International Journal of Mathematics and Mathematical Sciences

International Journal of Mathematics and Mathematical Sciences / 2009 / Article

Research Article | Open Access

Volume 2009 |Article ID 582181 | https://doi.org/10.1155/2009/582181

Basudeb Dhara, "Vanishing Power Values of Commutators with Derivations on Prime Rings", International Journal of Mathematics and Mathematical Sciences, vol. 2009, Article ID 582181, 8 pages, 2009. https://doi.org/10.1155/2009/582181

Vanishing Power Values of Commutators with Derivations on Prime Rings

Academic Editor: Howard Bell
Received30 Aug 2009
Accepted14 Dec 2009
Published26 Jan 2010

Abstract

Let 𝑅 be a prime ring of char 𝑅2, 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,𝑦𝜌, where 𝑛0, 𝑚0, 𝑡1 are fixed integers. If [𝜌,𝜌]𝜌0, then 𝑑(𝜌)𝜌=0.

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,𝑏𝑅, let [𝑎,𝑏]0=𝑎 and let [𝑎,𝑏]1 be the usual commutator 𝑎𝑏𝑏𝑎, and inductively for 𝑛>1, [𝑎,𝑏]𝑛=[[𝑎,𝑏]𝑛1,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,𝑦𝑅, 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 𝑥𝑈,𝑦𝑅, 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 𝑦𝑅, 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 𝑅2 and [𝑑(𝑥),𝑑(𝑦)]=0 for all 𝑥,𝑦𝑅, 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]𝜌0 and [[𝑑(𝑥),𝑥],[𝑑(𝑦),𝑦]]=0 for all 𝑥,𝑦𝜌, 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,𝑦𝜌, 𝑛,𝑚0,𝑡1 are fixed integers. We have studied this identity in the present paper.

It is well known that any derivation of a prime ring 𝑅 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 by 𝑄𝐶𝐶{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. Let 𝜌 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 𝑏𝑄 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]). Let 𝑅 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,,𝑟𝑛𝐼, then either (i) 𝐼 satisfies the generalized polynomial identity 𝑓𝑟1,𝑟2,,𝑟𝑛,𝑥1,𝑥2,,𝑥𝑛=0,(2.1) or (ii) 𝑑 is 𝑄-inner, that is, for some 𝑞𝑄,𝑑(𝑥)=[𝑞,𝑥] and 𝐼 satisfies the generalized polynomial identity 𝑓𝑟1,𝑟2,,𝑟𝑛,𝑞,𝑟1,𝑞,𝑟2,,𝑞,𝑟𝑛=0.(2.2)

Theorem 2.3. Let 𝑅 be a prime ring of char 𝑅2 and 𝑑 a derivation of 𝑅 such that [[𝑑(𝑥),𝑥]𝑛,[[𝑦,𝑑(𝑦)]𝑚]𝑡=0 for all 𝑥,𝑦𝑅, where 𝑛0,𝑚0,𝑡1 are fixed integers. Then 𝑅 is commutative or 𝑑=0.

