# Generalized Derivations on Prime Near Rings

## Full text

(1)

Volume 2013, Article ID 170749,5pages http://dx.doi.org/10.1155/2013/170749

## Generalized Derivations on Prime Near Rings

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2

### and Phool Miyan

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1

Department of Mathematics, Aligarh Muslim University, Aligarh 202002, India

2

Department of Mathematics, Brock University, St. Catharines, ON, Canada L2S 3A1

Correspondence should be addressed to Asma Ali; asma.ali345@gmail.com

Received 8 October 2012; Accepted 2 January 2013

Copyright Β© 2013 Asma Ali et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Letπbe a near ring. An additive mappingπ : π β πis said to be a right generalized (resp., left generalized) derivation with associated derivationπonπifπ(π₯π¦) = π(π₯)π¦ + π₯π(π¦)(resp.,π(π₯π¦) = π(π₯)π¦ + π₯π(π¦)) for allπ₯, π¦ β π. A mappingπ : π β πis said to be a generalized derivation with associated derivationπonπifπis both a right generalized and a left generalized derivation with associated derivationπonπ. 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,πdenotes a zero-symmetric left near ring with multiplicative centerπ, and for any pair of elements

π₯, π¦ β π,[π₯, π¦]denotes the commutatorπ₯π¦ β π¦π₯, while the symbol(π₯, π¦)denotes the additive commutatorπ₯+π¦βπ₯βπ¦. A near ringπis called zero-symmetric if0π₯ = 0, for allπ₯ β π (recall that left distributivity yields thatπ₯0 = 0). The near ring

πis said to be 3-prime ifπ₯ππ¦ = {0}forπ₯, π¦ β πimplies that

π₯ = 0orπ¦ = 0. A near ringπis called 2-torsion-free if(π, +) has no element of order 2. A nonempty subsetπ΄ofπis called a semigroup right (resp., semigroup left) ideal ifπ΄π β π΄ (resp.,ππ΄ β π΄), and ifπ΄is both a semigroup right ideal and a semigroup left ideal, it is called a semigroup ideal. An additive mappingπ : π β πis a derivation onπifπ(π₯π¦) = π(π₯)π¦+

π₯π(π¦), for allπ₯, π¦ β π. An additive mappingπ : π β π

is said to be a right (resp., left) generalized derivation with associated derivationπ if π(π₯π¦) = π(π₯)π¦ + π₯π(π¦) (resp.,

π(π₯π¦) = π(π₯)π¦ + π₯π(π¦)), for allπ₯, π¦ β π, andπis said to be a generalized derivation with associated derivationπonπif it is both a right generalized derivation and a left generalized derivation onπwith associated derivationπ. (Note that this definition differs from the one given by Hvala in [1]; his generalized derivations are our right generalized derivations.) Every derivation onπ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]). Letπbe a 3-prime near ring

andπbe a nonzero derivation onπ.

(i)Ifπis a nonzero semigroup right ideal or a nonzero

semigroup left ideal ofπ, thenπ(π) ΜΈ= {0}.

(ii)Ifπis a nonzero semigroup right ideal ofπandπ₯is

an element ofπwhich centralizesπ, thenπ₯ β π.

Lemma 2 (see [3, Lemma 1.2]). Letπbe a 3-prime near ring.

(i)Ifπ§ β π \ {0}, thenπ§is not a zero divisor.

(ii)Ifπ \ {0}contains an elementπ§for whichπ§ + π§ β π, then(π, +)is abelian.

(2)

Lemma 3 (see [3, Lemmas 1.3 and 1.4]). Letπbe a 3-prime

near ring andπbe a nonzero semigroup ideal ofπ. Letπbe a

nonzero derivation onπ.

(i)Ifπ₯, π¦ β πandπ₯ππ¦ = {0}, thenπ₯ = 0orπ¦ = 0.

(ii)Ifπ₯ β πandπ₯π = {0}orππ₯ = {0}, thenπ₯ = 0.

(iii)Ifπ₯ β πandπ(π)π₯ = {0}orπ₯π(π) = {0}, thenπ₯ = 0.

Lemma 4 (see [3, Lemma 1.5]). Ifπis a 3-prime near ring

andπcontains a nonzero semigroup left ideal or a semigroup

right ideal, thenπis a commutative ring.

