In this note we give some basic results on one sided(&#963;,&#964;)-Lie ideals of prime rings with characteristic not 2.</p

Güven, Evrim

Kaya, Kazim

Soytürk, Muharrem

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Some Results on (σ,τ)-Lie Ideals

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Mathematical Journal of OkayamaUniversityVolume 49, Issue 1 2007 Article 4JANUARY 2007Some Results on (σ,τ )-Lie IdealsEvrim Gu¨ven∗ Kazim Kaya†Muharrem Soytu¨rk‡∗Kocaeli University†C¸anakkale Onsekiz Mart University‡Kocaeli UniversityCopyright c©2007 by the authors. Mathematical Journal of Okayama University is produced byThe Berkeley Electronic Press (bepress). http://escholarship.lib.okayama-u.ac.jp/mjouSome Results on (σ,τ )-Lie IdealsEvrim Gu¨ven, Kazim Kaya, and Muharrem Soytu¨rkAbstractIn this note we give some basic results on one sided(σ,τ )-Lie ideals of prime rings with char-acteristic not 2.KEYWORDS: Prime ring, (sigma, tau)-LIe ideal, (sigma, tau)-derivationMath. J. Okayama Univ. 49 (2007), 59–64SOME RESULTS ON (σ, τ)-LIE IDEALSEvrim GU¨VEN, Kazım KAYA and Muharrem SOYTU¨RKAbstract. In this note we give some basic results on one sided (σ, τ )−Lieideals of prime rings with characteristic not 2.1. IntroductionLet R be a ring and σ, τ be two mappings from R into itself. We write[x, y] = xy − yx, and [x, y]σ,τ = xσ(y) − τ(y)x for x, y ∈ R. For subsetsA,B ⊂ R, let [A,B] be the additive subgroup generated by all [a, b], and[A,B]σ,τ be the additive subgroup generated by all [a, b]σ,τ for a ∈ A andb ∈ B. We recall that a Lie ideal L is an additive subgroup of R such that[R,L] ⊂ L. We first introduce the generalized Lie ideal in [3] as follows. LetU be an additive subgroup of R. (i) U is called a (σ, τ)−right Lie ideal of Rif [U,R]σ,τ ⊂ U, (ii) U is called a (σ, τ)−left Lie ideal if [R,U ]σ,τ ⊂ U. (iii)U is called a (σ, τ)−Lie ideal if U is both a (σ, τ)−right and a (σ, τ)−leftLie ideal. An additive mapping d : R −→ R is called a (σ, τ)−derivationif d(xy) = d(x)σ(y) + τ(x)d(y) for all x, y ∈ R. We write Cσ,τ = {c ∈ R |cσ(r) = τ(r)c for r ∈ R}, and will make extensive use of the following basiccommutator identities:[xy, z]σ,τ = x[y, z]σ,τ + [x, τ(z)]y = x[y, σ(z)] + [x, z]σ,τ y[x, yz]σ,τ = τ(y)[x, z]σ,τ + [x, y]σ,τσ(z)Throughout the present paper, R will represent a prime ring (of charR 6=2, exclude Lemmas 1 and 2) and σ, τ, α, β, λ and µ will be automorphismsof R. In this note, we give the following proporties on prime rings and someresults on one sided (σ, τ)−Lie ideals. Let I be a nonzero ideal of R. (1) If[[I, a]σ,τ , b]α,β = 0 for a, b ∈ R, then [τ(a), β(b)] = 0. (2) If [[a, I]σ,τ , b]α,β = 0for a, b ∈ R, then b ∈ Z or [a, τ−1β(b)]σ,τ = 0. (3) If [b, [a,R]σ,τ ]α,β = 0for a, b ∈ R, then b ∈ Cα,β , a ∈ Cσ,τ or a + τσ−1(a) ∈ Cσ,τ . On theother hand, in [4] Park and Jung proved that if d : R −→ R is a nonzero(σ, τ)−derivation and a ∈ R such that d[R, a]σ,τ = 0, then σ(a) + τ(a) ∈ Z.We prove that if d : R −→ R is a nonzero (σ, τ)−derivation and a ∈ R suchthat d[a,R]α,β = 0, then a ∈ Cα,β or a + βα−1(a) ∈ Cα,β.2. ResultsThe following Lemmas 1 and 2 are generalizations of [1, Lemma 1.5].Mathematics Subject Classification. 16N10, 16W25, 16U80.Key words and phrases. Prime ring, (σ,τ )-Lie ideal, (σ,τ )-derivation.591Güven et al.: Some Results on (&#963;,&#964;)-Lie IdealsProduced by The Berkeley Electronic Press, 200760 E. GU¨VEN, K. KAYA AND M. SOYTU¨RKLemma 1. Let I be a nonzero ideal of R and a, b ∈ R. If [[I, a]σ,τ , b]α,β = 0,then [τ(a), β(b)] = 0.Proof. Let [[I, a]σ,τ , b]α,β = 0. Then we have , 0 = [[τ(a)y, a]σ,τ , b]α,β =[τ(a)[y, a]σ,τ + [τ(a), τ(a)]y, b]α,β = τ(a)[[y, a]σ,τ , b]α,β +[τ(a), β(b)][y, a]σ,τfor all y ∈ I.This gives that(2.1) [τ(a), β(b)][y, a]σ,τ = 0 for all y ∈ I.Replacing yr, r ∈ R by y in (2.1), we get 0 = [τ(a), β(b)]y[r, σ(a)] +[τ(a), β(b)][y, a]σ,τ r and so(2.2) [τ(a), β(b)]y[r, σ(a)] = 0 for all y ∈ I, r ∈ R.Since R is prime, we get(2.3) [τ(a), β(b)] = 0 or a ∈ Z.Thus, [τ(a), β(b)] = 0 is obtained for two cases in (2.3) Corollary 1. (1) If I is a nonzero ideal of R and a ∈ R such that [I, a]α,β ⊂Cλ,µ, then a ∈ Z.(2) Let U be a nonzero (σ, τ)−right(left) Lie ideal of R and I a nonzeroideal of R. If [[I, I]σ,τ , U ]α,β = 0 then U ⊂ Z.(3) If a ∈ R such that [[I, I]σ,τ , a]α,β = 0 then a ∈ Z.Proof. (1) [I, a]α,β ⊂ Cλ,µ implies that [[I, a]α,β , R]λ,µ = 0. By Lemma 1we obtain that [β(a), µ(R)] = 0. Since µ is onto, we have β(a) ∈ Z and soa ∈ Z.(2) By Lemma 1 we have [τ(I), β(U)] = 0 and so U ⊂ Z.(3) [[I, I]σ,τ , a]α,β = 0 implies that [τ(I), β(a)] = 0 by Lemma 1 and soa ∈ Z. Lemma 2. Let I be a nonzero ideal of R. If a, b ∈ R and [[a, I]σ,τ , b]α,β = 0,then b ∈ Z or [a, τ−1β(b)]σ,τ = 0.Proof. For any x, y ∈ I we have0 = [[a, xy]σ,τ , b]α,β= [τ(x)[a, y]σ,τ + [a, x]σ,τ σ(y), b]α,β= τ(x)[[a, y]σ,τ , b]α,β + [τ(x), β(b)][a, y]σ,τ + [a, x]σ,τ [σ(y), α(b)]+ [[a, x]σ,τ , b]α,βσ(y)and so(2.4) [τ(x), β(b)][a, y]σ,τ + [a, x]σ,τ [σ(y), α(b)] = 0 for all x, y ∈ I.2Mathematical Journal of Okayama University, Vol. 