# Characterization of Multiplicative Lie Triple Derivations on Rings

Authors: Xiaofei Qi
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

LetR be a ring having unit 1. Denote by ZR the center of R. Assume that the characteristic of R is not 2 and there is  an idempotent element e∈R such that  aRe=0⇒a=0andaR1-e=0⇒a=0. It is shown that, under some mild conditions, a map L:R→R is a multiplicative Lie triple derivation if and only if Lx=δx+hx for all x∈R, where δ:R→R is an additive derivation and h:R→ZR is a map satisfying ha,b,c=0 for all a,b,c∈R. As applications, all Lie (triple) derivations on prime rings and von Neumann algebras are characterized, which generalize some known results.


---

## Body

## 1. Introduction

LetR be an associative ring with the center Z(R). For any element a,b∈R, we set [a,b]=ab-ba. Recall that a map L:R→R is a multiplicative derivation or nonlinear derivation if L(ab)=L(a)b+aL(b), for all a,b∈R, is a multiplicative Lie derivation, if L([a,b])=[L(a),b]+[a,L(b)], for all a,b∈R, and is a multiplicative Lie triple derivation, if L([[a,b],c])=[[L(a),b],c]+[[a,L(b)],c]+[[a,b],L(c)], for all a,b,c∈R. Particularly, if L is additive (linear), then above maps are, respectively, additive (linear) derivations, additive (linear) Lie derivations, and additive (linear) Lie triple derivations. We often omit “linear” for “linear derivations.”The structure of additive (linear) derivations and additive (linear) Lie (triple) derivations on rings or algebras has been studied by many authors. Bresˇar in [1] proved that every additive Lie derivation on a prime ring R with characteristic not 2 can be decomposed as τ+ζ, where τ is an additive derivation from R into its central closure and ζ is an additive map of R into the extended centroid C sending commutators to zero. Mathieu and Villena [2] showed that every linear Lie derivation on a C*-algebra is standard, that is, can be decomposed as the form τ+h, where τ is a derivation and h is a central valued linear map vanishing at each commutator. In [3] Qi and Hou proved that the same is true for additive Lie derivations of nest algebras on Banach spaces. Miers [4] showed that every linear Lie triple derivation on M, a von Neumann algebra with no central summands of type I1, is of the form τ+h, where τ is a derivation and h is a central valued linear map vanishing at every Lie triple products [[A,B],C]. Recently, Wang and Lu [5] described the structure of linear Lie triple derivations on J-subspace lattice algebras. For other results, see [6–10] and the references therein.For the study of multiplicative derivations and multiplicative Lie (triple) derivations, Daif [11] initially proved that each multiplicative derivation on a 2-torsion free prime ring containing a nontrivial idempotent is additive. Yu and Zhang [12] showed that every multiplicative Lie derivation on triangular algebras is the sum of an additive derivation and a map into its center sending commutators to zero, and, later, Ji et al. [13] generalized this result to the case of multiplicative Lie triple derivations. Assume that N is a nontrivial nest on a Banach space X over the complex field which contains a nontrivial complemented element, and AlgN is the associated nest algebra. Li and Fang in [14] obtained the same result as the above for multiplicative Lie triple derivations on AlgN.The purpose of the present paper is to consider the problem of characterizing nonlinear Lie triple derivations on general rings.LetR be a ring having unit 1 and an idempotent element e, and let Z(R) denote the center of R. Assume that the characteristic of R is not 2 and satisfies that aRe={0}⇒a=0 and aR(1-e)={0}⇒a=0. Let L:R→R be a multiplicative Lie triple derivation. We show that if eRe and (1-e)R(1-e) do not contain nonzero central ideals, then L(a+b)=L(a)+L(b)+λa,b, for all a,b∈R, where λa,b∈Z(R) is a central element depending on a and b (Theorem 1); furthermore, if R also satisfies that, for a∈R, [eae,eRe]⊆Z(eRe)⇒eae∈Z(eRe), and [(1-e)a(1-e),(1-e)R(1-e)]⊆Z((1-e)R(1-e))⇒(1-e)a(1-e)∈Z((1-e)R(1-e)), then L(a)=δ(a)+h(a), for all a∈R, where δ:R→R is an additive derivation and h:R→Z(R) is a map satisfying h([[a,b],c])=0, for all a,b,c∈R (Theorem 2). As applications, some characterizations of multiplicative (additive) Lie (triple) derivations on prime rings and von Neumann algebras are obtained, respectively (Corollaries 3–7).


## 2. Main Results and Corollaries

The following are our main results in this paper.Theorem 1.
LetR be a ring having unit 1 and an idempotent element e. Assume that the characteristic of R is not 2 and R satisfies the following two conditions:(i)
aRe={0}⇒a=0 and aR(1-e)={0}⇒a=0;(ii)
eRe and (1-e)R(1-e) do not contain nonzero central ideals.