Lemma 5. Ifπis a generalized derivation onπwith

associ-ated derivationπ, then(π(π₯)π¦ + π₯π(π¦))π§ = π(π₯)π¦π§ + π₯π(π¦)π§,

for allπ₯, π¦, π§ β π.

Proof. We prove only (ii), since (i) is proved in [2]. For all

π₯, π¦, π§ β π we have π((π₯π¦)π§) = π(π₯π¦)π§ + π₯π¦π(π§) =

(π(π₯)π¦+π₯π(π¦))π§+π₯π¦π(π§)andπ(π₯(π¦π§)) = π(π₯)π¦π§+π₯π(π¦π§) =

π(π₯)π¦π§ + π₯π(π(π¦)π§ + π¦π(π§)) = π(π₯)π¦π§ + π₯π(π¦)π§ + π₯π¦π(π§). Comparing the two expressions forπ(π₯π¦π§)gives(ii).

Lemma 6. Letπbe a 3-prime near ring andπa generalized

derivation with associated derivationπ.

(i)π(π₯)π¦ + π₯π(π¦) = π₯π(π¦) + π(π₯)π¦for allπ₯, π¦ β π.

(ii)π(π₯)π¦ + π₯π(π¦) = π₯π(π¦) + π(π₯)π¦for allπ₯, π¦ β π.

Proof. (i)π(π₯(π¦ + π¦)) = π(π₯)(π¦ + π¦) + π₯π(π¦ + π¦) = π(π₯)π¦ +

π(π₯)π¦ + π₯π(π¦) + π₯π(π¦), andπ(π₯π¦ + π₯π¦) = π(π₯)π¦ + π₯π(π¦) +

π(π₯)π¦ + π₯π(π¦). Comparing these two equations gives the desired result.

(ii) Again, calculate π(π₯(π¦ + π¦)) and π(π₯π¦ + π₯π¦) and compare.

Lemma 7. Letπbe a 3-prime near ring andπa generalized

derivation with associated derivationπ. Thenπ(π) β π.

Proof. Letπ§ β πandπ₯ β π. Thenπ(π§π₯) = π(π₯π§); that is,

π(π§)π₯ + π§π(π₯) = π(π₯)π§ + π₯π(π§). ApplyingLemma 6(i), we get

π§π(π₯) + π(π§)π₯ = π(π₯)π§ + π₯π(π§). It follows thatπ(π§)π₯ = π₯π(π§) for allπ₯ β π, soπ(π§) β π.

Lemma 8. Letπbe a 3-prime near ring and πa nonzero

semigroup ideal ofπ. Ifπis a nonzero right generalized

deri-vation ofπwith associated derivationπ, thenπ(π) ΜΈ= {0}.

Proof. Supposeπ(π) = {0}. Thenπ(π’π₯) = π(π’)π₯ + π’π(π₯) =

0 = π’π(π₯) for all π’ β πand π₯ β π, and it follows by Lemma 3(ii) thatπ = 0. Thereforeπ(π₯π’) = π(π₯)π’ = 0for allπ’ β πandπ₯ β π, and another appeal toLemma 3(ii) givesπ = 0, which is a contradiction.

Lemma 9 (see [2, Theorem 2.1]). Letπbe a 3-prime near ring

with a nonzero right generalized derivationπwith associated

derivationπ. Ifπ(π) β πthen(π, +)is abelian. Moreover, if

πis 2-torsion-free, thenπis a commutative ring.

Lemma 10 (see [2, Theorem 4.1]). Letπbe a 2-torsion-free

3-prime near ring andπa nonzero generalized derivation on

πwith associated derivationπ. If[π(π), π(π)] = {0}, thenπ

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. Letπbe a 3-prime near ring andπbe a nonzero

semigroup ideal of π. Let πbe a nonzero right generalized

derivation with associated derivationπ. Ifπ(π) β π, then

(π, +)is abelian. Moreover, ifπis 2-torsion-free, thenπis a commutative ring.

