Some theorems of commutativity on semiprime rings with mappings

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1 Some theorems of commutativity on semiprime rings with mappings S. K. Tiwari Department of Mathematics Indian Institute of Technology Delhi, New Delhi , INDIA R. K. Sharma Department of Mathematics Indian Institute of Technology Delhi, New Delhi , INDIA B. Dhara Department of Mathematics Belda College, Belda, Paschim Medinipur , INDIA basu April 10, 2016 Abstract Let R be a semiprime ring and F : R R a mapping such that F (xy) = F (y)x + yd(x) for all x, y R, where d is any map on R. In this paper, we investigate the commutativity of semiprime rings with a mapping F on R. Several theorems of commutativity of semiprime rings are obtained. Keywords: Semiprime ring, generalized derivation, multiplicative (generalized)- derivation, reverse derivation. AMS Subject Classification: 16U80, 16N60 and 16W25. 1 Introduction Throughout this paper, Z(R) will denote the center of an associative ring R. The symbols x y and [x, y], where x, y R, stand for the anti-commutator xy + yx and commutator xy yx respectively. A ring R is said to be a prime if for any a, b R, whenever arb = 0 implies a = 0 or b = 0 and is a semiprime if for any a R, ara = 0 implies a = 0. A mapping f is called an additive Corresponding Author, address: shaileshiitd84@gmail.com 1

2 Some theorems of commutativity of semiprime rings with mappings 2 mapping on R if f(x + y) = f(x) + f(y) holds for all x, y R. Let a mapping d : R R defined as d(xy) = d(x)y + xd(y) for all x, y R. If d is an additive mapping, then d is said to be a derivation on R. Recall that an additive mapping f on R is said to be left multiplier if f(xy) = f(x)y for all x, y R. The notion of generalized derivation is given by Brešar in [8]. More precisely, he defined as follows: An additive mapping F : R R is said to be a generalized derivation if there exists a derivation d : R R such that F (xy) = F (x)y + xd(y) for all x, y R. Obviously, every derivation is a generalized derivation but converse need not be true in general. Hence generalized derivation covers both the concepts of derivations and left multiplier maps. Let S be a non empty-subset of R. A mapping h : R R is called centralizing on S if [h(x), x] Z(R) for all x S and is called commuting on S if [h(x), x] = 0 for all x S. In this direction, Posner in [19] was first who investigated the commutativity of ring. More precisely, he proved that: if R is a prime ring with non-zero derivation d on R such that d centralizing on R, then R is commutative. Further, a number of authors have extended these theorems of Posner and have shown that derivations, automorphisms, and some related maps cannot be centralizing on certain subsets of noncommutative prime (and some other) rings. For these kind of results, we refer the reader to ([4], [5], [6], [7], [11], [13], [14], [15], [16], [17], [20], [21], [22], [23], [24], [25]) where further references can be found). On the other hand, the concept of multiplicative derivation was first time introduced by Daif in [10]. According to him, a mapping D : R R is said to be multiplicative derivation if it satisfies D(xy) = D(x)y + xd(y) for all x, y R. In case of multiplicative derivation the mappings are not supposed to be an additive. Further, Daif and Tammam El-Sayiad in [12] extended multiplicative derivation to multiplicative generalized derivation as follows: a mapping F on R is said to be multiplicative generalized derivation if there exists a derivation d on R such that F (xy) = F (x)y + xd(y) for all x, y R. In this definition, if we take d is any mapping not necessarily additive and not necessarily derivation then F is said to be multiplicative (generalized)-derivation which introduced by Dhara and Ali in [13]. Recently, Dhara and Ali in [13] gave a more precise definition of multiplicative (generalized)-derivation as follows: a mapping F : R R is said to be a multiplicative (generalized)-derivation if there exists a map g on R such that F (xy) = F (x)y + xg(y) for all x, y R, where g is any mapping on R (not necessarily additive). Hence the concepts of multiplicative (generalized)-derivations cover the concepts of multiplicative derivations and multiplicative generalized derivations. A multiplicative (generalized)-derivation associated with mapping g = 0 covers the concept of multiplicative centralizers (not necessarily additive). Recently, Dhara and Ali in [13] studied the following identities related on multiplicative (generalized)-derivation on semiprime ring: (i) F (xy) ± xy = 0, (ii) F (xy) ± yx = 0, (iii) F (x)f (y) ± xy Z(R), (iv) F (x)f (y) ± yx Z(R) for all x, y in some suitable subset of semiprime ring R.

