𝑓𝛽𝑧+𝑑𝛼𝑦𝛽𝑑₂(𝑥)𝛼𝑦+𝑑(𝑥)𝛼𝑑𝑦+𝑑(𝑥)𝛼𝑓𝑦𝛽𝑧+2𝑑(𝑥)𝛼𝑦𝛽𝑑(𝑧)∀𝑥,𝑦,𝑧∈𝑁,𝛼,𝛽∈Γ. (3.19) Using (3.18) and 𝑁 being 2-torsion free Γ-near-ring, we get 𝑑(𝑁)Γ𝑁Γ𝑑(𝑁)=0.
Thus we obtain that 𝑑=0. It contradicts by 𝑑≠0. The theorem is proved.

4. Generalized Derivations of Γ-Near-Rings

We denote a generalized derivation 𝑓∶𝑁→𝑁 determined by a derivation 𝑑 of 𝑁 by (𝑓,𝑑). We assume that 𝑑 is a nonzero derivation of 𝑁 unless stated otherwise.

Theorem 4.1. Let (𝑓,𝑑) be a generalized derivation of 𝑁. If 𝑓([𝑥,𝑦]_𝛼)=0  for all 𝑥,𝑦∈𝑁,𝛼∈Γ, then 𝑁 is a commutative Γ-ring.

Proof. Assume that 𝑓([𝑥,𝑦]_𝛼)=0 for all 𝑥,𝑦∈𝑁, 𝛼∈Γ. Substitute 𝑥𝛽𝑦 instead of 𝑦, obtaining 𝑓𝑥,𝑥𝛽𝑦_𝛼=𝑓𝑥𝛽𝑥,𝑦_𝛼=𝑑(𝑥)𝛽𝑥,𝑦_𝛼+𝑥𝛽𝑓𝑥,𝑦_𝛼=0. (4.1) Since the second term is zero, it is clear that 𝑑(𝑥)𝛼𝑥𝛽𝑦=𝑑(𝑥)𝛼𝑦𝛽𝑥∀𝑥,𝑦∈𝑁,𝛼,𝛽∈Γ. (4.2) Replacing 𝑦 by 𝑦𝛿𝑧 in (4.2) and using this equation, we get 𝑑(𝑥)𝛼𝑦𝛽[𝑥,𝑧]_𝛿=0∀𝑥,𝑦,𝑧∈𝑁,𝛼,𝛽,𝛿∈Γ. (4.3) Hence either 𝑥∈𝑍(𝑁) or 𝑑(𝑥)=0. Let 𝐿={𝑥∈𝑁∣𝑑(𝑥)=0}. Then 𝑍(𝑁) and 𝐿 are two additive subgroups of (𝑁,+)=𝑍(𝑁)∪𝐿. However, a group cannot be the union of proper subgroups, hence either 𝑁=𝑍(𝑁) or 𝑁=𝐿. Since 𝑑≠0, we are forced to conclude that 𝑁 is a commutative Γ-ring.

Theorem 4.2. Let (𝑓,𝑑) be a generalized derivation of 𝑁. If 𝑓([𝑥,𝑦]_𝛼)=±[𝑥,𝑦]_𝛼  for all 𝑥,𝑦∈𝑁, 𝛼∈Γ, then 𝑁 is a commutative Γ-ring.

Proof. Assume that 𝑓([𝑥,𝑦]_𝛼)=±[𝑥,𝑦]_𝛼 for all 𝑥,𝑦∈𝑁, 𝛼∈Γ. Replacing 𝑦 by 𝑥𝛽𝑦, 𝛽∈Γ, in the hypothesis, we have 𝑓𝑥,𝑥𝛽𝑦_𝛼=±𝑥𝛼𝑥𝛽𝑦−𝑥𝛼𝑦𝛽𝑥=±𝑥𝛽𝑥,𝑦_𝛼. (4.4) On the other hand, 𝑓𝑥,𝑥𝛽𝑦_𝛼=𝑓𝑥𝛽𝑥,𝑦_𝛼=𝑑(𝑥)𝛽𝑥,𝑦_𝛼+𝑥𝛽𝑓𝑥,𝑦_𝛼=𝑑(𝑥)𝛽𝑥,𝑦_𝛼+𝑥𝛽±𝑥,𝑦_𝛼. (4.5) It follows from the two expressions for 𝑓([𝑥,𝑥𝛽𝑦]_𝛼) that 𝑑(𝑥)𝛼𝑥𝛽𝑦=𝑑(𝑥)𝛼𝑦𝛽𝑥∀𝑥,𝑦∈𝑁,𝛼,𝛽∈Γ. (4.6) Using the same argument as in the proof of Theorem 4.1, we get that 𝑁 is a commutative Γ-ring.