Proof. We begin by showing that(π, +)is abelian, which by

Lemma 2(ii) is accomplished by producingπ§ β π \ {0}such thatπ§ + π§ β π. Letπbe an element ofπsuch thatπ(π) ΜΈ= 0. Then for allπ₯ β π,ππ₯ β πandππ₯ + ππ₯ = π(π₯ + π₯) β π, so thatπ(ππ₯) β πandπ(ππ₯)+π(ππ₯) β π; hence we need only to show that there existsπ₯ β πsuch thatπ(ππ₯) ΜΈ= 0. Suppose that this is not the case, so thatπ((ππ₯)π) = 0 = π(ππ₯)π+ππ₯π(π) =

ππ₯π(π), for allπ₯ β π. ByLemma 3(i) eitherπ = 0orπ(π) = 0. Ifπ(π) = 0, thenπ(π₯π) = π(π₯)π + π₯π(π); that is, π(π₯π) =

π(π₯)π β π, for allπ₯ β π. Thus[π(π’)π, π¦] = 0, for allπ¦ β π, andπ’ β π. This implies thatπ(π’)[π, π¦] = 0, for allπ’ β π andπ¦ β πandLemma 2(i) givesπ β π. Thus0 = π(ππ₯) =

π(π₯π) = π(π₯)π, for allπ₯ β π. Replacingπ₯byπ’ β π, we have

π(π)π = 0, and by Lemmas2(i) and8, we get thatπ = 0. Thus, we have a contradiction.

To complete the proof, we show that ifπis 2-torsion-free, thenπis commutative.

Consider first the caseπ = 0. This implies thatπ(π’π₯) =

π(π’)π₯ β πfor allπ’ β πandπ₯ β π. ByLemma 8, we have

π’ β πsuch thatπ(π’) β π \ {0}, soπis commutative by Lemma 2(iii).

Now consider the caseπ ΜΈ= 0. Letπ β π \ {0}. This implies that ifπ₯ β π,π(π₯π) = π(π₯)π + π₯π(π) β π. Thus(π(π₯)π + π₯π(π))π¦ = π¦(π(π₯)π + π₯π(π))for allπ₯, π¦ β πand π β π. Therefore byLemma 5(i),π(π₯)ππ¦+π₯π(π)π¦ = π¦π(π₯)π+π¦π₯π(π)

for allπ₯, π¦ β πandπ β π. Sinceπ(π) β πandπ(π₯) β π, we obtainπ(π)[π₯, π¦] = 0, for allπ₯, π¦ β πandπ β π. Letπ(π) ΜΈ= 0. Choosingπsuch thatπ(π) ΜΈ= 0and noting thatπ(π)is not a zero divisor, we have[π₯, π¦] = 0for allπ₯, π¦ β π. ByLemma 1(ii),

π β π; henceπis commutative byLemma 4.

The remaining case isπ ΜΈ= 0andπ(π) = {0}. Suppose we can show thatπ β© π ΜΈ= {0}. Takingπ§ β (π β© π) \ {0}andπ₯ β

π, we haveπ(π₯π§) = π(π₯)π§ β π; thereforeπ(π) β πby Lemma 2(iii) andπis commutative byLemma 9.

Assume, then, thatπ β© π = {0}. For eachπ’ β π,π(π’2) = π(π’)π’ + π’π(π’) = π’(π(π’) + π(π’)) β π β© π, soπ(π’2) = 0. Thus, for allπ’ β πandπ₯ β π,π(π’2π₯) = π(π’2)π₯ + π’2π(π₯) =

π’2π(π₯) β π β© π, soπ’2π(π₯) = 0, and byLemma 3(iii)π’2= 0.

Since π(π₯π’) = π(π₯)π’ + π₯π(π’) β π for allπ’ β π and

π₯ β π, we have (π(π₯)π’ + π₯π(π’))π’ = π’(π(π₯)π’ + π₯π(π’)), and right multiplying byπ’givesπ’π₯π(π’)π’ = 0. Consequently

π(π’)π’ππ(π’)π’ = {0}, so thatπ(π’)π’ = 0for allπ’ β π. Since

(3)

π(π’)π’ = 0for allπ’ β π. But byLemma 8, there existsπ’0β π for whichπ(π’0) ΜΈ= 0; and sinceπ(π’0) β π, we getπ’0 = 0βa contradiction. Thereforeπ β© π ΜΈ= {0}as required.

Theorem 12. Let πbe a 3-prime near ring with a nonzero

generalized derivationπwith associated nonzero derivationπ.

Letπbe a nonzero semigroup ideal ofπ. If[π(π), π(π)] = 0,

then(π, +)is abelian.

