On (σ, τ) Derivations in Prime Rings

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1 Int. J. Contemp. Math. Sciences, Vol. 3, 2008, no. 26, On (σ, τ) Derivations in Prime Rings Evrim Güven Kocaeli University Abstract. Let R be a prime ring with characteristic not 2 and U, a nonzero ideal of R. Let σ, τ, α, β, λ, μ : R R automorphisms. In this paper we proved that, (1) Let d : R R, a nonzero (σ, τ) derivation and a R. (i) If [a, d(u)] λ,μ = 0 then a C σ,τ.(ii) If [d(r),a] α,β = 0 then a Z or (dτ 1 β(a) = 0 and dσ 1 α(a) = 0). (iii) If U is a nonzero right ideal of R such that d(u) C λ,μ then R is commutative. (2) Let d 1 : R R be a(σ, τ) derivation and d 2 : R R, a(α, β) derivation such that d 2 α = αd 2,d 2 β = βd 2.(i) If [d 1 (U),d 2 (U)] λ,μ = 0 then R is commutative. (ii) If d 1 d 2 (U) = 0 then d 1 =0ord 2 =0. Mathematics Subject Classification: 16W25, 16U80 Keywords: Prime ring, (σ, τ) derivation, commutativity 1. Introduction In what follows, R is a prime ring with characteristic not 2, and α, β,σ,τ,λ,μ are automorphisms. An additive mapping d : R R is called a (σ, τ) derivation if d(xy) = d(x)σ(y)+ τ(x)d(y), for all x, y R. We write C σ,τ = {c R cσ(r) =τ(r)c for all r R}, [x, y] σ,τ = xσ(y) τ(y)x and will make extensive use of the following basic commutator identities: (A) [xy, z] σ,τ = x[y, z] σ,τ +[x, τ(z)]y = x[y, σ(z)] + [x, z] σ,τ y (B) [x, yz] σ,τ = τ(y)[x, z] σ,τ +[x, y] σ,τ σ(z) Suppose that a is an element of R such that ad(x) =d(x)a for all x R. Then by a theorem of Herstein [5] a must be central. In [3,Theorem 1], J.C.Chang extended this result by assuming that [a, δ(x)] = 0 for all x R, where δ is an (α, β) derivation of R such that δα = αδ, δβ = βδ. One of the aim of this paper is to generalize the above results as follows. Let d : R R be a nonzero (σ, τ) derivation, U a nonzero ideal of R and a R such that [a, d(u)] λ,μ = 0 then a C σ,τ. If [d(x),d(y)] = 0 for all x R then R is commutative. This result is proved in [4,Theorem 2] by Herstein. J.C.Chang extended this result in [3,Theorem-2(i)] by assuming that [δ(x),δ(y)] = 0 for all x, y R, where δ is

2 1290 E. Güven an (α, β) derivation of R such that δα = αδ, δβ = βδ. In [6] we proved that, if d 1 is a (σ, τ) derivation, d 2 an (α, β) derivation of R such that d 2 α = αd 2, d 2 β = βd 2, and [d 1 (R),d 2 (R)] σ,τ = 0 then R is commutative. Here this result is generalized as follows. Let d 1 : R R be a (σ, τ) derivation, d 2 : R R, a(α, β) derivation, U a nonzero ideal of R such that d 2 α = αd 2,d 2 β = βd 2 and [d 1 (U),d 2 (U)] λ,μ = 0 then R is commutative. On the other hand Ashraf and Rehman proved in [1] that, if d 1 and d 2 are two (σ, τ) derivations of R such that d 1 σ = σd 1, d 1 τ = τd 1, d 2 σ = σd 2, d 2 τ = τd 2 and d 1 d 2 (R) = 0 then d 1 =0ord 2 = 0. In [6] we proved that, if d 1 is a (σ, τ) derivation, d 2 an (α, β) derivation of R such that d 2 α = αd 2, d 2 β = βd 2 and d 1 d 2 (R) = 0 then d 1 =0ord 2 = 0. Here we generalized that, if d 1 : R R is a (σ, τ) derivation, d 2 : R R, an(α, β) derivation, U a nonzero ideal of R such that d 2 α = αd 2,d 2 β = βd 2 and d 1 d 2 (U) = 0 then d 1 =0ord 2 =0. 2. Results Lemma 1. [7, Lemma 1] Let U be a nonzero ideal of R and d : R R, a nonzero (σ, τ) derivation such that dσ = σd, dτ = τd. If d 2 (U) =0then d =0. Proof. For any u, v U we have 0 = d 2 (uv) =d(d(u)σ(v)+τ(u)d(v)) = τd(u)dσ(v)+ dτ(u)σd(v) =2τd(u)dσ(v) for all u, v U by hypothesis. Since σ, τ are automorphisms and σ(u) is a nonzero ideal of R, we have d(u) =0by[7, Lemma 3]. This implies that d =0by[2, Lemma 1]. Lemma 2. Let U be a nonzero ideal of R. d : R R is a nonzero (σ, τ) derivation such that d(u) C λ,μ. Then R is commutative. Proof. For any x, y U, r R we have, 0 = [d(xy),r] λ,μ = [d(x)σ(y) + τ(x)d(y),r] λ,μ = d(x)[σ(y),λ(r)]+[d(x),r] λ,μ σ(y)+τ(x)[d(y),r] λ,μ +[τ(x),μ(r)]d(y). And so (2.1) d(x)[σ(y),λ(r)] + [τ(x),μ(r)]d(y) = 0 for all x, y U, r R. Replacing λ 1 σ(y) byr in (2.1) we have (2.2) [τ(x),μλ 1 σ(y)]d(y) = 0 for all x, y U If we take xt, t U, instead of x in (2.2) we get 0 = [τ(xt),μλ 1 σ(y)]d(y) = τ(x)[τ(t),μλ 1 σ(y)]d(y) +[τ(x),μλ 1 σ(y)]τ(t)d(y) for all x, y, t U, r R. Thus (2.3) [τ(x),μλ 1 σ(y)]τ(t)d(y) for all x, y, t U is obtained. Since R is prime and τ(u) 0 an ideal of R then we have (2.4) [τ(x),μλ 1 σ(y)] = 0 for all x Uor d(y) =0