Theorem 4.3. Let (𝑓,𝑑) be a nonzero generalized derivation of 𝑁. If f acts as a homomorphism on 𝑁, then 𝑓 is the identity map.

Proof. Assume that 𝑓 acts as a homomorphism on 𝑁. Then one obtains 𝑓𝑥𝛼𝑦=𝑓(𝑥)𝛼𝑓𝑦=𝑑(𝑥)𝛼𝑦+𝑥𝛼𝑓𝑦∀𝑥,𝑦∈𝑁,𝛼∈Γ. (4.7) Replacing 𝑦 by 𝑦𝛽𝑧 in (4.7), we arrive at 𝑓(𝑥)𝛼𝑓𝑦𝛽𝑧=𝑑(𝑥)𝛼𝑦𝛽𝑧+𝑥𝛼𝑓𝑦𝛽𝑧. (4.8) Since (𝑓,𝑑) is a generalized derivation and 𝑓 acts as a homomorphism on 𝑁, we deduce that 𝑓𝑥𝛼𝑦𝛽𝑓(𝑧)=𝑑(𝑥)𝛼𝑦𝛽𝑧+𝑥𝛼𝑑𝑦𝛽𝑧+𝑥𝛼𝑦𝛽𝑓(𝑧). (4.9) By Lemma 3.9(ii), we get 𝑑(𝑥)𝛼𝑦𝛽𝑓(𝑧)+𝑥𝛼𝑓𝑦𝛽𝑓(𝑧)=𝑑(𝑥)𝛼𝑦𝛽𝑧+𝑥𝛼𝑑𝑦𝛽𝑧+𝑥𝛼𝑦𝛽𝑓(𝑧), (4.10) and so 𝑑(𝑥)𝛼𝑦𝛽𝑓(𝑧)+𝑥𝛼𝑓𝑦𝛽𝑧=𝑑(𝑥)𝛼𝑦𝛽𝑧+𝑥𝛼𝑑𝑦𝛽𝑧+𝑥𝛼𝑦𝛽𝑓(𝑧). (4.11) That is, 𝑑(𝑥)𝛼𝑦𝛽𝑓(𝑧)+𝑥𝛼𝑑𝑦𝛽𝑧+𝑥𝛼𝑦𝛽𝑓(𝑧)=𝑑(𝑥)𝛼𝑦𝛽𝑧+𝑥𝛼𝑑𝑦𝛽𝑧+𝑥𝛼𝑦𝛽𝑓(𝑧). (4.12) Hence, we deduce that 𝑑(𝑥)𝛼𝑦𝛽𝑓(𝑧)−𝑧=0∀𝑥,𝑦,𝑧∈𝑁,𝛼,𝛽∈Γ. (4.13) Because 𝑁 is prime and 𝑑≠0, we have 𝑓(𝑧)=𝑧 for all 𝑧∈𝑁. Thus, 𝑓 is the identity map.

Theorem 4.4. Let (𝑓,𝑑) be a nonzero generalized derivation of 𝑁. If f acts as an antihomomorphism on 𝑁, then 𝑓 is the identity map.