Proof. Let 𝑅 be noncommutative. If 𝑑 is not 𝑄-inner, then by Kharchenko's Theorem [9] 𝑔[](𝑥,𝑦,𝑢,𝑣)=𝑢,𝑥𝑛,[]𝑦,𝑣𝑚𝑡=0,(2.3) for all 𝑥,𝑦,𝑢,𝑣𝑅. This is a polynomial identity and hence there exists a field 𝐹 such that 𝑅𝑀𝑘(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 []0=𝑢,𝑥𝑛,[]𝑦,𝑣𝑚𝑡=(1)𝑡𝑛𝑒11+()𝑡𝑒22,(2.4) which is a contradiction.
Now, let 𝑑 be 𝑄-inner derivation, say 𝑑=𝑎𝑑(𝑎) for some 𝑎𝑄, that is, 𝑑(𝑥)=[𝑎,𝑥] for all 𝑥𝑅, then we have []𝑎,𝑥𝑛+1,[[𝑦,𝑎,𝑦]]𝑚𝑡=0,(2.5) for all 𝑥,𝑦𝑅. Since 𝑑0, 𝑎𝐶 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 𝑒𝑅𝐶. Moreover we may assume that 𝐻 is noncommutative; otherwise, 𝑅 must be commutative which is a contradiction.
Notice that 𝐻 satisfies [[𝑎,𝑥]𝑛+1,[𝑦,[𝑎,𝑦]]𝑚]𝑡=0 (see [10, Proof of Theorem 1]). For any idempotent 𝑒𝐻 and 𝑥𝐻, we have []0=𝑎,𝑒𝑛+1,[[𝑒𝑥(1𝑒),𝑎,𝑒𝑥(1𝑒)]]𝑚𝑡.(2.6) Right multiplying by 𝑒, we get []0=𝑎,𝑒𝑛+1,[[𝑒𝑥(1𝑒),𝑎,𝑒𝑥(1𝑒)]]𝑚𝑡𝑒=[]𝑎,𝑒𝑛+1,[[𝑒𝑥(1𝑒),𝑎,𝑒𝑥(1𝑒)]]𝑚𝑡1[]𝑎,𝑒𝑛+1[[𝑒𝑥(1𝑒),𝑎,𝑒𝑥(1𝑒)]]𝑚[[𝑒𝑒𝑥(1𝑒),𝑎,𝑒𝑥(1𝑒)]]𝑚[]𝑎,𝑒𝑛+1𝑒=[]𝑎,𝑒𝑛+1,[[𝑒𝑥(1𝑒),𝑎,𝑒𝑥(1𝑒)]]𝑚𝑡1[]𝑎,𝑒𝑛+1𝑚𝑗=0(1)𝑗𝑚𝑗[]𝑎,𝑒𝑥(1𝑒)𝑗[]𝑒𝑥(1𝑒)𝑎,𝑒𝑥(1𝑒)𝑚𝑗𝑒𝑚𝑗=0(1)𝑗𝑚𝑗[]𝑎,𝑒𝑥(1𝑒)𝑗[]𝑒𝑥(1𝑒)𝑎,𝑒𝑥(1𝑒)𝑚𝑗[]𝑎,𝑒𝑛+1𝑒=[]𝑎,𝑒𝑛+1,[[𝑒𝑥(1𝑒),𝑎,𝑒𝑥(1𝑒)]]𝑚𝑡10𝑚𝑗=0(1)𝑗𝑚𝑗(𝑒𝑥(1𝑒)𝑎)𝑗𝑒𝑥(1𝑒)(𝑎𝑒𝑥(1𝑒))𝑚𝑗[]𝑎𝑒=𝑎,𝑒𝑛+1,[[𝑒𝑥(1𝑒),𝑎,𝑒𝑥(1𝑒)]]𝑚𝑡1𝑚𝑗=0𝑚𝑗(𝑒𝑥(1𝑒)𝑎)𝑚+1𝑒=2𝑚[]𝑎,𝑒𝑛+1,[[𝑒𝑥(1𝑒),𝑎,𝑒𝑥(1𝑒)]]𝑚𝑡1(𝑒𝑥(1𝑒)𝑎)𝑚+1𝑒=()𝑡2𝑚𝑡(𝑒𝑥(1𝑒)𝑎)(𝑚+1)𝑡𝑒.(2.7)
This implies that 0=()𝑡2𝑚𝑡((1𝑒)𝑎𝑒𝑥)(𝑚+1)𝑡+1. Since char 𝑅2, ((1𝑒)𝑎𝑒𝑥)(𝑚+1)𝑡+1=0. By Levitzki's lemma [12, Lemma 1.1], (1𝑒)𝑎𝑒𝑥=0 for all 𝑥𝐻. Since 𝐻 is prime ring, (1𝑒)𝑎𝑒=0, that is, 𝑒𝑎𝑒=𝑎𝑒 for any idempotent 𝑒𝐻. Now replacing 𝑒 with 1𝑒, we get that 𝑒𝑎(1𝑒)=0, that is, 𝑒𝑎𝑒=𝑒𝑎. Therefore for any idempotent 𝑒𝐻, 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 𝑎𝐶 contradicting 𝑑0. In the last case, 𝐻 is a finite dimensional division algebra over 𝐶. This implies that 𝐻=𝑅𝐶=𝑄 and 𝑎𝐻. By [10,Lemma 2], there exists a field 𝐹 such that 𝐻𝑀𝑘(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 𝑎𝐶, that is, 𝑑=0. This completes the proof of the theorem.

