Southeast Asian Bulletin of Mathematics (2007) 31: 415–421
Southeast Asian Bulletin of Mathematics c SEAMS. 2007 °
Some Commutativity Theorems for Rings with Generalized Derivations Mohammad Ashraf, Asma Ali and Shakir Ali Department of Mathematics, Aligarh Muslim University, Aligarh-202002 India E-mail:
[email protected], asma−
[email protected],
[email protected]
AMS Mathematics Subject Classification (2000): 16W25, 16N60, 16U80 Abstract. Let R be a ring with center Z(R). The aim of this paper is to investigate the commutativity of R satisfying any one of the following properties: (i) F (xy) − xy ∈ Z(R), (ii) F (xy) + xy ∈ Z(R), (iii) F (xy) − yx ∈ Z(R), (iv) F (xy) + yx ∈ Z(R), (v) F (x)F (y) − xy ∈ Z(R) and (vi) F (x)F (y) + xy ∈ Z(R), for all x, y ∈ R, where F is a generalized derivation on R. Keywords: Prime rings; Derivations; Generalized derivations.
1. Introduction Throughout the paper, R will denote an associative ring with center Z(R). For any x, y ∈ R, we write [x, y] = xy − yx. Recall that R is prime if aRb = (0) implies that a = 0 or b = 0. An additive mapping d : R −→ R is called a derivation if d(xy) = d(x)y + xd(y), holds for all x, y ∈ R. Following Bre˘ sar [5], an additive mapping F : R −→ R is said to be a generalized derivation on R if there exists a derivation d : R −→ R such that F (xy) = F (x)y + xd(y) holds for all x, y ∈ R. However, generalized derivation covers the concept of derivation. Also with d = 0, a generalized derivation covers the concept of left multiplier that is, an additive mapping F satisfying F (xy) = F (x)y, for all x, y ∈ R. Very recently, Hvala [12] initiated the algebraic study of generalized derivation and extended some results concerning derivation to generalized derivation. There has been ongoing interest concerning the relationship between the commutativity of a rings and the existence of certain specific types of derivations of R. Recently, many authors viz. [2], [3],[4], [8] and [9] etc. have obtained commutativity of prime and semiprime rings with derivations satisfying certain polynomial constants. The first author in 2001 together with Nadeem [2] established that a prime ring R with a non-zero ideal I must be commutative if it
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admits a derivation d satisfying either of the properties d(xy) + xy ∈ Z(R) or d(xy) − xy ∈ Z(R) for all x, y ∈ I. Inspired by the above result, we shall explore the commutativity of a prime ring R in which the generalized derivation F satisfies any one of the properties: (i) F (xy) − xy ∈ Z(R), (ii) F (xy) + xy ∈ Z(R), (iii) F (xy) − yx ∈ Z(R), (iv) F (xy) + yx ∈ Z(R), (v) F (x)F (y) − xy ∈ Z(R) and (vi) F (x)F (y) + xy ∈ Z(R), for all x, y ∈ R.
2. Main Results We begin with the following theorem. Theorem 2.1. Let R be a prime ring and I be a non-zero ideal of R. If R admits a generalized derivation F associated with a non-zero derivation d such that F (xy) − xy ∈ Z(R) for all x, y ∈ I, then R is commutative. Proof. If F = 0, then xy ∈ Z(R) for all x, y ∈ I. In particular [xy, x] = 0, for all x, y ∈ I and hence x[y, x] = 0. Replacing y by yz, we get xy[z, x] = 0, for all x, y, z ∈ I. Hence, it follows that xRI[z, x] = (0), for all x, z ∈ I. Thus the primeness of R forces that for each x ∈ I, either x = 0 or I[x, z] = (0). But x = 0 also implies that I[x, z] = (0). Hence in both the cases we find that I[x, z] = (0), for all z ∈ I i.e., IR[x, z] = (0). Since I 6= (0) and R is prime, the above expression yields that [x, z] = 0 for all x, z ∈ I. Now replace x by xr, to get x[r, z] = 0. Again replacing x by xs, we get xs[r, z] = 0 for all x, z ∈ I and r, s ∈ R. i.e., xR[r, z] = (0), and the primeness of R forces that either x = 0 or [r, z] = 0, but I 6= (0), we have [r, z] = 0. Replace z by zs to get z[r, s] = 0 for all z ∈ I and r, s ∈ R, this implies that zR[r, s] = (0). The primeness of R forces that either z = 0 or [r, s] = 0, but I 6= (0), we have [r, s] = 0 for all r, s ∈ R i.e., R is commutative. Hence, onward we assume that F 6= 0. For any x, y ∈ I, we have F (xy)−xy ∈ Z(R). This can be rewritten as F (x)y + xd(y) − xy ∈ Z(R). Replacing y by yz, we obtain F (x)yz + xd(y)z + xyd(z) − xyz ∈ Z(R), for all x, y, z ∈ I. Thus, in particular [(F (x)y + xd(y) − xy)z + xyd(z), z] = 0, for all x, y, z ∈ I.