Assume thatL:R→R is a multiplicative Lie triple derivation. Then L(a+b)=L(a)+L(b)+λa,b,  for all a,b∈R, where λa,b∈Z(R) is a central element depending on a and b.Moreover, if the ringR in Theorem 1 also satisfies that, for a∈R, [eae,eRe]⊆Z(eRe)⇒eae∈Z(eRe), and [(1-e)a(1-e),(1-e)R(1-e)]⊆Z((1-e)R(1-e))⇒(1-e)a(1-e)∈Z((1-e)R(1-e)), then L has more concrete form.Theorem 2.
LetR be a ring having unit 1 and an idempotent element e. Assume that the characteristic of R is not 2 and R satisfies the following two conditions:(i)
aRe={0}⇒a=0 and aR(1-e)={0}⇒a=0;(ii)
eRe and (1-e)R(1-e) do not contain nonzero central ideals;(iii)
fora∈R, [eae,eRe]⊆Z(eRe)⇒eae∈Z(eRe), and [(1-e)a(1-e),(1-e)R(1-e)]⊆Z((1-e)R(1-e))⇒(1-e)a(1-e)∈Z((1-e)R(1-e)).
Then a mapL:R→R is a multiplicative Lie triple derivation if and only if L(a)=δ(a)+h(a), for all a∈R, where δ:R→R is an additive derivation and h:R→Z(R) is a map satisfying h([[a,b],c])=0, for all a,b,c∈R.Recall that a ringR is prime if, for any a,b∈R, aRb={0} implies a=0 or b=0.Applying Theorem2 to prime rings, we have the following result.Corollary 3.
LetR be a prime ring having unit 1 and a nontrivial idempotent, and let L:R→R be a map. If the characteristic of R is not 2 and eRe, (1-e)R(1-e) are noncommutative, then the following two statements are equivalent.(1)
L is a multiplicative Lie triple derivation.(2)
There exist an additive derivationδ:R→R and a map h:R→Z(R) satisfying h([[a,b],c])=0 for all a,b,c∈R such that L(x)=δ(x)+h(x) for all x∈R.Proof.
Lete∈R be a nontrivial idempotent. It is obvious that R satisfies the condition (i) in Theorem 2.
Claim. If U is a central ideal of a noncommutative prime ring R′, then U={0}.
Take anyu∈U. Since U⊆R′ is central, we have ua=au∈U for all a∈R′. Thus, for any a,b∈R′, one gets
(1)u[a,b]=uab-uba=uab-a(ub)=uab-(au)b=uab-(ua)b=0,
and so
(2)uc[a,b]=u[ca,b]-u[c,b]a=0∀a,b,c∈R′.
Since R′ is noncommutative, there exist two elements a0,b0∈R′ such that [a0,b0]≠0. It follows from the primeness of R′ that u=0. The claim holds.
SinceR is prime, both eRe and (1-e)R(1-e) are prime. Thus, by the above claim, the condition (ii) in Theorem 2 is satisfied.
Now, for any fixeda∈R, define two maps δ1:eRe→eRe and δ2:(1-e)R(1-e)→(1-e)R(1-e), respectively, by
(3)δ1(exe)=[eae,exe],δ2((1-e)x(1-e))=[(1-e)a(1-e),(1-e)x(1-e)],
for all x∈R. It is clear that both δ1 and δ2 are derivations. Posner in [9, Theorem 2] proved that, if d is a derivation of a noncommutative prime R′ such that, for all a∈R′, ad(a)-d(a)a is in the center of R′, then d is the zero derivation. Thus, by [9, Theorem 2], for a∈R, if eRe is noncommutative and [eae,eRe]⊆Z(eRe), then [eae,eRe]=0; that is, eae∈Z(eRe). Similarly, if (1-e)R(1-e) is noncommutative and [(1-e)a(1-e),(1-e)R(1-e)]⊆Z((1-e)R(1-e)), then (1-e)a(1-e)∈Z((1-e)R(1-e)). Hence the condition (iii) in Theorem 2 is also satisfied.
Now, by Theorem2, the corollary is true.LetB(H) be the algebra of all bounded linear operators acting on a complex Hilbert space H. Recall that a von Neumann algebra M is a subalgebra of some B(H) satisfying M′′={M′}′=M, where M′={T:T∈B(H)andTA=AT∀A∈M} ([15]). For A∈M, the central carrier of A, denoted by A¯, is the intersection of all central projections P such that PA=0. If A is self-adjoint, then the core of A, denoted by A_, is sup{S∈Z(M):S=S*,S≤A}. Particularly, if A=P is a projection, it is clear that P_ is the largest central projection ≤P. A projection P is core-free if P_=0. It is easy to see that P_=0 if and only if I-P¯=I ([16]).Applying Theorem2 to von Neumann algebras, we have the following corollary.Corollary 4.
LetM be a von Neumann algebra without central summands of type I1 and L:M→M a map. Then the following statements are equivalent.(1)
L is a multiplicative Lie triple derivation.(2)
There exist an additive derivationδ:M→M and a map h:M→Z(M) vanishing at each Lie triple product [[A,B],C] such that L(A)=δ(A)+h(A), for all A∈M.Proof.
