# Vanishing Power Values of Commutators with Derivations on Prime Rings

Authors: Basudeb Dhara
Journal: International Journal of Mathematics and Mathematical Sciences
 (2009)

---

## Abstract

LetR be a prime ring of char R≠2, d a nonzero derivation of R and ρ a nonzero right ideal of R such that [[d(x),x]n,[y,d(y)]m]t=0 for all x,y∈ρ, where n≥0, m≥0, t≥1 are fixed integers. If [ρ,ρ]ρ≠0, then d(ρ)ρ=0.


---

## Body

## 1. Introduction

Throughout this paper, unless specifically stated,R always denotes a prime ring with center Z(R) and extended centroid C, Q the Martindale quotients ring. Let n be a positive integer. For given a,b∈R, let [a,b]0=a and let [a,b]1 be the usual commutator ab-ba, and inductively for n>1, [a,b]n=[[a,b]n-1,b]. By d we mean a nonzero derivation in R.A well-known result proven by Posner [1] states that if [[d(x),x],y]=0 for all x,y∈R, then R is commutative. In [2], Lanski generalized this result of Posner to the Lie ideal. Lanski proved that if U is a noncommutative Lie ideal of R such that [[d(x),x],y]=0 for all x∈U,y∈R, then either R is commutative or char R=2 and R satisfies S4, the standard identity in four variables. Bell and Martindale III [3] studied this identity for a semiprime ring R. They proved that if R is a semiprime ring and [[d(x),x],y]=0 for all x in a non-zero left ideal of R and y∈R, then R contains a non-zero central ideal. Clearly, this result says that if R is a prime ring, then R must be commutative.Several authors have studied this kind of Engel type identities with derivation in different ways. In [4], Herstein proved that if char R≠2 and [d(x),d(y)]=0 for all x,y∈R, then R is commutative. In [5], Filippis showed that if R is of characteristic different from 2 and ρ a non-zero right ideal of R such that [ρ,ρ]ρ≠0 and [[d(x),x],[d(y),y]]=0 for all x,y∈ρ, then d(ρ)ρ=0.In continuation of these previous results, it is natural to consider the situation when[[d(x),x]n,[y,d(y)]m]t=0 for all x,y∈ρ,  n,m≥0,t≥1 are fixed integers. We have studied this identity in the present paper.It is well known that any derivation of a prime ringR can be uniquely extended to a derivation of Q, and so any derivation of R can be defined on the whole of Q. Moreover Q is a prime ring as well as R and the extended centroid C of R coincides with the center of Q. We refer to [6, 7] for more details.Denote byQ*CC{X,Y} the free product of the C-algebra Q and C{X,Y}, the free C-algebra in noncommuting indeterminates X,Y.


