Wang Zheng-ping,Rehman Ur Nadeemand Huang Shu-liang
(1.Supervising Section,Chuzhou Vocational and Technical College,Chuzhou,Anhui,239000)
(2.Department of Mathematics,Aligarh Muslim University,Aligarh,202002,India)
(3.School of Mathematics and Finance,Chuzhou University,Chuzhou,Anhui,239000)
Communicated by Du Xian-kun
An Identity with Skew Derivations on Lie Ideals
Wang Zheng-ping1,Rehman Ur Nadeem2and Huang Shu-liang3,*
(1.Supervising Section,Chuzhou Vocational and Technical College,Chuzhou,Anhui,239000)
(2.Department of Mathematics,Aligarh Muslim University,Aligarh,202002,India)
(3.School of Mathematics and Finance,Chuzhou University,Chuzhou,Anhui,239000)
Communicated by Du Xian-kun
Let R be a 2-torsion free prime ring and L a noncommutative Lie ideal of R.Suppose that(d,σ)is a skew derivation of R such that xsd(x)xt=0 for all x∈L,where s,t are fixed non-negative integers.Then d=0.
skew derivation,generalized polynomial identity,Lie ideal,prime ring
2010 MR subject classification:16N20,16W25,16N55
Document code:A
Article ID:1674-5647(2016)01-0083-05
10.13447/j.1674-5647.2016.01.06
Throughout this paper,unless specifically stated,R always denotes a prime ring with center Z(R),Q its Martindale quotient ring.Note that Q is also a prime ring and the center C of Q,which is called the extended centroid of R,is a field(we refer the readers to[1] for the definitions and related properties of these notions).For any x,y∈R,the symbol [x,y]stands for the commutator xy−yx.For subsets A,B of R,[A,B]is the additive subgroup generated by all[a,b]with a∈A and b∈B.An additive subgroup L of R is said to be a Lie ideal of R if[l,r]∈L for all l∈L and r∈R.A Lie ideal L is called noncommutative if[L,L]≠0.Let L be a noncommutative Lie ideal of R.It is well known that[R[L,L]R,R]⊆L(see the proof of Lemma 1.3 in[2]).Since[L,L]≠0,we have o≠=[I,R]⊆L for I=R[L,L]R a nonzero ideal of R.Recall that a ring R is called prime if for any x,y∈R,xRy=0 implies that either x=0 or y=0.An additive mapping d:R−→R is called a derivation if d(xy)=d(x)y+xd(y)holds for all x,y∈R.Given anyautomorphism σ of R,an additive mapping d:R→R satisfying
is called a σ-derivation of R,or a skew derivation of R with respect to σ,denoted by(d,σ). It is easy to see that if σ=1R,the identity map of R,then a σ-derivation is merely an ordinary derivation.And if σ≠1R,then σ−1Ris a skew derivation.Thus the concept of skew derivations can be regarded as a generalization of derivations.When d(x)=σ(x)b−bx for some b∈Q,then(d,σ)is called an inner skew derivation,and otherwise it is outer.Any skew derivation(d,σ)extends uniquely to a skew derivation of Q(see[3])via extensions of both maps to Q.Thus we may assume that any skew derivation of R is the restriction of a skew derivation of Q.Recall that σ is called an inner automorphism if when acting on Q,σ(q)=uqu−1for some invertible u∈Q.When σ is not inner,then it is called an outer automorphism.The skew derivations have been extensively studied by many researchers from various views(see for instance[4]–[7]where further references can be found).
A well-known paper of Herstein[2]states that if I is a right ideal of R such that xn=0 for all x∈I,then I=0.Chang and Lin[8]studied a more general case when d(x)xn=0 and xnd(x)=0 for all x∈I,where d is a nonzero derivation and I is a nonzero right ideal of a prime ring R.Dhara and De Filippis[9]proved the following:Let R be a prime ring,F a generalized derivation of R and L a noncommutative Lie ideal of R.Suppose that xsF(x)xt=0 for all x∈L,where s≥0,t≥0 are fixed integers,then F=0 except when charR=2 and R satisfies s4.
In this paper,we continue to investigation on Lie ideals of prime rings,involving a skew derivation(d,σ)with a nontrivial associated automorphism σ.Here we examine what happens replacing the generalized derivation F by a skew derivation(d,σ)in the result of [9].