Proof. By the hypothesis, we have 𝑓𝑥𝛼𝑦=𝑓𝑦𝛼𝑓(𝑥)=𝑑(𝑥)𝛼𝑦+𝑥𝛼𝑓𝑦∀𝑥,𝑦∈𝑁,𝛼∈Γ. (4.14) Replacing 𝑦 by 𝑥𝛽𝑦 in the last equation, we obtain 𝑓𝑥𝛽𝑦𝛼𝑓(𝑥)=𝑑(𝑥)𝛽𝑥𝛼𝑦+𝑥𝛽𝑓𝑥𝛼𝑦. (4.15) Since (𝑓,𝑑) is a generalized derivation and 𝑓 acts as an antihomomorphism on 𝑁, we get 𝑑(𝑥)𝛽𝑦+𝑥𝛽𝑓𝑦𝛼𝑓(𝑥)=𝑑(𝑥)𝛼𝑥𝛽𝑦+𝑥𝛼𝑓𝑦𝛽𝑓(𝑥). (4.16) By Lemma 3.9(ii), we conclude that 𝑑(𝑥)𝛼𝑦𝛽𝑓(𝑥)+𝑥𝛼𝑓𝑦𝛽𝑓(𝑥)=𝑑(𝑥)𝛼𝑥𝛽𝑦+𝑥𝛼𝑓𝑦𝛽𝑓(𝑥), (4.17) and so 𝑑(𝑥)𝛼𝑦𝛽𝑓(𝑥)=𝑑(𝑥)𝛼𝑥𝛽𝑦∀𝑥,𝑦∈𝑁,𝛼,𝛽∈Γ. (4.18) Replacing 𝑦 by 𝑦𝛿𝑧 and using this equation, we have 𝑑(𝑥)𝛼𝑦𝛽𝑓(𝑥),𝑧_𝛼=0∀𝑥,𝑧∈𝑁,𝛼,𝛽∈Γ. (4.19) Hence we obtain the following alternatives: 𝑑(𝑥)=0 or 𝑓(𝑥)∈𝑍(𝑁), for all 𝑥∈𝑁. By a standard argument, one of these must hold for all 𝑥∈𝑁. Since 𝑑≠0, the second possibility gives that 𝑁 is commutative Γ-ring by Theorem 3.12, and so we deduce that 𝑓 is the identity map by Theorem 4.3.

Theorem 4.5. Let (𝑓,𝑑) be a generalized derivation of 𝑁 such that 𝑑(𝑍(𝑁))≠0, and 𝑎∈𝑁. If [𝑓(𝑥),𝑎]_𝛼=0  for all 𝑥∈𝑁, 𝛼,𝛽∈Γ, then 𝑎∈𝑍(𝑁).

Proof. Since 𝑑(𝑍(𝑁))≠0, there exists 𝑐∈𝑍(𝑁) such that 𝑑(𝑐)≠0. Furthermore, as 𝑑 is a derivation, it is clear that 𝑑(𝑐)∈𝑍(𝑁). Replacing 𝑥 by 𝑐𝛽𝑥, 𝛽∈Γ, in the hypothesis and using Lemma 3.9(ii), we have 𝑓,𝑑(𝑐)(𝑥)(𝑐)(𝑥).𝑐𝛽𝑥𝛼𝑎=𝑎𝛼𝑓𝑐𝛽𝑥𝛼𝑥𝛽𝑎+𝑐𝛼𝑓𝛽𝑎=𝑎𝛼𝑑𝛽𝑥+𝑎𝛼𝑐𝛽𝑓 (4.20) Since 𝑐∈𝑍(𝑁) and 𝑑(𝑐)∈𝑍(𝑁), we get 𝑑(𝑐)𝛼𝑥𝛽𝑦,𝑎_𝛼=0∀𝑦∈𝑁,𝛼,𝛽,𝛿∈Γ. (4.21) By the primeness of 𝑁 and 0≠𝑑(𝑐)∈𝑍(𝑁), we obtain that 𝑎∈𝑍(𝑁).

Theorem 4.6. Let (𝑓,𝑑) be a generalized derivation of 𝑁, and 𝑎∈𝑁. If [𝑓(𝑥),𝑎]_𝛼=0  for all 𝑥∈𝑁, then 𝑑(𝑎)∈𝑍(𝑁).