Proof. Assume thatπ₯ β πis such that [π₯, π(π)] = [π₯ +

π₯, π(π)] = 0. For all π’, π£ β π such that π’ + π£ β π,

[π₯ + π₯, π(π’ + π£)] = 0. This implies that

(π₯ + π₯) π (π’ + π£) = π (π’ + π£) (π₯ + π₯) , (π₯ + π₯) π (π’) + (π₯ + π₯) π (π£)

= π (π’ + π£) π₯ + π (π’ + π£) π₯, π (π’) (π₯ + π₯) + π (π£) (π₯ + π₯)

= π₯π (π’ + π£) + π₯π (π’ + π£) , π (π’) π₯ + π (π’) π₯ + π (π£) π₯ + π (π£) π₯

= π₯π (π’) + π₯π (π£) + π₯π (π’) + π₯π (π£) , π₯π (π’) + π₯π (π’) + π₯π (π£) + π₯π (π£)

= π₯π (π’) + π₯π (π£) + π₯π (π’) + π₯π (π£) , π₯ (π (π’) + π (π£) β π (π’) β π (π£)) = 0,

π₯π (π’ + π£ β π’ β π£) = 0.

(1)

This equation may be restated asπ₯π(π) = 0, whereπ = (π’, π£). Letπ, π β π. Thenππ β πandππ + ππ = π(π + π) β π, so[π(ππ), π(π)] = {0} = [π(ππ) + π(ππ), π(π)], and by the argument in the previous paragraph,π(ππ)π(π) = 0. We now haveπ(π2)π(π’, π£) = 0for allπ’, π£ β πsuch thatπ’ + π£ β π. Taking π€ β π2 and π₯ β π, we get π(π₯π€)π((π’, π£)) =

(π(π₯)π€ + π₯π(π€))π((π’, π£)) = 0 = π(π₯)π€π((π’, π£)), and since

π2is a nonzero semigroup ideal by Lemma 3(ii) andπ ΜΈ= 0,

Lemma 3(i) gives

π ((π’, π£)) = 0 βπ’, π£ β πsuch thatπ’ + π£ β π. (2)

Takeπ’ = ππ¦andπ£ = ππ§, whereπ β πandπ¦, π§ β π, so that

π’ + π£ = ππ¦ + ππ§ = π(π¦ + π§) β π. By (2) we have

π ((ππ¦, ππ§)) = π (π (π¦, π§)) = 0 β π β π, π¦, π§ β π. (3)

Replacingπbyππ€,π€ β π, we obtainπ(π(π€π¦, π€π§)) = 0 =

π(π)(π€(π¦, π§)) + ππ((π€π¦, π€π§)); so by (3)π(π)π(π¦, π§) = 0for all

π β πandπ¦, π§ β π. It follows immediately by Lemmas1(i) and3(i) that(π¦, π§) = 0for allπ¦, π§ β π; that is,(π, +)is abelian.

Theorem 13. Let π be a 2-torsion-free 3-prime near ring

andπbe a nonzero semigroup ideal ofπ. Ifπis a nonzero

generalized derivation with associated derivationπsuch that

[π(π), π(π)] = 0, thenπis a commutative ring if it satisfies

one of the following: (i)π(π) ΜΈ= {0}; (ii)π β© π ΜΈ= {0}; (iii)π = 0

andπ(π) ΜΈ= {0}.

Proof. (i) Letπ β πcentralizes π(π), and letπ§ β π such

thatπ(π§) ΜΈ= 0. Then a centralizesπ(π’π§)for allπ’ β π, so that

π(π(π’)π§ + π’π(π§)) = (π(π’)π§ + π’π(π§))πandππ’π(π§) = π’π(π§)π. Sinceπ(π§) β π \ {0},π(π§)[π, π’] = 0 = [π, π’]for allπ’ β π. Therefore a centralizesπ, and byLemma 1(ii),π β π. Since

π(π)centralizesπ(π),π(π) β πand our result follows by Theorem 11.

(ii) We may assumeπ(π) = {0}. Letπ§ β (π β© π) \ {0}. Then for allπ₯, π¦ β π,π(π₯π§) = π(π₯)π§andπ(π¦π§) = π(π¦)π§ commute; hence π§2[π(π₯), π(π¦)] = 0 = [π(π₯), π(π¦)]. Our result now follows fromLemma 10.

