Let M 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.
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 results
Let M 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
Let M 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:
For any a,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:[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[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.
Further M 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 ifT(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.
Let M 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.
Let x∈M. Then
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.
Let M 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 replace x 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
θ(x+u,y)αwβφ(x+u,y)=0⟹θ(x,y)αwβφ(u,y)=-θ(u,y)αwβφ(x,y).
Then
(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 replace y by y+v and obtain the assertion of the lemma with the similar observation as above.
Lemma 3.3.
Let M 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 (*),
[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.
Let M 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) We have D(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:
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) Replace y by xβy+yβx in the above relation (3.5), then we get,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
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)
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,
⟹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,
2D(xαyβx)=2{D(x)αyβx+xαyβd(x)+xαd(y)βx}.
Since M is 2-torsion-free, it gives
D(xαyβx)=D(x)αyβx+xαd(yβx).
(iii) Replace x for x+z in (3.12), we get,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.
Let M 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 function Gα(x,y) has the following properties:
Gα(x,y)+Gα(y,x)=0.
Gα(x,y+z)=Gα(x,y)+Gα(x,z).
Gα(x+y,z)=Gα(x,z)+Gα(y,z).
Gα+β(x,y)=Gα(x,y)+Gβ(x,y).
Proof.
The results easily follow from Lemma 3.4(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.
Let M be a 2-torsion-free semiprime Γ-ring satisfying the condition (*). Let x,y,z∈M and β,δ∈Γ. Then
Gα(x,y)βzδ[x,y]α=0 for all z∈M,
Gα(x,y)β[x,y]α=0.
Proof.
(i) From Lemma 3.4(iii) we get
D(xαyβz+zβyαx)=D(x)αyβz+D(z)βyαx+xαd(yβz)+zβd(yαx).
We set
A=xαyβzδyαx+yαxβzδxαy.
Since
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
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,
(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 Lemma 3.4(ii) we have,D(xαyβz)=D(x)αyβz+xαd(yβz).
Replace z by xαy in (3.18) we find
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,
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 Theorem 3.8(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.
Let M 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
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,
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
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,
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
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
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
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
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 have4D(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
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
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,
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
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
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 derivation d to be the zero derivation in the above theorem, we get the following result.
Theorem 3.11.
Let M 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 centralizer
Proof.
We have
D(xαx)=D(x)αx.
If we replace x by x+y, we get
D(xαy+yαx)=D(x)αy+D(y)αx.
By replacing y with xαy+yαx and using (3.5), we arrive at
D(xα(xαy+yαx)+(xαy+yαx)αx)=D(x)αxαy+D(x)αyαx+D(x)αyαx+D(y)αxαx.
But on the other hand,
D(xαxαy+yαxαx)+2D(xαyαx)=D(x)αxαy+D(y)αxαx+2D(xαyαx).
Comparing (3.37) and (3.38) we obtain
D(xαyαx)=D(x)αyαx.
If we linearize (3.39) in x, we get
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
D(xαyαzαyαx+yαxαzαxαy)=D(x)αyαzαyαx+D(y)αxαzαxαy.
But if we use (3.40) we have
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
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
Gα(x,y)αzα[x,y]α=0.
Using first Lemma 3.2 and then Lemma 3.1 we have
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 givesxβGα(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),
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
D(x)ΓGα(x,y)Γy=D(y)ΓGα(x,y)ΓGα(x,y)Γx.
On the other hand, we also have
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,
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.
Acknowledgments
The paper was supported by Grant 01-12-10-978FR MOHE (Malaysia). The authors are thankful to the referee for valuable comments.
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