Generalized Derivations on Prime Near Rings

Abstract

Let 𝑁 be a near ring. An additive mapping 𝑓∶𝑁→𝑁 is said to be a right generalized (resp., left generalized) derivation with associated derivation 𝑑 on 𝑁 if 𝑓(𝑥𝑦)=𝑓(𝑥)𝑦+𝑥𝑑(𝑦) (resp., 𝑓(𝑥𝑦)=𝑑(𝑥)𝑦+𝑥𝑓(𝑦)) for all 𝑥,𝑦∈𝑁. A mapping 𝑓∶𝑁→𝑁 is said to be a generalized derivation with associated derivation 𝑑 on 𝑁 if 𝑓 is both a right generalized and a left generalized derivation with associated derivation 𝑑 on 𝑁. The purpose of the present paper is to prove some theorems in the setting of a semigroup ideal of a 3-prime near ring admitting a generalized derivation, thereby extending some known results on derivations.

1. Introduction

Throughout the paper, 𝑁 denotes a zero-symmetric left near ring with multiplicative center 𝑍, and for any pair of elements 𝑥,𝑦∈𝑁, [𝑥,𝑦] denotes the commutator 𝑥𝑦−𝑦𝑥, while the symbol (𝑥,𝑦) denotes the additive commutator 𝑥+𝑦−𝑥−𝑦. A near ring 𝑁 is called zero-symmetric if 0𝑥=0, for all 𝑥∈𝑁 (recall that left distributivity yields that 𝑥0=0). The near ring 𝑁 is said to be 3-prime if 𝑥𝑁𝑦={0} for 𝑥,𝑦∈𝑁 implies that 𝑥=0 or 𝑦=0. A near ring 𝑁 is called 2-torsion-free if (𝑁,+) has no element of order 2. A nonempty subset 𝐴 of 𝑁 is called a semigroup right (resp., semigroup left) ideal if 𝐴𝑁⊆𝐴 (resp., 𝑁𝐴⊆𝐴), and if 𝐴 is both a semigroup right ideal and a semigroup left ideal, it is called a semigroup ideal. An additive mapping 𝑑∶𝑁→𝑁 is a derivation on 𝑁 if 𝑑(𝑥𝑦)=𝑑(𝑥)𝑦+𝑥𝑑(𝑦), for all 𝑥,𝑦∈𝑁. An additive mapping 𝑓∶𝑁→𝑁 is said to be a right (resp., left) generalized derivation with associated derivation 𝑑 if 𝑓(𝑥𝑦)=𝑓(𝑥)𝑦+𝑥𝑑(𝑦) (resp., 𝑓(𝑥𝑦)=𝑑(𝑥)𝑦+𝑥𝑓(𝑦)), for all 𝑥,𝑦∈𝑁, and 𝑓 is said to be a generalized derivation with associated derivation 𝑑 on 𝑁 if it is both a right generalized derivation and a left generalized derivation on 𝑁 with associated derivation 𝑑. (Note that this definition differs from the one given by Hvala in [1]; his generalized derivations are our right generalized derivations.) Every derivation on 𝑁 is a generalized derivation.

In the case of rings, generalized derivations have received significant attention in recent years. We prove some theorems in the setting of a semigroup ideal of a 3-prime near ring admitting a generalized derivation and thereby extend some known results [2, Theorem 2.1], [3, Theorem 3.1], [4, Theorem 3], and [5, Theorem 3.3].

2. Preliminary Results

We begin with several lemmas, most of which have been proved elsewhere.

Lemma 1 (see [3, Lemma 1.3]). Let 𝑁 be a 3-prime near ring and 𝑑 be a nonzero derivation on 𝑁.(i)If𝑈 is a nonzero semigroup right ideal or a nonzero semigroup left ideal of 𝑁, then 𝑑(𝑈)≠{0}.(ii)If𝑈 is a nonzero semigroup right ideal of 𝑁 and 𝑥 is an element of 𝑁 which centralizes 𝑈, then 𝑥∈𝑍.

Lemma 2 (see [3, Lemma 1.2]). Let 𝑁 be a 3-prime near ring.(i) If𝑧∈𝑍⧵{0}, then 𝑧 is not a zero divisor.(ii) If𝑍⧵{0} contains an element 𝑧 for which 𝑧+𝑧∈𝑍, then (𝑁,+) is abelian.(iii) If𝑧∈𝑍⧵{0} and 𝑥 is an element of 𝑁 such that 𝑥𝑧∈𝑍, then 𝑥∈𝑍.

Lemma 3 (see [3, Lemmas 1.3 and 1.4]). Let 𝑁 be a 3-prime near ring and 𝑈 be a nonzero semigroup ideal of 𝑁. Let 𝑑 be a nonzero derivation on 𝑁.(i) If 𝑥,𝑦∈𝑁 and 𝑥𝑈𝑦={0}, then 𝑥=0 or 𝑦=0.(ii) If 𝑥∈𝑁 and 𝑥𝑈={0} or 𝑈𝑥={0}, then 𝑥=0.(iii) If 𝑥∈𝑁 and 𝑑(𝑈)𝑥={0} or 𝑥𝑑(𝑈)={0}, then 𝑥=0.

