Archivum Mathematicum Mohammad Ashraf; Nadeem-ur-Rehman On (σ, τ )-derivations in prime rings Archivum Mathematicum, Vol. 38 (2002), No. 4, 259--264 Persistent URL: http://dml.cz/dmlcz/107839 Terms of use: © Masaryk University, 2002 Institute of Mathematics of the Academy of Sciences of the Czech Republic provides access to digitized documents strictly for personal use. Each copy of any part of this document must contain these Terms of use. This paper has been digitized, optimized for electronic delivery and stamped with digital signature within the project DML-CZ: The Czech Digital Mathematics Library http://project.dml.cz ARCHIVUM MATHEMATICUM (BRNO) Tomus 38 (2002), 259 – 264 ON (σ, τ )-DERIVATIONS IN PRIME RINGS MOHAMMAD ASHRAF AND NADEEM-UR-REHMAN Abstract. Let R be a 2-torsion free prime ring and let σ, τ be automorphisms of R. For any x, y ∈ R, set [x, y] σ,τ = xσ(y) − τ (y)x. Suppose that d is a (σ, τ )-derivation defined on R. In the present paper it is shown that (i) if R satisfies [d(x), x] σ,τ = 0, then either d = 0 or R is commutative (ii) if I is a nonzero ideal of R such that [d(x), d(y)] = 0, for all x, y ∈ I, and d commutes with both σ and τ , then either d = 0 or R is commutative. (iii) if I is a nonzero ideal of R such that d(xy) = d(yx), for all x, y ∈ I, and d commutes with τ , then R is commutative. Finally a related result has been obtain for (σ, τ )-derivation. 1. Introduction Throughout the present paper R will denote an associative ring with center Z(R). For any x, y ∈ R the symbol [x, y] represents commutator xy − yx and for a non-empty subset S of R, we put CR (S) = {x ∈ R | [x, s] = 0, for all s ∈ S}. The set of all commutators of elements of S will be written as [S, S]. Recall that R is prime if aRb = (0) implies that a = 0 or b = 0. Let σ and τ be any two automorphisms of R. For any a, b ∈ R we set [a, b] σ,τ = aσ(b) − τ (b)a. An additive mapping d : R → R is called a derivation if d(xy) = d(x)y + xd(y), for all x, y ∈ R. An additive mapping d : R → R is called a (σ, τ )-derivation if d(xy) = d(x)σ(y) + τ (x)d(y) holds for all x, y ∈ R. Of course a (1, 1)-derivation where 1 is the identity map on R is a derivation. A mapping F : R → R is said to be centralizing if [F (x), x] ∈ Z(R), for all x ∈ R, in the special case when [F (x), x] = 0, the mapping F is said to be commuting on R. Mapping F : R → R is said to be (σ, τ )-centralizing (resp. (σ, τ )-commuting) if [F (x), x] σ,τ ∈ Z(R) (resp. [F (x), x]σ,τ = 0) holds for all x ∈ R. Of course a (1, 1)-centralizing (resp. (1, 1)-commuting) mapping is a centralizing (resp. commuting) on R. There are several results in the existing literature dealing with centralizing and commuting mappings in rings. The study of centralizing mappings was initiated by Posner [11] which states that the existence of a nonzero centralizing derivation on a prime 2000 Mathematics Subject Classification: 16W25, 16N60, 16U80. Key words and phrases: prime rings, (σ, τ )-derivations, ideals, torsion free rings and commutativity. Received January 9, 2001. 