3 Some theorems of commutativity of semiprime rings with mappings 3 The concept of reverse derivation was first time introduced by Herstein in [18]. According to him: an additive mapping d on R is said to be a reverse derivation if d(xy) = d(y)x+yd(x) holds for all x, y R. After that Brešar and Vukman in [9] studied the reverse derivation. Recently many authors have studied on reverse derivation, generalized reverse derivation, reverse -derivation, generalized reverse -derivation on different kind of rings. For further study in this direction we refer the reader to ([1], [2], [3], [9], [18]; where further references can be found). The natural question arises what will happen if we replace multiplicative (generalized)-derivation with F : R R, where F (xy) = F (y)x + yg(x) for all x, y R and g : R R is any mapping. In this regards, we shall introduce the notion of multiplicative (generalized)-reverse derivation as follows. Definition 1.1. Let R be a ring and let d be any mappings on R. Assume that θ and ϕ are automorphisms on R. A mapping F : R R is said to be multiplicative (generalized) (θ, ϕ)-reverse derivation of R associated with mapping d on R if F (xy) = F (y)θ(x) + ϕ(y)d(x) for all x, y R. When θ = 1 and ϕ = 1, where 1 is an identity mapping on R, then we will call it multiplicative (generalized)-reverse derivation on R. We observe that every reverse derivation, generalized (1, 1)-reverse derivation is a multiplicative (generalized)-reverse derivation but converse need not to be true in general. The following examples explore the existence of such mappings and justify the fact: Example 1.1. Suppose the ring R = 0 x y z y x x, y, z R, where R is the set of real numbers. Define the mappings F and d : R R respectively such that 0 x y z 0 x y z F y x = y x and d 0 x y z y x = z 2

4 Some theorems of commutativity of semiprime rings with mappings 4 respectively on R. Then it is easy to verify that F is a multiplicative (generalized)- reverse derivation on R but not multiplicative (generalized)-derivation on R. Example 1.2. Consider the ring R as in Example 1.1. Define the mappings F, d : R R as follows: 0 x y z 0 x y z F y x = y x and d 0 x y z y x = z 3 respectively on R. Here, we notice that F is a multiplicative (generalized)- derivation on R but not multiplicative (generalized)-reverse derivation on R. In present paper our main aim is to study the commutativity of semiprime ring with the help of multiplicative (generalized)-reverse derivation. Moreover, we study the following situations: (i) F (x)f (y) ± xy = 0 for all x, y I; (ii) F (x)f (y) ± yx = 0 for all x, y I; (iii) F (x)f (y) ± d(y)f (x) = 0 for all x, y I; (iv) F ([x, y]) ± x y = 0 for all x, y I; (v) F (xy) ± x y = 0 for all x, y I; (vi) F (xy) ± [x, y] = 0 for all x, y I; (vii) [F (x), d(y)] ± xy = 0 for all x, y I; (viii) F (xy) ± F (x)f (y) = 0 for all x, y I; (ix) F (xy) ± F (y)f (x) = 0 for all x, y I. The following Lemmas are used in the proof of main results: Lemma 1.3. [19, Lemma 3] Let R be a prime ring and d a derivation of R such that ad(a) d(a)a = 0 for all a R. Then R is commutative or d is zero. Lemma 1.4. [11, Lemma 2] (a) If R is a semiprime ring, then the center of a non-zero one-sided ideal is contained in the center of R. In particular, any commutative one-sided ideal is contained in the center of R. (b) If R is a prime with a non-zero central ideal, then R must be commutative. 2 Main Results Theorem 2.1. Let R be a semiprime ring, I a non-zero ideal of R. Suppose F is a multiplicative (generalized)-reverse derivation associated with the mapping d on R. If F (x)f (y) ± xy = 0 for all x, y I, then R is commutative.