Proof. If 𝑎=0, then there is nothing to prove. Hence, we assume that 𝑎≠0.
Replacing 𝑥 by 𝑎𝛽𝑥 in the hypothesis, we have 𝑓,𝑑(𝑎)(𝑥)(𝑎)(𝑥).𝑎𝛽𝑥𝛼𝑎=𝑎𝛼𝑓𝑎𝛽𝑥𝛼𝑥𝛽𝑎+𝑎𝛼𝑓𝛽𝑎=𝑎𝛼𝑑𝛽𝑥+𝑎𝛼𝑎𝛽𝑓 (4.22) Using 𝑓(𝑥)𝛼𝑎=𝑎𝛼𝑓(𝑥), we have 𝑑(𝑎)𝛼𝑥𝛽𝑎=𝑎𝛼𝑑(𝑎)𝛽𝑥∀𝑥∈𝑁,𝛼,𝛽∈Γ. (4.23) Taking 𝑥𝛿𝑦 instead of 𝑥 in the last equation and using this, we conclude that 𝑑(𝑎)𝛼𝑁𝛽𝑎,𝑦_𝛼=0∀𝑦∈𝑁,𝛼,𝛽∈Γ. (4.24) Since 𝑁 is a prime Γ-near-ring, we have either 𝑑(𝑎)=0 or 𝑎∈𝑍(𝑁). If 0≠𝑎∈𝑍(𝑁), then (𝑁,+) is abelian by Lemma 3.2(ii). Thus 𝑓()()𝑓(𝑥)(𝑎)(𝑎)(𝑥)𝑥𝛼𝑎=𝑓𝑎𝛼𝑥𝛼𝑎+𝑥𝛼𝑑=𝑑𝛼𝑥+𝑎𝛼𝑓 (4.25) and so [𝑑(𝑎),𝑥]_𝛼=0∀𝑥∈𝑁,𝛼∈Γ. (4.26)
That is, 𝑑(𝑎)∈𝑍(𝑁). Hence in either case we have 𝑑(𝑎)∈𝑍(𝑁). This completes the proof.

Theorem 4.7. Let (𝑓,𝑑) be a generalized derivation of 𝑁. If 𝑁 is a 2-torsion free Γ-near-ring and 𝑓²(𝑁)⊂𝑍(𝑁), then 𝑁 is a commutative Γ-ring.