## 2. The Case:R Prime Ring

We need the following lemma.Lemma 2.1.
Letρ be a non-zero right ideal of R and d a derivation of R. Then the following conditions are equivalent: (i) d is an inner derivation induced by some b∈Q such that bρ=0; (ii) d(ρ)ρ=0 (for its proof refer to [8, Lemma]).We mention an important result which will be used quite frequently as follows.Theorem 2.2 (see Kharchenko [9]).
LetR be a prime ring,  d a derivation on  R and  I a non-zero ideal of  R. If  I satisfies the differential identity f(r1,r2,…,rn,d(r1),d(r2),…,d(rn))=0foranyr1,r2,…,rn∈I, then either (i) I satisfies the generalized polynomial identity
(2.1)f(r1,r2,…,rn,x1,x2,…,xn)=0,
or (ii) d is Q-inner, that is, for some q∈Q,d(x)=[q,x] and I satisfies the generalized polynomial identity
(2.2)f(r1,r2,…,rn,[q,r1],[q,r2],…,[q,rn])=0.Theorem 2.3.
LetR be a prime ring of char  R≠2 and  d a derivation of  R such that  [[d(x),x]n,[[y,d(y)]m]t=0 for all  x,y∈R, where  n≥0,m≥0,t≥1 are fixed integers. Then  R is commutative or  d=0.Proof.
LetR be noncommutative. If d is not Q-inner, then by Kharchenko's Theorem [9]
(2.3)g(x,y,u,v)=[[u,x]n,[y,v]m]t=0,
for all x,y,u,v∈R. This is a polynomial identity and hence there exists a field F such that R⊆Mk(F) with k>1, and R and Mk(F) satisfy the same polynomial identity [10,Lemma 1]. But by choosing u=e12,x=e11,v=e11 and y=e21, we get
(2.4)0=[[u,x]n,[y,v]m]t=(-1)tn(e11+(-)te22),
which is a contradiction.
Now, letd be Q-inner derivation, say d=ad(a) for some a∈Q, that is, d(x)=[a,x] for all x∈R, then we have
(2.5)[[a,x]n+1,[y,[a,y]]m]t=0,
for all x,y∈R. Since d≠0, a∉C and hence R satisfies a nontrivial generalized polynomial identity (GPI). By [11], it follows that RC is a primitive ring with H=Soc(RC)≠0, and eHe is finite dimensional over C for any minimal idempotent e∈RC. Moreover we may assume that H is noncommutative; otherwise, R must be commutative which is a contradiction.
Notice thatH satisfies [[a,x]n+1,[y,[a,y]]m]t=0 (see [10, Proof of Theorem 1]). For any idempotent e∈H and x∈H, we have
(2.6)0=[[a,e]n+1,[ex(1-e),[a,ex(1-e)]]m]t.
Right multiplying by e, we get
(2.7)0=[[a,e]n+1,[ex(1-e),[a,ex(1-e)]]m]te=[[a,e]n+1,[ex(1-e),[a,ex(1-e)]]m]t-1·{[a,e]n+1([ex(1-e),[a,ex(1-e)]]m)e-([ex(1-e),[a,ex(1-e)]]m)[a,e]n+1e}=[[a,e]n+1,[ex(1-e),[a,ex(1-e)]]m]t-1·{[a,e]n+1(∑j=0m(-1)j(mj)[a,ex(1-e)]jex(1-e)[a,ex(1-e)]m-j)e-(∑j=0m(-1)j(mj)[a,ex(1-e)]jex(1-e)[a,ex(1-e)]m-j)[a,e]n+1e}=[[a,e]n+1,[ex(1-e),[a,ex(1-e)]]m]t-1·{0-(∑j=0m(-1)j(mj)(-ex(1-e)a)jex(1-e)(aex(1-e))m-j)ae}=-[[a,e]n+1,[ex(1-e),[a,ex(1-e)]]m]t-1(∑j=0m(mj)(ex(1-e)a)m+1)e=-2m[[a,e]n+1,[ex(1-e),[a,ex(1-e)]]m]t-1(ex(1-e)a)m+1e=(-)t2mt(ex(1-e)a)(m+1)te.
This implies that0=(-)t2mt((1-e)aex)(m+1)t+1. Since char R≠2, ((1-e)aex)(m+1)t+1=0. By Levitzki's lemma [12, Lemma 1.1], (1-e)aex=0 for all x∈H. Since H is prime ring, (1-e)ae=0, that is, eae=ae for any idempotent e∈H. Now replacing e with 1-e, we get that ea(1-e)=0, that is, eae=ea. Therefore for any idempotent e∈H, we have [a,e]=0. So a commutes with all idempotents in H. Since H is a simple ring, either H is generated by its idempotents or H does not contain any nontrivial idempotents. The first case gives a∈C contradicting d≠0. In the last case, H is a finite dimensional division algebra over C. This implies that H=RC=Q and a∈H. By [10,Lemma 2], there exists a field F such that H⊆Mk(F) and Mk(F) satisfies [[a,x]n+1,[y,[a,y]]m]t. Then by the same argument as earlier, a commutes with all idempotents in Mk(F), again giving the contradiction a∈C, that is, d=0. This completes the proof of the theorem.Theorem 2.4.
LetR be a prime ring of char  R≠2,  d a non-zero derivation of  R and  ρ a non-zero right ideal of  R such that  [[d(x),x]n,[y,d(y)]m]t=0 for all  x,y∈ρ, where  n≥0,m≥0,t≥1 are fixed integers. If  [ρ,ρ]ρ≠0, then  d(ρ)ρ=0.We begin the proof by proving the following lemma.Lemma 2.5.
Ifd(ρ)ρ≠0 and  [[d(x),x]n,[y,d(y)]m]t=0 for all  x,y∈ρ,m,n≥0,t≥1 are fixed integers, then  R satisfies nontrivial generalized polynomial identity (GPI).Proof.
