Jordan k Derivations on Lie Ideals of Prime Γ Rings

Full text

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Jordan

k

-Derivations on Lie Ideals of Prime

Γ-Rings

A.C. Paul

,

Ayesha Nazneen

Department of Mathematics,Rajshahi University, Rajshahi - 6205, Bangladesh Corresponding Author: acpaulrubd math@yahoo.com

Copyright © 2014 Horizon Research Publishing All rights reserved.

Abstract

Let M be a Γ- ring and U a Lie ideal of M. Let d : M M and k : Γ Γ be additive mappings. Then d is a k- derivation on U ofM if

d(uαv) =d(u)αv+uk(α)v+uαd(v) is satisfied for all

u, v∈U andα∈Γ. And d is a Jordank- derivation on

U ofM ifd(uαu) =d(u)αu+uk(α)u+uαd(u) holds for all u∈U andα∈Γ. It is well-known that every k -derivation on U of M is a Jordan k- derivation on U

of M but the converse is not true in general. In this article we prove that every Jordan k- derivation on U

ofM is ak- derivation onU ofM if ,M is a 2- torsion free prime Γ- ring andU is a Lie ideal ofM such that

uαu∈U for allu∈U andα∈Γ.

Keywords

Lie ideal, Jordan k- derivation, k -derivation, Prime Γ- ring

AMS(2010)subject classification: Primary 16N60, secondary 03E72,54A40,54B15

1

Introduction

The notion of a Γ- ring was introduced as an extensive generalisation of the concept of a classical ring. N. Nobusawa [10] introduced the notion of a Γ- ring (which is presently known as a ΓN- ring) and afterwords it was

generalised by W.E. Bernes [1] as a more broadsense (that served us now a days to call it as a Γ- ring generally). It is wellknown that every ring is a Γ- ring and also that every ΓN- ring is a Γ- ring. We begin

with the following necessary preliminary definitions. Let M and Γ be additive abelian groups. If there is a mapping (x, α, y) xαy of M ×Γ×M M which satisfies the conditions

(a) (x + y)αz = xαz +yαz, x(α + β)y = xαy +

xβy, xα(y+z) =xαy+xαz and

(b)(xαy)βz=(yβz) for allx, y, z∈M andα, β∈Γ, thenM is said to be a Γ- ring in the sense of Bernes [1]. In addition to the definition given above, if there is a mapping (α, x, β)→αxβ of Γ× ΓΓ satisfying the conditions

(a∗) (α+β) = αxγ +βxγ, α(x+y)β = αxβ +

αyβ, αx(β+γ) =αxβ+αxγ

(b∗) (xαy)βz = x(αyβ)z = (yβz)

(c∗)xαy = 0, for allx, y, z∈M impliesα= 0, thenM

is called a Γ- ring in the sense of nobusawa [12] as simply a ΓN- ring. It is clear thatM is a ΓN- ring implies that

Γ is aM- ring.

LetM be a Γ- ring . Then M is called 2-torsion free if 2x= 0 impliesx= 0 for allx∈M. BesidesM is called a prime Γ- ring if for all x, y ∈M, xΓMΓy = 0 implies

x = 0 or y = 0. And M is called a semiprime Γ- ring if for all x∈M, xΓMΓx= 0 impliesx= 0. Note that every prime Γ- ring is clearly semiprime.

The concept of derivation and Jordan derivation of a Γ-ring was first introduced by M. Sapanci and A. Nakajima in [14], whereas the notion ofk- derivation of a Γ- ring was used and developed by H. Kandamar [9]. The notion of Jordank- derivation of a Γ- ring was first initiated by S. Chakraborty and A.C. Paul [3].

The definition ofk- derivation and Jordank- derivation are given as follows :

Let M be a Γ- ring. Let d : M M and k : Γ Γ be additive mappings. If d(xαy) =d(x)αy+xk(α)y+

xαd(y) is satisfied for all x, y M and α Γ, then d is said to be a k- derivation of M. And if d(xαx) =

d(x)αx+xk(α)x+xαd(x) holds for allx∈M andα∈Γ, then dis said to be a Jordank- derivation ofM. Note that everyk-derivation is a Jordank- derivation but the converse is not in general true.