5 Some theorems of commutativity of semiprime rings with mappings 5 Proof. First we consider the case F (x)f (y) + xy = 0 (1) for all x, y I. Substituting zy in place of y, we obtain F (x)f (y)z+f (x)yd(z)+ xzy = 0. Using (1), it gives F (x)yd(z) + x[z, y] = 0 (2) for all x, y, z I. Let r R, substituting ry instead of y in (2), we get F (x)ryd(z) + xr[z, y] + x[z, r]y = 0. (3) Now replacing x with rx in (2) and then subtracting from (3), we obtain [x, r][z, y] + x[z, r]y xd(r)yd(z) = 0 (4) for all x, y, z I, r R. Substituting tx instead of x in (4), where t R, we get t[x, r][z, y] + [t, r]x[z, y] + tx[z, r]y txd(r)yd(z) = 0 (5) for all x, y, z I, t, r R. Left multiplying (4) by t and then subtracting from (5), we obtain [t, r]x[z, y] = 0 for all y, z I, t, r R. In particular t = z, r = y, it follows that [z, y]x[z, y] = 0 for all x, y, z I. By semiprimeness of R, it implies that [y, z] = 0 for all y, z I. By Lemma 1.4, we conclude that R is commutative. Using similar approach we can prove that the same result holds for F (x)f (y) xy = 0 for all x, y I. Theorem 2.2. Let R be a semiprime ring, I a non-zero ideal of R. Suppose F is a multiplicative (generalized)-reverse derivation associated with the mapping d on R. If F (x)f (y) ± yx = 0 for all x, y I, then R is commutative. Proof. First we consider the case F (x)f (y) + yx = 0 (6) for all x, y I. Substituting zy in place of y, we obtain F (x)f (y)z+f (x)yd(z)+ zyx = 0. Using (6), it gives F (x)yd(z) + [z, y]x + y[z, x] = 0 (7) for all x, y, z I. Let r R, substituting ry instead of y in (7), we get F (x)ryd(z) + r[z, y]x + [z, r]yx + ry[z, x] = 0. (8) Now replacing x with rx in (7) and then subtracting from (8), we obtain z[r, y]x + [y, r]xz xd(r)yd(z) = 0 (9)

6 Some theorems of commutativity of semiprime rings with mappings 6 for all x, y, z I, r R. Substituting ux instead of x in (9), where u R, we get z[r, y]ux + [y, r]uxz uxd(r)yd(z) = 0 (10) for all x, y, z I, u, r R. Left multiplying (9) by u and then subtracting from (10), we obtain [z[r, y], u]x + [[y, r], u]xz = 0 (11) for all x, y, z I, u, r R. Again substituting xt in place of x in (11), we get [z[r, y], u]xt + [[y, r], u]xtz = 0 (12) for all x, y, z I, u, r, t R. Right multiplying (11) by t and then subtracting from (12), we obtain [[y, r], u]x[t, z] = 0 (13) for all x, y, z I, r, t, u R. Putting t = [y, r] and u = z in (13), we obtain [[y, r], z]x[[y, r], z] = 0 for all x, y, z I, r R. By semiprimeness of R, it implies that [[y, r], z] = 0 for all y, z I, r R. Replacing ry in place of y in last expression, we obtain [r, z][y, r] = 0 for all y, z I, r R. By putting z = yz to obtain [r, y]z[y, r] = 0 for all y, z I, r R. By semiprimeness of R, its gives I Z(R) which implies that R is commutative, by Lemma 1.4. By similar manner, the same conclusion holds for F (x)f (y) yx = 0 for all x, y I. Theorem 2.3. Let R be a semiprime ring, I a non-zero ideal of R. Suppose F is a multiplicative (generalized)-reverse derivation associated with the mapping d on R. If F (x)f (y) ± d(y)f (x) = 0 for all x, y I, then d(i) Z(R). Proof. First we shall begin with hypothesis F (x)f (y) + d(y)f (x) = 0 (14) for all x, y I. Substituting zx in place of x in (14), we obtain F (x)zf (y) + xd(z)f (y) + d(y)f (x)z + d(y)xd(z) = 0 (15) for all x, y, z I. Replacing ux in place of x, we get F (x)uzf (y) + xd(u)zf (y) + uxd(z)f (y) + d(y)f (x)uz + d(y)xd(u)z Substituting uz in place of z in (15), we obtain +d(y)uxd(z) = 0. (16) F (x)uzf (y) + xd(uz)f (y) + d(y)f (x)uz + d(y)xd(uz) = 0 (17) for all x, y, z, u, v I. Subtracting (17) from (16), we obtain xd(u)zf (y) + uxd(z)f (y) + d(y)xd(u)z + d(y)uxd(z) xd(uz)f (y) d(y)xd(uz) = 0. (18)

