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Generalize derivations on semiprime rings.

[section]1. Introduction

Throughout R will represent an associative ring with Z(R). R is said to be 2-torsion free if 2x = 0, x [member of] R implies x = 0. As usual the commutator xy - yx will be denoted by [x, y]. We shall use the basic commutator identites [xy, z] = [x, z]y + x[y, z] and is [x, yz] = y[x, y] + [x, y]z. Recall that a ring R is prime if aRb = 0 implies that either a = 0 or b = 0, and R is semi prime if aRa = 0 implies a = 0. An additive mapping d: R [right arrow] R is called derivation if d(ab) = d(a)b + ad(b) for all a,b [member of] R. And d is called Jordan derivation if d([a.sup.2]) = d(a)a + ad(a) for all a [member of] R. In [3], M. Bresar introduced the definition of generalize derivation on rings as follows: An additive map g: R [right arrow] R is called a generalize derivation if there exists a derivation d: R [right arrow] R such that g(ab) = g(a)b + ad(b) for all a,b [member of] R (we will call it of type 1). It is clear that every derivation is generalize derivation, and an additive map J: R [right arrow] R is called Jordan generalize if there exists a derivation d: R [right arrow] R such that J ([a.sup.2]) = J (a) a + ad(a) for all a [member of] R (we will call it of type 1). An additive map [partial derivative]: R x R [right arrow] R be called hochschild 2-cocylc if x(y,z) - [partial derivative] (xy,z) + [partial derivative] (x,yz) - [partial derivative] (x,y)z = 0 for all x,y,z [member of] R, the map [partial derivative] is called symmetric if [partial derivative] (x, y) = [partial derivative] (y, x) for all x,y [member of] R. It is clear that every Jordan derivation is Jordan generalize derivation, and every generalize derivation is Jordan generalize derivation, and since Jordan derivation may be not derivation, so Jordan generalize derivation, may be not generalize derivation in general. The properties of this type of mapping were discussed in many papers ([1].[2].[7]), especially, in [2], M. Ashraf and N. Rehman showed that if R is a 2-torsion free ring which has a commtator non zero divisor, then every Jordan generalize derivation on R is generalize derivation. In [7], Nurcan Argac, Atsushi Nakajima and Emine Albas introduced the notion of orthogonality for a pair f, g of generalize derivation, and they gave several necessary and sufficient conditions for f, g to be orthogonal. Nakajima [1] introduced another definition of generalize derivation as following: An additive mapping f: R [right arrow] R is called generalize derivation if there exist a 2 -coycle d: R x R [right arrow] R such that f (ab) = f (a)b + af (b) + [partial derivative](a, b) for all a, b [member of] R (we will call it of type 2). And is called Jordan generalize derivation if f ([a.sup.2]) = f (a)a + af (a) + [partial derivative] (a, a) for all a [member of] R (we will call it of type 2). In this paper we work on the generalize derivation (of type 2), we will extend the result of ([1], Theorem 1) and also we give several sufficient and necessary conditions which makes two generalize derivation (of type 2) to be orthogonal. For a ring R, let U be a subset of R, then the left annihilator of U (rep, right annihilator of U) is the set a [member of] R such that aU = 0 (res, is the set a [member of] R such that Ua = 0). We denote the annihilator of U by Ann(U). Note that U [intersection] Ann(U) = 0 and U [direct sum] Ann(U) is essential ideal of R.

[section]2. Preliminaries and examples

In this section, we give some examples and some well-known lemmas we are needed in our work.

Example 2.1. Let R be a ring d : R [right arrow] R be a derivation and [partial derivative] : R x R [right arrow] R defined by [partial derivative] (a,b) = 2d(a)d(b) and g : R [right arrow] R is defined as follows g(a) = d(d(a)) for all a [member of] R. g(a + b) = d(d(a + b)) = d(d(a) + d(b)) = d(d(a)) + d(d(b)) = g(a) + g(b) for all a, b [member of] R.

And, g(ab) = d(d(ab)) = d(d(a)b + ad(b)) = d(d(a)b) + d(ad(b)) = d(d(a))b + d(a)d(b) + d(a)d(b) + ad(d(b)) = g(a)b + ag(b) + 2d(a)d(b) = g(a)b + ag(b) + [partial derivative] (a, b) for all a, b [member of] R. Hence, g is generalize derivation (of type 2) on R.

The following Example explains that the definition of generalize derivation (of type 2) is more generalizing than the generalize derivation (of type 1).

Example 2.2. Let (f, d) be a generalize derivation on a ring R then the map (f, [partial derivative]) is generalize derivation, where [partial derivative] (a, b) = a(d - f)(b) for all a, b [member of] R.

