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.