Lemma 4 (see [3, Lemma 1.5]). If 𝑁 is a 3-prime near ring and 𝑍 contains a nonzero semigroup left ideal or a semigroup right ideal, then 𝑁 is a commutative ring.

Lemma 5. If 𝑓 is a generalized derivation on 𝑁 with associated derivation 𝑑, then (𝑑(𝑥)𝑦+𝑥𝑓(𝑦))𝑧= 𝑑(𝑥)𝑦𝑧+𝑥𝑓(𝑦)𝑧, for all 𝑥,𝑦,𝑧∈𝑁.

Proof. We prove only (ii), since (i) is proved in [2]. For all 𝑥,𝑦,𝑧∈𝑁 we have 𝑓((𝑥𝑦)𝑧)= 𝑓(𝑥𝑦)𝑧+𝑥𝑦𝑑(𝑧)=(𝑑(𝑥)𝑦+𝑥𝑓(𝑦))𝑧+𝑥𝑦𝑑(𝑧) and 𝑓(𝑥(𝑦𝑧))=𝑑(𝑥)𝑦𝑧+𝑥𝑓(𝑦𝑧)= 𝑑(𝑥)𝑦𝑧+𝑥𝑓(𝑓(𝑦)𝑧+ 𝑦𝑑(𝑧))= 𝑑(𝑥)𝑦𝑧+𝑥𝑓(𝑦)𝑧+𝑥𝑦𝑑(𝑧). Comparing the two expressions for 𝑓(𝑥𝑦𝑧) gives (ii).

Lemma 6. Let 𝑁 be a 3-prime near ring and 𝑓 a generalized derivation with associated derivation 𝑑.(i)𝑓(𝑥)𝑦+𝑥𝑑(𝑦)=𝑥𝑑(𝑦)+𝑓(𝑥)𝑦 for all 𝑥,𝑦∈𝑁.(ii)𝑑(𝑥)𝑦+𝑥𝑓(𝑦)=𝑥𝑓(𝑦)+𝑑(𝑥)𝑦 for all 𝑥,𝑦∈𝑁.

Proof. (i) 𝑓(𝑥(𝑦+𝑦))=𝑓(𝑥)(𝑦+𝑦)+𝑥𝑑(𝑦+𝑦)=𝑓(𝑥)𝑦+𝑓(𝑥)𝑦+𝑥𝑑(𝑦)+𝑥𝑑(𝑦), and 𝑓(𝑥𝑦+𝑥𝑦)= 𝑓(𝑥)𝑦+𝑥𝑑(𝑦)+𝑓(𝑥)𝑦+𝑥𝑑(𝑦). Comparing these two equations gives the desired result.
(ii) Again, calculate 𝑓(𝑥(𝑦+𝑦)) and 𝑓(𝑥𝑦+𝑥𝑦) and compare.

Lemma 7. Let 𝑁 be a 3-prime near ring and 𝑓 a generalized derivation with associated derivation 𝑑. Then 𝑓(𝑍)⊆𝑍.

Proof. Let 𝑧∈𝑍 and 𝑥∈𝑁. Then 𝑓(𝑧𝑥)=𝑓(𝑥𝑧); that is, 𝑓(𝑧)𝑥+𝑧𝑑(𝑥)=𝑑(𝑥)𝑧+𝑥𝑓(𝑧). Applying Lemma 6(i), we get 𝑧𝑑(𝑥)+𝑓(𝑧)𝑥=𝑑(𝑥)𝑧+𝑥𝑓(𝑧). It follows that 𝑓(𝑧)𝑥=𝑥𝑓(𝑧) for all 𝑥∈𝑁, so 𝑓(𝑧)∈𝑍.

Lemma 8. Let 𝑁 be a 3-prime near ring and 𝑈 a nonzero semigroup ideal of 𝑁. If 𝑓 is a nonzero right generalized derivation of 𝑁 with associated derivation 𝑑, then 𝑓(𝑈)≠{0}.

Proof. Suppose 𝑓(𝑈)={0}. Then 𝑓(𝑢𝑥)=𝑓(𝑢)𝑥+𝑢𝑑(𝑥)=0=𝑢𝑑(𝑥) for all 𝑢∈𝑈 and 𝑥∈𝑁, and it follows by Lemma 3(ii) that 𝑑=0. Therefore 𝑓(𝑥𝑢)=𝑓(𝑥)𝑢=0 for all 𝑢∈𝑈 and 𝑥∈𝑁, and another appeal to Lemma 3(ii) gives 𝑓=0, which is a contradiction.

Lemma 9 (see [2, Theorem 4.2]). Let 𝑁 be a 3-prime near ring with a nonzero right generalized derivation 𝑓 with associated derivation 𝑑. If 𝑓(𝑁)⊆𝑍 then (𝑁,+) is abelian. Moreover, if 𝑁 is 2-torsion-free, then 𝑁 is a commutative ring.

