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Hindawi Publishing CorporationInternational Journal of Mathematics and Mathematical SciencesVolume 2013, Article ID 170749, 5 pageshttp://dx.doi.org/10.1155/2013/170749
Research ArticleGeneralized Derivations on Prime Near Rings
Asma Ali,1 Howard E. Bell,2 and Phool Miyan1
1 Department of Mathematics, Aligarh Muslim University, Aligarh 202002, India2Department of Mathematics, Brock University, St. Catharines, ON, Canada L2S 3A1
Correspondence should be addressed to Asma Ali; [email protected]
Received 8 October 2012; Accepted 2 January 2013
Academic Editor: Christian Corda
Copyright Β© 2013 Asma Ali et al.This is an open access article distributed under the Creative CommonsAttribution License, whichpermits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Let π be a near ring. An additive mapping π : π β π is said to be a right generalized (resp., left generalized) derivation withassociated derivation π onπ if π(π₯π¦) = π(π₯)π¦ + π₯π(π¦) (resp., π(π₯π¦) = π(π₯)π¦ + π₯π(π¦)) for all π₯, π¦ β π. A mapping π : π β π issaid to be a generalized derivation with associated derivation π onπ ifπ is both a right generalized and a left generalized derivationwith associated derivation π onπ. The purpose of the present paper is to prove some theorems in the setting of a semigroup idealof 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 nearring withmultiplicative centerπ, and for any pair of elementsπ₯, π¦ β π, [π₯, π¦] denotes the commutator π₯π¦ β π¦π₯, while thesymbol (π₯, π¦) denotes the additive commutatorπ₯+π¦βπ₯βπ¦. Anear ringπ is called zero-symmetric if 0π₯ = 0, for all π₯ β π(recall that left distributivity yields that π₯0 = 0).The near ringπ is said to be 3-prime if π₯ππ¦ = {0} for π₯, π¦ β π implies thatπ₯ = 0 or π¦ = 0. A near ringπ is called 2-torsion-free if (π, +)has no element of order 2. A nonempty subsetπ΄ ofπ is calleda semigroup right (resp., semigroup left) ideal if π΄π β π΄(resp.,ππ΄ β π΄), and ifπ΄ is both a semigroup right ideal and asemigroup left ideal, it is called a semigroup ideal. An additivemapping π : π β π is a derivation onπ if π(π₯π¦) = π(π₯)π¦+π₯π(π¦), for all π₯, π¦ β π. An additive mapping π : π β πis said to be a right (resp., left) generalized derivation withassociated derivation π if π(π₯π¦) = π(π₯)π¦ + π₯π(π¦) (resp.,π(π₯π¦) = π(π₯)π¦ + π₯π(π¦)), for all π₯, π¦ β π, and π is said to bea generalized derivation with associated derivation π onπ ifit is both a right generalized derivation and a left generalizedderivation onπ with associated derivation π. (Note that thisdefinition differs from the one given by Hvala in [1]; hisgeneralized derivations are our right generalized derivations.)Every derivation onπ is a generalized derivation.
In the case of rings, generalized derivations have receivedsignificant attention in recent years.We prove some theorems
in the setting of a semigroup ideal of a 3-prime near ringadmitting a generalized derivation and thereby extend someknown 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 beenproved elsewhere.
Lemma 1 (see [3, Lemma 1.3]). Letπ be a 3-prime near ringand π be a nonzero derivation onπ.
(i) If π is a nonzero semigroup right ideal or a nonzerosemigroup left ideal ofπ, then π(π) ΜΈ= {0}.
(ii) If π is a nonzero semigroup right ideal of π and π₯ isan element ofπ which centralizes π, then π₯ β π.
Lemma 2 (see [3, Lemma 1.2]). Letπ be a 3-prime near ring.
(i) If π§ β π \ {0}, then π§ is not a zero divisor.
(ii) If π \ {0} contains an element π§ for which π§ + π§ β π,then (π, +) is abelian.
(iii) If π§ β π\{0} andπ₯ is an element ofπ such thatπ₯π§ β π,then π₯ β π.
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Lemma 3 (see [3, Lemmas 1.3 and 1.4]). Let π be a 3-primenear ring and π be a nonzero semigroup ideal ofπ. Let π be anonzero derivation onπ.
(i) If π₯, π¦ β π and π₯ππ¦ = {0}, then π₯ = 0 or π¦ = 0.(ii) If π₯ β π and π₯π = {0} or ππ₯ = {0}, then π₯ = 0.(iii) If π₯ β π and π(π)π₯ = {0} or π₯π(π) = {0}, then π₯ = 0.