Theorem 2.1Let R be a 2-torsion free prime ring and L be a noncommutative Lie ideal of R.Suppose that(d,σ)is a skew derivation of R such that xsd(x)xt=0 for all x∈L, where s,t are fixed non-negative integers.Then d=0.
Proof.Suppose that d≠0.We divide the proof into two cases.
Case 1.Suppose that(d,σ)is X-outer.Set I=R[L,L]R.Then o≠=[I,R]⊆L.By the assumption,we have[x,y]s(d([x,y]))[x,y]t=0 for all x,y∈I and also for all x,y∈Q by Theorem 2 in[10].By Theorem 1 in[11],we get
Subcase 1.1.If σ is X-inner,that is,σ(x)=gxg−1for some g∈Q−C since σ is nontrivial.This implies that
Letting z=0 and replacing w by gw in(2.2),we find thatand,in particular,when w=y,we have
Set F(x)=gx for all x∈R.It is easy to see that F is a generalized derivation of R.Using Theorem 1 in[9],we find that F=0,that is,g=0,a contradiction.
Subcase 1.2.Suppose that σ is X-outer.By(2.1),we find that
Letting z=0 and replacing m by x in(2.4),we get
In particular,
Using Theorem 1 in[9]again,we conclude that 1=0,a contradiction.
Case 2.Suppose that d is X-inner,that is,d(x)=σ(x)b−bx with o≠=b∈Q.
Subcase 2.1.If σ is X-inner,then there exists an invertible element q∈Q such that σ(x)=qxq−1,where q∈Q−C.So Q satisfies the generalized polynomial identity
Let dimCV=∞and recall that as Lemma 2 in[12].The set[Q,Q]is dense in Q and so from
we have
Let v∈V such that{v,q−1bv}is linearly C-independent.Therefore there exist v1,···,vt, w∈V such that{v,q−1bv,v1,···,vt,w}is linearly C-independent.By the density of Q, there exists an r∈Q such that
Thus we get the contradiction
Hence{v,q−1bv}is linearly C-dependent for all v∈V and a standard argument shows that q−1b∈C,that is,d=0.
Let dimCV=k be a finite integer,that is,Q=Mk(C)for k≥2.Denote p=q−1b.Let i≠j and choose[x,y]=eii−ejjin(2.5).Both left multiplying by ejjand right multiplying by eiiit follows
in particular,
Let ϕ and ξ be the following automorphisms of Mk(C):
Since ϕ(q),ϕ(p),ξ(q)and ξ(p)satisfy the same property of q and p,it follows that
and also
which means that
and also
Comparing these last two relations we get 4qjipji=0,that is,
If k≥3,then by Proposition 1 in[13],it follows that either p∈C or q∈C.In the first case we get d=0.On the other hand,if q∈C then d(x)=[x,b]and the result follows as an application of main theorem in[14].Let k=2,that is,Q∼=M2(C).Assume that neither q nor p is a diagonal matrix in M2(C).Without loss of generality,we consider p21≠0.Thus, by(2.8),it follows q21=0,then q12≠0 and so p12=0.Moreover,since q is invertible,we also have q22≠0.Notice that,if u∈[Q,Q]is an invertible matrix,then by(2.6)it follows that
For u=e11−e22and by computations it follows that the(2,1)-entry of the matrix X is
On the other hand,for u=e12+e21,it follows that the(2,2)-entry of the matrix X is b21=0.Thus,by(2.9)we get the contradiction q22p21=0.The previous contradiction means that either q or p is diagonal.In this case,a standard argument shows that either q or p is central and we are done as above.
Subcase 2.2.If σ is X-outer,then
So by Kharchenko[15]we find that
Setting m=0,we get
Repeating the same argument already used after(2.3)we get b=0,which is a contradiction.
AcknowledgmentThe authors sincerely thank to Professor Asma Ali for her suggestions in the preparation of the manuscript.
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date:July 23,2014.
The NSF(1408085QA08)of Anhui Provincial,the Key University Science Research Project (KJ2014A183)of Anhui Province of China,and the Training Program(2014PY06)of Chuzhou University of China.
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E-mail address:598438243@qq.com(Wang Z P),shulianghuang@sina.com(Huang S L).
Communications in Mathematical Research2016年1期