(2.1)
This gives that [xyd(z), z] = 0, for all x, y, z ∈ I and hence xy[d(z), z] + x[y, z]d(z) + [x, z]yd(z) = 0, for all x, y, z ∈ I.
(2.2)
For any y1 ∈ I, replace x by y1 x in (2.2) and use (2.2), to get [y1 , z]xyd(z) = 0 for all x, y, z ∈ I and hence [y1 , z]xRId(z) = (0). Thus, the primeness of R implies that for each z ∈ I, either Id(z) = (0) or [y1 , z]x = 0. The set of z ∈ I
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for which these two properties hold are additive subgroups of I whose union is I. Therefore either Id(z) = (0), for all z ∈ I or [y1 , z]x = 0, for all x, y1 , z ∈ I. If Id(z) = (0), for all z ∈ I, then IRd(z) = (0), for all z ∈ I. Since I 6= (0), and R is prime, the above expression gives that d(z) = 0, for all z ∈ I. This implies that d(zr) = 0, for all z ∈ I and r ∈ R. Hence it follows that zd(r) = 0 that is IRd(r) = (0). Since I 6= (0), the primeness of R yields that d(r) = 0 for all r ∈ R, a contradiction. On the other hand if [y1 , z]x = 0, for all x, y1 , z ∈ I, then [y1 , z]RI = (0), for all y1 , z ∈ I. The primeness of R implies that [y1 , z] = 0, for all y1 , z ∈ I and hence, we get the required result.
Theorem 2.2. Let R be a prime ring and I be a non-zero ideal of R. If R admits a generalized derivation F associated with a non-zero derivation d such that F (xy) + xy ∈ Z(R), for all x, y ∈ I, then R is commutative. Proof. If F is a generalized derivation satisfying the property F (xy)+xy ∈ Z(R), for all x, y ∈ I, then the generalized derivation (−F ) satisfies the condition (−F )(xy) − xy ∈ Z(R), for all x, y ∈ I and hence by Theorem 2.1, R is commutative.
Theorem 2.3. Let R be a prime ring and I be a non-zero ideal of R. If R admits a generalized derivation F associated with a non-zero derivation d such that F (xy) − yx ∈ Z(R), for all x, y ∈ I, then R is commutative. Proof. Given that F (xy)−yx ∈ Z(R), for all x, y ∈ I. If F = 0, then yx ∈ Z(R), for all x, y ∈ I. Using the same arguments as used in the beginning of the proof of Theorem 2.1, we get the required result. Hence, onward we shall assume that F 6= 0. For any x, y ∈ I, we have F (xy) − yx ∈ Z(R). This can be rewritten as F (x)y + xd(y) − yx ∈ Z(R), for all x, y ∈ I that is, [F (x)y + xd(y) − yx, r] = 0, for all x, y ∈ I, r ∈ R.
(2.3)
The above expression implies that [F (x), r]y + F (x)[y, r] + [x, r]d(y) + x[d(y), r] = [y, r]x + y[x, r].
(2.4)
Replacing y by yr in (2.4), we obtain ([F (x), r]y + F (x)[y, r] + [x, r]d(y) + x[d(y), r])r + [x, r]yd(r) +xy[d(r), r] + x[y, r]d(r) = [y, r]rx + yr[x, r].
(2.5)
Using (2.4) in (2.5), we find that [y, r][x, r] + y[[x, r], r] + [x, r]yd(r) + x[y, r]d(r) + xy[d(r), r] = 0, for all x, y ∈ I, r ∈ R.
(2.6)
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Now, replace y by xy in (2.6) and use (2.6), to get [x, r]xyd(r) + [x, r]y[x, r] = 0, for all x, y ∈ I, r ∈ R.
(2.7)
Further, replacing r by r + x in (2.7) and using (2.7), we get [x, r]xyd(x) = 0, for all x, y ∈ I, r ∈ R. This yields that [x, r]xRId(x) = (0). Thus primeness of R shows that for each x ∈ I either [x, r]x = 0 or Id(x) = (0). If [x, r]x = 0, for all r ∈ R, then replace r by rs, to get [x, r]sx = (0) for all r, s ∈ R that is, [x, r]Rx = (0). Again primeness of R gives that x = 0 or [x, r] = 0. But x = 0 also implies that [x, r] = 0, for all r ∈ R. Thus, it remains only to dispose of the case when for each x ∈ I either [x, r] = 0 or d(x) = 0. The sets of elements of I for which these two conditions hold are additive subgroups of I whose union is I; consequently, we must have either [x, r] = 0, for all x ∈ I, r ∈ R or d(I) = (0). But if d(I) = (0) then d = 0, a contradiction. Thus, [x, r] = 0, for all x ∈ I, r ∈ R. Therefore I is central and hence R is commutative.
Using similar arguments as above, we can prove the following: Theorem 2.4. Let R be a prime ring and I be a non-zero ideal of R. If R admits a generalized derivation F associated with a non-zero derivation d such that F (xy) + yx ∈ Z(R), for all x, y ∈ I, then R is commutative.