Lemma 10 (see [2, Theorem 4.1]). Let 𝑁 be a 2-torsion-free 3-prime near ring and 𝑓 a nonzero generalized derivation on 𝑁 with associated derivation 𝑑. If [𝑓(𝑁),𝑓(𝑁)]={0}, then 𝑁 is a commutative ring.

3. Main Results

The theorems that we prove in this section extend the results proved in [2, Theorems 2.1 and 3.1], [3, Theorems 2.1, 3.1, and 3.3], and [5, Theorem 3.3].

Theorem 11. Let 𝑁 be a 3-prime near ring and 𝑈 be a nonzero semigroup ideal of 𝑁. Let 𝑓 be a nonzero right generalized derivation with associated derivation 𝑑. If 𝑓(𝑈)⊆𝑍, then (𝑁,+) is abelian. Moreover, if 𝑁 is 2-torsion-free, then 𝑁 is a commutative ring.

Proof. We begin by showing that (𝑁,+) is abelian, which by Lemma 2(ii) is accomplished by producing 𝑧∈𝑍⧵{0} such that 𝑧+𝑧∈𝑍. Let 𝑎 be an element of 𝑈 such that 𝑓(𝑎)≠0. Then for all 𝑥∈𝑁, 𝑎𝑥∈𝑈 and 𝑎𝑥+𝑎𝑥=𝑎(𝑥+𝑥)∈𝑈, so that 𝑓(𝑎𝑥)∈𝑍 and 𝑓(𝑎𝑥)+𝑓(𝑎𝑥)∈𝑍; hence we need only to show that there exists 𝑥∈𝑁 such that 𝑓(𝑎𝑥)≠0. Suppose that this is not the case, so that 𝑓((𝑎𝑥)𝑎)=0= 𝑓(𝑎𝑥)𝑎+𝑎𝑥𝑑(𝑎)= 𝑎𝑥𝑑(𝑎), for all 𝑥∈𝑁. By Lemma 3(i) either 𝑎=0 or 𝑑(𝑎)=0.
If 𝑑(𝑎)=0, then 𝑓(𝑥𝑎)=𝑓(𝑥)𝑎+𝑥𝑑(𝑎);thatis,𝑓(𝑥𝑎)=𝑓(𝑥)𝑎∈𝑍, for all 𝑥∈𝑁. Thus [𝑓(𝑢)𝑎,𝑦]=0, for all 𝑦∈𝑁, and 𝑢∈𝑈. This implies that 𝑓(𝑢)[𝑎,𝑦]=0, for all 𝑢∈𝑈 and 𝑦∈𝑁 and Lemma 2(i) gives 𝑎∈𝑍. Thus 0=𝑓(𝑎𝑥)=𝑓(𝑥𝑎)=𝑓(𝑥)𝑎, for all 𝑥∈𝑁. Replacing 𝑥 by 𝑢∈𝑈, we have 𝑓(𝑈)𝑎=0, and by Lemmas 2(i) and 8, we get that 𝑎=0. Thus, we have a contradiction.
To complete the proof, we show that if 𝑁 is 2-torsion-free, then 𝑁 is commutative.
Consider first the case 𝑑=0. This implies that 𝑓(𝑢𝑥)=𝑓(𝑢)𝑥∈𝑍 for all 𝑢∈𝑈 and 𝑥∈𝑁. By Lemma 8, we have 𝑢∈𝑈 such that 𝑓(𝑢)∈𝑍⧵{0}, so 𝑁 is commutative by Lemma 2(iii).
Now consider the case 𝑑≠0. Let 𝑐∈𝑍⧵{0}. This implies that if 𝑥∈𝑈, 𝑓(𝑥𝑐)=𝑓(𝑥)𝑐+𝑥𝑑(𝑐)∈𝑍. Thus (𝑓(𝑥)𝑐+𝑥𝑑(𝑐))𝑦=𝑦(𝑓(𝑥)𝑐+𝑥𝑑(𝑐)) for all 𝑥,𝑦∈𝑈 and 𝑐∈𝑍. Therefore by Lemma 5(i), 𝑓(𝑥)𝑐𝑦+𝑥𝑑(𝑐)𝑦=𝑦𝑓(𝑥)𝑐+𝑦𝑥𝑑(𝑐) for all 𝑥,𝑦∈𝑈 and 𝑐∈𝑍. Since 𝑑(𝑐)∈𝑍 and 𝑓(𝑥)∈𝑍, we obtain 𝑑(𝑐)[𝑥,𝑦]=0, for all 𝑥,𝑦∈𝑈 and 𝑐∈𝑍. Let 𝑑(𝑍)≠0. Choosing 𝑐 such that 𝑑(𝑐)≠0 and noting that 𝑑(𝑐) is not a zero divisor, we have [𝑥,𝑦]=0 for all 𝑥,𝑦∈𝑈. By Lemma 1(ii), 𝑈⊆𝑍; hence 𝑁 is commutative by Lemma 4.
The remaining case is 𝑑≠0 and 𝑑(𝑍)={0}. Suppose we can show that 𝑈∩𝑍≠{0}. Taking 𝑧∈(𝑈∩𝑍)⧵{0} and 𝑥∈𝑁, we have 𝑓(𝑥𝑧)=𝑓(𝑥)𝑧∈𝑍; therefore 𝑓(𝑁)⊆𝑍 by Lemma 2(iii) and 𝑁 is commutative by Lemma 9.
Assume, then, that 𝑈∩𝑍={0}. For each 𝑢∈𝑈, 𝑓(𝑢²)=𝑓(𝑢)𝑢+𝑢𝑑(𝑢)=𝑢(𝑓(𝑢)+𝑑(𝑢))∈𝑈∩𝑍, so 𝑓(𝑢²)=0. Thus, for all 𝑢∈𝑈 and 𝑥∈𝑁, 𝑓(𝑢²𝑥)=𝑓(𝑢²)𝑥+𝑢²𝑑(𝑥)=𝑢²𝑑(𝑥)∈𝑈∩𝑍, so 𝑢²𝑑(𝑥)=0, and by Lemma 3(iii)  𝑢²=0. Since 𝑓(𝑥𝑢)=𝑓(𝑥)𝑢+𝑥𝑑(𝑢)∈𝑍 for all 𝑢∈𝑈 and 𝑥∈𝑁, we have (𝑓(𝑥)𝑢+𝑥𝑑(𝑢))𝑢=𝑢(𝑓(𝑥)𝑢+𝑥𝑑(𝑢)), and right multiplying by 𝑢 gives 𝑢𝑥𝑑(𝑢)𝑢=0. Consequently 𝑑(𝑢)𝑢𝑁𝑑(𝑢)𝑢={0}, so that 𝑑(𝑢)𝑢=0 for all 𝑢∈𝑈. Since 𝑢²=0, 𝑑(𝑢²)=𝑑(𝑢)𝑢+𝑢𝑑(𝑢)=0 for all 𝑢∈𝑈, so 𝑓(𝑢)𝑢=0 for all 𝑢∈𝑈. But by Lemma 8, there exists 𝑢₀∈𝑈 for which 𝑓(𝑢₀)≠0; and since 𝑓(𝑢₀)∈𝑍, we get 𝑢₀=0—a contradiction. Therefore 𝑈∩𝑍≠{0} as required.

