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 | 8 pages | 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 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.

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 ğ‘Ž,𝑏∈𝑅, let [ğ‘Ž,𝑏]0=ğ‘Ž and let [ğ‘Ž,𝑏]1 be the usual commutator ğ‘Žğ‘âˆ’ğ‘ğ‘Ž, and inductively for 𝑛>1, [ğ‘Ž,𝑏]𝑛=[[ğ‘Ž,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 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 𝑄∗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. 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 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]). 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,…,𝑟𝑛∈𝐼, then either (i) 𝐼 satisfies the generalized polynomial identity 𝑓𝑟1,𝑟2,…,𝑟𝑛,𝑥1,𝑥2,…,𝑥𝑛=0,(2.1) or (ii) 𝑑 is 𝑄-inner, that is, for some ğ‘žâˆˆğ‘„,𝑑(𝑥)=[ğ‘ž,𝑥] and 𝐼 satisfies the generalized polynomial identity 𝑓𝑟1,𝑟2,…,𝑟𝑛,î€ºğ‘ž,𝑟1,î€ºğ‘ž,𝑟2,…,ğ‘ž,𝑟𝑛=0.(2.2)

Theorem 2.3. Let 𝑅 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. Let 𝑅 be noncommutative. If 𝑑 is not 𝑄-inner, then by Kharchenko's Theorem [9] 𝑔[](𝑥,𝑦,𝑢,𝑣)=𝑢,𝑥𝑛,[]𝑦,𝑣𝑚𝑡=0,(2.3) 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 []0=𝑢,𝑥𝑛,[]𝑦,𝑣𝑚𝑡=(−1)𝑡𝑛𝑒11+(−)𝑡𝑒22,(2.4) which is a contradiction.
Now, let 𝑑 be 𝑄-inner derivation, say 𝑑=ğ‘Žğ‘‘(ğ‘Ž) for some ğ‘Žâˆˆğ‘„, that is, 𝑑(𝑥)=[ğ‘Ž,𝑥] for all 𝑥∈𝑅, then we have []ğ‘Ž,𝑥𝑛+1,[[𝑦,ğ‘Ž,𝑦]]𝑚𝑡=0,(2.5) for all 𝑥,𝑦∈𝑅. Since 𝑑≠0, ğ‘Žâˆ‰ğ¶ 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 that 𝐻 satisfies [[ğ‘Ž,𝑥]𝑛+1,[𝑦,[ğ‘Ž,𝑦]]𝑚]𝑡=0 (see [10, Proof of Theorem 1]). For any idempotent 𝑒∈𝐻 and 𝑥∈𝐻, we have []0=ğ‘Ž,𝑒𝑛+1,[[𝑒𝑥(1−𝑒),ğ‘Ž,𝑒𝑥(1−𝑒)]]𝑚𝑡.(2.6) Right multiplying by 𝑒, we get []0=ğ‘Ž,𝑒𝑛+1,[[𝑒𝑥(1−𝑒),ğ‘Ž,𝑒𝑥(1−𝑒)]]𝑚𝑡𝑒=[]ğ‘Ž,𝑒𝑛+1,[[𝑒𝑥(1−𝑒),ğ‘Ž,𝑒𝑥(1−𝑒)]]𝑚𝑡−1⋅[]ğ‘Ž,𝑒𝑛+1[[𝑒𝑥(1−𝑒),ğ‘Ž,𝑒𝑥(1−𝑒)]]𝑚[[𝑒−𝑒𝑥(1−𝑒),ğ‘Ž,𝑒𝑥(1−𝑒)]]𝑚[]ğ‘Ž,𝑒𝑛+1𝑒=[]ğ‘Ž,𝑒𝑛+1,[[𝑒𝑥(1−𝑒),ğ‘Ž,𝑒𝑥(1−𝑒)]]𝑚𝑡−1⋅[]ğ‘Ž,𝑒𝑛+1𝑚𝑗=0(−1)𝑗𝑚𝑗[]ğ‘Ž,𝑒𝑥(1−𝑒)𝑗[]𝑒𝑥(1−𝑒)ğ‘Ž,𝑒𝑥(1−𝑒)𝑚−𝑗𝑒−𝑚𝑗=0(−1)𝑗𝑚𝑗[]ğ‘Ž,𝑒𝑥(1−𝑒)𝑗[]𝑒𝑥(1−𝑒)ğ‘Ž,𝑒𝑥(1−𝑒)𝑚−𝑗[]ğ‘Ž,𝑒𝑛+1𝑒=[]ğ‘Ž,𝑒𝑛+1,[[𝑒𝑥(1−𝑒),ğ‘Ž,𝑒𝑥(1−𝑒)]]𝑚𝑡−1⋅0−𝑚𝑗=0(−1)𝑗𝑚𝑗(−𝑒𝑥(1−𝑒)ğ‘Ž)𝑗𝑒𝑥(1−𝑒)(ğ‘Žğ‘’ğ‘¥(1−𝑒))𝑚−𝑗[]ğ‘Žğ‘’=âˆ’ğ‘Ž,𝑒𝑛+1,[[𝑒𝑥(1−𝑒),ğ‘Ž,𝑒𝑥(1−𝑒)]]𝑚𝑡−1𝑚𝑗=0𝑚𝑗(𝑒𝑥(1−𝑒)ğ‘Ž)𝑚+1𝑒=−2𝑚[]ğ‘Ž,𝑒𝑛+1,[[𝑒𝑥(1−𝑒),ğ‘Ž,𝑒𝑥(1−𝑒)]]𝑚𝑡−1(𝑒𝑥(1−𝑒)ğ‘Ž)𝑚+1𝑒=(−)𝑡2𝑚𝑡(𝑒𝑥(1−𝑒)ğ‘Ž)(𝑚+1)𝑡𝑒.(2.7)