In [2], Y. Ceven proved that every Jordan left derivation of a 2- torsion free completly prime Γ- ring is a Jordan left derivation. Paul and Halder [5] extended these re-sults to a Lie ideal of a Prime Γ- ring. S. Chakraborty and A.C. Paul [3] worked on a Jordan k-derivation and proved that every Jordank-derivation of a 2- torsion free Prime ΓN- ring is a k-derivation.

We shall use the notation [x, y]α for the commutator x

andywith respect toα, defined by [x, y]α=xαy−yαx.

IfAis a subset ofM, byZ(A) we shall mean the centre of A with respect to M, defined byZ(A) = {x∈ M : [x, y]α= 0 for alla∈A andα∈Γ}. The centre of a Γ

ringM is denoted byZ(M) which is defined byZ(M) = {x M : [x, y]α = 0 for ally M andα Γ}. A

Γ-ringM is commutative if and only ifM =Z(M). Throughout the paper, we shall use the condition (*)

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[aαb, x]β = [a, x]βαb + a[α, β]xb + [b, x]β and

[x, aαb]β=[x, b]β+a[α, β]xb+[x, a]βαbgiven in [7]

re-duce to [aαb, x]β=[b, x]β+ [a, x]βαband [x, aαb]β= [x, b]β+ [x, a]βαb.

which are used in [7]. F. Hoque and A. C. Paul also used the condition (*) in [8].

In this present article , we introduce the concept of Jor-dan k-derivation on a Lie ideal U of a Γ-ring M. We prove that every Jordank-derivation on a Lie idealU of a 2- torsion free Prime Γ- ring is ak-derivation.

2

Lie

ideals

and

Jordan

k

-derivations

LetM be a Γ-ring. An additive subgroupU ofM is called a Lie ideal of M if [u, m]α ∈U for every u∈ U

and m∈ M. Note that every ideal of a Γ-ring M is a Lie ideal ofM but the converse is not true in general.

Example 2.1.

Let R be a commutative ring with a unity 1 having characteristic 2. DefineM =M1,2(R)

and Γ =

{(

n.1

n.1

)

: n∈Zandnis not divisible by 2

}

. ThenM is a Γ- ring . Define N ={(a, a) :a∈R}. It is clear that N is an additive subgroup ofM. Now for

u= (a, a) ∈N, m= (x, y) ∈M and α=

(

n n

)

Γ we have ,

uαm−mαu= (a, a)

(

n n

)

(x, y)(x, y)

(

n n

)

(a, a) = (anx−yna, any−xna)

= (anx−2yna+yna, any−2xna+xna) = (anx+yna, any+xna)

= (anx+any, anx+any)∈N

Therefore , uαm−mαu N and N is a Lie ideal of

M. It is clear thatN is not an ideal ofM.

In [11] and [12] , Paul and Sabur Uddin worked on Lie and Jordan structure of a 2- torsion free simple Γ- ring and they developed a number of significant results of classical ring thories in Γ- rings.

Now we introduce the concepts of a k-derivation , a Jordank- derivation of Lie ideals in a Γ- ring and then build up a relationship between these two concepts in a concrete manner .

Let M be a Γ- ring and U be a Lie ideal of M. Let d : M M and k : Γ Γ be additive map-pings. If d(uαv) = d(u)αv +uk(α)v +uαd(v) is satisfied for every u, v U and α Γ, then d is called a k -derivation on a Lie ideal U of M. And, if

d(uαu) = d(u)αu +uk(α)u+uαd(u) holds for all

u U and α Γ , then d is said to be a Jordan

k-derivation on a Lie idealU ofM.

It is clear that every k-derivation on a Lie ideal is a Jordan k-derivation on a Lie ideal but the converse may not be true. Now we make an example of a Jordan

k- derivation for the case of a Lie ideal which ensures that Jordan k-derivation on a Lie ideal exists and it is evidently not ak- derivation on a Lie ideal.

Example 2.2.

LetM be a Γ- ring and let U be a Lie ideal of M. Let d: M M be a k- derivation on a Lie ideal U ofM. Define M1 ={(x, x) : x∈M}

and Γ1 = {(α, α) : α Γ}. Define addition and

multiplication onM1 as follows :

(x, x) + (y, y) = (x+y, x+y),

(x, x)(α, α)(y, y) = (xαy, xαy).