Theorem 12. Let 𝑁 be a 3-prime near ring with a nonzero generalized derivation 𝑓 with associated nonzero derivation 𝑑. Let 𝑈 be a nonzero semigroup ideal of 𝑁. If [𝑓(𝑈),𝑓(𝑈)]=0, then (𝑁,+) is abelian.

Proof. Assume that 𝑥∈𝑁 is such that [𝑥,𝑓(𝑈)]=[𝑥+𝑥,𝑓(𝑈)]=0. For all 𝑢,𝑣∈𝑈 such that 𝑢+𝑣∈𝑈, [𝑥+𝑥,𝑓(𝑢+𝑣)]=0.
This implies that ()𝑓()(),()𝑓(𝑢)+()𝑓(𝑣)()()𝑓(𝑢)(𝑣)()(),𝑓(𝑢)(𝑢)(𝑣)(𝑣)𝑥(𝑢)(𝑣)(𝑢)(𝑣),(𝑢)(𝑢)(𝑣)(𝑣)(𝑢)(𝑣)(𝑢)(𝑣),𝑥𝑓(𝑢)(𝑣)(𝑢)(𝑣)()𝑥+𝑥𝑢+𝑣=𝑓𝑢+𝑣)(𝑥+𝑥𝑥+𝑥𝑥+𝑥=𝑓𝑢+𝑣𝑥+𝑓𝑢+𝑣𝑥,)(𝑥+𝑥+𝑓)(𝑥+𝑥=𝑥𝑓𝑢+𝑣+𝑥𝑓𝑢+𝑣𝑥+𝑓𝑥+𝑓𝑥+𝑓=𝑥𝑓+𝑥𝑓+𝑥𝑓+𝑥𝑓𝑥𝑓+𝑥𝑓+𝑥𝑓+𝑥𝑓=𝑥𝑓+𝑥𝑓+𝑥𝑓+𝑥𝑓+𝑓−𝑓−𝑓=0,𝑥𝑓𝑢+𝑣−𝑢−𝑣=0. (1) This equation may be restated as 𝑥𝑓(𝑐)=0, where 𝑐=(𝑢,𝑣).
Let 𝑎,𝑏∈𝑈. Then 𝑎𝑏∈𝑈 and 𝑎𝑏+𝑎𝑏=𝑎(𝑏+𝑏)∈𝑈, so [𝑓(𝑎𝑏),𝑓(𝑈)]={0}=[𝑓(𝑎𝑏)+𝑓(𝑎𝑏),𝑓(𝑈)], and by the argument in the previous paragraph, 𝑓(𝑎𝑏)𝑓(𝑐)=0. We now have 𝑓(𝑈²)𝑓(𝑢,𝑣)=0 for all 𝑢,𝑣∈𝑈 such that 𝑢+𝑣∈𝑈. Taking 𝑤∈𝑈² and 𝑥∈𝑁, we get 𝑓(𝑥𝑤)𝑓((𝑢,𝑣))=(𝑑(𝑥)𝑤+ 𝑥𝑓(𝑤))𝑓((𝑢,𝑣))=0=𝑑(𝑥)𝑤𝑓((𝑢,𝑣)), and since 𝑈² is a nonzero semigroup ideal by Lemma 3(ii) and 𝑑≠0, Lemma 3(i) gives 𝑓((𝑢,𝑣))=0∀𝑢,𝑣∈𝑈suchthat𝑢+𝑣∈𝑈. (2) Take 𝑢=𝑟𝑦 and 𝑣=𝑟𝑧, where 𝑟∈𝑈 and 𝑦,𝑧∈𝑁, so that 𝑢+𝑣=𝑟𝑦+𝑟𝑧=𝑟(𝑦+𝑧)∈𝑈. By (2) we have 𝑓𝑟𝑦,𝑟𝑧=𝑓𝑟𝑦,𝑧=0∀𝑟∈𝑈,𝑦,𝑧∈𝑁. (3) Replacing 𝑟 by 𝑟𝑤, 𝑤∈𝑈, we obtain 𝑓(𝑟(𝑤𝑦,𝑤𝑧))=0=𝑑(𝑟)(𝑤(𝑦,𝑧))+𝑟𝑓((𝑤𝑦,𝑤𝑧)); so by (3) 𝑑(𝑟)𝑈(𝑦,𝑧)=0 for all 𝑟∈𝑈 and 𝑦,𝑧∈𝑁. It follows immediately by Lemmas 1(i) and 3(i) that (𝑦,𝑧)=0 for all 𝑦,𝑧∈𝑁; that is, (𝑁,+) is abelian.