(iii) Letπ’, π£ β πand π§ β πsuch thatπ(π§) ΜΈ= 0. Then

[π(π§π’), π(π’)] = 0 = [π(π§)π’, π(π£)], and sinceπ(π§) β π \ {0},

π(π§)[π’, π(π’)] = 0 = [π’, π(π£)]. Thus π(π) centralizes π, and byLemma 1(ii),π(π) β π. Our result now follows by Theorem 11.

We have already observed that if π is a generalized derivation withπ = 0, thenπ(π₯)π¦ = π₯π(π¦)for allπ₯, π¦ β π. For 3-prime near rings, we have the following converse.

Theorem 14. Letπbe a 3-prime near ring andπbe a nonzero

semigroup ideal of π. If π is a nonzero right generalized

derivation of πwith associated derivation π andπ(π₯)π¦ =

π₯π(π¦), for allπ₯, π¦ β π, thenπ = 0.

Proof. We are given thatπ(π₯)π¦ = π₯π(π¦)for allπ₯, π¦ β π.

Substitutingπ¦π§forπ¦, we getπ(π₯)π¦π§ = π₯π(π¦π§) = π₯(π(π¦)π§ + π¦π(π§))for all π₯, π¦, π§ β π. It follows thatπ₯π¦π(π§) = 0 for allπ₯, π¦, π§ β π; that is,π₯ππ(π§) = {0}for allπ₯, π§ β π. By Lemma 3(i),π(π) = 0, and henceπ = 0byLemma 1(i).

### an Antihomomorphism

In [4], Bell and Kappe proved that ifπis a semiprime ring and

πis a derivation onπwhich is either an endomorphism or an antiendomorphism onπ, thenπ = 0. Of course, derivations which are not endomorphisms or antiendomorphisms onπ may behave as such on certain subsets ofπ; for example, any derivationπbehaves as the zero endomorphism on the subring πΆ consisting of all constants (i.e., the elements π₯ for which π(π₯) = 0). In fact in a semiprime ring π, π may behave as an endomorphism on a proper ideal ofπ. However as noted in [4], the behaviour ofπ is somewhat restricted in the case of a prime ring. Recently the authors in [6] considered(π, π)-derivationπacting as a homomorphism or an antihomomorphism on a nonzero Lie ideal of a prime ring and concluded thatπ = 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. Letπbe a 3-prime near ring andπbe a

non-zero semigroup ideal of π. Let πbe a nonzero generalized

(4)

homomorphism onπ, then πis the identity map onπand

π = 0.

Proof. By the hypothesis

π (π₯π¦) = π (π₯) π¦ + π₯π (π¦) = π (π₯) π (π¦) βπ₯, π¦ β π.

(4)

Replacingπ¦byπ¦π§in the above relation, we get

π (π₯π¦π§) = π (π₯) π¦π§ + π₯π (π¦π§) βπ₯, π¦, π§ β π, (5)

or

π (π₯π¦) π (π§) = π (π₯) π¦π§ + π₯ (π (π¦) π§ + π¦π (π§)) βπ₯, π¦, π§ β π. (6)

This implies that

(π (π₯) π¦ + π₯π (π¦)) π (π§) = π (π₯) π¦π§ + π₯π (π¦) π§ + π₯π¦π (π§) βπ₯, π¦, π§ β π.

(7)

UsingLemma 5(ii), we get

π (π₯) π¦π (π§) + π₯π (π¦) π (π§)

= π (π₯) π¦π§ + π₯π (π¦) π§ + π₯π¦π (π§) βπ₯, π¦, π§ β π, (8)

or

π (π₯) π¦π (π§) + π₯π (π¦π§)

= π (π₯) π¦π§ + π₯π (π¦) π§ + π₯π¦π (π§) βπ₯, π¦, π§ β π. (9)

This implies that

π (π₯) π¦π (π§) + π₯π (π¦) π§ + π₯π¦π (π§)

= π (π₯) π¦π§ + π₯π (π¦) π§ + π₯π¦π (π§) βπ₯, π¦, π§ β π. (10)

That is,

π (π₯) π¦π (π§) = π (π₯) π¦π§ βπ₯, π¦, π§ β π. (11)

Therefore

π (π₯) π¦ (π (π§) β π§) = 0 βπ₯, π¦, π§ β π, (12)

which implies that

π (π₯) π (π (π§) β π§) = {0} βπ₯, π§ β π. (13)

It follows byLemma 3(i) that eitherπ(π) = 0orπ(π§) = π§for allπ§ β π.