Under these addition and multiplication, M1 is a

Γ1- ring. Define U1 = {(u, u) : u U}. Now we

show that U1 is a Lie ideal of M as follows. For

(u, u)∈U1,(α, α)Γ1 and (x, x)∈M1, we have,

(u, u)(α, α)(x, x) (x, x)(α, α)(u, u) = (uαx, uαx) (xαu, xαu) = (uαx−xαu, uαx −xαu) U1, since

uαx−xαu∈U.

Now let d1 : M1 M1 and k1 : Γ1 Γ1 be the

mappings defined by d1((u, u)) = (d(u), d(u)) for

all u U, andk1((α, α)) = ((k(α), k(α)) for all

α∈Γ.Thend1andk1 are additive mappings. If we say

that (u, u) =u1∈U1for allu∈U and (α, α) =γ∈Γ1

for allα∈Γ, then we have

d1(u1γu1) =d1((u, u)(α, α)(u, u))

=d1((uαu, uαu))

= (d(uαu), d(uαu))

= (d(u)αu+uk(α)u+uαd(u), d(u)αu+uk(α)u+

uαd(u))

= (d(u)αu, d(u)αu) + ((uk(α)u, uk(α)u) + (uαd(u), uαd(u)))

= (d(u), d(u))(α, α)(u, u) + (u, u)(k(α), k(α))(u, u) + (u, u)(α, α)(d(u), d(u))

= d1(u, u)(α, α)(u, u) + (u, u)k1(α, α)(u, u) +

(u, u)(α, α)d1(u, u)

=d1(u1)γu1+u1k1(γ)u1+u1γd1(u1).

Hence it follows that d1 is a Jordan k1- derivation on

a Lie ideal U1 of M1. It is obvious that d1 is not a

k1-derivation on a Lie idealU ofM.

Now we begin with the following results:

Lemma 2.3

. Let M be a 2- torsion free Γ- ring satisfying the condition (*) and let U be a Lie ideal of

M such that uαu U for all u U and α Γ. Let

d:M →M be a Jordank-derivation onU ofM. Then for allu, v, w∈U andα, β∈Γ , we have the following : (i)d(uαv+vαu) =d(u)αv+uk(α)v+uαd(v) +d(v)αu+

vk(α)u+vαd(u)

(ii)d(uαvβu) = d(u)αvβu+uk(α)vβu+uαd(v)βu+

uαvk(β)u+uαvβd(u)

(iii)d(uαvβw+wαvβu) =d(u)alphavβw+uk(α)vβw+

uαd(v)βw + uαvk(β)w + uαvβd(w) + d(w)αvβu +

wk(α)vβu+wαd(v)βu+wαvk(β)u+wαvβd(u).

Proof.

(i) Since uαv+vαu = (u+v)α(u+v)−uαu−vαv

and the right side is in U, we have the left side of the identity is inU. Hence

d(uαv+vαu) =d((u+v)α(u+v)−uαu−vαv) =d(u+v)α(u+v)+(u+v)k(α)(u+v)+(u+v)αd(u+v) (d(u)αu+uk(α)u+uαd(u) +d(v)αv+vk(α)v+vαd(v)) = (d(u) +d(v))α(u+v) + (u+v)k(α)(u+v) + (u+

v)α(d(u) +d(v))−d(u)αu−uk()u−uαd(u)−d(v)αv− vk(α)v−vαd(v)

=d(u)αu+d(u)αv+d(v)αu+d(v)αv+uk(α)u+uk(α)v+

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=d(u)αv+uk(α)v+uαd(v)+d(v)αu+vk(α)u+vαd(u).

(ii) Replacev byuβv+vβuin (i) we have

d((uβv+vβu) + (uβv+vβu)αu)

=d(u)α(uβv+vβu) +uk(α)(uβv+vβu) +uαd(uβv+

vβu) +d(uβv+vβu)αu+ (uβv+vβu)k(α)u+ (uβv+

vβu)αd(u)

