# Generalized Derivations in Semiprime Gamma Rings

Authors: Kalyan Kumar Dey; Akhil Chandra Paul; Isamiddin S. Rakhimov
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

LetM be a 2-torsion-free semiprime Γ-ring satisfying the condition aαbβc=aβbαc for all a,b,c∈M,α,β∈Γ, and let D:M→M be an additive mapping such that D(xαx)=D(x)αx+xαd(x) for all x∈M,α∈Γ  and for some derivation d of M. We prove that D is a generalized derivation.


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## Body

## 1. Introduction

Hvala [1] first introduced the generalized derivations in rings and obtained some remarkable results in classical rings. Daif and Tammam El-Sayiad [2] studied the generalized derivations in semiprime rings. The authors consider an additive mapping G:R→R of a ring R with the property G(x2)=G(x)x+xD(x) for some derivation D of R. They prove that G is a Jordan generalized derivation.Aydin [3] studied generalized derivations of prime rings. The author proved that if I is an ideal of a noncommutative prime ring R, a is a fixed element of R and F is an generalized derivation on R associated with a derivation d then the condition F([x,a])=0 or [F(x),a]=0 for all x∈I implies d(x)=λ[x,a].Çeven and Öztürk [4] have dealt with Jordan generalized derivations in Γ-rings and they proved that every Jordan generalized derivation on some Γ-rings is a generalized derivation.Generalized derivations of semiprime rings have been treated by Ali and Chaudhry [5]. The authors proved that if F is a commuting generalized derivation of a semiprime ring R associated with a derivation d then d(x)[y,z]=0 for all x,y,z∈R and d is central. They characterized a decomposition of R relative to the generalized derivations.Atteya [6] proved that if U is nonzero ideal of a semiprime ring R and R admits a generalized derivation D such that D(xy)-xy∈Z(R) then R contains a nonzero central ideal.Rehman [7, 8] studied the commutativity of a ring R by means of generalized derivations acting as homomorphisms and antihomomorphisms.In this paper, we prove the following resultsLetM be a 2-torsion-free semiprime Γ-ring satisfying the following assumption:(*)aαbβc=aβbαc∀a,b,c∈M,α,β∈Γ,
and D:M→M be an additive mapping. If there exists a derivation d of M such that D(xαx)=D(x)αx+xαd(x) for all x∈M, α∈Γ, then D is a Jordan generalized derivation.


## 2. Preliminaries

LetM and Γ be additive abelian groups. M is called a Γ-ring if there exists a mapping M×M×M→M such that for all a,b,c∈M, α,β∈Γ the following conditions are satisfied:(i)
aβb∈M,(ii)
(a+b)αc=aαc+bαc, a(α+β)b=aαb+aβb, aα(b+c)=aαb+aαc, (aαb)βc=aα(bβc).For anya,b∈M and for α,β∈Γ the expressions aαb-bαa is denoted by [a,b]α and αaβ-βaα are denoted by [α,β]a. Then one has the following identities:(2.1)[aβb,c]α=aβ[b,c]α+[a,c]αβb+a[β,α]cb,[a,bβc]α=bβ[a,c]α+[a,b]αβc+b[β,α]ac,
for all a,b,c∈M and for all α,β∈Γ. Using the assumption (*) the above identities reduce to(2.2)[aβb,c]α=aβ[b,c]α+[a,c]αβb,[a,bβc]α=bβ[a,c]α+[a,b]αβc,
for all a,b,c∈M and α, β∈Γ.FurtherM stands for a prime Γ-ring with center Z(M). The ring M is n-torsion-free if nx=0, x∈M implies x=0, where n is a positive integer, M is prime if aΓMΓb=0 implies a=0 or b=0, and it is semiprime if aΓMΓa=0 implies a=0. An additive mapping T:M→M is called a left (right) centralizer if T(xαy)=T(x)αy(T(xαy)=xαT(y)) for x,y∈M, α∈Γ and it is called a Jordan left (right) centralizer if(2.3)T(xαx)=T(x)αx(T(xαx)=xαT(x))∀x∈M,α∈Γ.