260 M. ASHRAF AND NADEEM-UR-REHMAN ring forces the ring to be commutative (Posner’s second theorem). In an attempt to generalize the above result Vukman [12] proved that if R is a 2-torsion free prime ring and d : R → R a nonzero derivation such that the map x 7→ [d(x), x] is commuting on R, then R is commutative. In the present paper it is shown that the conclusion of the above theorem holds if for a (σ, τ )-derivation d the mapping x 7→ d(x) is (σ, τ )-commuting. In fact we have proved the following. Theorem 1. Let R be a 2-torsion free prime ring. Suppose there exists a (σ, τ )derivation d : R → R such that [d(x), x] σ,τ = 0, for all x ∈ R. Then either d = 0 or R is commutative. A famous result due to Herstein [9] states that if R is prime ring of characteristic not 2 which admits a nonzero derivation d such that [d(x), d(y)] = 0, for all x, y ∈ R, then R is commutative. Motivated by this result, recently Bell and Daif [5] studied derivation d satisfying d(xy) = d(yx), for all x, y ∈ R. Now our object is to generalize these two results for (σ, τ )-derivations as follows: Theorem 2. Let R be a 2-torsion free prime ring, and I a nonzero ideal of R. If R admits a (σ, τ )-derivation d such that [d(x), d(y)] = 0, for all x, y ∈ I and d commutes with both σ, τ , then either d = 0 or R is commutative. Theorem 3. Let R be a 2-torsion free prime ring, and I a nonzero ideal of R. If R admits a nonzero (σ, τ )-derivation d such that d(xy) = d(yx), for all x, y ∈ I and d commutes with τ , then R is commutative. 2. Proof of the Main Results Throughout the present paper, we shall make extensive use of the following basic commutator identities: [xy, z]σ,τ = x[y, z]σ,τ + [x, τ (z)]y = x[y, σ(z)] + [x, z]σ,τ y and [x, yz]σ,τ = τ (y)[x, z]σ,τ + [x, y]σ,τ σ(z) . To facilitate our discussion, we begin with the following lemmas. Lemma 2.1 ([1, Lemma 3]). Let R be a prime ring, I a nonzero ideal of R and a ∈ R. If R admits a (σ, τ )-derivation d such that ad(I) = (0) (or d(I)a = (0)), then either d = 0 or a = 0. Lemma 2.2. Let R be a 2-torsion free prime ring, I be a nonzero ideal of R. If R admits a (σ, τ )-derivation d such that d2 (I) = (0) and d commutes with both σ, τ , then d = 0. Proof. For any x ∈ I, we have d2 (x) = 0. Replacing x by xy, we get d2 (x)σ2 (y) + τ (d(x))d(σ(y)) + d(τ (x))σ(d(y)) + τ 2 (x)d2 (y) = 0, for all x, y ∈ I and hence using the fact that d2 (I) = (0) and d commutes with both σ, τ , the above relation yields that τ (d(x))σ(d(y)) = 0, for all x, y ∈ I i.e. σ −1 (τ (d(x)))d(y) = 0, for all x, y ∈ I. ON (σ, τ )-DERIVATIONS IN PRIME RINGS 261 Thus application of Lemma 2.1 gives that either d = 0 or σ −1 (τ (d(x))) = 0. If σ−1 (τ (d(x))) = 0, for all x ∈ I, then d(x) = 0, for all x ∈ I. For any r ∈ R, replace x by xr, to get d(x)σ(r) + τ (x)d(r) = 0, for all x ∈ I and hence xτ −1 (d(r)) = 0, for all x ∈ I, r ∈ R i.e. IRτ −1 (d(r)) = (0). Since I is a nonzero ideal of R and R is prime the above relation yields that τ −1 (d(r)) = 0, for all r ∈ R and hence d = 0. Proof of Theorem 1. Let us introduce a mapping B(·, ·) : R × R → R by the relation B(x, y) = [d(x), y]σ,τ + [y, d(x)]σ,τ , for all x, y ∈ R. Obviously B(·, ·) is symmetric (that is