Suppose on the contrary thatR does not satisfy any nontrivial GPI. We may assume that R is noncommutative; otherwise, R satisfies trivially a nontrivial GPI. We consider two cases.Case 1.
Suppose thatd is Q-inner derivation induced by an element a∈Q. Then for any x∈ρ,(2.8)[[a,xX]n+1,[xY,[a,xY]]m]t
is a GPI for R, so it is the zero element in Q*CC{X,Y}. Expanding this, we get
(2.9)([a,xX]n+1∑j=0m(-1)j(mj)[a,xY]jxY[a,xY]m-j-∑j=0m(-1)j(mj)[a,xY]jxY[a,xY]m-j[a,xX]n+1)A(X,Y)=0,
where A(X,Y)=[[a,xX]n+1,[xY,[a,xY]]m]t-1. If ax and x are linearly C-independent for some x∈ρ, then
(2.10)((axX)n+1∑j=0m(-1)j(mj)[a,xY]jxY[a,xY]m-j-∑j=0m(-1)j(mj)(axY)jxY[a,xY]m-j[a,xX]n+1)A(X,Y)=0.
Again, since ax and x are linearly C-independent, above relation implies that
(2.11)(-xY[a,xY]m[a,xX]n+1)A(X,Y)=0,
and so
(2.12)(-xY(axY)m(axX)n+1)A(X,Y)=0.
Repeating the same process yields
(2.13)(-xY(axY)m(axX)n+1)t=0
in Q*CC{X,Y}. This implies that ax=0, a contradiction. Thus for any x∈ρ, ax and x are C-dependent. Then (a-α)ρ=0 for some α∈C. Replacing a with a-α, we may assume that aρ=0. Then by Lemma 2.1, d(ρ)ρ=0, contradiction.
Case 2.
Suppose thatd is not Q-inner derivation. If for all x∈ρ, d(x)∈xC, then [d(x),x]=0 which implies that R is commutative (see [13]). Therefore there exists x∈ρ such that d(x)∉xC, that is, x and d(x) are linearly C-independent.
By our assumption, we have thatR satisfies
(2.14)[[d(xX),xX]n,[xY,d(xY)]m]t=0.
By Kharchenko's Theorem [9],
(2.15)[[d(x)X+xr1,xX]n,[xY,d(x)Y+xr2]m]t=0,
for all X,Y,r1,r2∈R. In particular for r1=r2=0,
(2.16)[[d(x)X,xX]n,[xY,d(x)Y]m]t=0,
which is a nontrivial GPI for R, because x and d(x) are linearly C-independent, a contradiction.We are now ready to prove our main theorem.Proof of Theorem2.4.
Suppose thatd(ρ)ρ≠0, then we derive a contradiction. By Lemma 2.5, R is a prime GPI ring, so is also Q by [14]. Since Q is centrally closed over C, it follows from [11] that Q is a primitive ring with H=Soc(Q)≠0.
By our assumption and by [7], we may assume that
(2.17)[[d(x),x]n,[y,d(y)]m]t=0
is satisfied by ρQ and hence by ρH. Let e=e2∈ρH and y∈H. Then replacing x with e and y with ey(1-e) in (2.17), then right multiplying it by e, we obtain that
(2.18)0=[[d(e),e]n,[ey(1-e),d(ey(1-e))]m]te=[[d(e),e]n,[ey(1-e),d(ey(1-e))]m]t-1·{[d(e),e]n∑j=0m(-1)j(mj)d(ey(1-e))jey(1-e)d(ey(1-e))m-je-∑j=0m(-1)j(mj)d(ey(1-e))jey(1-e)d(ey(1-e))m-j[d(e),e]ne}.
Now we have the fact that for any idempotente, d(y(1-e))e=-y(1-e)d(e), ed(e)e=0 and so
(2.19)0=[[d(e),e]n,[ey(1-e),d(ey(1-e))]m]t-1·{0-∑j=0m(-1)j(mj)e(-y(1-e)d(e))jy(1-e)d(ey(1-e))m-jd(e)e}.
Now since for any idempotent e and for any y∈R, (1-e)d(ey)=(1-e)d(e)y, above relation gives
(2.20)0=[[d(e),e]n,[ey(1-e),d(ey(1-e))]m]t-1·{-e∑j=0m(mj)(y(1-e)d(e))jy(1-e)(d(e)y(1-e))m-jd(e)e}=[[d(e),e]n,[ey(1-e),d(ey(1-e))]m]t-1{-e∑j=0m(mj)(y(1-e)d(e))m+1e}=[[d(e),e]n,[ey(1-e),d(ey(1-e))]m]t-1{-2me(y(1-e)d(e))m+1e}={-2me(y(1-e)d(e))m+1}te.
This implies that 0=(-1)t2mt((1-e)d(e)ey)(m+1)t+1 for all y∈H. Since char R≠2, we have by Levitzki's lemma [12,Lemma 1.1] that (1-e)d(e)ey=0 for all y∈H. By primeness of H, (1-e)d(e)e=0. By [15,Lemma 1], since H is a regular ring, for each r∈ρH, there exists an idempotent e∈ρH such that r=er and e∈rH. Hence (1-e)d(e)e=0 gives (1-e)d(e)=(1-e)d(e2)=(1-e)d(e)e=0 and so d(e)=ed(e)∈eH⊆ρH and d(r)=d(er)=d(e)er+ed(er)∈ρH. Hence for each r∈ρH, d(r)∈ρH. Thus d(ρH)⊆ρH. Set J=ρH. Then J¯=J/(J∩lH(J)), a prime C-algebra with the derivation d¯ such that d¯(x¯)=d(x)¯, for all x∈J. By assumption, we have that
(2.21)[[d¯(x¯),x¯]n,[y¯,d¯(y¯)]m]t=0,
for all x¯,y¯∈J¯. By Theorem 2.3, we have either d¯=0 or ρH¯ is commutative. Therefore we have that either d(ρH)ρH=0 or [ρH,ρH]ρH=0. Now d(ρH)ρH=0 implies that 0=d(ρρH)ρH=d(ρ)ρHρH and so d(ρ)ρ=0. [ρH,ρH]ρH=0 implies that 0=[ρρH,ρH]ρH=[ρ,ρH]ρHρH and so [ρ,ρH]ρ=0, then 0=[ρ,ρρH]ρ=[ρ,ρ]ρHρ implying that [ρ,ρ]ρ=0. Thus in all the cases we have contradiction. This completes the proof of the theorem.