=⇒d(uαuβv+uαvβu+uβvαu+vβuαu) =d(u)α(uβv+

vβu)+uk(α)(uβv+vβu)+(d(u)βv+uk(β)v+uβd(v)+

d(v)βu+vk(β)u+vβd(u) + (d(u)βv+uk(β)v+uβd(v) +

d(v)βu+vk(β)u+vβd(u))αu+(uβv+vβu)k(α)u+(uβv+

vβu)αd(u) Here,

d((uαu)βv +(uαu)) = d(uαu)βv + (uαu)k(β)v + (uαu)βd(v) +d(v)β(uαu) +vk(β)(uαu) +vβd(uαu) = d(u)αuβv +uk(α)uβv +uαd(u)βv + uαuk(β)v +

uαuβd(v) + d(v)βuαu + vk(β)uαu + vβd(u)αu +

vβuk(α)u+vβuαd(u). Therefore we have,

d(uαvβu+uβvαu)+d(u)αuβv+uk(α)uβv+uαd(u)βv+

uαuk(β)v + uαuβd(v) + d(v)βuαu + vk(β)uαu +

vβd(u)αu + vβuk(α)u + vβuαd(u) = d(u)αuβv +

d(u)αvβu + uk(α)uβv + uk(α)vβu + uαd(u)βv +

uαuk(β)v + uαuβd(v) + uαd(v)βu + uαvk(β)u +

uαvβd(u) + d(u)βvαu + uk(β)vαu + uβd(v)αu +

d(v)βuαu + vk(β)uαu + vβd(u)αu + uβvk(α)u +

vβuk(α)u+uβvαd(u) +vβuαd(u)

= d(uαvβu+uβvαu) = d(u)αvβu +uk(α)vβu+

uαd(v)βu + uαvk(β)u + uαvβd(u) + d(u)βvαu +

uk(β)vαu+uβd(v)αu+uβvk(α)u+uβvαd(u). Putuαvβu=uβvαu, we have

d(2uαvβu) = d(u)αvβu + uk(α)vβu + uαd(v)βu +

uαvk(β)u + uαvβd(u) + d(u)αvβu + uαvk(β)u +

uαd(v)βu+uk(α)vβu+uαvβd(u)

=2d(uαvβu) = 2(d(u)αvβu+uk(α)vβu+uαd(v)βu+

uαvk(β)u+uαvβd(u))

SinceM is 2- torsion free, hence

d(uαvβu) = d(u)αvβu + uk(α)vβu + uαd(v)βu +

uαvk(β)u+uαvβd(u).

(iii) Replaceu+wfor u in (ii) we have ,

d((u + w)αvβ(u + w)) = d(u + w)αvβ(u + w) + (u+w)k(α)(u+w) + (u+w)αd(v)β(u+w) + (u+

w)αvk(β)(u+w) + (u+w)αvβd(u+w).

=⇒d(uαvβu+uαvβw+wαvβu+wαvβw) = (d(u) +

d(w))αvβ(u + w) + (u + w)k(α)(u + w) + (u +

w)αd(v)β(u + w) + (u + w)αvk(β)(u + w) + (u +

w)αvβ(d(u) +d(w)) Here we have

d(uαvβu+wαvβw) =d(uαvβu) +d(wαvβw)

= d(u)αvβu +uk(α)vβu +uαd(v)βu+ uαvk(β)u+

uαvβd(u) + d(w)αvβw + wk(α)vβw + wαd(v)βw +

wαvk(β)w+wαvβd(w). And also

d(uαvβu+wαvβu+uαvβw+wαvβw) = d(uαvβw+

wαvβu) +d(uαvβu+wαvβw). Hence we have

d(uαvβw+wαvβu)+d(u)αvβu+uk(α)vβu+uαd(v)βu+

uαvk(β)u + uαvβd(u) + d(w)αvβw + wk(α)vβw +

wαd(v)βw +wαvk(β)w+ wαvβd(w) = d(u)αvβu +

d(w)αvβu + d(u)αvβw + d(w)αvβw + uk(α)vβu +

wk(α)vβu + uk(α)vβw + wk(α)vβw + uαd(v)βu +

wαd(v)βu + uαd(v)βw + wαd(v)βw + uαvk(β)u +

wαvk(β)u + uαvk(β)w + wαvk(β)w + uαvβd(u) +

wαvβd(u) +uαvβd(w) +wαvβd(w).

= d(uαvβw+wαvβu) = d(u)αvβw +uk(α)vβw+

uαd(v)βw + uαvk(β)w + uαvβd(w) + d(w)αvβu +

wk(α)vβu+wαd(v)βu+wαvk(β)u+wαvβd(u).