Theorem 2.4. Let 𝑅 be a prime ring of char 𝑅2, 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,𝑦𝜌, where 𝑛0,𝑚0,𝑡1 are fixed integers. If [𝜌,𝜌]𝜌0, then 𝑑(𝜌)𝜌=0.

We begin the proof by proving the following lemma.

Lemma 2.5. If 𝑑(𝜌)𝜌0 and [[𝑑(𝑥),𝑥]𝑛,[𝑦,𝑑(𝑦)]𝑚]𝑡=0 for all 𝑥,𝑦𝜌,𝑚,𝑛0,𝑡1 are fixed integers, then 𝑅 satisfies nontrivial generalized polynomial identity (GPI).

Proof. Suppose on the contrary that 𝑅 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 that 𝑑 is 𝑄-inner derivation induced by an element 𝑎𝑄. Then for any 𝑥𝜌,[]𝑎,𝑥𝑋𝑛+1,[[𝑥𝑌,𝑎,𝑥𝑌]]𝑚𝑡(2.8) is a GPI for 𝑅, so it is the zero element in 𝑄𝐶𝐶{𝑋,𝑌}. Expanding this, we get []𝑎,𝑥𝑋𝑚𝑛+1𝑗=0(1)𝑗𝑚𝑗[]𝑎,𝑥𝑌𝑗[]𝑥𝑌𝑎,𝑥𝑌𝑚𝑗𝑚𝑗=0(1)𝑗𝑚𝑗[]𝑎,𝑥𝑌𝑗[]𝑥𝑌𝑎,𝑥𝑌𝑚𝑗[]𝑎,𝑥𝑋𝑛+1𝐴(𝑋,𝑌)=0,(2.9) where 𝐴(𝑋,𝑌)=[[𝑎,𝑥𝑋]𝑛+1,[𝑥𝑌,[𝑎,𝑥𝑌]]𝑚]𝑡1. If 𝑎𝑥 and 𝑥 are linearly 𝐶-independent for some 𝑥𝜌, then (𝑎𝑥𝑋)𝑚𝑛+1𝑗=0(1)𝑗𝑚𝑗[]𝑎,𝑥𝑌𝑗[]𝑥𝑌𝑎,𝑥𝑌𝑚𝑗𝑚𝑗=0(1)𝑗𝑚𝑗(𝑎𝑥𝑌)𝑗[]𝑥𝑌𝑎,𝑥𝑌𝑚𝑗[]𝑎,𝑥𝑋𝑛+1𝐴(𝑋,𝑌)=0.(2.10) Again, since 𝑎𝑥 and 𝑥 are linearly 𝐶-independent, above relation implies that []𝑥𝑌𝑎,𝑥𝑌𝑚[]𝑎,𝑥𝑋𝑛+1𝐴(𝑋,𝑌)=0,(2.11) and so 𝑥𝑌(𝑎𝑥𝑌)𝑚(𝑎𝑥𝑋)𝑛+1𝐴(𝑋,𝑌)=0.(2.12) Repeating the same process yields 𝑥𝑌(𝑎𝑥𝑌)𝑚(𝑎𝑥𝑋)𝑛+1𝑡=0(2.13) in 𝑄𝐶𝐶{𝑋,𝑌}. This implies that 𝑎𝑥=0, a contradiction. Thus for any 𝑥𝜌, 𝑎𝑥 and 𝑥 are 𝐶-dependent. Then (𝑎𝛼)𝜌=0 for some 𝛼𝐶. Replacing 𝑎 with 𝑎𝛼, we may assume that 𝑎𝜌=0. Then by Lemma 2.1, 𝑑(𝜌)𝜌=0, contradiction.
Case 2. Suppose that 𝑑 is not 𝑄-inner derivation. If for all 𝑥𝜌, 𝑑(𝑥)𝑥𝐶, then [𝑑(𝑥),𝑥]=0 which implies that 𝑅 is commutative (see [13]). Therefore there exists 𝑥𝜌 such that 𝑑(𝑥)𝑥𝐶, that is, 𝑥 and 𝑑(𝑥) are linearly 𝐶-independent.
By our assumption, we have that 𝑅 satisfies
