Orthogonal Derivations on an Ideal. of Semiprime Γ-Rings

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1 International Mathematical Forum, Vol. 7, 2012, no. 28, Orthogonal Derivations on an Ideal of Semiprime Γ-Rings Nishteman N. Suliman Department of Mathematics College of Education, Scientific Departments Salahaddin University Kurdistan Region Erbil, Iraq Abdul-Rahman H. Majeed Baghdad University/College of Science Department of Mathematics, Iraq Abstract In this paper, we generalized some results concerning orthogonal derivations for a nonzero ideal of a semiprime Γ-ring. These results which are related to some results concerning product derivations on a Γ-rings. Mathematics Subject Classification: 16W25, 16Y30 Keywords: Semiprime Γ- rings, derivations, orthogonal derivations. 1. Introduction Nobusawa [5] introduced the notion of a Γ-ring, more general than a ring. Burnes [2] weakened the conditions in the definition of Γ-ring in the sense of Nobusawa. After these two authors, many mathematicians made works on Γ-ring in the sense of Barnes and Nobusawa, which are parallel to the results in the ring theory. The gamma ring is defined by Barnes in [2] as follows:

2 1406 N. N. Suliman and A.-R. H. Majeed A Γ-ring is a pair (M, Γ) where M and Γ are additive abelian groups for which there exists a map from MXΓXM M (the image of (x, α, y) was denoted by xαy) for all x, y z M and α Γ satisfying the following conditions: (i) xαy M (ii) (x + y)αz = xαz + yαz, x(α + β)y = xαy + xβy, xα(y + z) = xαy + xαz, (iii) (xαy)βz = xα(yβz). We may note that it follows from (i) (iii) that 0αx = x0y =0αx = 0, for all x, y M and α Γ. A Γ-ring M is said to be 2-torsion free if 2x = 0 implies x = 0 for x M. M is called a prime if for any two elements x, y M, xγmγy = 0 implies either x = 0 or y = 0, and M is called semiprime if xγmγx = 0 with x M implies x = 0. Note that every prime Γ-ring is obviously semiprime. An additive subgroup U of M is called a left (right) ideal of M if MΓU U (UΓM U). If U is both left and right ideal of M, then we say U is an ideal of M.. Following [6] a subset U of M, Ann l (U) = {a M aγu = <0> }is called the left annihilator of U. A right annihilator Ann r (U) can be defined similarly. It is known that the right and left annihilators of an ideal U of a semiprime Γ-ring M coincide, it will be denoted by Ann(U). Not that U Ann l (U) = {0} (U Ann r (U) = {0}). Jing in [4], defined the derivation of Γ-ring as follows: An additive mapping d: M M is called a derivation if d(xαy) = d(x)αy + xαd(y) for all x, y M and α Γ. Beršar and Vukman in [ 3] introduced the notion of orthogonality for a pair of derivations (d, g) of a semiprime ring, and they gaves several necessary and sufficient conditions for d, g to be orthogonal on a semiprime Γ-ring. Asharf and Jamal in [1] they study the concepts of orthogonal derivation in Γ-ring M as follows: Two mappings f and g of a Γ-ring M are said to be orthogonal on M if f(x)γmγg(y) = 0 = g(y)γmγf(x) for all x, y M and α Γ, and obtained some results analogous to obtained by Brešar and Vukman [3]. In this paper we extend the results Ashraf and Jamal in [1] to orthogonal derivation on a nonzero ideal of semiprime Γ-ring M and M satisfying xαyβz = xβyαz, for all x, y, z M and α, β Γ, and it will be represented by (*).