Lemma 4 (see [3, Lemma 1.5]). If π is a 3-prime near ringand π contains a nonzero semigroup left ideal or a semigroupright ideal, thenπ is a commutative ring.
Lemma 5. If π is a generalized derivation on π with associ-ated derivation π, then (π(π₯)π¦ +π₯π(π¦))π§ = π(π₯)π¦π§+π₯π(π¦)π§,for all π₯, π¦, π§ β π.
Proof. We prove only (ii), since (i) is proved in [2]. For allπ₯, 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 for π(π₯π¦π§) gives (ii).
Lemma 6. Let π be a 3-prime near ring and π a generalizedderivation with associated derivation π.
(i) π(π₯)π¦ + π₯π(π¦) = π₯π(π¦) + π(π₯)π¦ for all π₯, π¦ β π.(ii) π(π₯)π¦ + π₯π(π¦) = π₯π(π¦) + π(π₯)π¦ for all π₯, 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¦), and π(π₯π¦ + π₯π¦) = π(π₯)π¦ + π₯π(π¦) +π(π₯)π¦ + π₯π(π¦). Comparing these two equations gives thedesired result.
(ii) Again, calculate π(π₯(π¦ + π¦)) and π(π₯π¦ + π₯π¦) andcompare.
Lemma 7. Let π be a 3-prime near ring and π a generalizedderivation with associated derivation π. Then π(π) β π.
Proof. Let π§ β π and π₯ β π. Then π(π§π₯) = π(π₯π§); that is,π(π§)π₯+π§π(π₯) = π(π₯)π§+π₯π(π§). Applying Lemma 6(i), we getπ§π(π₯) +π(π§)π₯ = π(π₯)π§+π₯π(π§). It follows that π(π§)π₯ = π₯π(π§)for all π₯ β π, so π(π§) β π.
Lemma 8. Let π be a 3-prime near ring and π a nonzerosemigroup ideal ofπ. If π is a nonzero right generalized deri-vation ofπ with associated derivation π, then π(π) ΜΈ= {0}.
Proof. Suppose π(π) = {0}. Then π(π’π₯) = π(π’)π₯ + π’π(π₯) =0 = π’π(π₯) for all π’ β π and π₯ β π, and it follows byLemma 3(ii) that π = 0. Therefore π(π₯π’) = π(π₯)π’ = 0 forall π’ β π and π₯ β π, and another appeal to Lemma 3(ii)gives π = 0, which is a contradiction.
Lemma9 (see [2,Theorem 2.1]). Letπ be a 3-prime near ringwith a nonzero right generalized derivation π with associatedderivation π. If π(π) β π then (π, +) is abelian. Moreover, ifπ is 2-torsion-free, thenπ is a commutative ring.
Lemma 10 (see [2, Theorem 4.1]). Let π be a 2-torsion-free3-prime near ring and π a nonzero generalized derivation on
π with associated derivation π. If [π(π), π(π)] = {0}, thenπis a commutative ring.
3. Main Results
The theorems that we prove in this section extend the resultsproved in [2, Theorems 2.1 and 3.1], [3, Theorems 2.1, 3.1, and3.3], and [5, Theorem 3.3].
Theorem 11. Letπ be a 3-prime near ring andπ be a nonzerosemigroup ideal of π. Let π be a nonzero right generalizedderivation with associated derivation π. If π(π) β π, then(π, +) is abelian. Moreover, if π is 2-torsion-free, then π isa commutative ring.
Proof. We begin by showing that (π, +) is abelian, which byLemma 2(ii) is accomplished by producing π§ β π \ {0} suchthat π§ + π§ β π. Let π be an element of π such that π(π) ΜΈ= 0.Then for all π₯ β π, ππ₯ β π and ππ₯ + ππ₯ = π(π₯ + π₯) β π, sothatπ(ππ₯) β π andπ(ππ₯)+π(ππ₯) β π; hence we need only toshow that there existsπ₯ β π such thatπ(ππ₯) ΜΈ= 0. Suppose thatthis is not the case, so thatπ((ππ₯)π) = 0 = π(ππ₯)π+ππ₯π(π) =ππ₯π(π), for all π₯ β π. By Lemma 3(i) either π = 0 or π(π) = 0.