Assume thatM is a von Neumann algebra without central summands of type I1. Then, by [16], there exists a nonzero core-free projection P∈M with P¯=I. Fix such P and note that P¯=I-P¯=I. It follows from the definition of the central carrier that both span{TP(x):T∈M,x∈H} and span{T(I-P)(x):T∈M,x∈H} are dense in H. So AMP={0}⇒A=0 and AM(I-P)={0}⇒A=0.
Bresˇar and Miers [17] proved that if Z∈Z(M) such that ZM⊆Z(M), then Z=0. This implies that M has no nonzero central ideals. Note that PMP and (I-P)M(I-P) are also von Neumann algebras without central summands of type I1. So both PMP and (I-P)M(I-P) have no nonzero central ideals.
Finally, forA∈M, if [PAP,PMP]⊆Z(PMP) and [(I-P)A(I-P),(I-P)M(I-P)]⊆Z((I-P)M(I-P)), by the Kleinecke-Shirokov theorem ([18]), both [PAP,PTP] and [(I-P)A(I-P),(I-P)T(I-P)] are central quasinilpotent for all T∈M. Hence [PAP,PTP]=[(I-P)A(I-P),(I-P)T(I-P)]=0 for all T∈M; that is, PAP∈Z(PMP) and (I-P)A(I-P)∈Z((I-P)M(I-P)).
Thus, ifM has no central summands of type I1, by what the above stated, M satisfies the corresponding assumptions (i)–(iii) in Theorem 2. By Theorem 2, the corollary is true.Note that a multiplicative Lie derivation must be a multiplicative Lie triple derivation. So the following corollary is immediate.Corollary 5.
LetM be a von Neumann algebra without central summands of type I1 and L:M→M a map. Then the following statements are equivalent.(1)
L is a multiplicative Lie derivation.(2)
There exist an additive derivationδ:M→M and a map h:M→Z(M) vanishing at each commutator [A,B] such that L(A)=δ(A)+h(A), for all A∈M.IfL in Corollaries 4 and 5 is additive, more can be said. In fact, a complete characterization of additive Lie (triple) derivations on any von Neumann algebras can be obtained, which is a slight generalization of the corresponding result in [4].Corollary 6.
LetM be a von Neumann algebra and L:M→M an additive map. Then the following statements are equivalent.(1)
L is a Lie triple derivation.(2)
There exist an additive derivationδ:M→M and an additive map h:M→Z(M) vanishing at each Lie triple product [[A,B],C] such that L(A)=δ(A)+h(A), for all A∈M.Proof.
Clearly, one only needs to check (1)⇒(2). In the following assume that L is an additive Lie triple derivation.
Take the central projectionQ∈M⊆B(H), so that, with respect to the space decomposition H=QH⊕(I-Q)H, M=M1⊕M2, where M1 is of type I1 and M2 has no central summands of type I1. Note that we may have Q=0. Then L(A) can be decomposed as
(4)L(A)=QL(A)+(I-Q)L(QA)+(I-Q)L((I-Q)A)hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh∀A∈M.
Claim.
L
(
Z
(
M
)
)
⊆
Z
(
M
).
For anyZ∈Z(M) and any A,B∈M, we have 0=L(0)=L([[Z,A],B])=[[L(Z),A],B], and so [L(Z),A]∈Z(M) for all A∈M. Hence [L(Z),A] is central quasinilpotent, which implies [L(Z),A]=0 for all A∈M. It follows that L(Z)∈Z(M). The claim holds.
SinceQ is a central projection and QM is of type I1, we have
(5)QM⊆Z(M).
By the above claim, L(QA)∈Z(M), for all A∈M, and so
(6)(I-Q)L(QA)=L(QA)-QL(QA)∈Z(M)∀A∈M.
Define a mapΨ:M2→M2 by
(7)Ψ(B)=L(0⊕B)|(I-Q)H=L((I-Q)(0⊕B))|(I-Q)Hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh∀B∈M2.
Then, for any A=A1⊕A2∈M=M1⊕M2, we have
(8)L(A)=QL(A)+(I-Q)L(QA)+(0⊕Ψ(A2)).
It is easy to prove that Ψ is an additive Lie triple derivation on M2. So, by Corollary 4, there exist an additive derivation δ2:M2→M2 and an additive map h2:M2→Z(M2) vanishing at each Lie triple product such that
(9)Ψ(B)=δ2(B)+h2(B)∀B∈M2.
Let δ=0⊕δ2 and h(A)=QL(A)+(I-Q)L(QA)+(0⊕h2(A|(I-Q)H)). By (4)–(9), one gets that δ is an additive derivation on M and h:M→Z(M) is an additive map satisfying h([[A,B],C])=0 for all A,B,C such that L(A)=δ(A)+h(A) holds for all A∈M. Hence (2) is true and the proof is finished.Particularly, for additive Lie derivations, we have the following corollary.Corollary 7.
LetM be a von Neumann algebra and L:M→M an additive map. Then the following statements are equivalent.(1)
L is a Lie derivation.(2)
There exist an additive derivationδ:M→M and an additive map h:M→Z(M) vanishing at each commutator such that L(A)=δ(A)+h(A) for all A∈M.