This implies that 0=(−)𝑡2𝑚𝑡((1−𝑒)ğ‘Žğ‘’ğ‘¥)(𝑚+1)𝑡+1. Since char 𝑅≠2, ((1−𝑒)ğ‘Žğ‘’ğ‘¥)(𝑚+1)𝑡+1=0. By Levitzki's lemma [12, Lemma 1.1], (1−𝑒)ğ‘Žğ‘’ğ‘¥=0 for all 𝑥∈𝐻. Since 𝐻 is prime ring, (1−𝑒)ğ‘Žğ‘’=0, that is, ğ‘’ğ‘Žğ‘’=ğ‘Žğ‘’ for any idempotent 𝑒∈𝐻. Now replacing 𝑒 with 1−𝑒, we get that ğ‘’ğ‘Ž(1−𝑒)=0, that is, ğ‘’ğ‘Žğ‘’=ğ‘’ğ‘Ž. Therefore for any idempotent 𝑒∈𝐻, we have [ğ‘Ž,𝑒]=0. So ğ‘Ž 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 ğ‘Žâˆˆğ¶ contradicting 𝑑≠0. In the last case, 𝐻 is a finite dimensional division algebra over 𝐶. This implies that 𝐻=𝑅𝐶=𝑄 and ğ‘Žâˆˆğ». By [10,Lemma 2], there exists a field 𝐹 such that 𝐻⊆𝑀𝑘(𝐹) and 𝑀𝑘(𝐹) satisfies [[ğ‘Ž,𝑥]𝑛+1,[𝑦,[ğ‘Ž,𝑦]]𝑚]𝑡. Then by the same argument as earlier, ğ‘Ž commutes with all idempotents in 𝑀𝑘(𝐹), again giving the contradiction ğ‘Žâˆˆğ¶, that is, 𝑑=0. This completes the proof of the theorem.

Theorem 2.4. Let 𝑅 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. If 𝑑(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 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 ğ‘Žâˆˆğ‘„. Then for any 𝑥∈𝜌,[]ğ‘Ž,𝑥𝑋𝑛+1,[[𝑥𝑌,ğ‘Ž,𝑥𝑌]]𝑚𝑡(2.8) is a GPI for 𝑅, so it is the zero element in 𝑄∗𝐶𝐶{𝑋,𝑌}. Expanding this, we get []ğ‘Ž,𝑥𝑋𝑚𝑛+1𝑗=0(−1)𝑗𝑚𝑗[]ğ‘Ž,𝑥𝑌𝑗[]ğ‘¥ğ‘Œğ‘Ž,𝑥𝑌𝑚−𝑗−𝑚𝑗=0(−1)𝑗𝑚𝑗[]ğ‘Ž,𝑥𝑌𝑗[]ğ‘¥ğ‘Œğ‘Ž,𝑥𝑌𝑚−𝑗[]ğ‘Ž,𝑥𝑋𝑛+1𝐴(𝑋,𝑌)=0,(2.9) where 𝐴(𝑋,𝑌)=[[ğ‘Ž,𝑥𝑋]𝑛+1,[𝑥𝑌,[ğ‘Ž,𝑥𝑌]]𝑚]𝑡−1. If ğ‘Žğ‘¥ and 𝑥 are linearly 𝐶-independent for some 𝑥∈𝜌, then (ğ‘Žğ‘¥ğ‘‹)𝑚𝑛+1𝑗=0(−1)𝑗𝑚𝑗[]ğ‘Ž,𝑥𝑌𝑗[]ğ‘¥ğ‘Œğ‘Ž,𝑥𝑌𝑚−𝑗−𝑚𝑗=0(−1)𝑗𝑚𝑗(ğ‘Žğ‘¥ğ‘Œ)𝑗[]ğ‘¥ğ‘Œğ‘Ž,𝑥𝑌𝑚−𝑗[]ğ‘Ž,𝑥𝑋𝑛+1𝐴(𝑋,𝑌)=0.(2.10) Again, since ğ‘Žğ‘¥ and 𝑥 are linearly 𝐶-independent, above relation implies that []âˆ’ğ‘¥ğ‘Œğ‘Ž,𝑥𝑌𝑚[]ğ‘Ž,𝑥𝑋𝑛+1𝐴(𝑋,𝑌)=0,(2.11) and so −𝑥𝑌(ğ‘Žğ‘¥ğ‘Œ)𝑚(ğ‘Žğ‘¥ğ‘‹)𝑛+1𝐴(𝑋,𝑌)=0.(2.12) Repeating the same process yields −𝑥𝑌(ğ‘Žğ‘¥ğ‘Œ)𝑚(ğ‘Žğ‘¥ğ‘‹)𝑛+1𝑡=0(2.13) in 𝑄∗𝐶𝐶{𝑋,𝑌}. This implies that ğ‘Žğ‘¥=0, a contradiction. Thus for any 𝑥∈𝜌, ğ‘Žğ‘¥ and 𝑥 are 𝐶-dependent. Then (ğ‘Žâˆ’ğ›¼)𝜌=0 for some 𝛼∈𝐶. Replacing ğ‘Ž with ğ‘Žâˆ’ğ›¼, we may assume that ğ‘ŽğœŒ=0. Then by Lemma 2.1, 𝑑(𝜌)𝜌=0, contradiction.