A mapping θ:M×M→M is called biadditive if it is additive in both arguments. An additive mapping D:M→M is called a derivation if D(xαy)=D(x)αy+xαD(y) for all x,y∈M, α∈Γ and it is called a Jordan derivation if D(xαx)=D(x)αx+xαD(x) for all x∈M, α∈Γ.A derivation D is inner if there exists a∈M, such that D(x)=aαx-xαa holds for all x∈M, α∈Γ. Every derivation is a Jordan derivation. The converse is in general not true. An additive mapping D:M→M is said to be a generalized derivation if there exists a derivation d:M→M such that D(xαy)=D(x)αy+xαd(y) for all x,y∈M, α∈Γ. The maps of the form x→aαx+xαb where a, b are fixed elements in M and for all α∈Γ called the generalized inner derivation. An additive mapping D:M→M is said to be a Jordan generalized derivation if there exists a derivation d:M→M such that D(xαx)=D(x)αx+xαd(x) for all x∈M, α∈Γ. Hence the concept of a generalized derivation covers both the concepts of a derivation and a left centralizers and the concept of a Jordan generalized derivation covers both the concepts of a Jordan derivation and a left Jordan centralizers. An example of a generalized derivation and a Jordan generalized derivation is given in [4].


## 3. Main Results

We start from the following subsidiary results.Lemma 3.1.
LetM be a semiprime Γ-ring. If a,b∈M are such that aαxβb=0 for all x∈M, α,β∈Γ,  then aαb=bαa=0.Proof.
Letx∈M. Then
(3.1)aαbβxγaαb=aα(bβxγa)αb=0,bαaβxγbαa=bα(aβxγb)αa=0.
By semiprimeness of M with respect to β,γ∈Γ, it follows that aαb=bαa=0.Lemma 3.2.
LetM be a semiprime Γ-ring and θ,φ:M×M→M biadditive mappings. If θ(x,y)αwβφ(x,y)=0 for all x,y,w∈M,  then θ(x,y)αwβφ(u,v)=0 for all x,y,u,v,w∈M, α,β∈Γ.Proof.
First we replacex with x+u in the relation θ(x,y)αwβφ(x,y)=0, and use the biadditivity of the θ and φ. Then we have
(3.2)θ(x+u,y)αwβφ(x+u,y)=0⟹θ(x,y)αwβφ(u,y)=-θ(u,y)αwβφ(x,y).
Then
(3.3)(θ(x,y)αwβφ(u,y))γzδ(θ(x,y)αwβφ(u,y))=-(θ(u,y)αwβφ(x,y))γzδ(θ(u,y)αwβφ(u,y))=0.
Hence θ(x,y)αwβφ(u,y)=0 by semiprimeness of M with respect to γ,δ∈Γ.
Now we replacey by y+v and obtain the assertion of the lemma with the similar observation as above.Lemma 3.3.
LetM be a semiprime Γ-ring satisfying the assumption (*) and a be a fixed element of M. If aβ[x,y]α=0 for all x,y∈M, α,β∈Γ,  then a∈Z(M).Proof.
First we calculate the following expressions using the assumption (*),
(3.4)[z,a]αβxδ[z,a]α=zαaβxδ[z,a]α-aαzβxδ[z,a]α=zαaβ[z,xαa]δ-zαaβ[z,x]δαa-aα[z,zδxαa]β+aα[z,zδx]βαa.
Since aβ[x,y]α=0 for all x,y∈M, α,β∈Γ, we get [z,a]αβxδ[z,a]α=0. By the semiprimeness of M we get [z,a]α=0 for all α∈Γ. Hence a∈Z(M).Lemma 3.4.
LetM be a Γ-ring satisfying the condition (*) and D:M→M be a Jordan generalized derivation with the associated derivation d. Let x,y,z∈M and α,β∈Γ. Then(i)
D(xαy+yαx)=D(x)αy+D(y)αx+xαd(y)+yαd(x).In particular, ifM is 2-torsion-free, then(ii)
D(xαyβx)=D(x)αyβx+xβd(yβx),(iii)
D(xαyβz+zβyαx)=D(x)αyβz+D(z)βyαx+xαd(yβz)+zβd(yαx).Proof.