Theorem 13. Let 𝑁 be a 2-torsion-free 3-prime near ring and 𝑈 be a nonzero semigroup ideal of 𝑁. If 𝑓 is a nonzero generalized derivation with associated derivation 𝑑 such that [𝑓(𝑈),𝑓(𝑈)]=0, then 𝑁 is a commutative ring if it satisfies one of the following: (i)  𝑑(𝑍)≠{0}; (ii) 𝑈∩𝑍≠{0}; (iii)𝑑=0 and 𝑓(𝑍)≠{0}.

Proof. (i) Let 𝑎∈𝑁 centralizes 𝑓(𝑈), and let 𝑧∈𝑍 such that 𝑑(𝑧)≠0. Then a centralizes 𝑓(𝑢𝑧) for all 𝑢∈𝑈, so that 𝑎(𝑓(𝑢)𝑧+𝑢𝑑(𝑧))=(𝑓(𝑢)𝑧+𝑢𝑑(𝑧))𝑎 and 𝑎𝑢𝑑(𝑧)=𝑢𝑑(𝑧)𝑎. Since 𝑑(𝑧)∈𝑍⧵{0}, 𝑑(𝑧)[𝑎,𝑢]=0=[𝑎,𝑢] for all 𝑢∈𝑈. Therefore a centralizes 𝑈, and by Lemma 1(ii), 𝑎∈𝑍. Since 𝑓(𝑈) centralizes 𝑓(𝑈), 𝑓(𝑈)⊆𝑍 and our result follows by Theorem 11.
(ii) We may assume 𝑑(𝑍)={0}. Let 𝑧∈(𝑈∩𝑍)⧵{0}. Then for all 𝑥,𝑦∈𝑁, 𝑓(𝑥𝑧)=𝑓(𝑥)𝑧 and 𝑓(𝑦𝑧)=𝑓(𝑦)𝑧 commute; hence 𝑧²[𝑓(𝑥),𝑓(𝑦)]=0=[𝑓(𝑥),𝑓(𝑦)]. Our result now follows from Lemma 10.
(iii) Let 𝑢,𝑣∈𝑈 and 𝑧∈𝑍 such that 𝑓(𝑧)≠0. Then [𝑓(𝑧𝑢),𝑓(𝑢)]=0=[𝑓(𝑧)𝑢,𝑓(𝑣)], and since 𝑓(𝑧)∈𝑍⧵{0}, 𝑓(𝑧)[𝑢,𝑓(𝑢)]=0=[𝑢,𝑓(𝑣)]. Thus 𝑓(𝑈) centralizes 𝑈, and by Lemma 1(ii), 𝑓(𝑈)⊆𝑍. Our result now follows by Theorem 11.

We have already observed that if 𝑓 is a generalized derivation with 𝑑=0, then 𝑓(𝑥)𝑦=𝑥𝑓(𝑦) for all 𝑥,𝑦∈𝑁. For 3-prime near rings, we have the following converse.

Theorem 14. Let 𝑁 be a 3-prime near ring and 𝑈 be a nonzero semigroup ideal of 𝑁. If 𝑓 is a nonzero right generalized derivation of 𝑁 with associated derivation 𝑑 and 𝑓(𝑥)𝑦=𝑥𝑓(𝑦), for all 𝑥,𝑦∈𝑈, then 𝑑=0.