49 [2007], Iss. 1, Art. 4http://escholarship.lib.okayama-u.ac.jp/mjou/vol49/iss1/4SOME RESULTS ON (σ, τ)-LIE IDEALS 61Replacing x by rx, r ∈ R in (2.4) we get0 = [τ(rx), β(b)][a, y]σ,τ + [a, rx]σ,τ [σ(y), α(b)]= τ(r)[τ(x), β(b)][a, y]σ,τ + [τ(r), β(b)]τ(x)[a, y]σ,τ + τ(r)[a, x]σ,τ [σ(y), α(b)]+ [a, r]σ,τσ(x)[σ(y), α(b)].That is(2.5)[τ(r), β(b)]τ(x)[a, y]σ,τ + [a, r]σ,τσ(x)[σ(y), α(b)] = 0 for all x, y ∈ I, r ∈ R.If we take τ−1β(b) instead of r in (2.5) then we have(2.6) [a, τ−1β(b)]σ,τσ(I)[σ(I), α(b)] = 0.Since σ(I) 6= 0 an ideal of R and R is prime we get(2.7) [a, τ−1β(b)]σ,τ = 0 or [σ(I), α(b)] = 0.Since R is prime, [σ(I), α(b)] = 0 implies that b ∈ Z. Thus [a, τ−1β(b)]σ,τ =0 or b ∈ Z is obtained. Lemma 3. Let U be a nonzero (σ, τ)−right Lie ideal of R and a ∈ R. If[U, a]α,β = 0, then a ∈ Z or U ⊂ Cσ,τ .Proof. Since [[U,R]σ,τ , a]α,β ⊂ [U, a]α,β = 0 then we havea ∈ Z or [U, τ−1β(a)]σ,τ = 0by Lemma 2. If [U, τ−1β(a)]σ,τ = 0 then a ∈ Z or U ⊂ Cσ,τ by [6, Lemma2]. Theorem 1. Let U be a nonzero (σ, τ)−right Lie ideal of R and I 6= 0 anideal of R.(1) If a ∈ R and [[U, I]α,β , a]λ,µ = 0, then a ∈ Z or U ⊂ Cσ,τ .(2) If [U, I]α,β ⊂ Cλ,µ, then U ⊂ Cσ,τ or R is commutative.Proof. (1) [[U, I]α,β , a]λ,µ = 0 implies that a ∈ Z or [U, β−1µ(a)]α,β = 0, byLemma 2. If [U, β−1µ(a)]α,β = 0 then a ∈ Z or U ⊂ Cσ,τ by Lemma 3.(2) Let [U, I]α,β ⊂ Cλ,µ then we have [[U, I]α,β , R]λ,µ = 0. If we use (1)we get R ⊂ Z or U ⊂ Cσ,τ and so U ⊂ Cσ,τ or R is commutative. Theorem 2. Let d be a nonzero (σ, τ)−derivation on R and a ∈ R. Ifd[a,R]α,β = 0, then a ∈ Cα,β or a + βα−1(a) ∈ Cα,β.Proof. For any x, y ∈ R we have0 = d[a, xy]α,β = d(β(x)[a, y]α,β + [a, x]α,βα(y))= dβ(x)σ[a, y]α,β + τ [a, x]α,βdα(y)3Güven et al.: Some Results on (&#963;,&#964;)-Lie IdealsProduced by The Berkeley Electronic Press, 200762 E. GU¨VEN, K. KAYA AND M. SOYTU¨RKReplacing x by β−1[a, z]α,β in the last relation we get[a, β−1[a, z]α,β ]α,βdα(y) = 0 for all y, z ∈ Rand so(2.8) [a, β−1[a, z]α,β ]α,β = 0 for all z ∈ Rby [5,Lemma 3]. Taking zy for z in (2.8) we get0 = [a, β−1[a, zy]α,β ]α,β = [a, β−1(β(z)[a, y]α,β + [a, z]α,βα(y))]α,β= [a, zβ−1[a, y]α,β + β−1[a, z]α,ββ−1α(y)]α,β= [a, z]α,βαβ−1[a, y]α,β + [a, z]α,β [a, β−1α(y)]α,βwhich leads to(2.9) [a, z]α,β(αβ−1[a, y]α,β + [a, β−1α(y)]α,β) = 0 for all z, y ∈ R.Replacing z by zt in (2.9), we get(2.10)[a, z]α,β = 0,∀ z ∈ R or αβ−1[a, y]α,β + [a, β−1α(y)]α,β = 0 for all y ∈ R.Hence a ∈ Cα,β or 0 = αβ−1[a, y]α,β + aαβ−1α(y) − α(y)a for all y ∈R. If we apply α−1 and