(i) We haveD(xαx)=D(x)αx+xαd(x), for all x∈M, α∈Γ. Then replacing x by x+y, and following the series of implications below we get the result:
(3.5)D((x+y)α(x+y))=D(x+y)α(x+y)+(x+y)αd(x+y)⟹D(xαx+xαy+yαx+yαy)=D(x)αx+D(y)αx+D(x)αy+D(y)αx+xαd(y)+xαd(x)+yαd(y)+yαd(x)⟹D(xαx+yαy)+D(xαy+yαx)=(D(x)αx+xαd(x)+D(y)αy+yαd(y))+D(x)αy+D(y)αx+xαd(y)+yαd(x)⟹D(xαy+yαx)=D(x)αy+D(y)αx+xαd(y)+yαd(x),∀x,y∈M,α∈Γ.
(ii) Replacey by xβy+yβx in the above relation (3.5), then we get,(3.6)D(xα(xβy+yβx)+(xβy+yβx)αx)=D(x)α(xβy+yβx)+D(xβy+yβx)αx+xαd(xβy+yβx)+(xβy+yβx)αd(x),∀x,y,z∈M,α,β∈Γ.⟹D(xαxβy+yβxαx)+D(xαyβx)+D(xβyαx)=D(x)αxβy+D(x)αyβx+D(x)βyαx+D(y)βxαx+xβd(y)αx+yβd(x)αx+xαd(xβy+yβx)+(xβy+yβx)αd(x),∀x,y,z∈M,α,β∈Γ.
Using the assumption (*), we conclude that
(3.7)D(xαxβy+yβxαx)+2D(xαyβx)=D(x)αxβy+D(x)αyβx+D(x)βyαx+D(y)βxαx+xβd(y)αx+yβd(x)αx+xαd(xβy+yβx)+(xβy+yβx)αd(x),∀x,y,z∈M,α,β∈Γ.
Again, replacing x by xβx in (3.5)
(3.8)D(xβxαy+yαxβx)=D(xβx)αy+D(y)αxβx+xβxαd(y)+yαd(xβx),∀x,y∈M,α,β∈Γ,⟹D(xβxαy+yαxβx)=D(x)βxαy+xβd(x)αy+D(y)αxβx+xβxαd(y)+yαd(x)βx+yαxβd(x),∀x,y∈M,α,β∈Γ.
Adding both sides 2D(xαyβx), we get,
(3.9)⟹D(xβxαy+yαxβx)+2D(xαyβx)=D(x)βxαy+xβd(x)αy+D(y)αxβx+xβxαd(y)+yαd(x)βx+yαxβd(x)+2D(xαyβx),∀x,y∈M,α,β∈Γ.
Comparing (3.7) and (3.9) we obtain,
(3.10)2D(xαyβx)=2{D(x)αyβx+xαyβd(x)+xαd(y)βx}.
Since M is 2-torsion-free, it gives
(3.11)D(xαyβx)=D(x)αyβx+xαd(yβx).
(iii) Replacex for x+z in (3.12), we get,(3.12)D((x+z)αyβ(x+z))=D(x+z)αyβ(x+z)+(x+z)αd(yβ(x+z))⟹D(xαyβx+xαyβz+zαyβx+zαyβz)=D(x)αyβx+D(z)αyβx+D(x)αyβz+D(z)αyβz+xαd(yβx)+xαd(yβz)+zαd(yβx+yβz)⟹D(xαyβx+zαyβz)+D(xαyβz+zαyβx)=(D(x)αyβx+D(z)αyβz+xαd(yβx)+zαd(yβz))+(D(z)αyβx+D(x)αyβz+xαd(yβz)+zαd(yβx))⟹D(xαyβz+zαyβx)=D(x)αyβz+D(z)αyβx+xαd(yβz)+zαd(yβx)⟹D(xαyβz+zβyαx)=D(x)αyβz+D(z)αyβx+xαd(yβz)+zαd(yβx),∀x,y∈M,α,β∈Γ.Definition 3.5.