[𝑑](𝑥𝑋),𝑥𝑋𝑛,[]𝑥𝑌,𝑑(𝑥𝑌)𝑚𝑡=0.(2.14) By Kharchenko's Theorem [9], 𝑑(𝑥)𝑋+𝑥𝑟1,𝑥𝑋𝑛,𝑥𝑌,𝑑(𝑥)𝑌+𝑥𝑟2𝑚𝑡=0,(2.15) for all 𝑋,𝑌,𝑟1,𝑟2𝑅. In particular for 𝑟1=𝑟2=0, [𝑑](𝑥)𝑋,𝑥𝑋𝑛,[]𝑥𝑌,𝑑(𝑥)𝑌𝑚𝑡=0,(2.16) 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 Theorem 2.4. Suppose that 𝑑(𝜌)𝜌0, 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 [𝑑](𝑥),𝑥𝑛,[]𝑦,𝑑(𝑦)𝑚𝑡=0(2.17) is satisfied by 𝜌𝑄 and hence by 𝜌𝐻. Let 𝑒=𝑒2𝜌𝐻 and 𝑦𝐻. Then replacing 𝑥 with 𝑒 and 𝑦 with 𝑒𝑦(1𝑒) in (2.17), then right multiplying it by 𝑒, we obtain that [𝑑]0=(𝑒),𝑒𝑛,[]𝑒𝑦(1𝑒),𝑑(𝑒𝑦(1𝑒))𝑚𝑡𝑒=[𝑑](𝑒),𝑒𝑛,[]𝑒𝑦(1𝑒),𝑑(𝑒𝑦(1𝑒))𝑚𝑡1[]𝑑(𝑒),𝑒𝑛𝑚𝑗=0(1)𝑗𝑚𝑗𝑑(𝑒𝑦(1𝑒))𝑗𝑒𝑦(1𝑒)𝑑(𝑒𝑦(1𝑒))𝑚𝑗𝑒𝑚𝑗=0(1)𝑗𝑚𝑗𝑑(𝑒𝑦(1𝑒))𝑗𝑒𝑦(1𝑒)𝑑(𝑒𝑦(1𝑒))𝑚𝑗[]𝑑(𝑒),𝑒𝑛𝑒.(2.18)
Now we have the fact that for any idempotent 𝑒, 𝑑(𝑦(1𝑒))𝑒=𝑦(1𝑒)𝑑(𝑒), 𝑒𝑑(𝑒)𝑒=0 and so []0=𝑑(𝑒),𝑒𝑛,[]𝑒𝑦(1𝑒),𝑑(𝑒𝑦(1𝑒))𝑚𝑡10𝑚𝑗=0(1)𝑗𝑚𝑗𝑒(𝑦(1𝑒)𝑑(𝑒))𝑗𝑦(1𝑒)𝑑(𝑒𝑦(1𝑒))𝑚𝑗.𝑑(𝑒)𝑒(2.19) Now since for any idempotent 𝑒 and for any 𝑦𝑅, (1𝑒)𝑑(𝑒𝑦)=(1𝑒)𝑑(𝑒)𝑦, above relation gives []0=𝑑(𝑒),𝑒𝑛,[]𝑒𝑦(1𝑒),𝑑(𝑒𝑦(1𝑒))𝑚𝑡1𝑒𝑚𝑗=0𝑚𝑗(𝑦(1𝑒)𝑑(𝑒))𝑗𝑦(1𝑒)(𝑑(𝑒)𝑦(1𝑒))𝑚𝑗=[]𝑑(𝑒)𝑒𝑑(𝑒),𝑒𝑛,[]𝑒𝑦(1𝑒),𝑑(𝑒𝑦(1𝑒))𝑚𝑡1𝑒𝑚𝑗=0𝑚𝑗(𝑦(1𝑒)𝑑(𝑒))𝑚+1𝑒=[]𝑑(𝑒),𝑒𝑛,[]𝑒𝑦(1𝑒),𝑑(𝑒𝑦(1𝑒))𝑚𝑡12𝑚𝑒(𝑦(1𝑒)𝑑(𝑒))𝑚+1𝑒=2𝑚𝑒(𝑦(1𝑒)𝑑(𝑒))𝑚+1𝑡𝑒.(2.20) This implies that 0=(1)𝑡2𝑚𝑡((1𝑒)𝑑(𝑒)𝑒𝑦)(𝑚+1)𝑡+1 for all 𝑦𝐻. Since char 𝑅2, we have by Levitzki's lemma [12,Lemma 1.1] that (1𝑒)𝑑(𝑒)𝑒𝑦=0 for all 𝑦𝐻. By primeness of 𝐻, (1𝑒)𝑑(𝑒)𝑒=0. By [15,Lemma 1], since 𝐻 is a regular ring, for each 𝑟𝜌𝐻, there exists an idempotent 𝑒𝜌𝐻 such that 𝑟=𝑒𝑟 and 𝑒𝑟𝐻. Hence (1𝑒)𝑑(𝑒)𝑒=0 gives (1𝑒)𝑑(𝑒)=(1𝑒)𝑑(𝑒2)=(1𝑒)𝑑(𝑒)𝑒=0 and so 𝑑(𝑒)=𝑒𝑑(𝑒)𝑒𝐻𝜌𝐻 and 𝑑(𝑟)=𝑑(𝑒𝑟)=𝑑(𝑒)𝑒𝑟+𝑒𝑑(𝑒𝑟)𝜌𝐻. Hence for each 𝑟𝜌𝐻, 𝑑(𝑟)𝜌𝐻. Thus 𝑑(𝜌𝐻)𝜌𝐻. Set 𝐽=𝜌𝐻. Then 𝐽=𝐽/(𝐽𝑙𝐻(𝐽)), a prime 𝐶-algebra with the derivation 𝑑 such that 𝑑(𝑥)=𝑑(𝑥), for all 𝑥𝐽. By assumption, we have that 𝑑𝑥,𝑥𝑛,𝑦,𝑑𝑦𝑚𝑡=0,(2.21) for all 𝑥,𝑦𝐽. 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 Theorem 2.3 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]). Let 𝑅 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, {𝑃𝑈𝑃isamaximalidealof𝐶with𝑈/𝑃𝑈2-torsionfree}=0.