## 3. The Proof of Main Results

In this section, we will give proofs of our main results, Theorems1 and 2.In the sequel, assume thatR is a unital ring and containing an idempotent e satisfying aRe={0}⇒a=0 and aR(1-e)={0}⇒a=0. It is clear that e≠0,1. Write e1=e and e2=1-e. Then R can be written as R=R11+R12+R21+R22, where Rij=eiRej (i,j∈{1,2}).We first give several lemmas, which are needed to prove the main results.Lemma 8 (see [3, Lemma 3.1]).
The center ofR is
(10)Z(R)={z11+z22:z11∈R11,z22∈R22,mmmz11a12=a12z22,mmmz22a21=a21z11∀a12∈R12,a21∈R21}.By Lemma8, it is easily seen that if z11+z22∈Z(R), then z11∈Z(R11) and z22∈Z(R22).Lemma 9.
Leta∈R. If ab12=b12a and ab21=b21a hold for all b12∈R12 and all b21∈R21, then a∈Z(R).Proof.
Writea=a11+a12+a21+a22. Since ab12=b12a for all b12∈R12, we have
(11)a11b12+a21b12=b12a21+b12a22.
Multiplying e1 and e2 from the left and right in the equation, one gets a11b12=b12a22. Multiplying e2 from both sides in the equation, one has a21b12=0; that is, a21be2=0 for all b∈R. This implies a21=0.
Similarly, from the relationab21=b21a for all b21∈R21, one can prove a22b21=b21a11 and a12=0. It follows from Lemma 8 that a=a11+a22∈Z(R).Lemma 10.
Leta∈R. If  [[a,tij],ej]=0 holds for all tij∈Rij (1≤i≠j≤2), then e1ae1+e2ae2∈Z(R).Proof.
For1≤i≠j≤2, since [[a,tij],ej]=0, we have eiaeitij=tijejaej for all tij∈Rij. It follows from Lemma 8 that e1ae1+e2ae2∈Z(R).Lemma 11.
Fora∈R, if a∈Z(R)∩Rii (i=1,2), then a=0.Proof.
Fori∈{1,2}, assume that a∈Z(R)∩Rii. So aeibej=eibeja=0 for all b∈R, where j≠i. It follows from the assumption on R that a=aei=0.Lemma 12 (see [19, Lemma 4]).
LetR′ be a ring having unit 1 and an idempotent element e. Assume that R′ satisfies that aR′e={0}, which implies that a=0. For a∈R′, if exeae=eaexe for all x∈R′, then there exists an element λ∈Z(R′) such that eae=λe.Applying Lemma12 to our ring R, we get that, for any aii∈Rii, if eixaii=aiixei holds for all x∈R, then there exists λi∈Z(R) such that aii=λiei, i=1,2.Now, we are in the position to give the proofs of Theorems1 and 2.Proof of Theorem1.
We will prove the theorem by a series of claims.
Claim
1.
L
(
0
)
=
0.
By the definition ofL, we have
(12)L(0)=L([[0,0],0])=[[L(0),0],0]+[[0,L(0)],0]+[[0,0],L(0)]=0.
Claim
2.
e
1
L
(
e
i
)
e
1
+
e
2
L
(
e
i
)
e
2
∈
Z
(
R
),  i=1,2.
Let1≤i≠j≤2. For any aij∈Rij and ei, we have
(13)L(aij)=L([[aij,ei],ei])=[[L(aij),ei],ei]+[[aij,L(ei)],ei]+[[aij,ei],L(ei)]=L(aij)ei-2eiL(aij)ei+eiL(aij)+aijL(ei)ei-aijL(ei)+eiL(ei)aij-aijL(ei)+L(ei)aij.
Multiplying by ei and ej from the left and the right in the above equation, respectively, one gets 2eiL(ei)eiaij=2aijejL(ei)ej for all aij∈Rij, which implies
(14)eiL(ei)eiaij=aijejL(ei)ej∀aij∈Rij,
since the characteristic of R is not 2.
Similarly, for anyaji∈Rji, by the relation L(aji)=L([[aji,ei],ei]), one can check that
(15)ajieiL(ei)ei=ejL(ei)ejaji∀aji∈Rji.
Applying Lemma 8 to (14) and (15), one obtains eiL(ei)ei+ejL(ei)ej∈Z(R) for 1≤i≠j≤2. The claim holds.
Now letδ(a)=L(a)-[a,e1L(e1)e2-e2L(e1)e1] for all a∈R. Then δ:R→R is also a multiplicative Lie triple derivation and satisfies δ(e1)∈Z(R) as Claim 2. Thus, without loss of generality, we may assume
(16)L(e1)∈Z(R).
Claim
3.
L
(
e
2
)
∈
Z
(
R
).
By Claim 2,e1L(e2)e1+e2L(e2)e2∈Z(R). So we only need to check e1L(e2)e2=e2L(e2)e1=0. In fact, by (16), we have
(17)0=L([[e2,e1],e1])=[[L(e2),e1],e1]+[[e2,L(e1)],e1]+[[e2,e1],L(e1)]=[[L(e2),e1],e1]=e2L(e2)e1+e1L(e2)e2,
which implies e1L(e2)e2=e2L(e2)e1=0. Hence L(e2)=e1L(e2)e1+e2L(e2)e2∈Z(R).