Lemma 2.1. [5] Let R be a 2-torsion free ring, (f, [partial derivative]) : R [right arrow] R be generalize Jordan derivation, then f (ab + ba) = f (ab) + f (ba) = f (a)b + af (b) + d(a, b) + f (b)a + bf (a) + d(b, a) for all a, b [member of] R.

Lemma 2.2. [5] Let R be a 2-torsion free ring and (f, [partial derivative]) : R [right arrow] R be generalize Jordan derivation the map S: R x R [right arrow] R defined as follows: S(a, b) = f (ab) - (f (a)b + af (b) + [partial derivative] (a, b) for all a, b [member of] R. And the map [] : R x R [right arrow] R defined by [a, b] = ab - ba for all a, b [member of] R, then the following relations hold

(1) S(a, b)c[a, b] + [a, b]cS(a, b) = 0 for all a, b [member of] R.

(2) S (a, b)[a, b] = 0 for all a, b [member of] R.

Lemma 2.3. Let R be a 2-torsion free ring, (f, [partial derivative]) : R [right arrow] R be generalize Jordan derivation the map S : R x R [right arrow] R defined as follows: S(a, b) = f (ab) - (f (a)b + af (b) + [partial derivative] (a, b)) for all a, b [member of] R, then the following relations hold

(1) S([a.sub.1] + [a.sub.2], b) = S([a.sub.1], b) + S([a.sub.2], b) for all [a.sub.1], [a.sub.2], b [member of] R.

(2) S (a, [b.sub.1] + [b.sub.2]) = S(a, [b.sub.1]) + S (a, [b.sub.2]) for all a, [b.sub.1], [b.sub.2] [member of] R.

Proof.

(1) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

(2) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Lemma 2.4. [1] If R is a 2-torsion free semi prime ring and a, b are elements in R then the following are equivalent.

(i) a x b = 0 for all x in R.

(ii) b x a = 0 for all x in R.

(iii) a x b + b x a = 0 for all x in R. If one of them fulfilled, then ab = ba = 0.

[section]3. Generalize Jordan derivations on a semi prime rings

In this section, we extend the result proved by Nakajima ([1], Theorem 1(1), (2))by adding condition.

Theorem 3.1. If R is a commutative 2-torsion free ring and (f, [partial derivative]) : R [right arrow] R be a generalize Jordan derivation then (f, [partial derivative]) is a generalize derivation.

Proof. Let S(a, b) = f (ab) - f (a)b - af (b) - [partial derivative] (a, b) for all a, b [member of] R. And by Lemma (2.1) we have S(a, b) + S(b, a) = 0. So

S(a, b) = -S(b, a). (1)

And, f (ab - ba) = f (a)b + af (b) + [partial derivative](a, b) - f(b)a - bf(a) - [partial derivative] ((b, a). Since R is commutative.

Then

S (a, b) = f (ab) - f (a)b - af (b) - [partial derivative](a, b) = f (ba) - bf (a) - f (b)a - [partial derivative] (b, a) = S(b, a). (2)

From (1) and (2), we get 2S(a, b) = 0. And since R is 2-torsion free ring, so S(a, b) = 0. Hence f is a generalize derivation.

Theorem 3.2. Let R be a non commutative 2-torsion free semi prime ring and (f, [partial derivative]) : R [right arrow] R be a generalize Jordan derivation then (f, [partial derivative]) is a generalize where [partial derivative] is symmetric.

Proof. Let a, b [member of] R, then by Lemma (2.2,(1)) S(a, b)c[a, b] + [a, b]cS(a, b) = 0, for all c [member of] R. And since R is semi prime and by Lemma (2.4), we get S(a, b)c[a, b] = 0 for all c [member of] R, it is clear that S are additive maps on each argument, then:

S(a, b)c[x, y] = 0 for all x,y,c [member of] R. (1)

Now,

2S(x,y)wS(x,y) = S(x,y)w(S (x,y) + S(x,y))S (x,y)w(S (x,y) - S (y,x)) = S(x, y)w(f (x, y) - (f (x)y + xf (y) + [partial derivative] (x, y)) - (f (yx) - (f (y)x - yf (x) - [partial derivative](y,x))

Since [partial derivative] is symmetric, so [partial derivative] (x, y) = [partial derivative] (y, x), then

2S (x,y)wS (x,y) = S(x,y)w(f (x,y) - f (x)y - xf (y) - f (yx) + f (y)x + yf (x)) = S(x, y)w((f (xy) - f (yx) + [f (y), x] + [y, f (x)]) = S(x, y)w(f (xy - yx) + [f (y), x] + [y, f (x)]) = S(x, y)wf (xy - yx) + S(x, y)w[f (y), x] + S(x, y)w[y, f (x)].