7 Some theorems of commutativity of semiprime rings with mappings 7 In (18), replacing x with d(y)x, we get d(y)xd(u)zf (y) + ud(y)xd(z)f (y) + d(y)d(y)xd(u)z + d(y)ud(y)xd(z) d(y)xd(uz)f (y) d(y)d(y)xd(uz) = 0 (19) for all x, y, z, u, v I. Left multiplying (18) by d(y) and then subtracting from (19), we get [u, d(y)]xd(z)f (y) + d(y)[u, d(y)]xd(z) = 0 (20) for all x, y, z, u, v I. In (20), substituting vu in place of u and using (20), we get v[u, d(y)]xd(z)f (y) + d(y)v[u, d(y)]xd(z) = 0 (21) for all x, y, z, u, v I. Left multiplying (20) by v and then subtracting from (21), we get [d(y), v][u, d(y)]xd(z) = 0 (22) for all x, y, z, u, v I. Replacing v with vw, w I and using (22), we get Substituting x with xv in (23), we obtain [d(y), v]w[u, d(y)]xd(z) = 0. (23) [d(y), v]w[u, d(y)]xvd(z) = 0 (24) for all x, y, z, u, v, w I. Right multiply (23) by v and then subtracting from (24), we get [d(y), v]w[u, d(y)]x[v, d(z)] = 0. In particular for z = y and u = v, we obtain [d(y), v]w[v, d(y)]x[v, d(y)] = 0 for all x, y, v, w I. That is ([d(x), y]z) 3 = 0 for all x, y, z I. By semiprimeness of R it implies that [d(x), y]z = 0, which gives [d(x), y] = 0 for all x, y I. Replacing y with yr, where r R and using the fact that [d(x), y] = 0, we obtain [d(x), r] = 0 for all x I and r R that is d(i) Z(R). By similar approach, we can prove the same conclusion holds for F (x)f (y) d(y)f (x) = 0 for all x, y I. Corollary 2.4. Let R be a prime ring and let F be a multiplicative (generalized)- reverse derivation associated with a derivation d 0 on R. If F (x)f (y) ± d(y)f (x) = 0 for all x, y R, then R is commutative. Proof. By Theorem 2.3, it implies that d(r) Z(R). conclude desired result. By Lemma 1.3, we Theorem 2.5. Let R be a semiprime ring, I a non-zero ideal of R. Suppose F is a multiplicative (generalized)-reverse derivation associated with the mapping d on R. If F ([x, y])±(x y) = 0 for all x, y I, then [d(x), x] = 0 for all x I. Proof. We consider the hypothesis F ([x, y]) (x y) = 0 (25)

8 Some theorems of commutativity of semiprime rings with mappings 8 for all x, y I. Since (x yz) = (x y)z + y[z, x], substituting xy in place of y, we have 0 = F ([x, xy]) (x xy) = F (x[x, y]) (x x)y x[y, x] = F ([x, y])x + [x, y]d(x) x 2 y xyx = (F ([x, y]) (x y))x + [x, y]d(x) + [y, x 2 ] = [x, y]d(x) + [y, x 2 ] (26) for all x, y I. In (26), substituting ry, r R, in place of y we obtain r[x, y]d(x) + [x, r]yd(x) + r[y, x 2 ] + [r, x 2 ]y = 0 (27) for all x, y I and r R. Left multiplying (26) by r and then subtracting from (27), we get [x, r]yd(x) + [r, x 2 ]y = 0. In particular r = z, we get [x, z]yd(x) + [z, x 2 ]y = 0 (28) for all x, y, z I. By (26), we have [z, x 2 ]y = [x, z]d(x)y. Substituting the value of [z, x 2 ]y in (28), we obtain 0 = [x, z]yd(x) [x, z]d(x)y = [x, z][y, d(x)] (29) for all x, y, z I. Replacing z with d(x)z in (29) and using (29), we get [x, d(x)]z[y, d(x)] = 0. In particular y = x, we have ([x, d(x)]z) 2 = 0 for all x, z I which implies that ([d(x), x]i) 2 = 0 for all x I. By semiprimeness of R, it yields that [d(x), x]i = 0. Again by semiprimeness of R, it forces that [d(x), x] = 0 for all x I. Using similar technique with some necessary variations, we can prove that same conclusion holds for F ([x, y]) + (x y) = 0 for all x, y I. Corollary 2.6. Let R be a semiprime ring and let F be a multiplicative (generalized)- reverse derivation associated with the mapping d on R. If F ([x, y]) ± (x y) = 0 for all x, y R, then [d(x), x] = 0 for all x R. Theorem 2.7. Let R be a semiprime ring, I a non-zero ideal of R. Suppose F is a multiplicative (generalized)-reverse derivation associated with the mapping d on R. If F ([x, y]) = 0 for all x, y I, then [d(x), x] = 0 for all x I. Proof. Proof is follows as Theorem 2.5. Theorem 2.8. Let R be a semiprime ring and I a non-zero ideal of R. Suppose that F is a multiplicative (generalized)-reverse derivation associated with the mapping d on R. If F (xy) ± x y = 0 for all x, y I, then R is commutative. Proof. First, we consider the case F (xy) x y = 0