LetM be a Γ-ring and D:M→M be a Jordan generalized derivation with the associated derivation d. Define Gα(x,y)=D(xαy)-D(x)αy-xαd(y) for all x,y∈M and α∈Γ.Lemma 3.6.
The functionGα(x,y) has the following properties:(i)
Gα(x,y)+Gα(y,x)=0.(ii)
Gα(x,y+z)=Gα(x,y)+Gα(x,z).(iii)
Gα(x+y,z)=Gα(x,z)+Gα(y,z).(iv)
Gα+β(x,y)=Gα(x,y)+Gβ(x,y).Proof.
The results easily follow from Lemma3.4(i).Remark 3.7.
D is a generalized derivation if and only if Gα(x,y)=0 for all x,y∈M, α∈Γ.Theorem 3.8.
LetM be a 2-torsion-free semiprime Γ-ring satisfying the condition (*). Let x,y,z∈M and β,δ∈Γ. Then(i)
Gα(x,y)βzδ[x,y]α=0 for all z∈M,(ii)
Gα(x,y)β[x,y]α=0.Proof.
(i) From Lemma3.4(iii) we get
(3.13)D(xαyβz+zβyαx)=D(x)αyβz+D(z)βyαx+xαd(yβz)+zβd(yαx).
We set
(3.14)A=xαyβzδyαx+yαxβzδxαy.
Since
(3.15)D(A)=D(xαyβzδyαx+yαxβzδxαy)=D(xα(yβzδyαx)+yα(xβzδxαy))=D(x)αyβzδyαx+xαd(yβzδyαx)+D(y)αxβzδxαy+yαd(xβzδxαy)=D(x)αyβzδyαx+xαd(y)βzδyαx+xαyβd(zδyαx)+D(y)αxβzδxαy+yαd(x)βzδxαy+yαxβd(zδxαy),∀x,y,z∈M,α,β,δ∈Γ.
Again
(3.16)D(A)=D(xαyβzδyαx+yαxβzδxαy)=D((xαy)βzδ(yαx)+(yαx)βzδ(xαy))=D(xαy)βzδyαx+D(yαx)βzδxαy+xαyβd(zδyαx)+yαxβd(zδxαy),∀x,y,z∈M,α,β,δ∈Γ.
From (3.15) and (3.16) we find,
(3.17)(D(xαy)-D(x)αy-xαd(y))βzδyαx+(D(yαx)-D(y)αx-yαd(x))βzδxαy=0.⟹Gα(x,y)βzδyαx+Gα(y,x)βzδxαy=0,⟹Gα(x,y)βzδyαx-Gα(x,y)βzδxαy=0,⟹Gα(x,y)βzδ[y,x]α=0,⟹Gα(x,y)βzδ[x,y]α=0,forx,y∈M,α,β,δ∈Γ.
(ii) According to Lemma3.4(ii) we have,(3.18)D(xαyβz)=D(x)αyβz+xαd(yβz).
Replace z by xαy in (3.18) we find
(3.19)D(xαyβxαy)=D(x)αyβxαy+xαd(yβxαy)⟹D((xαy)β(xαy))=D(x)αyβxαy+xαd(yβxαy)⟹D(xαy)βxαy+xαyβd(xαy)-D(x)αyβxαy-xαd(yβxαy)=0⟹(D(xαy)-D(x)αy-xαd(y))βxαy)+xαyβd(xαy)-xαyβd(xαy)=0⟹Gα(x,y)βxαy=0.
Similarly Gα(x,y)βyαx=0. Therefore,
(3.20)Gα(x,y)βxαy-Gα(x,y)βyαx=Gα(x,y)β[x,y]α=0.Lemma 3.9.
Gα(x,y)∈Z(M), for all x,y∈M, α∈Γ.Proof.
From Theorem3.8(ii), we have Gα(x,y)β[x,y]α=0. Therefore due to Lemma 3.4(ii) Gα(x,y)∈Z(M).Theorem 3.10.
LetM be a 2-torsion-free semiprime Γ-ring satisfying the assumption (*) and D:M→M be a Jordan generalized derivation with associated derivation d on M. Then D is a generalized derivation.Proof.