Theorem 3.2. Let 𝑅 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,𝑦𝑅, 𝑛,𝑚0,𝑡1 fixed are integers. Then 𝑑 maps 𝑅 into its center.

Proof. Since any derivation 𝑑 can be uniquely extended to a derivation in 𝑈, and 𝑅 and 𝑈 satisfy the same differential identities [7, Theorem 3], we have [𝑑](𝑥),𝑥𝑛,[]𝑦,𝑑(𝑦)𝑚𝑡=0,(3.1) for all 𝑥,𝑦𝑈. 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 𝑥𝑈. Therefore, 𝑑𝑥,𝑥𝑛,𝑦,𝑑𝑦𝑚𝑡=0,(3.2) for all 𝑥,𝑦𝑈. 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, {𝑃𝑈𝑃isamaximalidealof𝐶with𝑈/𝑃𝑈2-torsionfree}=0. Thus 𝑑(𝑈)[𝑈,𝑈]=0. Without loss of generality, we have 𝑑(𝑅)[𝑅,𝑅]=0. This implies that 𝑅0=𝑑2[][][][].𝑅,𝑅=𝑑(𝑅)𝑅𝑅,𝑅+𝑅𝑑(𝑅)𝑅,𝑅=𝑑(𝑅)𝑅𝑅,𝑅(3.3) 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.

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Copyright © 2009 Basudeb Dhara. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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