#### Abstract

Let be a prime ring of char , a nonzero derivation of and a nonzero right ideal of such that for all , where , , are fixed integers. If , then .

#### 1. Introduction

Throughout this paper, unless specifically stated, always denotes a prime ring with center and extended centroid , the Martindale quotients ring. Let be a positive integer. For given , let and let be the usual commutator and inductively for , . By we mean a nonzero derivation in .

A well-known result proven by Posner [1] states that if for all , then is commutative. In [2], Lanski generalized this result of Posner to the Lie ideal. Lanski proved that if is a noncommutative Lie ideal of such that for all , then either is commutative or char and satisfies , the standard identity in four variables. Bell and Martindale III [3] studied this identity for a semiprime ring . They proved that if is a semiprime ring and for all in a non-zero left ideal of and , then contains a non-zero central ideal. Clearly, this result says that if is a prime ring, then 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 and for all , then is commutative. In [5], Filippis showed that if is of characteristic different from and a non-zero right ideal of such that and for all , then .

In continuation of these previous results, it is natural to consider the situation when for all , are fixed integers. We have studied this identity in the present paper.

It is well known that any derivation of a prime ring can be uniquely extended to a derivation of , and so any derivation of can be defined on the whole of . Moreover is a prime ring as well as and the extended centroid of coincides with the center of . We refer to [6, 7] for more details.

Denote by the free product of the -algebra and , the free -algebra in noncommuting indeterminates .

#### 2. The Case: Prime Ring

We need the following lemma.

Lemma 2.1. *Let be a non-zero right ideal of and a derivation of . Then the following conditions are equivalent: (i) is an inner derivation induced by some such that ; (ii) (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 be a prime ring, a derivation on and a non-zero ideal of . If satisfies the differential identity then either (i) satisfies the generalized polynomial identity
**
or (ii) is -inner, that is, for some and satisfies the generalized polynomial identity
*

Theorem 2.3. * Let be a prime ring of char and a derivation of such that for all , where are fixed integers. Then is commutative or .*

*Proof. *Let be noncommutative. If is not -inner, then by Kharchenko's Theorem [9]
for all . This is a polynomial identity and hence there exists a field such that with and and satisfy the same polynomial identity [10,Lemma ]. But by choosing and , we get
which is a contradiction.

Now, let be -inner derivation, say for some , that is, for all , then we have
for all . Since , and hence satisfies a nontrivial generalized polynomial identity (GPI). By [11], it follows that is a primitive ring with and is finite dimensional over for any minimal idempotent . Moreover we may assume that is noncommutative; otherwise, must be commutative which is a contradiction.

Notice that satisfies (see [10, Proof of Theorem ]). For any idempotent and we have
Right multiplying by , we get

This implies that . Since char , . By Levitzki's lemma [12, Lemma ], for all . Since is prime ring, , that is, for any idempotent . Now replacing with , we get that that is, . Therefore for any idempotent , we have . So commutes with all idempotents in . Since is a simple ring, either is generated by its idempotents or does not contain any nontrivial idempotents. The first case gives contradicting . In the last case, is a finite dimensional division algebra over . This implies that and . By [10,Lemma ], there exists a field such that and satisfies . Then by the same argument as earlier, commutes with all idempotents in , again giving the contradiction , that is, . This completes the proof of the theorem.

Theorem 2.4. * Let be a prime ring of char , a non-zero derivation of and a non-zero right ideal of such that for all , where are fixed integers. If , then .*

We begin the proof by proving the following lemma.

Lemma 2.5. * If and for all are fixed integers, then satisfies nontrivial generalized polynomial identity (GPI).*

*Proof. *Suppose on the contrary that does not satisfy any nontrivial GPI. We may assume that is noncommutative; otherwise, satisfies trivially a nontrivial GPI. We consider two cases.*Case 1. *Suppose that is -inner derivation induced by an element . Then for any
is a GPI for , so it is the zero element in . Expanding this, we get
where . If and are linearly -independent for some then
Again, since and are linearly -independent, above relation implies that
and so
Repeating the same process yields
in . This implies that , a contradiction. Thus for any , and are -dependent. Then for some . Replacing with , we may assume that . Then by Lemma 2.1, , contradiction.*Case 2. *Suppose that is not -inner derivation. If for all , , then which implies that is commutative (see [13]). Therefore there exists such that , that is, and are linearly -independent.

By our assumption, we have that satisfies

By Kharchenko's Theorem [9],
for all . In particular for ,
which is a nontrivial GPI for , because and are linearly -independent, a contradiction.

We are now ready to prove our main theorem.

*Proof of Theorem 2.4. *Suppose that then we derive a contradiction. By Lemma 2.5, is a prime GPI ring, so is also by [14]. Since is centrally closed over , it follows from [11] that is a primitive ring with .

By our assumption and by [7], we may assume that
is satisfied by and hence by . Let and . Then replacing with and with in (2.17), then right multiplying it by we obtain that

Now we have the fact that for any idempotent , , and so
Now since for any idempotent and for any , , above relation gives
This implies that for all . Since char , we have by Levitzki's lemma [12,Lemma ] that for all . By primeness of , . By [15,Lemma ], since is a regular ring, for each , there exists an idempotent such that and . Hence gives and so and . Hence for each , . Thus . Set . Then a prime -algebra with the derivation such that , for all . By assumption, we have that
for all . By Theorem 2.3, we have either or is commutative. Therefore we have that either or . Now implies that and so . implies that and so then implying that . Thus in all the cases we have contradiction. This completes the proof of the theorem.

#### 3. The Case: Semiprime Ring

In this section we extend Theorem 2.3 to the semiprime case. Let be a semiprime ring and be its right Utumi quotient ring. It is well known that any derivation of a semiprime ring can be uniquely extended to a derivation of its right Utumi quotient ring and so any derivation of can be defined on the whole of [7,Lemma ].

By the standard theory of orthogonal completions for semiprime rings, we have the following lemma.

Lemma 3.1 (see [16, Lemma and Theorem ] or [7,pages 31-32]). *Let be a -torsion free semiprime ring and a maximal ideal of . Then is a prime ideal of invariant under all derivations of . Moreover, .*

Theorem 3.2. * Let be a -torsion free semiprime ring and a non-zero derivation of such that for all , fixed are integers. Then maps into its center.*

*Proof. *Since any derivation can be uniquely extended to a derivation in and and satisfy the same differential identities [7, Theorem ], we have
for all . Let be any maximal ideal of such that is -torsion free. Then by Lemma 3.1, is a prime ideal of invariant under . Set Then derivation canonically induces a derivation on defined by for all . Therefore,
for all . By Theorem 2.3, either or , that is, or . In any case for any maximal ideal of . By Lemma 3.1, . Thus . Without loss of generality, we have . This implies that
Therefore . By semiprimeness of , we have , that is, . This completes the proof of the theorem.