Proof. We are given that 𝑓(𝑥)𝑦=𝑥𝑓(𝑦) for all 𝑥,𝑦∈𝑈. Substituting 𝑦𝑧 for 𝑦, we get 𝑓(𝑥)𝑦𝑧=𝑥𝑓(𝑦𝑧)=𝑥(𝑓(𝑦)𝑧+𝑦𝑑(𝑧)) for all 𝑥,𝑦,𝑧∈𝑈. It follows that 𝑥𝑦𝑑(𝑧)=0 for all 𝑥,𝑦,𝑧∈𝑈; that is, 𝑥𝑈𝑑(𝑧)={0} for all 𝑥,𝑧∈𝑈. By Lemma 3(i), 𝑑(𝑈)=0, and hence 𝑑=0 by Lemma 1(i).

4. Generalized Derivations Acting as a Homomorphism or an Antihomomorphism

In [4], Bell and Kappe proved that if 𝑅 is a semiprime ring and 𝑑 is a derivation on 𝑅 which is either an endomorphism or an antiendomorphism on 𝑅, then 𝑑=0. Of course, derivations which are not endomorphisms or antiendomorphisms on 𝑅 may behave as such on certain subsets of 𝑅; for example, any derivation 𝑑 behaves as the zero endomorphism on the subring 𝐶 consisting of all constants (i.e., the elements 𝑥 for which 𝑑(𝑥)=0). In fact in a semiprime ring 𝑅, 𝑑 may behave as an endomorphism on a proper ideal of 𝑅. However as noted in [4], the behaviour of 𝑑 is somewhat restricted in the case of a prime ring. Recently the authors in [6] considered (𝜃,𝜙)-derivation 𝑑 acting as a homomorphism or an antihomomorphism on a nonzero Lie ideal of a prime ring and concluded that 𝑑=0. In this section we establish similar results in the setting of a semigroup ideal of a 3-prime near ring admitting a generalized derivation.

Theorem 15. Let 𝑁 be a 3-prime near ring and 𝑈 be a nonzero semigroup ideal of 𝑁. Let 𝑓 be a nonzero generalized derivation on 𝑁 with associated derivation 𝑑. If 𝑓 acts as a homomorphism on 𝑈, then 𝑓 is the identity map on 𝑁 and 𝑑=0.

Proof. By the hypothesis 𝑓𝑥𝑦=𝑑(𝑥)𝑦+𝑥𝑓𝑦=𝑓(𝑥)𝑓𝑦∀𝑥,𝑦∈𝑈. (4) Replacing 𝑦 by 𝑦𝑧 in the above relation, we get 𝑓𝑥𝑦𝑧=𝑑(𝑥)𝑦𝑧+𝑥𝑓𝑦𝑧∀𝑥,𝑦,𝑧∈𝑈, (5) or 𝑓𝑥𝑦𝑓(𝑧)=𝑑(𝑥)𝑦𝑧+𝑥𝑑𝑦𝑧+𝑦𝑓(𝑧)∀𝑥,𝑦,𝑧∈𝑈. (6) This implies that 𝑑(𝑥)𝑦+𝑥𝑓𝑦𝑓(𝑧)=𝑑(𝑥)𝑦𝑧+𝑥𝑑𝑦𝑧+𝑥𝑦𝑓(𝑧)∀𝑥,𝑦,𝑧∈𝑈. (7) Using Lemma 5(ii), we get 𝑑(𝑥)(𝑧)𝑦𝑓(𝑧)(𝑥)𝑦(𝑧)𝑦𝑓+𝑥𝑓=𝑑𝑦𝑧+𝑥𝑑𝑧+𝑥𝑦𝑓∀𝑥,𝑦,𝑧∈𝑈, (8) or 𝑑(𝑥)(𝑧)(𝑥)𝑦(𝑧)𝑦𝑓+𝑥𝑓𝑦𝑧=𝑑𝑦𝑧+𝑥𝑑𝑧+𝑥𝑦𝑓∀𝑥,𝑦,𝑧∈𝑈. (9) This implies that 𝑑(𝑥)(𝑧)𝑦(𝑧)(𝑥)𝑦(𝑧)𝑦𝑓+𝑥𝑑𝑧+𝑥𝑦𝑓=𝑑𝑦𝑧+𝑥𝑑𝑧+𝑥𝑦𝑓∀𝑥,𝑦,𝑧∈𝑈. (10) That is, 𝑑(𝑥)𝑦𝑓(𝑧)=𝑑(𝑥)𝑦𝑧∀𝑥,𝑦,𝑧∈𝑈. (11) Therefore 𝑑(𝑥)𝑦𝑓(𝑧)−𝑧=0∀𝑥,𝑦,𝑧∈𝑈, (12) which implies that 𝑑(𝑥)𝑈𝑓(𝑧)−𝑧={0}∀𝑥,𝑧∈𝑈. (13) It follows by Lemma 3(i) that either 𝑑(𝑈)=0 or 𝑓(𝑧)=𝑧 for all 𝑧∈𝑈.
In fact, as we now show, both of these conditions hold.
Suppose that 𝑓(𝑢)=𝑢 for all 𝑢∈𝑈. Then for all 𝑢∈𝑈 and 𝑥∈𝑁, 𝑓(𝑥𝑢)=𝑥𝑢=𝑑(𝑥)𝑢+ 𝑥𝑓(𝑢)= 𝑑(𝑥)𝑢+𝑥𝑢; hence 𝑑(𝑥)𝑈={0} for all 𝑥∈𝑁, and thus 𝑑=0.
On the other hand, suppose that 𝑑(𝑈)={0}, so that 𝑑=0. Then for all 𝑥,𝑦∈𝑈, 𝑓(𝑥𝑦)=𝑓(𝑥)𝑦= 𝑓(𝑥)𝑓(𝑦), so that 𝑓(𝑥)(𝑦−𝑓(𝑦))=0. Replacing 𝑦 by 𝑧𝑦, 𝑧∈𝑁, and noting that 𝑓(𝑧𝑦)=𝑧𝑓(𝑦), we see that 𝑓(𝑥)𝑁(𝑦−𝑓(𝑦))={0} for all 𝑥,𝑦∈𝑈. Therefore, 𝑓(𝑈)={0} or 𝑓 is the identity on 𝑈. But 𝑓(𝑈)={0} contradicts Lemma 8, so 𝑓 is the identity on 𝑈.
We now know that 𝑓 is the identity on 𝑈 and 𝑓(𝑥𝑦)=𝑥𝑓(𝑦) for all 𝑥,𝑦∈𝑁. Consequently, 𝑓(𝑢𝑥)=𝑢𝑥=𝑢𝑓(𝑥) for all 𝑢∈𝑈 and 𝑥∈𝑁, so that 𝑈(𝑥−𝑓(𝑥))={0} for all 𝑥∈𝑁. It follows that 𝑓 is the identity on 𝑁.