3 On (σ, τ) derivations in prime rings 1291 Since τ(u) is a nonzero ideal, (2.4) implies that for any y R we have (2.5) y Z or d(y) =0 If d(u) = 0 then d = 0 by [7, Lemma 1]. So it must be d(u) 0 by hypothesis. If we consider Brauer s trick then we have U Z by (2.5). Since R is prime and U, a nonzero ideal of R then U Z implies that R is commutative. Theorem 1. Let U be a nonzero ideal of R and d : R R a nonzero (σ, τ) derivation such that dσ = σd, dτ = τd. If a R and [a, d(u)] λ,μ =0 then a C λ,μ. Proof. Let [a, d(x)] λ,μ = 0 for all x U. Then 0 = [a, d(xy)] λ,μ =[a, d(x)σ(y + τ(x)d(y)] λ,μ = μd(x)[a, σ(y)] λ,μ +[a, d(x)] λ,μ λσ(y)+μτ(x)[a, d(y)] λ,μ +[a, τ(x)] λ,μ λd(y) for all x, y U is obtained. That is (2.6) μd(x)[a, σ(y)] λ,μ +[a, τ(x)] λ,μ λd(y) = 0 for all x, y U Replacing yr, r R by y in (2.6) and applying (2.6) we have 0=μd(x)μσ(y)[a, σ(r)] λ,μ + μd(x)[a, σ(y)] λ,μ λσ(r) +[a, τ(x)] λ,μ λd(y)λσ(r)+ [a, τ(x)] λ,μ λτ(y)λd(r). = μd(x)μσ(y)[a, σ(r)] λ,μ +[a, τ(x)] λ,μ λτ(y)λd(r) for all x, y U, r R. If we take σ 1 d(v),v U instead of r in the last relation we get (2.7) [a, τ(x)] λ,μ λτ(y)λdσ 1 d(v) for all x, y, v U Here, λτ(u) is a nonzero ideal of R. Hence (2.8) [a, τ(u)] λ,μ =0ordσ 1 d(u) =0 is obtained by (2.7).Since dσ = σd and R is prime ring then dσ 1 d(u) =0 implies that d 2 (U) = 0 and so d = 0 by Lemma 1. This contradicts to hypothesis. Thus it must be [a, τ(u)] λ,μ = 0 by (2.8). Then we have 0 = [a, τ(xr)] λ,μ = [a, τ(x)τ(r)] λ,μ = μτ(x) [a, τ(r)] λ,μ +[a, τ(x)] λ,μ λτ(r) for all x U, r R. This gives that (2.9) μτ(u)[a, R] λ,μ =0 Since R is a prime ring and μτ(u) 0 an ideal of R then we have [a, R] λ,μ = 0 by (2.9). Thus a C λ,μ is obtained. Corollary 1. Let U be a nonzero ideal of R and d : R R, a nonzero (σ, τ) derivation such that dσ = σd, dτ = τd. If [U, d(u)] λ,μ =0then R is commutative. Proof. U C λ,μ is obtained by Theorem 1. Thus for any v U, r, x R we have 0 = [vr, x] λ,μ = v[r, λ(x)] + [v, x] λ,μ r. That is U[R, λ(r)] = 0. Since U is a nonzero ideal and R, a prime ring we obtain that R is commutative by the preceding relation.