Theorem 2.5. Let R be a prime ring and I be a non-zero ideal of R. If R admits a generalized derivation F associated with a non-zero derivation d such that F (x)F (y) − xy ∈ Z(R) for all x, y ∈ I, then R is commutative. Proof. By hypothesis, we have F (x)F (y) − xy ∈ Z(R), for all x, y ∈ I. If F = 0, then xy ∈ Z(R), for all x, y ∈ I. Using the same arguments as we have used in the proof of Theorem 2.1, we get the required result. Hence, onward we shall assume that F 6= 0. For any x, y ∈ I, we have F (x)F (y) − xy ∈ Z(R). Replacing y by yr, we find that F (x)(F (y)r + yd(r)) − xyr ∈ Z(R) i.e., (F (x)F (y) − xy)r + F (x)yd(r) ∈ Z(R), for all x, y ∈ I, r ∈ R.
(2.8)
This implies that [F (x)yd(r), r] = 0, for all x, y ∈ I, r ∈ R.
(2.9)
This can be rewritten as F (x)[yd(r), r] + [F (x), r]yd(r) = 0, for all x, y ∈ I, r ∈ R.
(2.10)
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Substituting F (x)y for y in (2.10) and using (2.10), we find that [F (x), r]F (x)yd(r) = 0, for all x, y ∈ I, r ∈ R.
(2.11)
That is, [F (x), r]F (x)RId(r) = (0). Thus for each r ∈ R, primeness of R forces that either [F (x), r]F (x) = 0 or Id(r) = (0). The set of r ∈ R for which these two properties hold form additive subgroups of R whose union is R. Hence either [F (x), r]F (x) = 0, for all x ∈ I, r ∈ R or Id(r) = (0), for all r ∈ R. If Id(r) = (0), for all r ∈ R, then IRd(r) = (0), for all r ∈ R. Since I 6= (0) and R is prime, the above relation yields that d = 0, which is a contradiction. Therefore, assume the remaining possibility that [F (x), r]F (x) = 0, for all x ∈ I, r ∈ R. For any s ∈ R, replace r by rs, to get [F (x), r]RF (x) = (0), for all x ∈ I, r ∈ R. The primeness of R implies that for each x ∈ I, either F (x) = 0 or [F (x), r] = 0. Thus in each case we have, [F (x), r] = 0, for all x ∈ I and r ∈ R. The above relation gives that F (x) ∈ Z(R) for all x ∈ I and hence F (x)F (y) ∈ Z(R), for all x, y ∈ I. Thus, our hypotheses yield that xy ∈ Z(R). Now using the same arguments as we have used in the beginning of the proof of the Theorem 2.1, we get the required result.
Application of similar arguments yields the following: Theorem 2.6. Let R be a prime ring and I be a non-zero ideal of R. If R admits a generalized derivation F associated with a non-zero derivation d such that F (x)F (y) + yx ∈ Z(R), for all x, y ∈ I, then R is commutative.
Theorem 2.7. Let R be a prime ring and I be a non-zero ideal of R. Then the following conditions are equivalent: (i) R admits a generalized derivation F associated with a non-zero derivation d such that F (xy) − xy ∈ Z(R) or F (xy) + xy ∈ Z(R) for all x, y ∈ I. (ii) R admits a generalized derivation F associated with a non-zero derivation d such that F (xy) − yx ∈ Z(R) or F (xy) + yx ∈ Z(R) for all x, y ∈ I. (iii) R admits a generalized derivation F associated with a non-zero derivation d such that F (x)F (y) − xy ∈ Z(R) or F (x)F (y) + xy ∈ Z(R) for all x, y ∈ I. (iv) R is commutative. Proof. Obviously, (iv) ⇒ (i), (iv) ⇒ (ii) and (iv) ⇒ (iii). (i) ⇒ (iv). For each x ∈ I, we put I1 = {y ∈ I | F (xy) − xy ∈ Z(R)}, I2 = {y ∈ I | F (xy) + xy ∈ Z(R)}. Then it can be easily seen that I1 and I2 are additive subgroups of I whose union is I. Thus by Brauer’s trick, either I1 = I
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or I2 = I. Further, using similar arguments we find that I = {x ∈ I | I = I1 } or I = {x ∈ I | I = I2 }. Therefore, R is commutative by Theorem 2.1 and Theorem 2.2. (ii) ⇒ (iv). For each x ∈ I, put I1 = {y ∈ I | F (xy) − yx ∈ Z(R)}, and I2 = {y ∈ I | F (xy) + yx ∈ Z(R)} and use the similar arguments as above, to get the required result. (iii) ⇒ (iv). For each x ∈ R, set I1 = {y ∈ I | F (x)F (y) − xy ∈ Z(R)} and I2 = {y ∈ I | F (x)F (y) + xy ∈ Z(R)}, use Theorems 2.5 and Theorem 2.6 , to get the required result. Finally, it is to remark that the above theorem is not true in the case of an arbitrary ring (see Example 2.14 of [1]). Acknowledgement. The third author gratefully acknowledges the financial support he received from U.G.C., India.
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