Theorem 16. Let 𝑁 be a 3-prime near ring and 𝑈 be a nonzero semigroup ideal of 𝑁. If 𝑓 is a nonzero generalized derivation on 𝑁 with associated derivation 𝑑. If 𝑓 acts as an antihomomorphism on 𝑈, then 𝑑=0, 𝑓 is the identity map on 𝑁, and 𝑁 is a commutative ring.

Proof. We begin by showing that 𝑑=0 if and only if 𝑓 is the identity map on 𝑁.
Clearly if 𝑓 is the identity map on 𝑁, 𝑥𝑑(𝑦)=0 for all 𝑥,𝑦∈𝑁, and hence 𝑑=0.
Conversely, assume that 𝑑=0, in which case 𝑓(𝑥𝑦)=𝑓(𝑥)𝑦=𝑥𝑓(𝑦) for all 𝑥,𝑦∈𝑁. It follows that for any 𝑥,𝑦,𝑧∈𝑈, 𝑓(𝑧)𝑓(𝑧)(𝑥)𝑓(𝑥)𝑦𝑓(𝑥).𝑦𝑥𝑧=𝑓𝑦𝑥=𝑓𝑦𝑓=𝑓𝑧𝑦=𝑧𝑓=𝑧𝑓𝑥𝑦 (14) On the other hand, 𝑓()𝑓𝑦(𝑥)𝑦(𝑥)𝑓(𝑥)𝑓𝑦𝑓(𝑧)𝑓(𝑧)𝑦(𝑧)𝑦𝑓()𝑦𝑓(𝑥)𝑦𝑥𝑧=𝑓𝑥𝑧=𝑓𝑧𝑓=𝑓𝑧𝑦=𝑓=𝑓𝑦𝑥=𝑓𝑥𝑓=𝑓𝑥𝑧=𝑓𝑧=𝑓𝑥𝑦𝑧. (15) Comparing (14) and (15) shows that 𝑓(𝑈²) centralizes 𝑈, so that 𝑓(𝑈²)⊆𝑍 by Lemma 1(ii).
Now 𝑈² is a nonzero semigroup ideal by Lemma 3(ii); hence 𝑓(𝑈²)≠0 by Lemma 8. Choosing 𝑥,𝑦∈𝑈 such that 𝑓(𝑥𝑦)≠0, we see that for any 𝑧∈𝑈, 𝑓(𝑥𝑦)𝑧=𝑓(𝑥𝑦𝑧)=𝑓(𝑦𝑧)𝑓(𝑥)= 𝑓(𝑦)𝑧𝑓(𝑥)= 𝑓(𝑦)𝑓(𝑧𝑥)= 𝑓(𝑦)𝑓(𝑥)𝑓(𝑧)= 𝑓(𝑥𝑦)𝑓(𝑧), and hence 𝑓(𝑥𝑦)(𝑧−𝑓(𝑧))=0. Since 𝑓(𝑥𝑦)∈𝑍⧵{0}, we conclude that 𝑓(𝑧)=𝑧 for all 𝑧∈𝑈, and it follows easily that 𝑓 is the identity map on 𝑁.
We note now that if the identity map on 𝑁 acts as an antihomomorphism on 𝑈, then 𝑈 is commutative, so that by Lemmas 1(ii) and 4𝑁 is a commutative ring.
To complete the proof of our theorem, we need only to argue that 𝑑=0. By our antihomomorphism hypothesis 𝑓𝑥𝑦=𝑑(𝑥)𝑦+𝑥𝑓𝑦=𝑓𝑦𝑓(𝑥)∀𝑥,𝑦∈𝑈. (16) Replacing 𝑦 by 𝑥𝑦 in the above relation, we get 𝑓𝑥𝑦𝑓(𝑥)=𝑓𝑥𝑥𝑦=𝑑(𝑥)𝑥𝑦+𝑥𝑓𝑥𝑦∀𝑥,𝑦∈𝑈. (17) This implies