β to the last relation we have aα(y) − β(y)a +βα−1(a)α(y) − β(y)βα−1(a) = 0 for all y ∈ R. This implies that (a +βα−1(a))α(y) − β(y)(a + βα−1(a)) = 0 and so a + βα−1(a) ∈ Cα,β for ally ∈ R. Thus we obtain a ∈ Cα,β or a + βα−1(a) ∈ Cα,β by (2.10). Corollary 2. If [b, [a,R]σ,τ ]α,β = 0, then a ∈ Cσ,τ or b ∈ Cα,β or a +τσ−1(a) ∈ Cσ,τ .Proof. d(x) = [b, x]α,β is a (α, β)−derivation on R. Furthermore d[a,R]σ,τ =0. This implies that a ∈ Cσ,τ , b ∈ Cα,β or a + τσ−1(a) ∈ Cσ,τ by Theorem2. Theorem 3. Let U be a nonzero (σ, τ)−right Lie ideal of R and d : R −→ Ra nonzero (λ, µ)−derivation.(1) If d(U) = 0, then v + τσ−1(v) ∈ Cσ,τ for all v ∈ U.(2) If d[U,R] = 0, then U ⊂ Z.Proof. (1) Suppose that d(U) = 0. Then d[U,R]σ,τ = 0. This implies thatU ⊂ Cσ,τ or v + τσ−1(v) ∈ Cσ,τ for all v ∈ U by Theorem 2.(2) Taking α = β = 1 in Theorem 2, we have U ⊂ Z. Theorem 4. Let U be a nonzero (σ, τ)−left Lie ideal of R and d : R −→ Ra nonzero (α, β)−derivation.(1) If d(U) = 0, then σ(v) + τ(v) ∈ Z for all v ∈ U.(2) If a ∈ R and [U, a] = 0, then a ∈ Z or σ(v) + τ(v) ∈ Z for all v ∈ U.4Mathematical Journal of Okayama University, Vol. 49 [2007], Iss. 1, Art. 4http://escholarship.lib.okayama-u.ac.jp/mjou/vol49/iss1/4SOME RESULTS ON (σ, τ)-LIE IDEALS 63(3) If a ∈ R and [U, a]α,β = 0, then a ∈ Z or σ(v) + τ(v) ∈ Z for allv ∈ U.(4) If [[R,U ]α,β , a]λ,µ = 0 then a ∈ Z or σ(v) + τ(v) ∈ Z for all v ∈ U.Proof. (1) Suppose that d(U) = 0. Then d[R, v]σ,τ = 0 for all v ∈ U. Thisimplies that σ(v) + τ(v) ∈ Z for all v ∈ U by [4, Corollary 5] for all v ∈ U .(2) Let d(x) = [x, a] for all x ∈ R. Then d is a derivation and furthermored(U) = 0. Thus we have a ∈ Z or σ(v) + τ(v) ∈ Z for all v ∈ U by (1).(3) Since [[R,U ]σ,τ , a]α,β ⊂ [U, a]α,β = 0 we have [τ(U), β(a)] = 0 byLemma 1. That is [U, τ−1β(a)] = 0. This implies that a ∈ Z or σ(v)+τ(v) ∈Z for all v ∈ U by (2).(4) By Lemma 1 and hypothesis, we have [β(U), µ(a)] = 0. That is[U, β−1µ(a)] = 0. This implies that a ∈ Z or σ(v) + τ(v) ∈ Z for all v ∈ Uby (2). Remark 1. Let U be a nonzero (σ, τ)−left Lie ideal of R such that [U,U ]α,β =0. Then we have σ(v) + τ(v) ∈ Z for all v ∈ U.Proof. By Theorem 4(3) we have σ(v) + τ(v) ∈ Z for all v ∈ U. Theorem 5. Let U be a nonzero (σ, τ)−left Lie ideal of R and a ∈ R.(1) If [a, U ]α,β = 0, then a ∈ Cα,β or σ(v) + τ(v) ∈ Z for all v ∈ U.(2) If [a, [R,U ]α,β ]λ,µ = 0, then a ∈ Cλ,µ or α(v)+β(v) ∈ Z for all v ∈ U.(3) If [R,U ]α,β ⊂ Cλ,µ, then R is commutative or σ(v) = τ(v) for allv ∈ U.