In fact, as we now show, both of these conditions hold. Suppose thatπ(π’) = π’for allπ’ β π. Then for allπ’ β π andπ₯ β π,π(π₯π’) = π₯π’ = π(π₯)π’ + π₯π(π’) = π(π₯)π’ + π₯π’; henceπ(π₯)π = {0}for allπ₯ β π, and thusπ = 0.

On the other hand, suppose thatπ(π) = {0}, so thatπ = 0. Then for allπ₯, π¦ β π,π(π₯π¦) = π(π₯)π¦ = π(π₯)π(π¦), so that

π(π₯)(π¦ β π(π¦)) = 0. Replacingπ¦byπ§π¦,π§ β π, and noting

thatπ(π§π¦) = π§π(π¦), we see thatπ(π₯)π(π¦ β π(π¦)) = {0}for all

π₯, π¦ β π. Therefore,π(π) = {0}orπis the identity onπ. But

π(π) = {0}contradictsLemma 8, soπis the identity onπ. We now know thatπis the identity onπandπ(π₯π¦) =

π₯π(π¦)for allπ₯, π¦ β π. Consequently,π(π’π₯) = π’π₯ = π’π(π₯)

for allπ’ β πandπ₯ β π, so thatπ(π₯ β π(π₯)) = {0}for all

π₯ β π. It follows thatπis the identity onπ.

Theorem 16. Let π be a 3-prime near ring and π be a

nonzero semigroup ideal ofπ. Ifπis a nonzero generalized

derivation onπwith associated derivationπ. Ifπacts as an

antihomomorphism onπ, thenπ = 0,πis the identity map on

π, andπis a commutative ring.

Proof. We begin by showing thatπ = 0if and only ifπis the

identity map onπ.

Clearly ifπis the identity map onπ,π₯π(π¦) = 0for all

π₯, π¦ β π, and henceπ = 0.

Conversely, assume thatπ = 0, in which caseπ(π₯π¦) =

π(π₯)π¦ = π₯π(π¦)for allπ₯, π¦ β π. It follows that for anyπ₯, π¦, π§ β

π,

π (π¦π₯π§) = π (π§) π (π¦π₯) = π (π§) π¦π (π₯)

= π (π§π¦) π (π₯) = π§π (π¦) π (π₯) = π§π (π₯π¦) . (14)

On the other hand,

π (π¦π₯π§) = π (π₯π§) π (π¦) = π (π₯) π§π (π¦) = π (π₯) π (π§π¦) = π (π₯) π (π¦) π (π§) = π (π¦π₯) π (π§) = π (π¦) π₯π (π§) = π (π¦) π (π₯π§) = π (π¦) π (π₯) π§ = π (π₯π¦) π§.

(15)

Comparing (14) and (15) shows thatπ(π2)centralizesπ, so thatπ(π2) β πbyLemma 1(ii).

Now π2 is a nonzero semigroup ideal by Lemma 3(ii); henceπ(π2) ΜΈ= 0byLemma 8. Choosingπ₯, π¦ β πsuch that

π(π₯π¦) ΜΈ= 0, we see that for anyπ§ β π,π(π₯π¦)π§ = π(π₯π¦π§) =

π(π¦π§)π(π₯) = π(π¦)π§π(π₯) = π(π¦)π(π§π₯) = π(π¦)π(π₯)π(π§) = π(π₯π¦)π(π§), and henceπ(π₯π¦)(π§ β π(π§)) = 0. Sinceπ(π₯π¦) β

π \ {0}, we conclude thatπ(π§) = π§for allπ§ β π, and it follows easily thatπis the identity map onπ.

We note now that if the identity map on πacts as an antihomomorphism onπ, thenπis commutative, so that by Lemmas1(ii) and4πis a commutative ring.

To complete the proof of our theorem, we need only to argue thatπ = 0. By our antihomomorphism hypothesis

π (π₯π¦) = π (π₯) π¦ + π₯π (π¦) = π (π¦) π (π₯) βπ₯, π¦ β π. (16)

Replacingπ¦byπ₯π¦in the above relation, we get

π (π₯π¦) π (π₯) = π (π₯π₯π¦) = π (π₯) π₯π¦ + π₯π (π₯π¦) βπ₯, π¦ β π.