Proof. Suppose that 𝑓²(𝑁)⊂𝑍(𝑁). Then we get 𝑓²𝑥𝛼𝑦=𝑓²(𝑥)𝛼𝑦+2𝑓(𝑥)𝛼𝑑𝑦+𝑥𝛼𝑑²𝑦∈𝑍(𝑁)∀𝑥,𝑦∈𝑁,𝛼∈Γ. (4.27) In particular, 𝑓²(𝑥)𝛼𝑐+2𝑓(𝑥)𝛼𝑑(𝑐)+𝑥𝛼𝑑²(𝑐)∈𝑍(𝑁) for all 𝑥∈𝑁, 𝑐∈𝑍(𝑁), 𝛼∈Γ. Since the first summand is an element of 𝑍(𝑁), we have 2𝑓(𝑥)𝛼𝑑(𝑐)+𝑥𝛼𝑑²(𝑐)∈𝑍(𝑁)∀𝑥∈𝑁,𝑐∈𝑍(𝑁),𝛼∈Γ. (4.28) Taking 𝑓(𝑥) instead of 𝑥 in (4.28), we obtain that 2𝑓²(𝑥)𝛼𝑑(𝑐)+𝑓(𝑥)𝛼𝑑²(𝑐)∈𝑍(𝑁)∀𝑥∈𝑁,𝑐∈𝑍(𝑁),𝛼∈Γ. (4.29) Since 𝑑(𝑐)∈𝑍(𝑁), 𝑓²(𝑥)∈𝑍(𝑁), and so 𝑓²(𝑥)𝛼𝑑(𝑐)∈𝑍(𝑁) for all 𝑥∈𝑁, 𝑐∈𝑍(𝑁), 𝛼∈Γ, we conclude 𝑓(𝑥)𝛼𝑑²(𝑐)∈𝑍(𝑁) for all 𝑥∈𝑁, 𝑐∈𝑍(𝑁), 𝛼∈Γ.
Since 𝑁 is prime, we get 𝑑²(𝑍(𝑁))=0 or 𝑓(𝑁)⊆𝑍(𝑁). If 𝑓(𝑁)⊆𝑍(𝑁), then 𝑁 is a commutative Γ-ring by Lemma 3.8. Hence, we assume 𝑑²(𝑍)=0. By (4.28), we get 2𝑓(𝑥)𝛼𝑑(𝑐)∈𝑍(𝑁) for all 𝑥∈𝑁, 𝑐∈𝑍(𝑁), 𝛼∈Γ.
Since 𝑁 is a 2-torsion free near-ring and 𝑑(𝑐)∈𝑍(𝑁), we obtain that either 𝑓(𝑁)⊂𝑍(𝑁) or 𝑑(𝑍(𝑁))=0. If 𝑓(𝑁)⊂𝑍(𝑁), then we are already done. So, we may assume that 𝑑(𝑍(𝑁))=0. Then 𝑓()(),𝑓(𝑐)(𝑥)(𝑥)(𝑐),𝑐𝛼𝑥=𝑓𝑥𝛼𝑐𝛼𝑥+𝑐𝛼𝑑=𝑓𝛼𝑐+𝑥𝛼𝑑 (4.30) and so 𝑓(𝑐)𝛼𝑥+𝑐𝛼𝑑(𝑥)=𝑓(𝑥)𝛼𝑐∀𝑥∈𝑁,𝑐∈𝑍(𝑁). (4.31) Now replacing 𝑥 by 𝑓(𝑥) in (4.31), and using the fact that 𝑓²(𝑁)⊂𝑍(𝑁), we get 𝑓(𝑐)𝛼𝑓(𝑥)+𝑐𝛼𝑑𝑓(𝑥)=𝑓²(𝑥)𝛼𝑐∀𝑥∈𝑁,𝑐∈𝑍(𝑁). (4.32) That is, 𝑓(𝑐)𝛼𝑓(𝑥)+𝑐𝛼𝑑𝑓(𝑥)∈𝑍∀𝑥∈𝑁,𝑐∈𝑍(𝑁),𝛼∈Γ. (4.33) Again taking 𝑓(𝑥) instead of 𝑥 in this equation, one can obtain 𝑓(𝑐)𝛼𝑓²(𝑥)+𝑐𝛼𝑑𝑓²(𝑥)∈𝑍∀𝑥∈𝑁,𝑐∈𝑍(𝑁),𝛼∈Γ. (4.34) The second term is equal to zero because of 𝑑(𝑍)=0. Hence we have 𝑓(𝑐)𝛼𝑓²(𝑥)∈𝑍(𝑁) for all 𝑥∈𝑁, 𝑐∈𝑍(𝑁), 𝛼∈Γ.
Since 𝑓²(𝑁)⊂𝑍(𝑁) by the hypothesis, we get either 𝑓²(𝑁)=0 or 𝑓(𝑍(𝑁))⊂𝑍(𝑁). If 𝑓²(𝑁)=0, then the theorem holds by Definition 3.7. If 𝑓(𝑍)⊂𝑍(𝑁), then 𝑓(𝑥𝛼𝑓(𝑐))=𝑓(𝑓(𝑐)𝛼𝑥) for all 𝑥∈𝑁, 𝑐∈𝑍(𝑁), and so 𝑑(𝑥)𝛼𝑓(𝑐)=𝑓(𝑐)𝛼𝑓(𝑥)∀𝑥∈𝑁,𝑐∈𝑍(𝑁). (4.35) Using 𝑓(𝑐)∈𝑍(𝑁), we now have 𝑓(𝑐)𝛼(𝑑(𝑥)−𝑓(𝑥))=0 for all 𝑥∈𝑁, 𝑐∈𝑍(𝑁), 𝛼∈Γ. Since 𝑓(𝑍(𝑁))⊂𝑍(𝑁), we have either 𝑓(𝑍(𝑁))=0 or 𝑑=𝑓. If 𝑑=𝑓, then 𝑓 is a derivation of 𝑁 and so 𝑁 is commutative Γ-ring by Lemma 3.11.
Now assume that 𝑓(𝑍(𝑁))=0. Returning to the equation (4.31), we have 𝑐𝛼𝑑(𝑥)−𝑓(𝑥)=0∀𝑥∈𝑁,𝑐∈𝑍(𝑁),𝛼∈Γ. (4.36) Since 𝑐∈𝑍(𝑁), we have either 𝑑=𝑓 or 𝑍(𝑁)=0. Clearly, 𝑑=𝑓 implies the theorem holds. If 𝑍(𝑁)=0, then 𝑓²(𝑁)=0 by the hypothesis, and so 𝑁 is a commutative Γ-ring by Lemma 3.4(iv). Hence, the proof is completed.