3 Orthogonal derivations on ideal The Results To prove the main result we need the following lemmas. Lemma 2.1. Let M be a 2-torsion free semiprime Γ-ring, U a nonzero ideal of M and a, b the elements of M. Then the following conditions are equivalent: (i) aγuγb = (0) (ii) bγuγa = (0) (iii) aγuγb + bγuγa = (0) If one of these conditions are satisfying and Ann l (U) = 0, then aγb = bγa = 0. Proof. (i) (ii) Suppose that aγuγb = 0. Then bγuγaγuγbγuγa = 0, since U is an ideal then bγuγaγuγmγbγuγaγu = 0. By semiprimeness, bγuγaγu = 0, hence bγuγa Ann l (U) = 0, we get bγuγa = 0. (ii) (iii) Suppose that bγuγa=0, that is aγuγb= 0, this implies aγuγb + bγuγa = 0. (iii) (i) Suppose that aγuγb + bγuγa = 0, that is aγuγb = - bγuγa. Let u and v be any two elements of U. Then by hypotheses we have (aγuγb)γvγ(aγuγb) = - aγuγaγvγbγuγb = aγuγbγvγ(bγuγa) = - aγuγbγvγaγuγb This implies 2(aΓuΓb)ΓvΓ(aΓuΓb) = 0. Since M 2-torsion free Γ-ring, we obtain (aγuγb)γvγ(aγuγb) = 0. Since U be an ideal, then (aγuγb)γuγmγ(aγuγb)γu = 0. By the semiprimeness we get (aγuγb)γu = 0, hence aγuγb Ann l (U) = 0, aγuγb = 0. For all u U. Hence we get aγuγb = bγuγa = 0. Lemma 2.2. Let M be a 2-torsion free semiprime Γ-ring, and U be a nonzero ideal of M such that Ann l (U) = 0. Suppose that additive mappings f and h of M into itself satisfy f (x)γuγh(x) = (0) for all x U. Then f (x)γuγh(y) = (0) for all x, y U. Proof. Suppose that f(x)αuβh(x) = 0 for all x, u U and α, β Γ. Linearizing we get Then we have f(x)αuβh(y) + f(y)αuβh(x) = 0 for all x, y, u U and α, β Γ

4 1408 N. N. Suliman and A.-R. H. Majeed f(x)αuβh(y)γvδf(x)αuβh(y) = - f(y)αuβh(x)γvβf(x)αuβh(y) = 0 Replacing v by vτm, we get f(x)αuβh(y)γvτmδf(x)αuβh(y)γv = 0, for all x, y,u,v U and α, β, γ, δ,τ Γ. By semiprimeness we obtain f(x)αuβh(y)γv = 0, that is f(x)αuβh(y) Ann l (U) = 0, this implies f(x)αuβh(y) = 0, for all x, y, u U and α, β Γ. Lemma 2.3. Let M be a 2-torsion free semiprime Γ-ring, d and g be derivations of M, and U be a nonzero ideal of M such that Ann l (U) = 0. If d(x)αg(y) + g(x)αd(y) = 0 for all x, y U and α Γ, then d and g are orthogonal. Proof. Suppose that d(x)αg(y) + g(x)αd(y) = 0 for all x, y U and α Γ. Replacing y by yβm we get 0 = d(x)αg(yβm) + g(x)αd(yβm) = d(x)αg(y)βm + d(x)αyβg(m) + g(x)αd(y)βm + g(x)αyβd(m) = d(x)αyβg(m) + g(x)αyβd(m), for all x, y U, m M and α Γ. Replacing x by mγx we get 0 = d(mγx)αyβg(m) + g(mγx)αyβd(m) = d(m)γxαyβg(m) + mγd(x)αyβg(m) + g(m)γxαyβd(m) + mγg(x)αyβd(m) = d(m)γxαyβg(m) + g(m)γxαyβd(m), for all x, y U, m M and α, β, γ Γ. Since xαy UΓU U. Put u = xαy, then by Lemma 2.1 we get d(m)γuβg(m) = 0. By Lemma 2.2 yield d(m)γuβg(s) = 0, for all u U, m, s M and β, γ Γ. Replacing u by tαg(s)βuγd(m)αt, we have d(m)γtαg(s)βuγd(m)αtβg(s) = 0. Replacing u by uτr we have d(m)γtαg(s)βuτrγd(m)αtβg(s)γu=0. By semiprimeness and (*) we get d(m)γtαg(s)βu=0, for all u U, m, t, s M and α, γ Γ, yields d(m)γtαg(s) Ann l (U), therefore d(m)γtαg(s) = 0, for all m, s, t M and γ, β Γ. Hence d and g are orthogonal. Theorem 2.4. [1, Theorem 2.1] Let M be a 2-torsion free semiprime Γ-ring. Suppose d and g are derivations of M. Then the following conditions are equivalent: (i) d and g are orthogonal. (ii) dg = 0. (iii) dg + gd = 0.