that 𝑑(𝑥)𝑦𝑓(𝑥)(𝑥)𝑦𝑓(𝑥)𝑦+𝑥𝑓=𝑑𝑥𝑦+𝑥𝑓∀𝑥,𝑦∈𝑈. (18) Using Lemma 5(ii), we get 𝑑(𝑥)(𝑥)𝑦𝑓(𝑥)(𝑥)𝑦𝑓(𝑥)𝑦𝑓+𝑥𝑓=𝑑𝑥𝑦+𝑥𝑓∀𝑥,𝑦∈𝑈. (19) Thus 𝑑(𝑥)𝑦𝑓(𝑥)=𝑑(𝑥)𝑥𝑦∀𝑥,𝑦∈𝑈. (20) Replacing 𝑦 by 𝑦𝑟 in (20) and using (20), we get 𝑑(𝑥)𝑦𝑟𝑓(𝑥)=𝑑(𝑥)𝑥𝑦𝑟,andso𝑑(𝑥)𝑦𝑟,𝑓(𝑥)=0∀𝑥,𝑦∈𝑈,𝑟∈𝑁. (21) Application of Lemma 3(i) yields that for each 𝑥∈𝑈 either 𝑑(𝑥)=0 or [𝑟,𝑓(𝑥)]=0; that is 𝑑(𝑥)=0 or 𝑓(𝑥)∈𝑍.
Suppose that there exists 𝑤∈𝑈 such that 𝑓(𝑤)∈𝑍⧵{0}. Then for all 𝑣∈𝑈 such that 𝑑(𝑣)=0, 𝑓(𝑤𝑣)=𝑓(𝑤)𝑣=𝑓(𝑣)𝑓(𝑤)=𝑓(𝑤)𝑓(𝑣), and hence 𝑓(𝑤)(𝑣−𝑓(𝑣))=0=𝑣−𝑓(𝑣). Now consider arbitrary 𝑥,𝑦∈𝑈. If one of 𝑓(𝑥),𝑓(𝑦) is in 𝑍, then 𝑓(𝑥𝑦)=𝑓(𝑥)𝑓(𝑦). If 𝑑(𝑥)=0=𝑑(𝑦), then 𝑑(𝑥𝑦)=𝑑(𝑥)𝑦+𝑥𝑑(𝑦)=0, so 𝑓(𝑥𝑦)=𝑥𝑦=𝑓(𝑥)𝑓(𝑦). Therefore 𝑓(𝑥𝑦)=𝑓(𝑥)𝑓(𝑦) for all 𝑥,𝑦∈𝑈, and by Theorem 15, 𝑓 is the identity map on 𝑁, and therefore 𝑑=0.
The remaining possibility is that for each 𝑥∈𝑈, either 𝑑(𝑥)=0 or 𝑓(𝑥)=0. Let 𝑢∈𝑈⧵{0}, and let 𝑈₁=𝑢𝑁. Then 𝑈₁ is a nonzero semigroup right ideal contained in 𝑈 and 𝑈₁ is an additive subgroup of 𝑁. The sets {𝑥∈𝑈₁|𝑑(𝑥)=0} and {𝑥∈𝑈₁|𝑓(𝑥)=0} are additive subgroups of 𝑈₁ with union equal to 𝑈₁, so 𝑑(𝑈₁)={0} or 𝑓(𝑈₁)={0}. If 𝑑(𝑈₁)={0}, then 𝑑=0 by Lemma 1(i). Suppose, then, that 𝑓(𝑈₁)={0}. Then for arbitrary 𝑥,𝑦∈𝑁  𝑓(𝑢𝑥𝑦)=𝑓(𝑢𝑥)𝑦+𝑢𝑥𝑑(𝑦)=0=𝑢𝑥𝑑(𝑦), so 𝑢𝑁𝑑(𝑦)={0}, and again 𝑑=0. This completes the proof.

Acknowledgments

A. Ali and P. Miyan gratefully acknowledge the support received by the University Grants Commission, India Grant F. no. 33–106 (2008) and Council of Scientific and Industrial Research, India Grant F. no. 9/112 (0475) 2 K12-EMR-I.