Claim
4.
L
(
R
i
j
)
⊆
R
i
j, 1≤i≠j≤2.
Here, we only give the proof forL(R12)⊆R12. The proof for the other inclusion L(R21)⊆R21 is similar.
For anya12∈R12, by (16), (13) can be reduced to
(18)L(a12)=L(a12)e1-2e1L(a12)e1+e1L(a12)=e1L(a12)e2+e2L(a12)e1.
Taking any b12∈R12 and any c∈R, one obtains
(19)0=L([[a12,b12],c])=[[L(a12),b12],c]+[[a12,L(b12)],c]+[[a12,b12],L(c)]=[[L(a12),b12]+[a12,L(b12)],c],
which implies [L(a12),b12]+[a12,L(b12)]∈Z(R). Note that
(20)[a12,L(b12)]=[[e1,a12],L(b12)]=L([[e1,a12],b12])-[[L(e1),a12],b12]-[[e1,L(a12)],b12]=-[[e1,L(a12)],b12].
So, by (18), one achieves
(21)[L(a12),b12]+[a12,L(b12)]=[e2L(a12)e1,b12]-[[e1,L(a12)],b12]=2[e2L(a12)e1,b12]∈Z(R).
It follows from char  R≠2 that [e2L(a12)e1,b12]=e2L(a12)e1b12-b12e2L(a12)e1∈Z(R) for all b12∈R12. By Lemma 8, one can get e2L(a12)e1b12∈Z(R22) for all b12∈R12. It is easily checked that e2L(a12)e1Re2 is an ideal of R22. So, by the assumption (ii), e2L(a12)e1Re2={0}, and so e2L(a12)e1=0 by the condition (i). This and (18) imply that L(a12)=e1L(a12)e2∈R12, as desired.
Claim
5. For any aij∈Rij, we have L(-aij)=-L(aij), 1≤i≠j≤2.
By (16) and Claims 3 and 4, it is clear that
(22)L(-aij)=L([[aij,ei],ej])=[[L(aij),ei],ej]=-L(aij).
Claim
6. For any aij∈Rij, we have L(ei+aij)-L(aij)∈Z(R), 1≤i≠j≤2.
Let1≤i≠j≤2 and aij∈Rij. For any bij∈Rij, by (16) and Claims 3 and 4, one has
(23)L(bij)=L([[ei+aij,bij],ej])=[[L(ei+aij),bij],ej]+L(bij);
that is, [[L(ei+aij),bij],ej]=0. Note that [[L(aij),bij],ej]=0 by Claim 4. So
(24)[[L(ei+aij)-L(aij),bij],ej]=0.
Similarly, for any bji∈Rji, by the relations -L(bji)=L(-bji)=L([[ei+aij,bji],ei]) and [[L(aij),bji],ei]=0, one can check
(25)[[L(ei+aij)-L(aij),bji],ei]=0.
Combining (24) and (25) and by Lemma 10, we achieve
(26)e1(L(ei+aij)-L(aij))e1+e2(L(ei+aij)-L(aij))e2∈Z(R).
Thus, to complete the proof of the claim, we still need to check e1(L(ei+aij)-L(aij))e2=e2(L(ei+aij)-L(aij))e1=0. In fact, by (16), we have
(27)L(aij)=L([[ei+aij,e1],e1])=[[L(ei+aij),e1],e1],L(aij)=L([[aij,e1],e1])=[[L(aij),e1],e1].
It follows that [[L(ei+aij)-L(aij),e1],e1]=0, which implies e1(L(ei+aij)-L(aij))e2=e2(L(ei+aij)-L(aij))e1=0.
Claim
7.
L is additive on Rij, 1≤i≠j≤2.
Take anyaij,bij∈Rij (1≤i≠j≤2). By (16) and Claims 3–6, one obtains
(28)L(aij+bij)=L([[ei-aij,ei+bij],ej])=[[L(ei-aij),ei+bij],ej]+[[ei-aij,L(ei+bij)],ej]=[[-L(aij),ei+bij],ej]+[[ei-aij,L(bij)],ej]=L(aij)+L(bij).
Claim
8. For 1≤i≠j≤2, we have L(Rii)⊆(Rii+Rjj), L(Rii+Rii)⊆(Rii+Rjj), and L(Rii+Rjj)⊆(Rii+Rjj).
Let1≤i≠j≤2. For any aii,bii∈Rii and ajj∈Rjj, by (16) and Claim 3, one has
(29)0=L([[aii+ajj,ej],ej])=[[L(aii+ajj),ej],ej],0=L([[aii,ej],ej])=[[L(aii),ej],ej],0=L([[aii+bii,ej],ej])=[[L(aii+bii),ej],ej].
A simple calculation yields e1L(aii+ajj)e2=e2L(aii+ajj)e1=0, e1L(aii)e2=e2L(aii)e1=0, and e1L(aii+bii)e2=e2L(aii+bii)e1=0. So the claim is true.