B(x, y) = B(y, x), for all x, y ∈ R) and additive in both the arguments. Notice that B(xy, z) = [d(xy), z]σ,τ + [d(z), xy]σ,τ (1) = B(x, z)σ(y) + τ (x)B(y, z) + d(x)σ([y, z]) + τ ([x, z])d(y) , for all x, y, z ∈ R. Now, introduce a mapping f from R into itself by f(x) = B(x, x), for all x ∈ R. We have f(x) = 2[d(x), x]σ,τ for all x ∈ R. The mapping f satisfies the relation f(x + y) = 2[d(x + y), x + y]σ,τ = 2[d(x), x]σ,τ + 2[d(y), x]σ,τ + 2[d(x), y]σ,τ + 2[d(y), y]σ,τ (2) = f(x) + f(y) + 2B(x, y) , for all x, y ∈ R. Throughout the proof we shall use the mappings B and f, as well as the relation (1) and (2) without specific references. The assumption of the theorem can be rewritten as (3) f(x) = 0 , for all x ∈ R . Linearization of (3) gives that f(x) + f(y) + 2B(x, y) = 0, for all x, y ∈ R and hence 2B(x, y) = 0, for all x, y ∈ R. Since char R 6= 2, we get B(x, y) = 0, for all x, y ∈ R. Replacing y by xy in the above relation, we obtain B(x, xy) = f(x)σ(x) + τ (x)B(x, y) + d(x)σ([x, y]) = 0 , for all x, y ∈ R and hence using (3) and the fact that B(x, y) = 0, we get d(x)σ([x, y]) = 0 , for all x, y ∈ R , i.e. σ (d(x))[x, y] = 0, for all x, y ∈ R. Again replace y by yz in the above expression, to get σ−1 (d(x))y[x, z] = 0, for all x, y, z ∈ R and hence σ −1 (d(x))R[x, z] = 0, for all x, z ∈ R. Thus for each x ∈ R, either σ −1 (d(x)) = 0 or [x, z] = 0, for all z ∈ R. This shows that additive group R is the union of two of its additive subgroups A = {x ∈ R | σ−1 (d(x)) = 0} and B = {x ∈ R | [x, z] = 0, for all z ∈ R}. This implies that either R = A or R = B. If R = A, then σ−1 (d(x)) = 0, for all x ∈ R, i.e. d = 0. On the other hand if R = B, then [x, z] = 0, for all x, z ∈ R, i.e. R is commutative. This completes the proof of the theorem. −1 Proof of Theorem 2. We have (4) [d(x), d(y)] = 0 , for all x, y ∈ I . 262 M. ASHRAF AND NADEEM-UR-REHMAN Replacing y by xy in (4) and using (4), we get d(x)[d(x), σ(y)] + [d(x), τ (x)]d(y) = 0 , for all x, y ∈ I . Now for any r ∈ R, replace y by yr in the above expression to get (5) d(x)σ(y)[d(x), σ(r)] + [d(x), τ (x)]τ (y)d(r) = 0 , for all x, y ∈ I, r ∈ R. In view of (4) for r = σ −1 (d(z)), for any z ∈ I (5) reduces to [d(x), τ (x)]τ (y)σ−1(d2 (z)) = 0 , for all x, y, z ∈ I . For any s ∈ R, replacing y by yτ −1 (s) in the above relation we get [d(x), τ (x)]τ (y)Rσ −1 (d2 (z)) = (0) , for all x, y, z ∈ I, s ∈ R. This implies that either σ −1 (d2 (z)) = 0 or [d(x), τ (x)] ·τ (y) = 0, for all x, y ∈ I. If σ −1 (d2 (z)) = 0, for all z ∈ I, then d2 (z) = 0 for all z ∈ I and hence by Lemma 2.2 we get the required result. On the other hand if [d(x), τ (x)]τ (y) = 0, for all x, y ∈ I, then τ −1 ([d(x), τ (x)])y = 0, for all x, y ∈ I and hence τ −1 ([d(x), τ (x)])RI = (0), for all x ∈ I. Since I is a nonzero ideal of R and R is prime the above relation yields that τ −1 ([d(x), τ (x)]) = 0, for all x ∈ I and hence (6) [d(x), τ (x)] = 0 , for all x ∈ I . Linearizing (6), we get (7) [d(x), τ (y)] + [d(y), τ (x)] = 0 for all x, y ∈ I . Now replacing y by yx in (7) and using (7), we get d(x)[σ(y), τ (x)] = 0, for all x, y ∈ I. For any r1 ∈ R, again