Corollary 4.8. Let 𝑁 be a 2-torsion free near-ring, and (𝑓,𝑑) a nonzero generalized derivation of 𝑁. If [𝑓(𝑁),𝑓(𝑁)]_𝛼=0, 𝛼∈Γ, then 𝑁 is a commutative Γ-ring.

Lemma 4.9. Let (𝑓,𝑑) and (𝑔,ℎ) be two generalized derivations of 𝑁. If h is a nonzero derivation on 𝑁 and 𝑓(𝑥)𝛼ℎ(𝑦)=−𝑔(𝑥)𝛼𝑑(𝑦) for all 𝑥,𝑦∈𝑁, then (𝑁,+) is abelian.

Proof. Suppose that 𝑓(𝑥)𝛼ℎ(𝑦)+𝑔(𝑥)𝛼𝑑(𝑦)=0 for all 𝑥,𝑦∈𝑁, 𝛼∈Γ.
We substitute 𝑦+𝑧 for 𝑦, thereby obtaining 𝑓(𝑥)𝛼ℎ𝑦+𝑓(𝑥)𝛼ℎ(𝑧)+𝑔(𝑥)𝛼𝑑𝑦+𝑔(𝑥)𝛼𝑑(𝑧)=0. (4.37) Using the hypothesis, we get 𝑓(𝑥)𝛼ℎ𝑦,𝑧=0∀𝑥,𝑦,𝑧∈𝑁,𝛼∈Γ. (4.38) It follows by Lemma 3.10(ii) that ℎ(𝑦,𝑧)=0 for all 𝑦,𝑧∈𝑁. For any 𝑤∈𝑁, we have ℎ(𝑤𝛼𝑦,𝑤𝛼𝑧)=ℎ(𝑤𝛼(𝑦,𝑧))=ℎ(𝑤)𝛼(𝑦,𝑧)+𝑤𝛼ℎ(𝑦,𝑧)=0 and so ℎ(𝑤)𝛼(𝑦,𝑧)=0 for all 𝑤,𝑦,𝑧∈𝑁, 𝛼∈Γ.
An appeal to Lemma 3.4(iii) yields that (𝑁,+) is abelian.

Theorem 4.10. Let (𝑓,𝑑) and (𝑔,ℎ) be two generalized derivations of 𝑁. If 𝑁 is 2-torsion free and 𝑓(𝑥)𝛼ℎ(𝑦)=−𝑔(𝑥)𝛼𝑑(𝑦) for all 𝑥,𝑦∈𝑁,𝛼∈Γ, then 𝑓=0 or 𝑔=0.