In particular,rγGα(x,y), Gα(x,y)γr∈Z(M) for all r∈M, α,γ∈Γ. This gives
(3.21)xαGα(x,y)δGα(x,y)βy=Gα(x,y)δGα(x,y)βyαx=yβGα(x,y)δGα(x,y)αx=yαGα(x,y)δGα(x,y)βx.
Then 4D(xαGα(x,y)δGα(x,y)βy)=4D(yβGα(x,y)δGα(x,y)αx).Now we will compute each side of this equality by using (3.15) and the above relation,
(3.22)4D(xαGα(x,y)δGα(x,y)βy)=2D(xαGα(x,y)δGα(x,y)βy+Gα(x,y)δGα(x,y)βyαx)=2D(x)αGα(x,y)δGα(x,y)βy+2xαd(Gα(x,y)δGα(x,y)βy)+2D(Gα(x,y)δGα(x,y)βy)αx+2Gα(x,y)δGα(x,y)βyαd(x)=2D(x)αGα(x,y)δGα(x,y)βy+D(Gα(x,y)δGα(x,y)βy+yβGα(x,y)δGα(x,y))αx+2xαd(Gα(x,y)δGα(x,y)βy)+2Gα(x,y)δGα(x,y)βyαd(x)=2D(x)αGα(x,y)δGα(x,y)βy+D(Gα(x,y))δGα(x,y)βyαx+Gα(x,y)δd(Gα(x,y))βyαx+D(y)βGα(x,y)δGα(x,y)αx+Gα(x,y)δGα(x,y)βd(y)αx+yβd(Gα(x,y)δGα(x,y))αx+2xαd(Gα(x,y)δGα(x,y)βy)+2Gα(x,y)δGα(x,y)βyαd(x).
So we get
(3.23)4D(xβGα(x,y)δGα(x,y)αy)=2D(x)βGα(x,y)δGα(x,y)αy+D(Gα(x,y))δGα(x,y)βyαx+Gα(x,y)δd(Gα(x,y))βyx+D(y)βGα(x,y)δGα(x,y)αx+Gα(x,y)δGα(x,y)βd(y)αx+yβd(Gα(x,y)δGα(x,y))αx+2xαd(Gα(x,y)δGα(x,y)βy)+2Gα(x,y)δGα(x,y)βyαd(x),x,y∈M,α,β,δ∈Γ.
Moreover,
(3.24)4D(yβGα(x,y)δGα(x,y)αx)=2D(yβGα(x,y)δGα(x,y)αx+Gα(x,y)δGα(x,y)βxαy)=2D(y)βGα(x,y)δGα(x,y)αx+2yβd(Gα(x,y)δGα(x,y)αx)+2D(Gα(x,y)δGα(x,y)βx)αy+2Gα(x,y)δGα(x,y)βxαd(y)=2D(y)βGα(x,y)δGα(x,y)αx+D(Gα(x,y)δGα(x,y)βx+xβGα(x,y)δGα(x,y))αy+2yβd(Gα(x,y)δGα(x,y)αx)+2Gα(x,y)δGα(x,y)βxαd(y)=2D(y)βGα(x,y)δGα(x,y)αx+D(Gα(x,y))δGα(x,y)βxαy+Gα(x,y)δd(Gα(x,y))βxαy+D(x)βGα(x,y)δGα(x,y)αy+Gα(x,y)δGα(x,y)αd(x)βy+xαd(Gα(x,y)δGα(x,y))βy+2yβd(Gα(x,y)δGα(x,y)αx)+2Gα(x,y)δGα(x,y)αxβd(y).
So we get
(3.25)4D(yβGα(x,y)δGα(x,y)αx)=2D(y)βGα(x,y)δGα(x,y)αx+D(Gα(x,y))δGα(x,y)αxβy+Gα(x,y)δd(Gα(x,y))αxβy+D(x)βGα(x,y)δGα(x,y)αy+Gα(x,y)δGα(x,y)αd(x)βy+xαd(Gα(x,y)δGα(x,y))βy+2yβd(Gα(x,y)δGα(x,y)αx)+2Gα(x,y)δGα(x,y)αxβd(y),x,y∈M,α,β,δ∈Γ.