By equation (1), we get 2S(x,y)wS(x,y) = S(x,y)wf(xy - yx) = S(x,y)w(f(x)y + xf(y) + [partial derivative](x,y) - f(y)x - yf(x) - [partial derivative](y,x)) Since [partial derivative] is symmetric. Thus 2S(x,y)wS(x,y) = S(x, y)w([f (x), y] + [x, f (y)]) = S(x, y)w[f (x), y] + S(x, y)w[x, f (y)]. Then by equation (1), we get 2S(x, y)wS(x, y) = 0 for all x, y, w [member of][member of] R, and since R is 2-torsion free, so S(x, y)wS(x, y) = 0 for all x, y, w [member of] R, since R is prime ring Thus, S(x, y) = 0 for all x, y [member of] R, then f(xy) = f (x)y + xf (y) + [partial derivative](x, y). Hence, f is generalize derivation on R.

[section]4. Orthogonal generalize derivations on semi prime rings

In this section, we gave some necessary and sufficient conditions for tow generalized derivation of type (2), to be orthogonal and also we show that the image of two orthogonal generalize derivations are different from each other except for both are is zero.

Definition 4.1. Two map f and g on a ring R are orthogonal if f (x)Rg(y) = 0 for all x, y [member of] R.

Lemma 4.1. If R is a ring [partial derivative]: R [right arrow] R is a hochschild 2-cocycle, S = R [direct sum] R, and [partial derivative] : S x S [right arrow] S defined by [partial derivative] : (([x.sub.1], [y.sub.1]), ([x.sub.2], [y.sub.2])) = ([partial derivative] ([x.sub.1], [x.sub.2]), 0) for all ([x.sub.1], [y.sub.1]), ([x.sub.2], [y.sub.2]) [member of] S x S is a hochschild 2-cocycle.

Proof. It is clear that [partial derivative] is an additive map in each argument and (([x.sub.1], [x.sub.2]) [partial derivative] (([y.sub.1], [y.sub.2]), ([z.sub.1], [z.sub.2])) - [partial derivative] (([x.sub.1], [x.sub.2])([y.sub.1], [y.sub.2]), ([z.sub.1], [z.sub.2])) + [partial derivative] (([x.sub.1], [x.sub.2]), ([y.sub.1], [y.sub.2])([z.sub.1], [z.sub.2])) - [partial derivative] (([x.sub.1], [x.sub.2]), ([y.sub.1], [y.sub.2])([z.sub.1], [z.sub.2])) = ([x.sub.1] [partial derivative] ([y.sub.1], [z.sub.1]) - [partial derivative] ([x.sub.1][y.sub.1], [z.sub.1]) + [partial derivative]([x.sub.1], [y.sub.1][z.sub.1]) -[partial derivative] ([x.sub.1], [y.sub.1]) [z.sub.1], 0) = (0,0) for all ([x.sub.1], [x.sub.2]), ([y.sub.1], [y.sub.2]), ([z.sub.1], [z.sub.2]) [member of] S x S. Thus, [partial derivative] is a hochschild 2-cocycle.

Theorem 4.1. For any generalize derivation (f, [[partial derivative].sub.1]) on a ring R there exist tow orthogonal generalize derivation (h, [[partial derivative].sub.1]), (g, [[partial derivative].sub.2]) on S = R [direct sum] R such that h(x,y) = (f (x), 0), g(x,y) = (0,f (y)) for all x, y [member of] R.

Proof. Let h : S [right arrow] S and g : S [right arrow] S defined as following h(x, y) = (f (x), 0) for all (x, y) [member of] S, g(x, y) = (0, f (y)) for all (x, y) [member of] S, and let [[partial derivative].sub.1] defined as in Lemma 4.1. Now it is clear that h, g are additive mapping and h((x, y)(z, w)) = h(xz, yz) = (f (xz), 0) =(f (x)z, 0) + (xf (z), 0) + ([partial derivative] (x, z), 0) =(f (x), 0)(z, w) + (x, y)(f (z), 0) + ([partial derivative] (x, z), 0) = h(x, y)(z, w) + (x, y)h(z, w) + [partial derivative] ((x, y) (z, w)). Thus (h, [partial derivative]) is a generalize derivation on S. In similar way (g, [partial derivative]) is a generalize derivation on S, and h(x, y)(m, n)g(z, w) = (f (x), 0)(m, n)(0, f (w)) = (0,0) for all (x, y), (m, n), (z, w) [member of] S. h(x, y)Sg(z, w) = 0 for all (x, y), (z, w) [member of] S. Hence h, g are orthogonal generalize derivation on S.

Theorem 4.2. If (f, [[partial derivative].sub.1]), (g, [[partial derivative].sub.2]) are a generalize derivation on commutative semi prime ring R with identity, then the following are equivalent:

(i) (f, [[partial derivative].sub.1]) and (g, [[partial derivative].sub.2]) are orthogonal.