5 Orthogonal derivations on ideal 1409 (iv) dg is a derivation. (v) there exists a, b M and α, β Γ such that (dg)(x) = aβx + xγb. Now we prove the main result. Theorem 2.5. Let M be a 2-torsion free semiprime Γ-ring, d and g are derivations of M, and U be a nonzero ideal of M such that Ann l (U) = 0. Then the following conditions are equivalent: (i) d and g are orthogonal of M. (ii) dg = 0 on U. (iii) dg + gd = 0 on U. (iv) dg is a derivation on U. (v) there exists a, b M and α, β Γ such that (dg)(x) = aβx + xγb, for all x U. Proof. (i) (ii), (iii), (iv)and (v) are clear by Theorem 2.4. (ii) (i) Suppose that dg = 0 on U, that is dg(x) = 0, for all x U. Replacing x by xαm, we get 0 = d(g(x)αm + xαg(m)) = d(g(x))αm + g(x)αd(m) + d(x)αg(m) + xαd(g(m)) = g(x)αd(m) + d(x)αg(m) + xαd(g(m)) Replacing x by mβx, we get 0 = g(m)βxαd(m) + mβg(x)αd(m) +d(m)βxαg(m) + mβd(x)αg(m) + mβxαd(g(m)) = g(m)βxαd(m) + d(m)βxαg(m) By Lemma 2.1 we have g(m)βxαd(m)=0. By Lemma 2.2 we get g(m)βxαd(s)=0, for all x U, m, s M and α, β Γ. Replacing x by tαd(s)γxδg(m)βt we have g(m)βtαd(s)γxδg(m)βtαd(s) = 0. Replacing x by xλr, we get g(m)βtαd(s)γxλrδg(m)βtαd(s)γx = 0, for all x U, m, s, t, r M and α, β, γ, δ, λ Γ. By semiprimeness we obtain g(m)βtαd(s)γx = 0, yields g(m)βtαd(s) Ann l (U) = 0, therefore g(m)βtαd(s) = 0, for all m, s, t M and α, β Γ. Hence d and g are orthogonal. (iii) (i) Suppose that dg + gd = 0, that is (dg + gd)(x) = 0 for all x U. Replacing x by xαm, we obtain

6 1410 N. N. Suliman and A.-R. H. Majeed 0 = d(g(x)αm + xαg(m)) + g(d(x)αm + xαd(m)) = d(g(x))αm+g(x)αd(m)+d(x)αg(m)+xαd(g(m))+g(d(x))αm+d(x)αg(m) +g(x)αd(m)+ xαg(d(m)) = 2d(x)αg(m) + 2g(x)αd(m) +xα{d(g(m)) + g(d(m))} Replacing x by mβx we get 0 = 2{d(m)βxαg(m) + mβd(x)αg(m) + g(m)βxαd(m) + mβg(x)αd(m)} + mβxα{d(g(m)) + g(d(m))} = 2{d(m)βxαg(m) + g(m)βxαd(m)} Since M is 2-torsion free Γ-ring, we have d(m)βxαg(m) + g(m)βxαd(m) = 0, for all x U, m, n M and α, β Γ. By Lemma 2.1 we get d(m)βxαg(m) = 0. By Lemma 2.2 we get d(m)βxαg(s) = 0, for all m, s M and α, β Γ. Replacing x by tαg(s)γxδd(m)βt we have d(m)βtαg(s)γxδd(m)βtαg(s) = 0. Replacing x by xλr, we get d(m)βtαg(s)γxλrδd(m)βtαg(s)γx = 0, for all x U, m, s, t, r M and α, β, γ, δ, λ Γ. By semiprimeness we obtain d(m)βtαg(s)γx = 0, yields d(m)βtαg(s) Ann l (U) = 0, therefore d(m)βtαg(s) = 0, for all m, s, t M and α, β Γ. Hence d and g are orthogonal. (iv) (i) Suppose that dg is derivation from U to M, we have In other hand dg(xαy) = d(g(x))αy + xβd(g(y)), for all x, y U and α Γ. (2.1) dg(xαy) = d(g(x))αy + g(x)αd(y) + d(x)αg(y) + xβd(g(y)) (2.2) Comparing (2.1) and (2.2) we get d(x)αg(y) + g(x)αd(y) = 0, for all x, y U and α Γ Hence by Lemma 2.3 we get d and g are orthogonal. (v) (i) Suppose that there exists a, b M and β, γ Γ such that dg(x) = aβx + xγb. Replacing x by xαm we get dg(xαm) = d(g(x)αm + xαg(m))