If π(π) = 0, then π(π₯π) = π(π₯)π + π₯π(π); that is, π(π₯π) =π(π₯)π β π, for all π₯ β π. Thus [π(π’)π, π¦] = 0, for all π¦ β π,and π’ β π. This implies that π(π’)[π, π¦] = 0, for all π’ β πand π¦ β π and Lemma 2(i) gives π β π. Thus 0 = π(ππ₯) =π(π₯π) = π(π₯)π, for all π₯ β π. Replacing π₯ by π’ β π, we haveπ(π)π = 0, and by Lemmas 2(i) and 8, we get that π = 0.Thus, we have a contradiction.
To complete the proof, we show that ifπ is 2-torsion-free,thenπ is commutative.
Consider first the case π = 0. This implies that π(π’π₯) =π(π’)π₯ β π for all π’ β π and π₯ β π. By Lemma 8, we haveπ’ β π such that π(π’) β π \ {0}, so π is commutative byLemma 2(iii).
Now consider the case π ΜΈ= 0. Let π β π \ {0}. This impliesthat if π₯ 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 all π₯, π¦ β π and π β π.Therefore by Lemma 5(i),π(π₯)ππ¦+π₯π(π)π¦ = π¦π(π₯)π+π¦π₯π(π)for all π₯, π¦ β π and π β π. Since π(π) β π and π(π₯) β π, weobtain π(π)[π₯, π¦] = 0, for all π₯, π¦ β π and π β π. Let π(π) ΜΈ= 0.Choosing π such thatπ(π) ΜΈ= 0 andnoting thatπ(π) is not a zerodivisor, we have [π₯, π¦] = 0 for all π₯, π¦ β π. By Lemma 1(ii),π β π; henceπ is commutative by Lemma 4.
The remaining case is π ΜΈ= 0 and π(π) = {0}. Suppose wecan show that π β© π ΜΈ= {0}. Taking π§ β (π β© π) \ {0} and π₯ βπ, we have π(π₯π§) = π(π₯)π§ β π; therefore π(π) β π byLemma 2(iii) andπ is commutative by Lemma 9.
Assume, then, that π β© π = {0}. For each π’ β π, π(π’2) =π(π’)π’ + π’π(π’) = π’(π(π’) + π(π’)) β π β© π, so π(π’2) = 0.Thus, for all π’ β π and π₯ β π, π(π’2π₯) = π(π’2)π₯ + π’2π(π₯) =π’2π(π₯) β π β© π, so π’2π(π₯) = 0, and by Lemma 3(iii) π’2 = 0.
Since π(π₯π’) = π(π₯)π’ + π₯π(π’) β π for all π’ β π andπ₯ β π, we have (π(π₯)π’ + π₯π(π’))π’ = π’(π(π₯)π’ + π₯π(π’)),and right multiplying by π’ gives π’π₯π(π’)π’ = 0. Consequentlyπ(π’)π’ππ(π’)π’ = {0}, so that π(π’)π’ = 0 for all π’ β π. Sinceπ’2= 0, π(π’2) = π(π’)π’ + π’π(π’) = 0 for all π’ β π, so
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π(π’)π’ = 0 for all π’ β π. But by Lemma 8, there exists π’0β π
for which π(π’0) ΜΈ= 0; and since π(π’
0) β π, we get π’
0= 0βa
contradiction. Therefore π β© π ΜΈ= {0} as required.
Theorem 12. Let π be a 3-prime near ring with a nonzerogeneralized derivation πwith associated nonzero derivation π.Let π be a nonzero semigroup ideal ofπ. If [π(π), π(π)] = 0,then (π, +) is abelian.
Proof. Assume that π₯ 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 as π₯π(π) = 0, where π = (π’, π£).Let π, π β π. Then ππ β π and ππ + ππ = π(π + π) β π,
so [π(ππ), π(π)] = {0} = [π(ππ) + π(ππ), π(π)], and by theargument in the previous paragraph, π(ππ)π(π) = 0. We nowhave π(π2)π(π’, π£) = 0 for all π’, π£ β π such that π’ + π£ β π.Taking π€ β π2 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 sinceπ2 is a nonzero semigroup ideal by Lemma 3(ii) and π ΜΈ= 0,
Lemma 3(i) gives
π ((π’, π£)) = 0 βπ’, π£ β π such that π’ + π£ β π. (2)
Take π’ = ππ¦ and π£ = ππ§, where π β π and π¦, π§ β π, so thatπ’ + π£ = ππ¦ + ππ§ = π(π¦ + π§) β π. By (2) we have
π ((ππ¦, ππ§)) = π (π (π¦, π§)) = 0 β π β π, π¦, π§ β π. (3)
Replacing π by ππ€, π€ β π, we obtain π(π(π€π¦, π€π§)) = 0 =π(π)(π€(π¦, π§)) + ππ((π€π¦, π€π§)); so by (3) π(π)π(π¦, π§) = 0 for allπ β π and π¦, π§ β π. It follows immediately by Lemmas 1(i)and 3(i) that (π¦, π§) = 0 for all π¦, π§ β π; that is, (π, +) isabelian.