9 Some theorems of commutativity of semiprime rings with mappings 9 for all x, y I. Putting zx in place of x, in above relation, we obtain 0 = F (zxy) zx y = F (xy)z + xyd(z) (x y)z + (x y)z zxy yzx = (F (xy) x y)z + xyd(z) + [xy, z] + y[x, z] (30) for all x, y, z I. Now using hypothesis, (30) gives xyd(z) + [xy, z] + y[x, z] = 0 (31) for all x, y, z I. Substituting xy instead of y in (31), we obtain x 2 yd(z) + x[xy, z] + [x, z]xy + xy[x, z] = 0 (32) for all x, y, z I. Right multiplying (31) by x and then subtracting from (32), we obtain [x, z]xy = 0 for all x, y, z I. Putting z = zu, u I, we get Replacing uz in place of u, we obtain [x, z]uxy = 0 for all x, y, z, u I. (33) [x, z]uzxy = 0 for all x, y, z, u I. (34) Lastly putting y = zy in (33) and subtracting from (34), we get [x, z]u[x, z]y = 0 for all x, y, z, u I. It implies that ([x, z]y) 2 = 0 for all x, y, z I. By semiprimeness of R, conclude that [x, z] = 0 for all x, z I. By Lemma 1.4, it implies that R is commutative. By using similar approach, we can prove that same conclusion holds for F (xy) + (x y) = 0 for all x, y I. Theorem 2.9. Let R be a semiprime ring and I a non-zero ideal of R. Suppose that F is a multiplicative (generalized)-reverse derivation associated with the mapping d on R. If F (xy) ± [x, y] = 0 for all x, y I, then R is commutative. Proof. Proof follows from Theorem 2.8 Theorem Let R be a semiprime ring and I a non-zero ideal of R. Suppose that F is a multiplicative (generalized)-reverse derivation associated with the mapping d on R. If [F (x), d(y)] ± yx = 0 for all x, y I, then [d(x), x] = 0 for all x I. Proof. We start with the hypothesis [F (x), d(y)] yx = 0 for all x, y I. (35) Substituting zx in place of x in (35), we obtain 0 = [F (zx), d(y)] yzx = [F (x)z + xd(z), d(y)] yzx = [F (x), d(y)]z + F (x)[z, d(y)] + [x, d(y)]d(z) + x[d(z), d(y)] yzx(36) for all x, y, z I. Using (35), (36) reduces as