(17)

This implies that

(π (π₯) π¦ + π₯π (π¦)) π (π₯)

(5)

UsingLemma 5(ii), we get

π (π₯) π¦π (π₯) + π₯π (π¦) π (π₯)

= π (π₯) π₯π¦ + π₯π (π¦) π (π₯) βπ₯, π¦ β π. (19)

Thus

π (π₯) π¦π (π₯) = π (π₯) π₯π¦ βπ₯, π¦ β π. (20)

Replacingπ¦byπ¦πin (20) and using (20), we get

π (π₯) π¦ππ (π₯) = π (π₯) π₯π¦π, and soπ (π₯) π¦ [π, π (π₯)] = 0

βπ₯, π¦ β π, π β π.

(21)

Application ofLemma 3(i) yields that for eachπ₯ β πeither

π(π₯) = 0or[π, π(π₯)] = 0; that isπ(π₯) = 0orπ(π₯) β π. Suppose that there existsπ€ β πsuch thatπ(π€) β π \ {0}. Then for allπ£ β πsuch thatπ(π£) = 0,π(π€π£) = π(π€)π£ =

π(π£)π(π€) = π(π€)π(π£), and henceπ(π€)(π£ β π(π£)) = 0 =

π£βπ(π£). Now consider arbitraryπ₯, π¦ β π. If one ofπ(π₯), π(π¦) is inπ, thenπ(π₯π¦) = π(π₯)π(π¦). Ifπ(π₯) = 0 = π(π¦), then

π(π₯π¦) = π(π₯)π¦ + π₯π(π¦) = 0, soπ(π₯π¦) = π₯π¦ = π(π₯)π(π¦). Therefore π(π₯π¦) = π(π₯)π(π¦) for all π₯, π¦ β π, and by Theorem 15,πis the identity map onπ, and thereforeπ = 0. The remaining possibility is that for eachπ₯ β π, either

π(π₯) = 0orπ(π₯) = 0. Letπ’ β π \ {0}, and letπ1= π’π. Then

π1is a nonzero semigroup right ideal contained inπandπ1 is an additive subgroup ofπ. The sets{π₯ β π1|π(π₯) = 0}and

{π₯ β π1|π(π₯) = 0}are additive subgroups ofπ1with union equal toπ1, soπ(π1) = {0}orπ(π1) = {0}. Ifπ(π1) = {0}, thenπ = 0byLemma 1(i). Suppose, then, thatπ(π1) = {0}. Then for arbitraryπ₯, π¦ β π π(π’π₯π¦) = π(π’π₯)π¦+π’π₯π(π¦) = 0 =

π’π₯π(π¦), soπ’ππ(π¦) = {0}, and againπ = 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. 33β106 (2008) and Council of Scientific and Industrial Research, India Grant F. no. 9/112 (0475) 2 K12-EMR-I. Let

π§ β πandπ₯ β π. Thenπ(π§π₯) = π(π₯π§); that is,π(π§)π₯ + π§π(π₯) = π(π₯)π§ + π₯π(π§).

### References

[1] B. Hvala, βGeneralized derivations in rings,βCommunications in

Algebra, vol. 26, no. 4, pp. 1147β1166, 1998.

[2] H. E. Bell, βOn prime near-rings with generalized derivation,β

International Journal of Mathematics and Mathematical

Sci-ences, vol. 2008, Article ID 490316, 5 pages, 2008.

[3] H. E. Bell, βOn derivations in near rings II,β inNearrings,

Nearfields and K-Loops, vol. 426 of Mathematical

Applica-tions, pp. 191β197, Kluwer Academic Publishers, Dordrecht, The

Netherlands, 1997.

[4] H. E. Bell and L. C. Kappe, βRings in which derivations satisfy certain algebraic conditions,βActa Mathematica Hungarica, vol. 53, no. 3-4, pp. 339β346, 1989.

[5] H. E. Bell and G. Mason, βOn derivations in near-rings,β in

Near-Rings and Near-Fields, vol. 137 of North-Holland

Math-ematical Studies, pp. 31β35, North-Holland, Amsterdam, The

Netherlands, 1987.

[6] A. Asma, N. Rehman, and A. Shakir, βOn Lie ideals with derivations as homomorphisms and antihomomorphisms,β

(6)

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