Case 2. Suppose that 𝑑 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 that 𝑅 satisfies
[𝑑](𝑥𝑋),𝑥𝑋𝑛,[]𝑥𝑌,𝑑(𝑥𝑌)𝑚𝑡=0.(2.14) By Kharchenko's Theorem [9], 𝑑(𝑥)𝑋+𝑥𝑟1,𝑥𝑋𝑛,𝑥𝑌,𝑑(𝑥)𝑌+𝑥𝑟2𝑚𝑡=0,(2.15) for all 𝑋,𝑌,𝑟1,𝑟2∈𝑅. In particular for 𝑟1=𝑟2=0, [𝑑](𝑥)𝑋,𝑥𝑋𝑛,[]𝑥𝑌,𝑑(𝑥)𝑌𝑚𝑡=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)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 [𝑑](𝑥),𝑥𝑛,[]𝑦,𝑑(𝑦)𝑚𝑡=0(2.17) 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 [𝑑]0=(𝑒),𝑒𝑛,[]𝑒𝑦(1−𝑒),𝑑(𝑒𝑦(1−𝑒))𝑚𝑡𝑒=[𝑑](𝑒),𝑒𝑛,[]𝑒𝑦(1−𝑒),𝑑(𝑒𝑦(1−𝑒))𝑚𝑡−1⋅[]𝑑(𝑒),𝑒𝑛𝑚𝑗=0(−1)𝑗𝑚𝑗𝑑(𝑒𝑦(1−𝑒))𝑗𝑒𝑦(1−𝑒)𝑑(𝑒𝑦(1−𝑒))𝑚−𝑗𝑒−𝑚𝑗=0(−1)𝑗𝑚𝑗𝑑(𝑒𝑦(1−𝑒))𝑗𝑒𝑦(1−𝑒)𝑑(𝑒𝑦(1−𝑒))𝑚−𝑗[]𝑑(𝑒),𝑒𝑛𝑒.(2.18)
Now we have the fact that for any idempotent 𝑒, 𝑑(𝑦(1−𝑒))𝑒=−𝑦(1−𝑒)𝑑(𝑒), 𝑒𝑑(𝑒)𝑒=0 and so []0=𝑑(𝑒),𝑒𝑛,[]𝑒𝑦(1−𝑒),𝑑(𝑒𝑦(1−𝑒))𝑚𝑡−1⋅0−𝑚𝑗=0(−1)𝑗𝑚𝑗𝑒(−𝑦(1−𝑒)𝑑(𝑒))𝑗𝑦(1−𝑒)𝑑(𝑒𝑦(1−𝑒))𝑚−𝑗.𝑑(𝑒)𝑒(2.19) Now since for any idempotent 𝑒 and for any 𝑦∈𝑅, (1−𝑒)𝑑(𝑒𝑦)=(1−𝑒)𝑑(𝑒)𝑦, above relation gives []0=𝑑(𝑒),𝑒𝑛,[]𝑒𝑦(1−𝑒),𝑑(𝑒𝑦(1−𝑒))𝑚𝑡−1⋅−𝑒𝑚𝑗=0𝑚𝑗(𝑦(1−𝑒)𝑑(𝑒))𝑗𝑦(1−𝑒)(𝑑(𝑒)𝑦(1−𝑒))𝑚−𝑗=[]𝑑(𝑒)𝑒𝑑(𝑒),𝑒𝑛,[]𝑒𝑦(1−𝑒),𝑑(𝑒𝑦(1−𝑒))𝑚𝑡−1−𝑒𝑚𝑗=0𝑚𝑗(𝑦(1−𝑒)𝑑(𝑒))𝑚+1𝑒=[]𝑑(𝑒),𝑒𝑛,[]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.(2.20) 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 𝑑𝑥,𝑥𝑛,𝑦,𝑑𝑦𝑚𝑡=0,(2.21) 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 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, ⋂{𝑃𝑈∣𝑃isamaximalidealof𝐶with𝑈/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,𝑦∈𝑅, 𝑛,𝑚≥0,𝑡≥1 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)𝑚𝑡=0,(3.1) 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, 𝑑𝑥,𝑥𝑛,𝑦,𝑑𝑦𝑚𝑡=0,(3.2) 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 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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