10 Some theorems of commutativity of semiprime rings with mappings 10 F (x)[z, d(y)] + [x, d(y)]d(z) + x[d(z), d(y)] + y[x, z] = 0 (37) for all x, y, z I. Putting x = ux, where u I in (37), we get F (x)u[z, d(y)] + xd(u)[z, d(y)] + u[x, d(y)]d(z) + [u, d(y)]xd(z) +ux[d(z), d(y)] + yu[x, z] + y[u, z]x = 0 (38) for all x, y, z, u I. Again putting x = ux in (38), we obtain F (x)u 2 [z, d(y)] + xd(u)u[z, d(y)] + uxd(u)[z, d(y)] + u 2 [x, d(y)]d(z) +u[u, d(y)]xd(z) + [u, d(y)]uxd(z) + u 2 x[d(z), d(y)] +yu 2 [x, z] + yu[u, z]x + y[u, z]ux = 0 (39) for all x, y, z, u I. Replacing u 2 in place of u in (38), we obtain F (x)u 2 [z, d(y)] + xd(u 2 )[z, d(y)] + u 2 [x, d(y)]d(z) +u[u, d(y)]xd(z) + [u, d(y)]uxd(z) + u 2 x[d(z), d(y)] + yu 2 [x, z] +yu[u, z]x + y[u, z]ux = 0 (40) for all x, y, z, u I. Subtracting (40) from (39), we get xd(u)u[z, d(y)] + uxd(u)[z, d(y)] xd(u 2 )[z, d(y)] = 0 (41) for all x, y, z, u I. Putting x = tx, t R, in (41), we get txd(u)u[z, d(y)] + utxd(u)[z, d(y)] txd(u 2 )[z, d(y)] = 0. (42) Left multiplying (41) by t and then subtracting from equation (42), we obtain [u, t]xd(u)[z, d(y)] = 0 for all x, y, z, u I and t R. Substituting zu with z in above relation and using [u, t]xd(u)[z, d(y)] = 0, we get [u, t]xd(u)z[u, d(y)] = 0 which implies that [u, t]x[u, d(u)]z[u, d(y)] = 0. In particular t = d(u), we get [u, d(u)]x[u, d(u)]z[u, d(y)] = 0, which gives {[x, d(x)]y} 3 = 0 for all x, y I. By semiprimeness of R it implies that [d(x), x]y = 0. Again by semiprimeness of R, we conclude desired result. By similar fashion, same conclusion holds for [F (x), d(y)] + yx = 0 for all x, y I. Theorem Let R be a semiprime ring and I a non-zero ideal of R. Suppose that F is a multiplicative (generalized)-reverse derivation associated with the mapping d on R. If [F (x), d(y)] ± xy = 0 for all x, y I, then [d(x), x] = 0 for all x I. Proof. Proof follows as Theorem Theorem Let R be a semiprime ring and I a non-zero ideal of R. Suppose that F is a multiplicative (generalized)-reverse derivation associated with the mapping d on R. If [F (x), d(y)] ± [x, y] = 0 for all x, y I, then [d(x), x] = 0 for all x I.

11 Some theorems of commutativity of semiprime rings with mappings 11 Proof. Proof follows as Theorem Theorem Let R be a semiprime ring and I a non-zero ideal of R. Suppose that F is a multiplicative (generalized)-reverse derivation associated with the mapping d on R. If one of the following condition holds: (i) F (xy) = F (x)f (y) for all x, y I; (ii) F (xy) = F (y)f (x) for all x, y I, then [d(x), x] = 0 for all x I. Proof. When F(xy)=F(x)F(y): Since F (xy) = F (y)x + yd(x) = F (x)f (y) for all x, y I. That is F (x)f (y) F (y)x yd(x) = 0 for all x, y I. (43) Putting y = zy in (43), we obtain F (x)f (y)z + F (x)yd(z) F (y)zx yd(z)x zyd(x) = 0 (44) for all x, y, z I. Right multiplying (43) by z and then subtracting from (44), we obtain F (x)yd(z) + F (y)[x, z] + [yd(x), z] yd(z)x = 0 (45) for all x, y, z I. Replacing zy in place of y in (45), we get F (x)zyd(z) + F (y)z[x, z] + yd(z)[x, z] + z[yd(x), z] zyd(z)x = 0 (46) for all x, y, z I. Again replacing zx in place of x in (45), we obtain F (x)zyd(z) + xd(z)yd(z) + F (y)z[x, z] + [yd(zx), z] yd(z)zx = 0 (47) for all x, y, z I. Now subtracting (46) from (47), we obtain xd(z)yd(z)+[yd(zx), z] yd(z)zx yd(z)[x, z] z[yd(x), z]+zyd(z)x = 0 (48) for all x, y, z I. Putting y = zy in (48), we get xd(z)zyd(z)+z[yd(zx), z] zyd(z)zx zyd(z)[x, z] z 2 [yd(x), z]+z 2 yd(z)x = 0 (49) for all x, y, z I. Left multiplying (48) by z and then subtracting from (49), we get [xd(z), z]yd(z) = 0 (50) for all x, y, z I. That is [xd(z), z]yd(z)z = 0 (51)