Comparing the results of (3.22) and (3.24) and using the above relations
(3.26)Gα(x,y)βyαx=Gα(x,y)βyαx=xαGα(x,y)βy=xαGα(x,y)βy=Gα(x,y)βxαy,Gα(x,y)δd(Gα(x,y))βyαx=d(Gα(x,y))δGα(x,y)βyαx=d(Gα(x,y))δGα(x,y)αxβy=Gα(x,y)δd(Gα(x,y))αxβy,xβGα(x,y)δd(Gα(x,y))αy=d(Gα(x,y))αxδGα(x,y)βy=d(Gα(x,y))δGα(x,y)βxαy=d(Gα(x,y))δGα(x,y)αyβx=Gα(x,y)βyδd(Gα(x,y))αx=yβGα(x,y)δd(Gα(x,y))αx,
we obtain
(3.27)D(x)αGα(x,y)δGα(x,y)βy+xαGα(x,y)δGα(x,y)βd(y)=D(y)αGα(x,y)δGα(x,y)βx+yαGα(x,y)δGα(x,y)βd(x),
which gives
(3.28)φα(x,y)βGα(x,y)δGα(x,y)=φα(y,x)αGα(x,y)δGα(x,y),
where φα(x,y) stands for D(x)αy+xαd(y).
On the other hand, we have(3.29)4D(xαyβGα(x,y)δGα(x,y))=4D(xαGα(x,y)δyβGα(x,y)).
Now we use (3.15) and the properties of Gα(x,y), to derive
(3.30)4D(xαyβGα(x,y)δGα(x,y))=2D(xαyβGα(x,y)δGα(x,y)+Gα(x,y)δGα(x,y)αxβy)=2D(xαy)βGα(x,y)δGα(x,y)+2xαyβd(Gα(x,y)δGα(x,y))+2D(Gα(x,y)δGα(x,y))αxβy+2Gα(x,y)δGα(x,y)βd(xαy),
which gives
(3.31)4D(xαyβGα(x,y)δGα(x,y))=2D(xαy)βGα(x,y)δGα(x,y)+2xαyβd(Gα(x,y)δGα(x,y))+2D(Gα(x,y)δGα(x,y))αxβy+2Gα(x,y)δGα(x,y)βd(xαy),x,y∈M,α,β,δ∈Γ.
Moreover,
(3.32)4D(xαGα(x,y)δyβGα(x,y))=2D(xαGα(x,y)δyβGα(x,y)+yβGα(x,y)δxαGα(x,y))=2D(Gα(x,y)αx)δGα(x,y)βy+2Gα(x,y)αxδd(Gα(x,y)βy)+2D(Gα(x,y)βy)δGα(x,y)αx+2Gα(x,y)βyδd(Gα(x,y)αx)=D(xαGα(x,y)+Gα(x,y)αx)δGα(x,y)βy+2Gα(x,y)αxδd(Gα(x,y)βy)+D(yβGα(x,y)+Gα(x,y)βy)δGα(x,y)αx+2Gα(x,y)βyδd(Gα(x,y)αx)=D(x)αGα(x,y)δGα(x,y)βy+D(Gα(x,y))δGα(x,y)αxβy+xαd(Gα(x,y))δGα(x,y)βy+Gα(x,y)αd(x)δGα(x,y)βy+2Gα(x,y)αxδd(Gα(x,y)βy)+D(y)βGα(x,y)δGα(x,y)αx+D(Gα(x,y))δGα(x,y)βyαx+yβd(Gα(x,y))δGα(x,y)αx+Gα(x,y)βd(y)δGα(x,y)αx+2Gα(x,y)βyδd(Gα(x,y)αx).
So we obtain
(3.33)4D(xαGα(x,y)δyβGα(x,y))=D(x)αGα(x,y)δGα(x,y)βy+D(Gα(x,y))δGα(x,y)βxαy+xαd(Gα(x,y))δGα(x,y)βy+Gα(x,y)αd(x)δGα(x,y)βy+2Gα(x,y)αxδd(Gα(x,y)βy)+D(y)βGα(x,y)δGα(x,y)αx+D(Gα(x,y))δGα(x,y)βyαx+yβd(Gα(x,y))δGα(x,y)αx+Gα(x,y)βd(y)δGα(x,y)αx+2Gα(x,y)βyδd(Gα(x,y)αx),x,y∈M,α,β,δ∈Γ.