Definition 2.4

We defineϕα(u, v) =d(uαv)−d(u)αv−uk(α)v−uαd(v)

for every u, v∈U andα∈Γ.

Remark

It is clear thatdis ak-derivation onU ofM if and only ifϕα(u, v) = 0.

Lemma 2.5

. Let M, U and d be as in Lemma 2.3.Then for all u, v, w∈U and α, β∈Γ, the following relations hold .

(i)ϕα(u, v) +ϕα(v, u) = 0

(ii)ϕα(u+w, v) =ϕα(u, v) +ϕα(w, v)

(iii)ϕα(u, v+w) =ϕα(u, v) +ϕα(u, w)

(iv)ϕα+β(u, v) =ϕα(u, v) +ϕβ(u, v)

Lemma 2.6

. Let M, U and d be as in Lemma 2.3, then for allu, v, w∈U andα, β, γ,∈Γ,

ϕα(u, v)βwγ[u, v]α+ [u, v]αβwγϕα(u, v) = 0.

Proof.

Consider A= (2uαv)βwγ(2vαu) + (2vαu)βwγ(2uαv) =⇒d(A) =d((2uαv)βwγ(2vαu) + (2vαu)βwγ(2uαv)) = d(2uαv)βwγ(2vαu) + 2uαvk(β)(2vαu) + (2uαv)βd(w)γ(2vαu) + (2uαv)βwk(γ)(2vαu) + (2uαv)βwγd(2vαu) + d(2vαu)βwγ(2uαv) + (2vαu)k(β)(2uαv) + (2vαu)βd(w)γ(2uαv) + (2vαu)βwk(γ)(2uαv) + (2vαu)βwγd(2uαv)

= 4d(uαv)βwγ(vαu) + 4(uαv)k(β)(vαu) + 4(uαv)βd(w)γ(vαu) + 4(uαv)βwk(γ)(vαu) + 4(uαv)βwγd(vαu) + 4d(vαu)βwγ(uαv) + 4(vαu)k(β)(uαv) + 4(vαu)βd(w)γ(uαv) + 4(vαu)βwk(γ)(uαv) + 4()βwγd(uαv)

AgainA= (2uαv)βwγ(2vαu) + (2vαu)βwγ(2uαv) =(4vβwγv)αu+(4uβwγu)αv

=⇒d(A) =d((4vβwγv)αu) +d((4uβwγu)αv) = d(u)α(4vβwγv)αu + uk(α)(4vβwγv)αu +

uαd(4vβwγv)αu + (4vβwγv)k(α)u +

(4vβwγv)αd(u) + d(v)α(4uβwγu)αv +

vk(α)(4uβwγu)αv + vαd(4uβwγu)αv +

(4uβwγu)k(α)v+(4uβwγu)αd(v)

= 4d(u)αvβwγvαu+4uk(α)vβwγvαu+4(d(v)βwγv+

vk(β)wγv +vβd(w)γv +vβwk(γ)v +vβwγd(v))αu + 4uαvβwγvk(α)u+4uαvβwγvαd(u)+4d(v)αuβwγuαv+ 4vk(α)uβwγuαv + 4(d(u)βwγu + uk(β)wγu +

uβd(w)γu + uβwk(γ)u + uβwγd(u))αv + 4vαuβwγuk(α)v+ 4vαuβwγuαd(v)

= 4d(u)αvβwγvαu + 4uk(α)vβwγvαu + 4uαd(v)βwγvαu+4uαvk(β)wγvαu+4uαvβd(w)γvαu+ 4uαvβwk(γ)vαu+4uαvβwγd(v)αu+4uαvβwγvk(α)u+ 4uαvβwγvαd(u)+4d(v)αuβwγuαv+4vk(α)uβwγuαv+ 4vαd(u)βwγuαv+4vαuk(β)wγuαv+4vαuβd(w)γuαv+ 4vαuβwk(γ)uαv + 4vαuβwγd(u))αv + 4vαuβwγuk(α)v+ 4vαuβwγuαd(v)

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4uαvβwγ(d(vαu d(v)αu vk(α)u vαd(u)) + 4(d(vαu) d(v)αu vk(α)u vαd(u))βwγuαv + 4vαuβwγ(d(uαv)−d(u)αv−uk(α)v−uαd(v)) = 0 = 4(ϕα(u, v)βwγvαu ϕα(u, v)βwγuαv uαvβwγϕα(v, u) +vαuβwγϕα(u, v)) = 0

SinceM is 2- torsion free, we have

−ϕα(u, v)βwγ(uαv−vαu)(uαv−vαu)βwγϕα(u, v) = 0

⇒ϕα(u, v)βwγ[u, v]α+ [u, v]αβwγϕα(u, v) = 0.