12 Some theorems of commutativity of semiprime rings with mappings 12 for all x, y, z I. By putting y = yx in above relation, it gives [xd(z), z]yxd(z)z = 0 (52) for all x, y, z I. Also, by putting y = yzx in (50), it implies that [xd(z), z]yzxd(z) = 0 (53) for all x, y, z I. Subtracting (52) from (53), we obtain [xd(z), z]y[xd(z), z] = 0. By semiprimeness of R, it implies that [xd(z), z] = 0 for all x, z I. Lastly substituting d(z)x in place of x in last expression, we get d(z)[xd(z), z] + [d(z), z]xd(z) = 0. Now using the fact that [xd(z), z] = 0, last relation gives [d(z), z]xd(z) = 0 for all x, z I. It follows that [d(z), z]x[d(z), z] = 0 for all x, z I. By semiprimeness of R, we conclude that [d(x), x] = 0 for all x I. When F(xy)=F(y)F(x): Since F (xy) = F (y)f (x) = F (y)x+yd(x) for all x, y I. This can be re-written as F (y){f (x) x} yd(x) = 0. (54) By putting y = zy, it gives F (y)z{f (x) x} + yd(z){f (x) x} zyd(x) = 0. (55) Substituting tz in place of z in (55), we obtain F (y)tz{f (x) x} + yd(tz){f (x) x} tzyd(x) = 0 (56) for all x, y, z I, t R. Replacing ty with y in (55), we get F (y)tz{f (x) x} + yd(t)z{f (x) x} + tyd(z){f (x) x} ztyd(x) = 0 (57) for all x, y, z I. By subtracting (56) from (57), it implies that {yd(t)z + tyd(z) yd(tz)}{f (x) x} + [t, z]yd(x) = 0. (58) In particular for t = z, (58) reduces to {yd(z)z + zyd(z) yd(z 2 )}{F (x) x} = 0 (59) for all x, y, z I. By putting x = rx, r R, (59) gives {yd(z)z + zyd(z) yd( 2 z)}{f (x)r + xd(r) rx} = 0. (60) Right multiplying (59) by r and then subtracting from (60), we get {yd(z)z + zyd(z) yd(z 2 )}{xd(r) [r, x]} = 0 (61) for all x, y, z I, r R. By substituting xr with x in (61), it gives {yd(z)z + zyd(z) yd(z 2 )}{xrd(r) [r, x]r} = 0 (62)

13 REFERENCES 13 for all x, y, z I, r R. Again right multiplying (61) by r and then subtracting from (62), we obtain {yd(z)z + zyd(z) yd(z 2 )}x[r, d(r)] = 0 (63) for all x, y, z I, r R. By replacing ty with y in (63), we get {tyd(z)z + ztyd(z) tyd(z 2 )}x[r, d(r)] = 0 (64) for all x, y, z I, r R. Left multiplying (63) by t and then subtracting from (64), we obtain [z, t]yd(z)x[r, d(r)] = 0 for all x, y, z I, r, t R. In particular for t = d(z) and r = z, above equation gives [z, d(z)]yd(z)x[z, d(z)] = 0 for all x, y, z I. Last expression implies that ([z, d(z)]y) 3 = 0 for y, z I. By semiprimeness of R, it implies that [d(x), x] = 0 for all x I. Hence proof of Theorem is complete. References [1] Aboubakr, A. and Gonzlez, S., Generalized reverse derivations on semiprime rings, Siberian Mathematical Journal 56(2) (2015), [2] Ashraf, M. and Ali, S., On (α, β) -derivation in H algebras, Advances in Algebra 2(1) (2009), [3] Ali, S., Ali, A., On generalized -derivations in -rings, Plastin Journal of Mathematics 1 (2012), [4] Ashraf, M., Ali, A., and Ali, S., Some commutative theorems for rings with generalized derivations, Southeast Asian Bull. Math. 31 (2007), [5] Ashraf, M. and Rehman, N., On derivations and commutativity in prime rings, East-West J. Math. 3(1) (2001), [6] Bell, H. E. and Daif, M. N., On derivations and commutativity in prime rings, Acta Math. Hungar. 66(4) (1995), [7] Bell, H. E. and Martindale, W. S. III, Centralizing mappings of semiprime rings, Canad. Math. Bull. 30(1) (1987), [8] Brešar, M., On the distance of the composition of two derivations to the generalized derivations, Glasgow Math. J. 33 (1991), [9] Brešar, M. and Vukman, J., On some additive mappings in rings with involution, Aequationes mathematicae 38 (1989), [10] Daif, M. N., When in a multiplicative derivation additive?, Int. J. Math. Math. Sci. 14(3) (1991),

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