Comparing (3.31) and (3.33), we derive
(3.34)2D(xαy)βGα(x,y)δGα(x,y)=φα(x,y)βGα(x,y)δGα(x,y)+φα(y,x)βGα(x,y)δGα(x,y),x,y∈M,α,β,δ∈Γ.
Finally using (3.31) we get D(xαy)βGα(x,y)δGα(x,y)=φα(x,y)βGα(x,y)δGα(x,y). But Gα(x,y)=D(xαy)-φα(x,y). By the semiprimeness of M, we have φα(x,y)βGα(x,y)=0. Again by the primeness of M, we get Gα(x,y). The proof is complete.It is clear that if we let the derivationd to be the zero derivation in the above theorem, we get the following result.Theorem 3.11.
LetM be a 2-torsion-free semiprime Γ-ring and D:M→M be an additive mapping which satisfies D(xαx)=D(x)αx for all x∈M, α∈Γ.  Then D is a left centralizerProof.
We have(3.35)D(xαx)=D(x)αx.
If we replace x by x+y, we get
(3.36)D(xαy+yαx)=D(x)αy+D(y)αx.
By replacing y with xαy+yαx and using (3.5), we arrive at
(3.37)D(xα(xαy+yαx)+(xαy+yαx)αx)=D(x)αxαy+D(x)αyαx+D(x)αyαx+D(y)αxαx.
But on the other hand,
(3.38)D(xαxαy+yαxαx)+2D(xαyαx)=D(x)αxαy+D(y)αxαx+2D(xαyαx).
Comparing (3.37) and (3.38) we obtain
(3.39)D(xαyαx)=D(x)αyαx.
If we linearize (3.39) in x, we get
(3.40)D(xαyαz+zαyαx)=D(x)αyαz+D(z)αyαx.
Now we shall compute D(xαyαzαyαx+yαxαzαxαy) in two different ways. If we use (3.39) we have
(3.41)D(xαyαzαyαx+yαxαzαxαy)=D(x)αyαzαyαx+D(y)αxαzαxαy.
But if we use (3.40) we have
(3.42)D(xαyαzαyαx+yαxαzαxαy)=D(xαy)αzαyαx+D(yαx)αzαxαy.
Comparing (3.41) and (3.42) and introducing a bi-additive mapping Gα(x,y)=D(xαy)-D(x)αy we arrive at
(3.43)Gα(x,y)αzαyαx+Gα(y,x)αzαxαy=0.
Equality (3.36) can be rewritten in this notation as Gα(x,y)=-Gα(y,x). Using this fact and (3.43) we obtain
(3.44)Gα(x,y)αzα[x,y]α=0.
Using first Lemma 3.2 and then Lemma 3.1 we have
(3.45)Gα(x,y)αzα[u,v]α=0.
Now fix some x,y∈M and using Lemma 3.3 we get Gα(x,y)∈Z.
In particular,rβGα(x,y), Gα(x,y)βr∈Z for all r∈M. This gives(3.46)xβGα(x,y)δGα(x,y)βy=Gα(x,y)ΓGα(x,y)ΓyΓx=yΓGα(x,y)ΓGα(x,y)Γx=yΓGα(x,y)ΓGα(x,y)Γx.