Theorem 13. Let π be a 2-torsion-free 3-prime near ringand π be a nonzero semigroup ideal of π. If π is a nonzerogeneralized derivation with associated derivation π such that[π(π), π(π)] = 0, then π is a commutative ring if it satisfies
one of the following: (i) π(π) ΜΈ= {0}; (ii) πβ©π ΜΈ= {0}; (iii) π = 0and π(π) ΜΈ= {0}.
Proof. (i) Let π β π centralizes π(π), and let π§ β π suchthat π(π§) ΜΈ= 0. Then a centralizes π(π’π§) for all π’ β π, so thatπ(π(π’)π§ + π’π(π§)) = (π(π’)π§ + π’π(π§))π and ππ’π(π§) = π’π(π§)π.Since π(π§) β π \ {0}, π(π§)[π, π’] = 0 = [π, π’] for all π’ β π.Therefore a centralizes π, and by Lemma 1(ii), π β π. Sinceπ(π) centralizes π(π), π(π) β π and our result follows byTheorem 11.
(ii) We may assume π(π) = {0}. Let π§ β (π β© π) \ {0}.Then for all π₯, π¦ β π, π(π₯π§) = π(π₯)π§ and π(π¦π§) = π(π¦)π§commute; hence π§2[π(π₯), π(π¦)] = 0 = [π(π₯), π(π¦)]. Ourresult now follows from Lemma 10.
(iii) Let π’, π£ β π and π§ β π such that π(π§) ΜΈ= 0. Then[π(π§π’), π(π’)] = 0 = [π(π§)π’, π(π£)], and since π(π§) β π \ {0},π(π§)[π’, π(π’)] = 0 = [π’, π(π£)]. Thus π(π) centralizes π,and by Lemma 1(ii), π(π) β π. Our result now follows byTheorem 11.
We have already observed that if π is a generalizedderivation with π = 0, then π(π₯)π¦ = π₯π(π¦) for all π₯, 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 nonzerosemigroup ideal of π. If π is a nonzero right generalizedderivation of π with associated derivation π and π(π₯)π¦ =π₯π(π¦), for all π₯, π¦ β π, then π = 0.
Proof. We are given that π(π₯)π¦ = π₯π(π¦) for all π₯, π¦ β π.Substituting π¦π§ for π¦, we get π(π₯)π¦π§ = π₯π(π¦π§) = π₯(π(π¦)π§ +π¦π(π§)) for all π₯, π¦, π§ β π. It follows that π₯π¦π(π§) = 0 forall π₯, π¦, π§ β π; that is, π₯ππ(π§) = {0} for all π₯, π§ β π. ByLemma 3(i), π(π) = 0, and hence π = 0 by Lemma 1(i).
4. Generalized DerivationsActing as a Homomorphism oran Antihomomorphism
In [4], Bell andKappe proved that ifπ is a semiprime ring andπ is a derivation on π which is either an endomorphism or anantiendomorphism on π , then π = 0. Of course, derivationswhich are not endomorphisms or antiendomorphisms on π may behave as such on certain subsets of π ; for example,any derivation π behaves as the zero endomorphism on thesubring πΆ consisting of all constants (i.e., the elements π₯for which π(π₯) = 0). In fact in a semiprime ring π , πmay behave as an endomorphism on a proper ideal of π .However as noted in [4], the behaviour of π is somewhatrestricted in the case of a prime ring. Recently the authors in[6] considered (π, π)-derivationπ acting as a homomorphismor an antihomomorphism on a nonzero Lie ideal of a primering and concluded that π = 0. In this section we establishsimilar results in the setting of a semigroup ideal of a 3-primenear ring admitting a generalized derivation.
Theorem 15. Let π be a 3-prime near ring and π be a non-zero semigroup ideal of π. Let π be a nonzero generalizedderivation on π with associated derivation π. If π acts as a
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homomorphism on π, then π is the identity map on π andπ = 0.