Lemma 2.7.

Let U ̸⊆ Z(M) be a Lie ideal of a 2-torsion free prime Γ-ringM , then Z(U) =Z(M).

Proof.

We haveZ(U) is both a sub Γring and a Lie ideal of M . Also we know that Z(U)cannot contain a nonzero ideal of M. So by [9, Lemma 3.7], Z(U) is contained inZ(M).Therefore,Z(U) =Z(M).

Lemma 2.8.

Let U be a Lie ideal of a 2-torsion free prime Γ-ring M satisfying the condition (*) and

a M. If a Z([U, U]Γ), then a Z(U). That is

Z([U, U]Γ) =Z(U).

Proof.

Obviously Z(u)⊆Z([U, U]Γ).

If Z([U, U]Γ) ̸⊆ Z(M), then by Lemma 2.7, a

Z(M)⇒a∈Z(U)

On the other hand if Z([U, U]Γ) Z(M), then for all

u U, m M, α, β Γ implies a = [u,[u, m]α]β Z(M).

Using the condition (*) we have

aγu= [u,[u, uγm]α]β ∈Z(M).

If =o, we get u∈Z(M) impliesa= 0 Thus [u,[u, uγm]α]β = 0 for allm∈M.

By the subLemma 3.8 of [9] u Z(M); hence U Z(M).

In both cases we see that a Z(U). This gives that

Z([U, U]Γ) =Z(U).

Lemma 2.9.

Let U ̸⊆Z(M) be a Lie ideal of a 2-torsion free Γ-ringM satisfying the condition (*) such that , uαu∈ U for all u∈U andα∈ Γ. Ifu∈ Z(U) thend(u)∈Z(M).

Proof.

Let u Z(U) = Z(M), then , uαv =

vαu, for every v∈U andα∈Γ.

From Lemma 2.3(i) we have,

d(uαv+vαu) =d(u)αv+uk(α)v+uαd(v) +d(v)αu+

vk(α)u+vαd(u)

= d(2uαv) = d(u)αv+vαd(u) +uk(α)v+uαd(v) +

uαd(v) +uk(α)v

=d(u)αv+vαd(u) + 2uk(α)v+ 2uαd(v) Replacevby (vβw+wβv),we have

d(2(vβw+wβv)) = d(u)α(vβw +wβv) + (vβw+

wβv)αd(u) + 2uk(α)(vβw+wβv) + 2uαd(vβw+wβv) = d(uαvβw+uαwβv) = d(u)αvβw +d(u)αwβv +

vβwαd(u) +wβvαd(u) + 2uk(α)vβw + 2uk(α)wβv + 2uαd(v)βw+ 2uαvk(β)w+ 2uαvβd(w) + 2uαd(w)βv+ 2uαwk(β)v+ 2uαwβd(v)

Now (2uαvβw+ 2uαwβv) = 2d(uαvβw+wαvβu) = 2d(u)αvβw+2uk(α)vβw+2uαd(v)βw+2uαvk(β)w+ 2uαvβd(w) + 2d(w)αvβu+ 2wk(α)vβu+ 2wαd(v)βu+ 2wαvk(β)u+ 2wαvβd(u)

So 2d(u)αvβw+2uk(α)vβw+2uαd(v)βw+2uαvk(β)w+ 2uαvβd(w) + 2d(w)αvβu+ 2wk(α)vβu+ 2wαd(v)βu+ 2wαvk(β)u+ 2wαvβd(u) = d(u)αvβw+d(u)αwβv +

vβwαd(u) +wβvαd(u) + 2uk(α)vβw+ 2uk(α)wβv+ 2uαd(v)βw+ 2uαvk(β)w+ 2uαvβd(w) + 2uαd(w)βv+ 2uαwk(β)u+ 2uαwβd(v)