7 Orthogonal derivations on ideal 1411 aβxαm + xαmγb = dg(x)αm + g(x)αd(m) + d(x)αg(m) + xαdg(m) aβxαm + xαmγb = aβxαm + xγbαm + g(x)αd(m) + d(x)αg(m) + xαdg(m), that is xγbαm + g(x)αd(m) + d(x)αg(m) + xαdg(m) - xαmγb = 0. Replacing x by mδx we get mδxγbαm + g(m)δxαd(m) + mδg(x)αd(m) + d(m)δxαg(m) + mδd(x)αg(m) + mδxαdg(m) - mδxαmγb = 0. Therefore g(m)δxαd(m)+ d(m)δxαg(m) = 0, for all x U m s, m M and α, δ Γ. By Lemma 2.1 we have g(m)βxαd(m)=0. By Lemma 2.2 we get g(m)βxαd(s)=0, for all x U, m, s M and α, β Γ. Replacing x by tαd(s)γxδg(m)βt we have g(m)βtαd(s)γxδg(m)βtαd(s) = 0. Replacing x by xλr, we get g(m)βtαd(s)γxλrδg(m)βtαd(s)γx = 0, for all x U, m, s, t, r M and α, β, γ, δ, λ Γ. By semiprimeness we obtain g(m)βtαd(s)γx = 0, yields g(m)βtαd(s) Ann l (U) = 0, therefore g(m)βtαd(s) = 0, for all m, s, t M and α, β Γ. Hence d and g are orthogonal. Corollary 2.6. Let M be a 2-torsion free semiprime Γ-ring, d and g are derivations from U to M, and U be a nonzero ideal such that Ann l (U) = 0. If dg is derivation from U to M. Then dg is derivation of M. Corollary 2.7. Let M be a 2-torsion free semiprime Γ-ring, d be a derivation from U to M, and U be a nonzero ideal such that Ann l (U) = 0. If d 2 is derivation, then d = 0. Proof. Suppose that d 2 is derivation on U, by Theorem 2.5 we get d and d are orthogonal. That is d(x)γmγd(x) = 0. Then by semiprimeness we get d = 0. References [1] M. Ashraf, and M. R. Jamal, Orthogonal Derivations in Γ-Rings. Advanced in Algebra, 3(1) (2010) 1-6. [2] W. E. Barness, On the Γ-Rings of Nobusawa. Pacific J. Math., 18(3)(1966),

8 1412 N. N. Suliman and A.-R. H. Majeed [3] M. Brešar and J. Vukman, Orthogonal derivation and an extension of a theorem of Posner, Radovi Mat., 5(1989) [4] Jing, FJ., On Derivations of Γ-Rings. Qu fu Shifan Xuebeo Ziran Kexue Ban, 13(4)(1987) [5] N. Nobusawa, On a Generlazetion of the Ring Theory, Osaka J. Math., 1 (1964), [6] M. A. Ӧztyrk and H. Yazarli, Modules Over The Generalized Centroid Of Semi- Prime Gamma Rings. Bull. Korean math. Soc. 44 (2) (2007) Received: December, 2011