(4) If U ⊂ Cλ,µ, then σ(v) = τ(v) for all v ∈ U or R is commutative.Proof. (1) Let d(x) = [a, x]α,β for all x ∈ R. Then d is (α, β)−derivation ofR. Since [a, [R,U ]σ,τ ]α,β ⊂ [a, U ]α,β = 0 then we have d[R,U ]σ,τ = 0. Thisimplies that a ∈ Cα,β or σ(v) + τ(v) ∈ Z for all v ∈ U by [4,Corollary 5].(2) Considering as in the proof (1) we obtain the result.(3) Suppose that [R,U ]α,β ⊂ Cλ,µ. Then we have [[R,U ]α,β , R]λ,µ = 0.This gives [β(U), µ(R)] = 0 by Lemma 1 and so U ⊂ Z. Thus [R,U ]σ,τ ⊂U ⊂ Z is obtained. For any r, s ∈ R, v ∈ U we have 0 = [[r, v]σ,τ , s] =[rσ(v)− τ(v)r, s] = [r(σ(v)− τ(v)), s] = r[σ(v)− τ(v), s]+ [r, s] (σ(v)− τ(v))which leads to(2.11) [r, s](σ(v) − τ(v)) = 0 for all r, s ∈ R, v ∈ U.Since R is prime and σ(v)− τ(v) ∈ Z we get(2.12) [r, s] = 0 for all r, s ∈ R or σ(v) = τ(v) for all v ∈ U.and so R is commutative or σ(v) = τ(v) for all v ∈ U.(4) If U ⊂ Cλ,µ, then [R,U ]σ,τ ⊂ Cλ,µ. This implies that R is commutativeor σ(v) = τ(v) for all v ∈ U by (3). 5Güven et al.: Some Results on (&#963;,&#964;)-Lie IdealsProduced by The Berkeley Electronic Press, 200764 E. GU¨VEN, K. KAYA AND M. SOYTU¨RKAcknowledgementThe authors wish to thank the referee for valuable suggestions.References[1] I.N.Herstein, Topics in Ring Theory, The University of Chicago Press,(1969).[2] J.C.Chang, On Fixed Power Central (α, β)−derivations, Bull.ofInst.Math.Acad.Scica.Vol.15, No.2,(1975),163-178.[3] K.Kaya,(σ, τ )−Lie ideals in Prime Rings,An.Univ.Timisoara,Stiinte Mat.30,No.2-3,251-255(1992).[4] Kyoo-Hong Park and Yong-Soo Jung, Some Results Concerning (θ, ϕ)−Derivations onPrime Rings, J.Korea Soc.Math.Educ.SerB, Pure Appl.Math.10(2003), no.4, 207-215.[5] N.Aydın and K.Kaya, Dog˘a-Tr.J.of Mathematics 16 (1992), 169-176.[6] N.Aydın and H.Kandamar, (σ, τ )−Lie Ideals in Prime Rings, Tr.J.Math. 18 (1994),143-148.Kocaeli UniversityFakulty of Art and SciencesDepartment of MathematicsKocaeli - TURKEYe-mail address: evrim@kou.edu.trURL: www.kou.edu.trC¸anakkale Onsekiz Mart UniversityFakulty of Art and SciencesDepartment of MathematicsCanakkale - TURKEYe-mail address: kkaya@comu.edu.trURL: www.comu.edu.trKocaeli UniversityFakulty of Art and SciencesDepartment of MathematicsKocaeli - TURKEYe-mail address: msoyturk@kou.edu.trURL: www.kou.edu.tr(Received November 28, 2005 )(Revised September 1, 2006 )6Mathematical Journal of Okayama University, Vol. 49 [2007], Iss. 1, Art. 4http://escholarship.lib.okayama-u.ac.jp/mjou/vol49/iss1/4

Some Results on (σ,τ)-Lie Ideals

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