Therefore 4D(xΓGα(x,y)ΓGα(x,y)Γy)=4D(yΓGα(x,y)ΓGα(x,y)Γx). Both sides of this equality will be computed in few steps using (3.36),
(3.47)2D(xΓGα(x,y)ΓGα(x,y)Γy+Gα(x,y)ΓGα(x,y)ΓyΓx)=2D(yΓGα(x,y)ΓGα(x,y)Γx+Gα(x,y)ΓGα(x,y)ΓxΓy),2D(x)ΓGα(x,y)ΓGα(x,y)Γy+2D(Gα(x,y)ΓGα(x,y)Γy)Γx=2D(y)ΓGα(x,y)ΓGα(x,y)Γx+2D(Gα(x,y)ΓGα(x,y)Γx)Γy,2D(x)ΓGα(x,y)ΓGα(x,y)Γy+D(Gα(x,y)ΓGα(x,y)Γy+yΓGα(x,y)ΓGα(x,y))Γx=2D(y)ΓGα(x,y)ΓGα(x,y)Γx+D(Gα(x,y))ΓGα(x,y)Γx+xΓGα(x,y)ΓGα(x,y)Γy,2D(x)ΓGα(x,y)ΓGα(x,y)Γy+D(Gα(x,y))ΓGα(x,y)ΓyΓx+D(y)ΓGα(x,y)ΓGα(x,y)Γx=2D(y)ΓGα(x,y)ΓGα(x,y)Γx+D(Gα(x,y))ΓGα(x,y)ΓxΓy+D(x)ΓGα(x,y)ΓGα(x,y)Γy,D(x)ΓGα(x,y)ΓGα(x,y)Γy+D(Gα(x,y))ΓGα(x,y)ΓyΓx=D(y)ΓGα(x,y)ΓGα(x,y)Γx+D(Gα(x,y))ΓGα(x,y)ΓxΓy.
Since Gα(x,y)ΓyΓx=Gα(x,y)ΓyΓx=xΓGα(x,y)Γy=xΓGα(x,y)Γy=Gα(x,y)ΓxΓy, we obtain
(3.48)D(x)ΓGα(x,y)Γy=D(y)ΓGα(x,y)ΓGα(x,y)Γx.
On the other hand, we also have
(3.49)4D(xΓyΓGα(x,y)ΓGα(x,y))=4D(xΓGα(x,y)ΓyΓGα(x,y)),2D(xΓyΓGα(x,y)ΓGα(x,y)+Gα(x,y)ΓGα(x,y)ΓxΓy)=2D(xΓGα(x,y)ΓyΓGα(x,y)+yΓGα(x,y)ΓxΓGα(x,y)),2D(xΓy)ΓGα(x,y)ΓGα(x,y)+2D(Gα(x,y))ΓGα(x,y)ΓxΓy=2D(Gα(x,y)Γx)ΓGα(x,y)Γy+2D(Gα(x,y)Γy)ΓGα(x,y)Γx,2D(xΓy)ΓGα(x,y)ΓGα(x,y)+2D(Gα(x,y))ΓGα(x,y)ΓxΓy=D(xΓGα(x,y)+Gα(x,y)Γx)ΓGα(x,y)Γy+D(yΓGα(x,y)+Gα(x,y)Γy)ΓGα(x,y)Γx,2D(xΓy)ΓGα(x,y)ΓGα(x,y)+2D(Gα(x,y))ΓGα(x,y)ΓxΓy=D(x)ΓGα(x,y)ΓGα(x,y)Γy+D(Gα(x,y))ΓGα(x,y)ΓxΓy+D(y)ΓGα(x,y)ΓGα(x,y)Γx+D(Gα(x,y))ΓGα(x,y)ΓxΓy,2D(xΓy)ΓGα(x,y)ΓGα(x,y)=D(x)ΓyΓGα(x,y)ΓGα(x,y)+D(y)ΓxΓGα(x,y)ΓGα(x,y).
Finally using (3.48) we arrive at D(xΓy)ΓGα(x,y)ΓGα(x,y)=D(x)ΓyΓGα(x,y)ΓGα(x,y), but this means that Gα(x,y)ΓGα(x,y)ΓGα(x,y)=0. Hence,
(3.50)Gα(x,y)ΓGα(x,y)ΓMΓGα(x,y)ΓGα(x,y)=Gα(x,y)ΓGα(x,y)ΓGα(x,y)ΓGα(x,y)ΓM=0,Gα(x,y)ΓMΓGα(x,y)=Gα(x,y)ΓGα(x,y)ΓM=0,
which implies Gα(x,y)=0. The proof is complete.



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*Source: 270132-2012-02-16.xml*