Proof. By the hypothesis
π (π₯π¦) = π (π₯) π¦ + π₯π (π¦) = π (π₯) π (π¦) βπ₯, π¦ β π.
(4)
Replacing π¦ by π¦π§ in the above relation, we get
π (π₯π¦π§) = π (π₯) π¦π§ + π₯π (π¦π§) βπ₯, π¦, π§ β π, (5)
or
π (π₯π¦) π (π§) = π (π₯) π¦π§ + π₯ (π (π¦) π§ + π¦π (π§))
βπ₯, 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§)
βπ₯, π¦, π§ β π.
(7)
Using Lemma 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§) βπ₯, 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§) βπ₯, 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§) βπ₯, π¦, π§ β π.
(10)
That is,
π (π₯) π¦π (π§) = π (π₯) π¦π§ βπ₯, π¦, π§ β π. (11)
Therefore
π (π₯) π¦ (π (π§) β π§) = 0 βπ₯, π¦, π§ β π, (12)
which implies that
π (π₯)π (π (π§) β π§) = {0} βπ₯, π§ β π. (13)
It follows by Lemma 3(i) that either π(π) = 0 or π(π§) = π§ forall π§ β π.
In fact, as we now show, both of these conditions hold.Suppose that π(π’) = π’ for all π’ β π. Then for all π’ β π
and π₯ β π, π(π₯π’) = π₯π’ = π(π₯)π’ + π₯π(π’) = π(π₯)π’ + π₯π’;hence π(π₯)π = {0} for all π₯ β π, and thus π = 0.
On the other hand, suppose thatπ(π) = {0}, so thatπ = 0.Then for all π₯, π¦ β π, π(π₯π¦) = π(π₯)π¦ = π(π₯)π(π¦), so thatπ(π₯)(π¦ β π(π¦)) = 0. Replacing π¦ by π§π¦, π§ β π, and noting
that π(π§π¦) = π§π(π¦), we see that π(π₯)π(π¦βπ(π¦)) = {0} for allπ₯, π¦ β π. Therefore, π(π) = {0} or π is the identity onπ. Butπ(π) = {0} contradicts Lemma 8, so π is the identity on π.
We now know that π is the identity on π and π(π₯π¦) =π₯π(π¦) for all π₯, π¦ β π. Consequently, π(π’π₯) = π’π₯ = π’π(π₯)for all π’ β π and π₯ β π, so that π(π₯ β π(π₯)) = {0} for allπ₯ β π. It follows that π is the identity onπ.
Theorem 16. Let π be a 3-prime near ring and π be anonzero semigroup ideal of π. If π is a nonzero generalizedderivation on π with associated derivation π. If π acts as anantihomomorphism onπ, then π = 0, π is the identity map onπ, andπ is a commutative ring.
Proof. We begin by showing that π = 0 if and only if π is theidentity map onπ.
Clearly if π is the identity map on π, π₯π(π¦) = 0 for allπ₯, π¦ β π, and hence π = 0.
Conversely, assume that π = 0, in which case π(π₯π¦) =π(π₯)π¦ = π₯π(π¦) for allπ₯, π¦ β π. It follows that for anyπ₯, π¦, π§ βπ,
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 that π(π2) centralizes π, sothat π(π2) β π by Lemma 1(ii).
Now π2 is a nonzero semigroup ideal by Lemma 3(ii);hence π(π2) ΜΈ= 0 by Lemma 8. Choosing π₯, π¦ β π such thatπ(π₯π¦) ΜΈ= 0, we see that for any 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(π§) =
π(π₯π¦)π(π§), and hence π(π₯π¦)(π§ β π(π§)) = 0. Since π(π₯π¦) βπ\ {0}, we conclude that π(π§) = π§ for all π§ β π, and it followseasily that π is the identity map onπ.
We note now that if the identity map on π acts as anantihomomorphism on π, then π is commutative, so that byLemmas 1(ii) and 4 π is a commutative ring.
To complete the proof of our theorem, we need only toargue that π = 0. By our antihomomorphism hypothesis
π (π₯π¦) = π (π₯) π¦ + π₯π (π¦) = π (π¦) π (π₯) βπ₯, π¦ β π. (16)
Replacing π¦ by π₯π¦ in the above relation, we get
π (π₯π¦)π (π₯) = π (π₯π₯π¦) = π (π₯) π₯π¦ + π₯π (π₯π¦) βπ₯, π¦ β π.