Proof. If ℎ=0 or 𝑑=0, then the proof of the theorem is obvious. So, we may assume that ℎ≠0 and 𝑑≠0. Therefore, we know that (𝑁,+) is abelian by Lemma 4.9.
Now suppose that 𝑓(𝑥)𝛼ℎ𝑦+𝑔(𝑥)𝛼𝑑𝑦=0∀𝑥,𝑦∈𝑁,𝛼∈Γ. (4.39) Replacing 𝑥 by 𝑢𝛽𝑣 in this equation and using the hypothesis, we get 𝑓𝑦𝑦(𝑣)𝑦(𝑢)𝑦(𝑣)𝑦(𝑢)𝑦𝑢𝛽𝑣𝛼ℎ+𝑔𝑢𝛽𝑣𝛼𝑑=𝑢𝛼𝑓𝛽ℎ+𝑑𝛼𝑣𝛽ℎ+𝑢𝛼𝑔𝛽𝑑+ℎ𝛼𝑣𝛽𝑑=0, (4.40) and so 𝑑(𝑢)𝛼𝑣𝛽ℎ𝑦=−ℎ(𝑢)𝛼𝑣𝛽𝑑𝑦∀𝑢,𝑣,𝑦∈𝑁,𝛼∈Γ. (4.41) Taking 𝑦𝛿𝑡 instead of 𝑦 in the above relation, we obtain 𝑑(𝑢)𝛼𝑣𝛽ℎ𝑦𝛿𝑡+𝑑(𝑢)𝛼𝑣𝑦𝛽𝛿ℎ(𝑡)=−ℎ(𝑢)𝛼𝑣𝛽𝑑𝑦𝛿𝑡−ℎ(𝑢)𝛼𝑣𝛽𝑦𝛿𝑑(𝑡). (4.42) That is, 𝑑(𝑢)𝛼𝑣𝛽𝑦𝛿ℎ(𝑡)=−ℎ(𝑢)𝛼𝑣𝛽𝑦𝛿𝑑(𝑡)∀𝑢,𝑣,𝑦,𝑡∈𝑁,𝛼,𝛽,𝛿∈Γ. (4.43) Replacing 𝑦 by ℎ(𝑦) in (4.43) and using this relation, we have ℎ(𝑢)𝛼𝑁𝛽𝑑𝑦𝛿ℎ(𝑡)−ℎ𝑦𝛼𝑑(𝑡)=0∀𝑢,𝑦,𝑡∈𝑁. (4.44) Since 𝑁 is a prime Γ-near-ring and ℎ≠0, we obtain that 𝑑𝑦𝛼ℎ(𝑡)=ℎ𝑦𝛼𝑑(𝑡),∀𝑦,𝑡∈𝑁. (4.45) Now again taking 𝑢𝜆𝑣 instead of 𝑥 in the initial hypothesis, we get 𝑓(𝑢)𝛼𝑣𝛽ℎ𝑦+𝑢𝛼𝑑(𝑣)𝛽ℎ𝑦+𝑔(𝑢)𝛼𝑣𝛽𝑑𝑦+𝑢𝛼ℎ(𝑣)𝛽𝑑𝑦=0. (4.46) Using (4.45) yields that 𝑓(𝑢)𝛼𝑣𝛽ℎ𝑦+2𝑢𝛼ℎ(𝑣)𝛽𝑑𝑦+𝑔(𝑢)𝛼𝑣𝛽𝑑𝑦=0∀𝑢,𝑣,𝑦∈𝑁, (4.47) Taking ℎ(𝑣) instead of 𝑣 in this equation, we arrive at 𝑓(𝑢)𝛼ℎ(𝑣)𝛽ℎ𝑦+2𝑢𝛼ℎ²(𝑣)𝛽𝑑𝑦+𝑔(𝑢)𝛼ℎ(𝑣)𝛽𝑑𝑦=0. (4.48) By the hypothesis and (4.45), we have 0=−𝑔(𝑢)𝛼𝑑(𝑣)𝛽ℎ𝑦+2𝑢𝛼ℎ²(𝑣)𝑦(𝑢)(𝑣)𝑦(𝑢)(𝑣)𝑦𝛽𝑑+𝑔𝛼ℎ𝛽𝑑=−𝑔𝛼ℎ𝛽𝑑+2𝑢𝛼ℎ²(𝑣)𝛽𝑑𝑦+𝑔(𝑢)𝛼ℎ(𝑣)𝛽𝑑𝑦, (4.49) and so 2𝑢𝛼ℎ²(𝑣)𝛽𝑑𝑦=0∀𝑢,𝑣,𝑦∈𝑁,𝛼,𝛽∈Γ. (4.50) Since 𝑁 is a 2-torsion free prime Γ-near-ring, we obtain that ℎ²(𝑁)Γ𝑑(𝑁)=0. An appeal to Lemma 3.4(iii) and (iv) gives that ℎ=0. This contradicts by our assumption. Thus the proof is completed.

Theorem 4.11. Let (𝑓,𝑑) and (𝑔,ℎ) be two generalized derivations of 𝑁. If (𝑓𝑔,𝑑ℎ) acts as a generalized derivation on 𝑁, then 𝑓=0 or 𝑔=0.

Proof. By calculating 𝑓𝑔(𝑥𝛼𝑦) in two different ways, we see that 𝑔(𝑥)𝛼𝑑(𝑦)+𝑓(𝑥)𝛼ℎ(𝑦)=0 for all 𝑥,𝑦∈𝑁, 𝛼∈Γ. The proof is completed by using Theorem 4.10.

Acknowledgments

The paper was supported by Grant 01-12-10-978FR MOHE Malaysia. The authors are thankful to the referee for valuable comments.