Orthogonal Derivations on an Ideal of Semiprime Γ-Rings

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

Full text

International Mathematical Forum, Vol. 7, 2012, no. 28, 1405 - 1412
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
vananesh@gmail.com
Abdul-Rahman H. Majeed
Baghdad University/College of Science
Department of Mathematics, Iraq
ahmajeed@yahoo.com
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:

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, Annl(U) = {a ∈ M|
aΓU = <0> }is called the left annihilator of U. A right annihilator Annr(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∩Annl(U) =
{0} (U∩Annr(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 (*).

Orthogonal derivations on ideal 1407
2. 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 Annl(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∈ Annl(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 ∈ Annl(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 Annl(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
f(x)αuβh(y) + f(y)αuβh(x) = 0 for all x, y, u∈ U and α, β∈ Γ
Then we have

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)∈ Annl(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 Annl(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)∈ Annl(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.

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 Annl(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)∈ Annl(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

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)∈ Annl(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
dg(xαy) = d(g(x))αy + xβd(g(y)), for all x, y∈ U and α∈ Γ. (2.1)
In other hand
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))

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)∈ Annl(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 Annl(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 Annl(U) = 0. If d2 is derivation, then d = 0.
Proof. Suppose that d2 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.
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1412 N. N. Suliman and A.-R. H. Majeed
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Received: December, 2011