replace y by yσ−1 (r1 ), to get d(x)σ(y)[r1 , τ (x)] = 0, for all x, y ∈ I, r 1 ∈ R and hence σ−1 (d(x))yσ−1 ([r1 , τ (x)]) = 0 i.e. σ −1 (d(x)) · IRσ −1 ([r1 , τ (x)]) = (0). The primeness of R implies that for each x ∈ I either σ−1 (d(x))I = (0) or σ−1 ([r1 , τ (x)]) = 0. If σ −1 (d(x))I = (0), then σ−1 (d(x))RI = (0). Since I is a nonzero ideal of R and R is prime the above relation yields that σ−1 (d(x)) = 0 and hence d(x) = 0. Thus for each x ∈ I, either d(x) = 0 or [r1 , τ (x)] = 0, for all r 1 ∈ R. Now let A = {x ∈ I | d(x) = 0}, B = {x ∈ I | [r1 , τ (x)] = 0, for all r 1 ∈ R}. Then A and B are additive subgroups of I and I = A ∪ B. But a group can not be a union of two its proper subgroups and hence I = A or I = B. If I = A, then d(x) = 0, for all x ∈ I. For any s1 ∈ R, replace x by xs1 , to get τ (x)d(s1 ) = 0, for all x ∈ I and hence IRτ −1 (d(s1 )) = (0). Again primeness of R implies that τ −1 (d(s1 )) = 0, for all s1 ∈ R, and hence d = 0. On the other hand if I = B, then that τ (x) ∈ Z(R), for all x ∈ I and hence x ∈ Z(R), for all x ∈ I i.e. I ⊆ Z(R). But if R is prime which has a nonzero central ideal, then R is commutative. Proof of Theorem 3. Let c ∈ I be a constant i.e. an element such that d(c) = 0 and let z be an arbitrary element of I. The condition that d(cz) = d(zc) yields that τ (c)d(z) = d(z)σ(c). Now for each x, y ∈ I, [x, y] is a constant and hence (8) τ ([x, y])d(z) = d(z)σ([x, y]) , for all x, y, z ∈ I . ON (σ, τ )-DERIVATIONS IN PRIME RINGS 263 We have d(xy) = d(yx), for all x, y ∈ I. This can be rewritten as (9) [d(x), y]σ,τ = [d(y), x]σ,τ , for all x, y ∈ I . Replacing x by x2 in (9) and using (9), we get (10) d(x)σ([x, y]) + τ ([x, y])d(x) = 0 , for all x, y ∈ I . In view of (8) the above yields that 2τ ([x, y])d(x) = 0, for all x, y ∈ I. This implies that (11) τ ([x, y])d(x) = 0 , for all x, y ∈ I . Now, replacing y by yz in(11) and using (11), we find that [x, y]zτ −1 (d(x)) = 0, for all x, y, z ∈ I and hence [x, y]IRτ −1 (d(x)) = (0), for all x, y ∈ I. Thus, primeness of R implies that for each x ∈ I, either [x, y]I = (0) or τ −1 (d(x)) = 0. Now, let A = {x ∈ I | [x, y]I = (0), for all y ∈ I}, B = {x ∈ I | τ −1 (d(x)) = 0}. Clearly, both A and B are additive subgroups of I whose union is I. By Brauer’s trick we have either I = A or I = B. If I = B, then τ−1 (d(x)) = 0, for all x ∈ I and hence d(x) = 0, for all x ∈ I. For any r ∈ R, replace x by xr, to get τ (x)d(r) = 0, for all x ∈ I. This implies that IRτ −1 (d(r)) = (0), for all r ∈ R. Since I 6= (0), and R is prime the above relation yields that τ −1 (d(r)) = 0, for all r ∈ R and hence d = 0, a contradiction. On the other hand if I = A, then [x, y]I = (0), for all x, y ∈ I i.e. [x, y]RI = (0). Again since I 6= (0), we get [x, y] = 0, for all x, y ∈ I and hence by the corollary of Lemma 1.1.5 of [10], R is commutative. The following example shows that the conclusion of the above theorem need not be true if I is a one sided ideal of R even in the case if d is assumed to be a derivation on R. Example. of 2 × 2 matrices over a field F ; let  Let Rbe a ring  1 0 a b a, b ∈ F . Let d be the inner