(17)
This implies that
(π (π₯) π¦ + π₯π (π¦)) π (π₯)
= π (π₯) π₯π¦ + π₯π (π¦) π (π₯) βπ₯, π¦ β π.
(18)
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International Journal of Mathematics and Mathematical Sciences 5
Using Lemma 5(ii), we get
π (π₯) π¦π (π₯) + π₯π (π¦) π (π₯)
= π (π₯) π₯π¦ + π₯π (π¦) π (π₯) βπ₯, π¦ β π.
(19)
Thus
π (π₯) π¦π (π₯) = π (π₯) π₯π¦ βπ₯, π¦ β π. (20)
Replacing π¦ by π¦π in (20) and using (20), we get
π (π₯) π¦ππ (π₯) = π (π₯) π₯π¦π, and so π (π₯) π¦ [π, π (π₯)] = 0
βπ₯, π¦ β π, π β π.
(21)
Application of Lemma 3(i) yields that for each π₯ β π eitherπ(π₯) = 0 or [π, π(π₯)] = 0; that is π(π₯) = 0 or π(π₯) β π.
Suppose that there exists π€ β π such that π(π€) β π \ {0}.Then for all π£ β π such that π(π£) = 0, π(π€π£) = π(π€)π£ =π(π£)π(π€) = π(π€)π(π£), and hence π(π€)(π£ β π(π£)) = 0 =π£βπ(π£). Now consider arbitrary π₯, π¦ β π. If one ofπ(π₯), π(π¦)is in π, then π(π₯π¦) = π(π₯)π(π¦). If π(π₯) = 0 = π(π¦), thenπ(π₯π¦) = π(π₯)π¦ + π₯π(π¦) = 0, so π(π₯π¦) = π₯π¦ = π(π₯)π(π¦).Therefore π(π₯π¦) = π(π₯)π(π¦) for all π₯, π¦ β π, and byTheorem 15, π is the identity map onπ, and therefore π = 0.
The remaining possibility is that for each π₯ β π, eitherπ(π₯) = 0 or π(π₯) = 0. Let π’ β π \ {0}, and let π
1= π’π. Then
π1is a nonzero semigroup right ideal contained in π and π
1
is an additive subgroup ofπ. The sets {π₯ β π1|π(π₯) = 0} and
{π₯ β π1|π(π₯) = 0} are additive subgroups of π
1with union
equal to π1, so π(π
1) = {0} or π(π
1) = {0}. If π(π
1) = {0},
then π = 0 by Lemma 1(i). Suppose, then, that π(π1) = {0}.
Then for arbitraryπ₯, π¦ β π π(π’π₯π¦) = π(π’π₯)π¦+π’π₯π(π¦) = 0 =π’π₯π(π¦), so π’ππ(π¦) = {0}, and againπ = 0.This completes theproof.
Acknowledgments
A. Ali and P. Miyan gratefully acknowledge the supportreceived by the University Grants Commission, India GrantF. no. 33β106 (2008) and Council of Scientific and IndustrialResearch, India Grant F. no. 9/112 (0475) 2 K12-EMR-I. Letπ§ β π and π₯ β π. Then π(π§π₯) = π(π₯π§); that is, π(π§)π₯ +π§π(π₯) = π(π₯)π§ + π₯π(π§).
References
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[2] H. E. Bell, βOn prime near-rings with generalized derivation,βInternational Journal of Mathematics and Mathematical Sci-ences, vol. 2008, Article ID 490316, 5 pages, 2008.
[3] H. E. Bell, βOn derivations in near rings II,β in Nearrings,Nearfields and K-Loops, vol. 426 of Mathematical Applica-tions, pp. 191β197, Kluwer Academic Publishers, Dordrecht, TheNetherlands, 1997.
[4] H. E. Bell and L. C. Kappe, βRings in which derivations satisfycertain algebraic conditions,βActaMathematica Hungarica, vol.53, no. 3-4, pp. 339β346, 1989.
[5] H. E. Bell and G. Mason, βOn derivations in near-rings,β inNear-Rings and Near-Fields, vol. 137 of North-Holland Math-ematical Studies, pp. 31β35, North-Holland, Amsterdam, TheNetherlands, 1987.
[6] A. Asma, N. Rehman, and A. Shakir, βOn Lie ideals withderivations as homomorphisms and antihomomorphisms,βActa Mathematica Hungarica, vol. 101, no. 1-2, pp. 79β82, 2003.
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