(ii) f(x)g(y) = 0 and [[partial derivative].sub.1] (x, y)g(w) = 0 for all x, y, w [member of] R.

(iii) f(x)g(y) = 0 and [[partial derivative].sub.2] (x, y)f(w) = 0 for all x, y, w [member of] R.

(iv) There exist tow ideal U, V of R such that

(a) U [intersection] V = 0.

(b) f (R) [subset or equal to] U and g(R), [[partial derivative].sub.2](R, R) [subset or equal to] V.

Proof. (i) [right arrow] (ii) Since f (x)Rg(y) = 0 for all x,y [member of] R, then by Lemma 2.4, we get f (x)g(y) = 0 for all x,y [member of] R. Now 0 = f (xy)g(w) = f (x)yg(w) + xf (y)g(w) + [[partial derivative].sub.1] (x, y)g(w). But f (x)yg(w) = 0 = xf (y)g(w), so [[partial derivative].sub.1](x,y)g(w) = 0 for all x,y,w [member of] R.

(i) [right arrow] (iii) In similar way of (i) [right arrow] (ii).

(i) [right arrow] (iv) Let U be an ideal of R generated by f (R) and V = Ann(U), then by Lemma

4.1, U [intersection] V = 0 and by (ii). [[partial derivative].sub.2](x, y)f(w) = 0 for all x, y, w [member of] R, and since R is commutative ring with identity, so it is clear that U = where [r.sub.i] [member of] R, [s.sub.i] [member of] f (R) and n is any positive integer=0, thus clear that [[partial derivative].sub.2](x, y)U = 0 and since f(x)g(y) = 0 for all x, y [member of] R, then where [r.sub.i] [member of] R, [s.sub.i] [member of] f(R) and n is any positive integer. g(y) = 0, for all y [member of] R. Ug(y) = 0 for all y [member of] R, then [[partial derivative].sub.2](R, R), g(R) [subset or equal to] Ann(U) = V.

(ii) [right arrow] (i) For all x,y,w [member of] R, f (xy)g(w) = 0 = f (x)yg(w) + xf (y)g(w) + [[partial derivative].sub.1](x,y)g(w). But f (y)g(w) = 0 = [[partial derivative].sub.1](x,y)g(w). So, f (x)yg(w) = 0, f (x)Rg(w) = 0. Thus f, g are orthogonal generalize derivations.

(iii) [right arrow] (i) In similar way of (ii) - (i).

(iv) [right arrow] (i) Since f(R) [subset or equal to] U, f(R)g(R) = 0, then g(R) [subset or equal to] V, Thus, f(R)Rg(R) [subset or equal to] U n V, then f(R)Rg(R) = 0. Hence f, g are orthogonal generalize derivations. This completes the proof.

Corollary 4.1. If (f, [[partial derivative].sub.1]), (g, [[partial derivative].sub.2]) are a generalize derivation on commutative semi prime ring R with unity, then f(R) [intersection] g(R) = 0.

Proof. By Theorem 4.2 there exists tow ideals U, V of R such that U [intersection] V = 0 And f (R) [subset or equal to] U and g(R) [subset or equal to] V. f (R)g(R) [subset or equal to] U n V, hence f (R) [intersection] g(R) = 0.

References

[1] A. Nakajima, On generalized Jordan derivation associate with hochschilds 2-cocyle on rings, Turk. J. math, 30(2006), 403-411.

[2] M. Ashraf and N. Rehman, On Jordan generalized derivation in rings, Math. J. Okayama Univ, 42(2002), 7-9.

[3] F. Harary, Structural duality, Behav. Sci., 2(1957), No. 4, 255-265.

[4] M. Bresar, On thedistance of the composition of two derivation to the generalized derivations, Glasgow. Math. J., 33(1991), 80-93.

[5] M. Bresar and J. Vukman, Orthogonal derivation and extension of a theorem of posner, Radovi Matematieki, 5(1989), 237-246.

[6] M. Ferero and C. Haetinger, Higher derivations and a theorem by Herstein, Quaestionnes Mathematicae, 25(2002), 1-9.

Arkan. Nawzad ([dagger]), Hiba Abdulla ([double dagger]) and A. H. Majeed (#)

([dagger]) Department of Mathematics, College of Science, Sulaimani University, Sulaimani-IRAQ

([double dagger], #) Department of Mathematics, College of Science, Baghdad University, Baghdad-IRAQ E-mail: hiba_abdulla@yahoo.com Ahmajeed6@yahoo.com
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Author:Nawzad, Arkan; Abdulla, Hiba; Majeed, A.H.
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