=⇒d(u)αvβw+2d(w)αvβu+2wk(α)vβu+2wαd(v)βu+ 2wαvk(β)u+ 2wβvαd(u) = d(u)αwβv +vβwαd(u) +

wβvαd(u) + 2wk(α)vβu+ 2d(w)αvβu+ 2wαvk(β)u+ 2wαd(v)βu,

=⇒d(u)vβw+ 2d(w)αvβu+ 2wk(α)vβu+ 2wαd(v)βu+ 2wαvk(β)u+ 2wαvβd(u) = d(u)αwβv +vβwαd(u) +

wβvαd(u) + 2uk(α)wβv+ 2uαd(w)βv+ 2uαwk(β)v+ 2uαwβd(v)

=⇒d(u)α(vβw−wβv) = (vβw−wβv)αd(u) =⇒d(u)∈Z([U, U]Γ)

But by Lemma 2.7 and Lemma 2.8 , we have,

Z([U, U]Γ) =Z(M).Hence d(u)∈Z(M).

To prove our main results we need the following two Lemmas .

Lemma 2.10 [ 13, Lemma 2.10]

Let U be a Lie ideal of a 2- torsion free Prime Γ- ring satisfying the condition (*) and U ̸⊆Z(M). Ifa, b∈ M (res. b ∈U

anda∈M ) such thataαU βb= 0 for allα, β∈Γ, then

a= 0 orb= 0.

Lemma 2.11

[13, Lemma 2.11 ]

Let

U ̸⊆ Z(M) be a 2- torsion free lie ideal of a prime Γ-ringM. Ifa, b∈M (res.a∈M andb∈U) such that

aαxβb+bαxβa= 0 for allx∈ U and α, β Γ, then

aαxβb=bαxβa= 0.

Now we have in position to prove our main result.

Theorem 2.12

. Let M be a 2 -torsion free prime Γ-ring satisfying the condition (*) and let U be a Lie ideal ofM such thatuαu∈U for allu∈U andα,∈Γ. If d : M −→ M is a Jordank-derivation on U of M , thendis ak-derivation onU ofM.

Proof.

IfU is a commutative Lie ideal ofM, then for all u, v ∈U and α∈Γ, [u, v]α = 0. Then uαv =vαu.

By Lemma 2.3(iii), we have

d(uαvβw + wαvβu) = d(u)αvβw + uk(α)vβw +

uαd(v)βw + uαvk(β)w + uαvβd(w) + d(w)αvβu +

wk(α)vβu+wαd(v)βu+wαvk(β)u+wαvβd(u).

By using (*) we obtain,

d(uαvβw+wαvβu) =d((uαv)βw+(uαv))

=d(uαv)βw+(uαv)k(β)w+(uαv)βd(w)+d(w)β(uαv)+

wk(β)(uαv) +wβd(uαv).

Comparing the above two expressions, we obtain

d(uαv)βw+ (uαv)k(β)w+ (uαv)βd(w) +d(w)β(uαv) +

wk(β)(uαv) +wβd(uαv) = d(u)αvβw +uk(α)vβw+

uαd(v)βw + (uαv)k(β)w + uαvβd(w) + d(w)βuαv +

wk(β)uαv+wβd(v)αu+wβvk(α)u+wβvαd(u) (d(uαv)−d(u)αv−uk(α)v−uαd(v))βw+(d(vαu)

d(v)αu−vk(α)u−vαd(u)) = 0 ⇒ϕα(u, v)βw+wβϕα(v, u) = 0 ⇒ϕα(u, v)βw−wβϕα(u, v) = 0

⇒ϕα(u, v)βw=wβϕα(u, v), for allw∈U, β∈Γ.

Thenϕα(u, v)∈Z(U) =Z(M) by Lemma 2.7.

(5)

sinceuαu∈U and(uαu)βv=(uαu) for all β∈Γ . Hence, d(uαuβv) d(uαu)βv (uαu)k(β)v

(uαu)βd(v)∈Z(M)

d(uαuβv) ((d(u)αu + uk(α)u + uαd(u))βv + (uαu)k(β)v+uαuβd(v))∈Z(M))...(1)

Also, 2uβv∈U and(2uβv) = (2uβv)αu, we get

d((2uβv)−d(u)αuβv)−uk(α)(2uβv)−uαd(2uβv)

Z(M)

i.e,2(d(uαuβv)−d(u)αuβv−uk(α)uβv−uαd(uβv))

Z(M).