Claim
9. For any a11∈R11 and a22∈R22, we have L(a11+a22)-L(a11)-L(a22)∈Z(R).
Take anya11∈R11anda22∈R22 and write L(a11+a22)-L(a11)-L(a22)=s11+s12+s21+s22. Note that, by Claim 8, one has L(a11+a22)-L(a11)-L(a22)=s11+s22. So, to prove the claim, one only needs to check s11+s22∈Z(R).
To do this, taking anyb12∈R12, by Claims 3–5 and 7, we have
(30)L([a11+a22,b12])=L(a11b12-b12a22)=L(a11b12)-L(b12a22)=L([[a11,b12],e2])-L([[b12,a22],e2])=[[L(a11),b12],e2]+[[a11,L(b12)],e2]-[[L(b12),a22],e2]-[[b12,L(a22)],e2]=[[L(a11),b12],e2]+[a11,L(b12)]+[a22,L(b12)]+[[L(a22),b12],e2],L([a11+a22,b12])=L([[a11+a22,b12],e2])=[[L(a11+a22),b12],e2]+[[a11+a22,L(b12)],e2]=[[L(a11+a22),b12],e2]+[a11,L(b12)]+[a22,L(b12)].
Combining the above two equations yields
(31)[[L(a11+a22)-L(a11)-L(a22),b12],e2]=0hhhhhhhhhhhhhhhhhhhhhhhhhh∀b12∈R12.
Similar to the above discussion, one can also prove that(32)[[L(a11+a22)-L(a11)-L(a22),b21],e1]=0hhhhhhhhhhhhhhhhhhhhhhhhhh∀b21∈R21.
Now, by Lemma 10, it follows that s11+s22∈Z(R), and the claim holds.
Claim
10. For any a∈R, any a12∈R12, and any a21∈R21, we have(i)
L(a+a12)-L(a)-e1(L(a+a12)-L(a))e2∈Z(R);(ii)
L(a+a21)-L(a)-e2(L(a+a21)-L(a))e1∈Z(R).
For anya,c∈R and any a12,b12∈R12, since [[a+a12,b12],c]=[[a,b12],c], by Claim 4, one can easily get [[L(a+a12),b12],c]=[[L(a),b12],c]; that is,
(33)[[L(a+a12)-L(a),b12],c]=0∀c∈R.
Write t=L(a+a12)-L(a)=t11+t12+t21+t22. Then (33) implies that
(34)[t,b12]=t11b12+t21b12-b12t21-b12t22∈Z(R).
Letting c=e2 in (33), a simple calculation yields
(35)t11b12=b12t22∀b12∈R12.
So (34) is reduced to t21b12-b12t21∈Z(R), which and Lemma 8 imply that t21b12∈Z(R22) holds for all b12∈R12. Note that t21e1Re2 is an ideal of R22. It follows from the assumption (ii) that t21e1Re2={0}, and so
(36)t21=0.
On the other hand, by using the equation [[a+a12,b21],e2]=[[a,b21],e2], one can also show
(37)t22b21=b21t11∀b21∈R21,t12=0.
Combining (35)–(37) and Lemma 8, we get that (i) holds.
The proof of (ii) is similar and we omit it here.
Claim
11. For any aij∈Rij (1≤i,j≤2), we have(i)
L(a12)=e1L(a11+a12+a22)e2 and e2L(a11+a12+a22)e1=0;(ii)
L(a21)=e2L(a11+a21+a22)e1 and e1L(a11+a21+a22)e2=0.
Here, we only give the proof of (i). The proof of (ii) is similar.
For anyaij∈Rij (1≤i,j≤2), by Claim 3, one gets
(38)L(a12)=L([[a11+a12+a22,e2],e2])=[[L(a11+a12+a22),e2],e2]=e1L(a11+a12+a22)e2+e2L(a11+a12+a22)e1,
which, together with Claim 4, implies L(a12)=e1L(a11+a12+a22)e2 and e2L(a11+a12+a22)e1=0, as desired.
Claim
12. For any aij∈Rij (1≤i,j≤2), we have(i)
L(a11+a12+a22)-L(a11)-L(a12)-L(a22)∈Z(R);(ii)
L(a11+a21+a22)-L(a11)-L(a21)-L(a22)∈Z(R).
Still, we only give the proof of (i).
Take anya11∈R11, a12∈R12, and a22∈R22. Firstly, by Claims 8, 9, and 11, there exists some z∈Z(R) such that
(39)L(a11+a12+a22)-L(a11)-L(a12)-L(a22)=L(a11+a12+a22)-L(a11+a22)-L(a12)+z=e1(L(a11+a12+a22)-L(a11+a22))e1+e2(L(a11+a12+a22)-L(a11+a22))e2+z.
Next, for anyb12∈R12, by Claims 3 and 4, one has
(40)L([[a11+a12+a22,b12],e2])=[[L(a11+a12+a22),b12],e2]+[[a11+a12+a22,L(b12)],e2]=[[L(a11+a12+a22),b12],e2]+[[a11+a22,L(b12)],e2],L([[a11+a12+a22,b12],e2])=L([[a11+a22,b12],e2])=[[L(a11+a22),b12],e2]+[[a11+a22,L(b12)],e2].
Comparing the above equations, we get
(41)[[L(a11+a12+a22)-L(a11+a22),b12],e2]=0.