derivation of R I = R = 0 0 0 0     0 1 0 1 given by d(x) = x − x, for all x ∈ R. It is readily verified 0 0 0 0 that d satisfies the property d(xy) = d(yx), for all x, y ∈ I. However, R is not commutative. Theorem 4. Let R be a 2-torsion free prime ring and σ, τ be automorphisms of R. Suppose that d1 and d2 are two (σ, τ )-derivations of R such that d1 σ = σd1 , d1 τ = τ d1 , d2 σ = σd2 and d2 τ = τ d2 . If d1 d2 (R) = 0, then either d1 = 0 or d2 = 0. Proof. We have (12) d1 d2 (x) = 0 , for all x ∈ R . Replacing x by xy in (12) and using (12), we get τ (d2 (x))σ(d1 (y)) + τ (d1 (x))σ(d2 (y)) = 0 , for all x, y ∈ R . 264 M. ASHRAF AND NADEEM-UR-REHMAN Again replace x by τ −1 (d2 (x)) in the above expression and use (12), to get d22 (x)σ(d1 (y)) = 0, for all x, y ∈ R and hence σ−1 (d22 (x))d1 (y) = 0, for all x, y ∈ R. Thus by Lemma 2.1 either σ −1 (d22 (x)) = 0, for all x ∈ R or d 1 = 0. If σ−1 (d22 (x)) = 0, for all x ∈ R, then d 22 (x) = 0, for all x ∈ R. Replacing x by xy and using the fact that d22 (R) = 0, we get 2τ (d2 (x))σ(d2 (y)) = 0, for all x, y ∈ R and hence τ (d2 (x))σ(d2 (y)) = 0. Again replace y by σ−1 (y), to get τ (d2 (x))d2 (y) = 0, for all x, y ∈ R and hence again application of Lemma 2.1 gives that d 2 = 0 or τ (d2 (x)) = 0, for all x ∈ R. If τ (d 2 (x)) = 0, for all x ∈ R, then d 2 = 0. This completes the proof of our theorem. References [1] Aydin, N. and Kaya, A., Some generalization in prime rings with (σ, τ )-derivation, Doga Turk. J. Math. 16 (1992), 169–176. [2] Bell, H. E. and Martindale, W. S., Centralizing mappings of semiprime rings, Canad. Math. Bull. 30 (1987), 92–101. [3] Bell, H. E. and Kappe, L. C., Ring in which derivations satisfy certain algebric conditions, Acta Math. Hungar. 53 (1989), 339–346. [4] Bell, H. E. and Daif, M. N., On commutativity and strong commutativity preserving maps, Canad. Math. Bull. 37 (1994), 443–447. [5] Bell, H. E. and Daif, M. N., On derivations and commutativity in prime rings, Acta Math. Hungar. 66 (1995), 337–343. [6] Bresar, M., On a generalization of the notion of centralizing mappings, Proc. Amer. Math. Soc. 114 (1992), 641–649. [7] Bresar, M., Centralizing mappings and derivations in prime rings, J. Algebra 156 (1993), 385–394. [8] Daif, M. N. and Bell, H. E., Remarks on derivations on semiprime rings, Int. J. Math. Math. Sci. 15 (1992), 205–206. [9] Herstein, I. N., A note on derivations, Canad. Math. Bull. 21 (1978), 369–370. [10] Herstein, I. N., Rings with involution, Univ. Chicago Press, Chicago 1976. [11] Posner, E. C., Derivations in prime rings, Proc. Amer. Math. Soc. 8 (1957), 1093–1100. [12] Vukman, J., Commuting and centralizing mappings in prime rings, Proc. Amer. Math. Soc. 109 (1990), 47–52. [13] Vukman, J., Derivations on semiprime rings, Bull. Austral. Math. Soc. 53 (1995), 353–359. Mohammad Ashraf Department of Mathematics Faculty of Science, King Abdul Aziz University P.O. Box. 80203, Jeddah 21589, Saudi-Arabia E-mail: mashraf80@hotmail.com Nadeem-ur-Rehman Department of Mathematics, University of Kaiserslautern P.O. Box 3049, 67653 Kaiserslautern, Germany E-mail: rehman100@postmark.net