SinceM is a 2-torsion free , we have

d(uαuβv) d(u)αuβv uk(α)uβv uαd(uβv)

Z(M)...(2)

From (1) and (2) we have d(uαuβv) −d(u)αuβv uk(α)uβv uαd(u)βv uαuk(β)v uαuβd(v)

d(uαuβv) +d(u)αuβv+uk(α)uβv+uαd(uβv) =uαd(uβv)−uαuk(β)v−uαuβd(v)−uαd(u)βv

=(d(uβv)−d(u)βv−uk(β)v−uβd(v)) =uαϕβ(u, v)∈Z(M).

Ifϕβ(u, v)̸= 0. SinceM is prime andϕβ(u, v)∈Z(M), thenu∈Z(M).

Sod(u)∈Z(M)

By Lemma 2.3(i), we have,

d(uαv+vαu) =d(u)αv+uk(α)v+uαd(v) +d(v)αu+

vk(α)u+vαd(u)

⇒d(2uαv) = 2(d(u)αv+uk(α)v+uαd(v)) 2(d(uαv)−d(u)αv−uk(α)v−uαd(v)) = 0 2ϕα(u, v) = 0⇒ϕα(u, v) = 0

Again, letU is not commutative. i.e,U ̸⊆Z(M) . Then by lemma 2.6 , we have

(i)...ϕα(u, v)βwγ[u, v]α+ [u, v]αβwγϕα(u, v) = 0.

Applying Lemma 2.11 in (i) , we obtain (ii)...ϕα(u, v)βwγ[u, v]α= 0 and

(iii)...[u, v]αβwγϕα(u, v) = 0.

In view of Lemma 2.10, we have from (ii) thatϕα(u, v) = 0 or [u, v]α= 0.

The same result follows from (iii) by applying Lemma 2.10 .

For everyv∈U, let us define

A={u∈U :ϕα(u, v) = 0} and B={u∈U : [u, v]α= 0}.

Then A and B are additive subgroup of U such that

A∪B=U, Therefore , by Brauer’s trick, eitherA=U

or B = U. By using the same argument,we have

U = {v U : U = A} and U = {v U : U = B}

For the later case , we haveU ⊆Z(M) which is a con-tradiction. So, we have ϕα(u, v) = 0, which completes the proof.

REFERENCES

[1] W.E. Barnes, On the Γ- rings of Nobusawa, Pacific J.Math.,18(1966), 411-422.

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[3] S. Chakraborty and A. C. Paul, On Jordank- deriva-tions of 2- torsion free Prime ΓN-rings , Punjab Univ.

J. Math. vol. 40 (2008), 97-101.

[4] S. Chakraborty and A. C. Paul, k- derivations andk -homomorphisms of Gamma rings, Lambert Academic Publishing CmbH Co, KG Heinrich -Beking- Str.6-8,66121,saarbrchen, Germany,2012.

[5] A.K. Halder and A.C. Paul, Jordan left derivations on Lie ideals of Prime Γrings, Punjab Univ. J. Math., (2012),23-29.

[6] I.N. Herstein, Topics in Ring Theory, The University of Chicago Press, Chicago, 111. London,1969.

[7] M. F. Hoque and A. C. Paul, On Centralizers of Semiprime Gamma Rings. Int. J. Math. Forum , vol. 6 (2011) no. 13, 627-638.

[8] M. F. Hoque and A. C. Paul, Centralizers on Semiprime Gamma Rings, Italian J. Pure and Applied Math. N-30, (2013), 289-302.

[9] H. Kandamar, The k- derivations of a gamma ring, Turkish J. Math., 24 (2000),221-231.

[10] N. Nobusawa, On the generalization of ring theory, Os-aka J. Math. 1(1964),81-89.

[11] A.C. Paul and Sabur Uddin, Lie and Jordan structure in simple gamma rings, J. Physical Sciences, 14(2010),77-86.

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[13] M.M. Rahman and A.C. Paul, Jordan derivations on Lie ideals of Prime Γ-rings. Mathematical Theory and Modeling (2013), Vol. 3(3); 128 - 135.

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