4 1292 E. Güven Corollary 2. Let d 1 : R R be a (σ, τ) derivation and d 2 : R R, a (α, β) derivation such that d 2 α = αd 2,d 2 β = βd 2. If U is a nonzero ideal of R and [d 1 (U),d 2 (U)] λ,μ =0then R is commutative. Proof. d 1 (U) C λ,μ is obtained by Theorem 1. This implies that R is commutative by Lemma 2. Theorem 2. Let U be a nonzero ideal of R and d 1 : R R, a nonzero (σ, τ) derivation and d 2 : R R, a (α, β) derivation such that d 2 α = αd 2,d 2 β = βd 2. If d 1 d 2 (U) =0then d 1 =0or d 2 =0. Proof. By hypothesis we have 0 = d 1 d 2 (xy) =d 1 (d 2 (x)α(y) +β(x)d 2 (y)) = τd 2 (x)d 1 α(y)+d 1 β(x)σd 2 (y) for all x, y U. Hence (2.10) τd 2 (x)d 1 α(y)+d 1 β(x)σd 2 (y) = 0 for all x, y U. is obtained. Let us substitute rx, r R for x in (2.10). Then we have 0 = τd 2 (rx)d 1 α(y)+d 1 β(rx)σd 2 (y)) = τd 2 (r)τα(x)d 1 α(y) +τβ(r)τd 2 (x)d 1 α(y)+ d 1 β(r)σβ(x)σd 2 (y)+τβ(r)d 1 β(x)σd 2 (y) whence (2.11) τd 2 (r)τα(x)d 1 α(y)+d 1 β(r)σβ(x)σd 2 (y) = 0 for all x, y U, r R. Putting r = β 1 d 2 (v),v U in (2.11) we get τd 2 β 1 d 2 (U)τα(U)d 1 α(u) =0. This gives that (2.12) d 2 β 1 d 2 (U) =0ord 1 =0. Since d 2 β = βd 2 ( that is d 2 β 1 = β 1 d 2 ) by hypothesis, d 2 β 1 d 2 (U) =0 implies that d 2 2(U) = 0 and so d 2 = 0 by Lemma 1. Theorem 3. Let d : R R be a nonzero (σ, τ) derivation and a R. If [d(r),a] α,β =0then a Z or {dτ 1 β(a) =0and dσ 1 α(a) =0}. Proof. If h : R R, h(x) =[x, a] α,β for all x R then (2.13) h(xy) =h(x)y + xd 1 (y) =d 2 (x)y + xh(y) for all x, y R. where d 1 (x) =[x, α(a)], x R and d 2 (x) =[x, β(a)], x R. Thus we have hd(r) = 0 by hypotesis. Now for any x, y R we have 0 = hd(xy) = h(d(x)σ(y)+τ(x)d(y)) = hd(x)σ(y)+d(x)d 1 σ(y)+d 2 τ(x)d(y)+τ(x)hd(y) = d(x)d 1 σ(y)+d 2 τ(x)d(y). This gives that (2.14) d(x)[σ(y),α(a)] + [τ(x),β(a)]d(y) = 0 for all x, y R. Replacing x by τ 1 β(a) in (2.14) we get dτ 1 β(a)[σ(y),α(a)] = 0 for all y R. If we take yz, z R instead of y in the last relation and using primeness of R we obtain a Z or dτ 1 β(a) =0. Another hand, if we take σ 1 α(a) instead of y in (2.14) we obtain that a Z or dσ 1 α(a) = 0 in the same way. Corollary 3. Let U be a nonzero ideal of R and d : R R, a nonzero (σ, τ) derivation such that dσ = σd, dτ = τd. If [d(r),u] α,β =0then R is commutative.

5 On (σ, τ) derivations in prime rings 1293 Proof. If [d(r),u] α,β = 0 then for any u U we have u Z or dσ 1 α(u) =0 by Theorem 3. Since σ 1 α(u) is a nonzero ideal it must be dσ 1 α(u) 0by hypotesis. Thus U Z is obtained. This implies that R is commutative in prime ring. References [1] Mohammad Ashraf and Nadeem-Ur-Rehmen, On (σ, τ) Derivations in Prime Rings, Archivum Mathematicum (BRNO) Tomus 38 (2002), [2] N.Aydın and K.Kaya, Some Generalizations in Prime Rings with (σ, τ) Derivation,Doğa-Tr.J.of Math.16(1992), [3] J.C.Chang, On (α, β) Derivations of Prime Rings,Chinese J.Math.(Taiwan,R.O.C.) 1994, Vol.22, No. 1, [4] I.N.Herstein, A Note On Derivations,Canad.Math.Bull.Vol.21(1978), [5] I.N.Herstein, A Note On Derivations II, Canad.Math.Bull.Vol.22(4),(1979). [6] K.Kaya, E.Güven and M.Soytürk, On (σ, τ) Derivations of Prime Rings, J.Korea Soc.Math.Educ.Ser.B: Pure Appl.Math.Vol.13,Number 3 (August 2006) [7] M.Soytürk, The Commutativity of Prime Rings with (σ, τ) Derivation, C.Ü.Fen Bil.Derg.,15-16(1993). Received: March 20, 2008