Finally, for anyb21∈R21, a similar argument to the above achieves
(42)[[L(a11+a12+a22)-L(a11+a22),b21],e1]=0.
Now, combining (39)–(42) and Lemma 10, the claim holds.
Claim
13. For any aij∈Rij(1≤i,j≤2), we have L(a11+a12+a21+a22)-L(a11)-L(a12)-L(a21)-L(a22)∈Z(R).
Take anyaij∈Rij(1≤i,j≤2) and write
(43)t=L(a11+a12+a21+a22)-L(a11)-L(a12)-L(a21)-L(a22).
For any b12∈R12 and any b21∈R21, by Claims 4, 10, and 12, one has
(44)[t,b12]=[L(a11+a12+a21+a22)-L(a11)-L(a21)-L(a22),b12]=[L(a11+a12+a21+a22)-L(a11+a21+a22),b12]=0,[t,b21]=[L(a11+a12+a21+a22)-L(a11)-L(a12)-L(a22),b21]=[L(a11+a12+a21+a22)-L(a11+a12+a22),b21]=0.
Now the above two equations and Lemma 9 imply t∈Z(R).
Claim
14. For any aii,bii∈Rii, we have L(aii+bii)-L(aii)-L(bii)∈Z(R),  i=1,2.
Let1≤i≠j≤2 and aii,bii∈Rii be arbitrary. For any cij∈Rij, by (16) and Claims 3 and 4, we have
(45)L([aii+bii,cij])=L([aii,cij]+[bii,cij])=L([aii,cij])+L([bii,cij])=L([[aii,cij],ej])+L([[bii,cij],ej])=[[L(aii),cij],ej]+[[aii,L(cij)],ej]+[[L(bii),cij],ej]+[[bii,L(cij)],ej],L([aii+bii,cij])=L([[aii+bii,cij],ej])=[[L(aii+bii),cij],ej]+[[aii+bii,L(cij)],ej].
It follows that
(46)[[L(aii+bii)-L(aii)-L(bii),cij],ej]=0hhhhhhhhhhhhhhhhhhhhhhhh∀cij∈Rij.
Similarly, one can check(47)[[L(aii+bii)-L(aii)-L(bii),cji],ei]=0hhhhhhhhhhhhhhhhhhhhhhhh∀cji∈Rji.
Now, together with Lemma10 and Claim 8, (46)-(47) imply that the claim holds.
Claim
15.
L
(
a
+
b
)
-
L
(
a
)
-
L
(
b
)
∈
Z
(
R
) for all a,b∈R. Therefore, Theorem 1 holds.
In fact, by Claim 7 and Claims 13 and 14, it is easily seen that the claim is true.
The proof of the theorem is complete.Proof of Theorem2.
The “if” part is clear. We will prove the “only if” part by several claims.
Claim
1. For any aii∈Rii, there exists a map fi:Rii→Z(R) such that L(aii)-fi(aii)∈Rii, i=1,2.
We only give the proof fora11 here. The proof for any a22 is similar.
For anya11∈R11, by Claim 8 in the proof of Theorem 1, we have
(48)L(a11)=e1L(a11)e1+e2L(a11)e2∈R11+R22.
Now, taking any b22∈R22 and any c∈R, one has
(49)0=L([[a11,b22],c])=[[L(a11),b22],c]+[[a11,L(b22)],c],
which implies [L(a11),b22]+[a11,L(b22)]∈Z(R). It follows that [L(a11),b22]∈Z(R22). By the assumption (iii) in the theorem, one obtains [L(a11),b22]=0 for all b22∈R22; that is, e2L(a11)e2be2=e2be2L(a11)e2 for all b∈R. It follows from Lemma 12 that e2L(a11)e2=f1(a11)e2 for some f1(a11)∈Z(R). Hence
(50)L(a11)=e1L(a11)e1+f1(a11)e2=e1L(a11)e1-f1(a11)e1+f1(a11)∈R11+Z(R).
The claim holds.
Now define two mapsδ:R→R and f:R→Z(R), respectively, by
(51)δ(a)=L(a11)+L(a12)+L(a21)+L(a22)-f1(a11)-f2(a22),f(a)=L(a)-δ(a)
for all a=a11+a12+a21+a22∈R. Then, by Claim 4 in the proof of Theorem 1 and Claim 1, we have that
(52)δ(Rij)=L(Rij)⊆Rij,δ(Rii)⊆Rii,1≤i≠j≤2.
Moreover, δ(ei)∈Z(R) for i=1,2.
Claim
2.
δ is additive on R.
By Claim 7 in the proof of Theorem1, δ is additive on R12 and R21.
Leti∈{1,2}. For any aii,bii∈Rii, by Claim 14 in the proof of Theorem 1, (52), and the definition of δ, we have
(53)δ(aii+bii)-δ(aii)-δ(bii)=L(aii+bii)-L(aii)-L(bii)-f(aii+bii)+f(aii)+f(bii)∈Z(R)∩Rii.
It follows from Lemma 11 that δ(aii+bii)-δ(aii)-δ(bii)=0; that is, δ is additive on Rii, i=1,2.