## 3. The Case:R Semiprime Ring

In this section we extend Theorem2.3 to the semiprime case. Let R be a semiprime ring and U be its right Utumi quotient ring. It is well known that any derivation of a semiprime ring R can be uniquely extended to a derivation of its right Utumi quotient ring U and so any derivation of R can be defined on the whole of U [7,Lemma 2].By the standard theory of orthogonal completions for semiprime rings, we have the following lemma.Lemma 3.1 (see [16, Lemma 1 and Theorem 1] or [7,pages 31-32]).
LetR be a 2-torsion free semiprime ring and P a maximal ideal of C. Then PU is a prime ideal of U invariant under all derivations of U. Moreover, ⋂{PU∣PisamaximalidealofCwithU/PU2-torsionfree}=0.Theorem 3.2.
LetR be a  2-torsion free semiprime ring and  d a non-zero derivation of  R such that  [[d(x),x]n,[y,d(y)]m]t=0 for all  x,y∈R,  n,m≥0,t≥1 fixed are integers. Then  d maps  R into its center.Proof.
Since any derivationd can be uniquely extended to a derivation in U, and R and U satisfy the same differential identities [7, Theorem 3], we have
(3.1)[[d(x),x]n,[y,d(y)]m]t=0,
for all x,y∈U. Let P be any maximal ideal of C such that U/PU is 2-torsion free. Then by Lemma 3.1, PU is a prime ideal of U invariant under d. Set U¯=U/PU. Then derivation d canonically induces a derivation d¯ on U¯ defined by d¯(x¯)=d(x)¯ for all x∈U. Therefore,
(3.2)[[d¯(x¯),x¯]n,[y¯,d¯(y¯)]m]t=0,
for all x¯,y¯∈U¯. By Theorem 2.3, either d¯=0 or [U¯,U¯]=0, that is, d(U)⊆PU or [U,U]⊆PU. In any case d(U)[U,U]⊆PU for any maximal ideal P of C. By Lemma 3.1, ⋂{PU∣PisamaximalidealofCwithU/PU2-torsionfree}=0. Thus d(U)[U,U]=0. Without loss of generality, we have d(R)[R,R]=0. This implies that
(3.3)0=d(R2)[R,R]=d(R)R[R,R]+Rd(R)[R,R]=d(R)R[R,R].
Therefore [R,d(R)]R[R,d(R)]=0. By semiprimeness of R, we have [R,d(R)]=0, that is, d(R)⊆Z(R). This completes the proof of the theorem.



---

*Source: 582181-2010-01-26.xml*