Now, for anya=∑i,j=12‍aij,b=∑i,j=12‍bij∈R, by what the above proved, one gets
(54)δ(a+b)=δ(∑i,j2‍(aij+bij))=L(a11+b11)+L(a12+b12)+L(a21+b21)+L(a22+b22)-f1(a11+b11)-f2(a22+b22)=δ(a11+b11)+δ(a12+b12)+δ(a21+b21)+δ(a22+b22)=∑i,j=12‍δ(aij)+∑i,j=12‍δ(bij)=δ(a)+δ(b).
Hence δ is additive on R.
Claim
3.
δ is a derivation.
We will divide the proof of the claim into five steps.
Step
1. For any aii∈Rii and any bij∈Rij, we have δ(aiibij)=δ(aii)bij+aiiδ(bij), 1≤i≠j≤2.
For anyaii∈Rii and bij∈Rij (1≤i≠j≤2), by the definition of δ and (33), one obtains
(55)δ(aiibij)=L(aiibij)=L([[aii,bij],ej])=[[L(aii),bij],ej]+[[aii,L(bij)],ej]=[[δ(aii),bij],ej]+[[aii,δ(bij)],ej]=δ(aii)bij+aiiδ(bij).
By a similar argument to that of Step 1, one can show the following.
Step
2. For any aij∈Rij and any bjj∈Rjj, we have δ(aijbjj)=δ(aij)bjj+aijδ(bjj), 1≤i≠j≤2.
Step
3. For any aii,bii∈Rii, we have δ(aiibii)=δ(aii)bii+aiiδ(bii), i=1,2.
Take anyaii,bii∈Rii and any cij∈Rij (1≤i≠j≤2). By Step 1, one has
(56)δ(aiibiicij)=δ(aii)biicij+aiiδ(biicij)=δ(aii)biicij+aiiδ(bii)cij+aiibiiδ(cij),δ(aiibiicij)=δ(aiibii)cij+aiibiiδ(cij).
The above two equations yield (δ(aiibii)-δ(aii)bii-aiiδ(bii))cij=0; that is, (δ(aiibii)-δ(aii)bii-aiiδ(bii))eicej=0 for all c∈R. Note that δ(aiibii)-δ(aii)bii-aiiδ(bii)∈Rii. It follows that δ(aiibii)-δ(aii)bii-aiiδ(bii)=0.
Step
4. For any aij∈Rij and any bji∈Rji, we have δ(aijbji)=δ(aij)bji+aijδ(bji), 1≤i≠j≤2.
For anyaij∈Rij and any bji∈Rji (1≤i≠j≤2), noting that L(ej)∈Z(R), by (52), Claim 2, and the definitions of δ and f, we have
(57)δ(aijbji)-δ(bjiaij)=δ(aijbji-bjiaij)=L(aijbji-bjiaij)-f(aijbji-bjiaij)=L([[aij,ej],bji])-f(aijbji-bjiaij)=[[L(aij),ej],bji]+[[aij,ej],L(bji)]-f(aijbji-bjiaij)=[δ(aij),bji]+[aij,δ(bji)]-f(aijbji-bjiaij)=δ(aij)bji+aijδ(bji)-bjiδ(aij)-δ(bji)aij-f(aijbji-bjiaij).
Multiplying ei from both sides in the above equation, and noting that (52), one obtains
(58)δ(aijbji)=δ(aij)bji+aijδ(bji)-f(aijbji-bjiaij)ei.
Define a set
(59)Ui={δ(aijbji)-δ(aij)bji-aijδ(bji):aij∈Rij,mmbji∈Rji,1≤i≠j≤2}.
It is easily seen that Ui⊆Z(R)ei=Z(Rii). Also note that, for any cii∈Rii,
(60)cii(δ(aijbji)-δ(aij)bji-aijδ(bji))=ciiδ(aijbji)+δ(cii)aijbji-δ(cii)aijbji-ciiδ(aij)bji-ciiaijδ(bji)=δ(ciiaijbji)-δ(ciiaij)bji-ciiaijδ(bji)∈Ui.
Thus, Ui is a central ideal of Rii. It follows from the assumption (ii) that Ui={0}. So δ(aijbji)=δ(aij)bji+aijδ(bji) for all aij∈Rij and bji∈Rji, 1≤i≠j≤2.
Step
5. For any a,b∈R, we have δ(ab)=δ(a)b+aδ(b); that is, δ is a derivation.
By Claim 2 and Steps 1–4, it is easily checked that the step is true.
Claim
4.
f satisfies f([[a,b],c])=0 for all a,b,c∈R.
In fact, for anya,b,c∈R, by the definition of f and Claim 3, we have
(61)f([[a,b],c])=L([[a,b],c])-δ([[a,b],c])=[[L(a),b],c]+[[a,L(b)],c]+[[a,b],L(c)]-[[δ(a),b],c]-[[a,δ(b)],c]-[[a,b],δ(c)]=[[f(a),b],c]+[[a,f(b)],c]+[[a,b],f(c)]=0.
The proof of the theorem is complete.



---

*Source: 739730-2014-07-09.xml*