Chinese Journal of MathematicsVolume 2013 (2013), Article ID 752928, 4 pageshttp://dx.doi.org/10.1155/2013/752928

Research Article

On Generalized Derivations in Nearrings

Mohd Rais Khan and Mohd Mueenul Hasnain

Department of Mathematics, Jamia Millia Islamia, New Delhi 110 025, India

Received 14 July 2013; Accepted 10 August 2013

Academic Editors: Z. Shi and Z. Wang

Copyright © 2013 Mohd Rais Khan and Mohd Mueenul Hasnain. 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.

Abstract.

The aim of this paper is to investigate some results of nearrings satisfying certain identities involving generalized derivations. Furthermore, we give some examples to demonstrate the restrictions imposed on the hypothesis of various results which are not superfluous.

1. Introduction

The study of derivations in rings goes back to 1957 when Posner [1] proved that the existence of a nonzero centralizing derivation on a prime ring forces the ring to be commutative. Many results in this vein were obtained by a number of authors [2–18] in several ways. In view of [19], the concept of generalized derivation is introduced by Hvala [20]. Familiar examples of generalized derivations are derivations and generalized inner derivations, and later includes left multiplier, that is, an additive mapping satisfying for all . Since the sum of two generalized derivations is a generalized derivation, every map of the form , where is fixed element of and a derivation of , is a generalized derivation; and if has , all generalized derivations have this form.

Throughout the paper, will denote a zero symmetric right abelian nearring with multiplicative center . For all , as usual and will denote the well-known Lie and Jordan products, respectively. A nonempty subset of will be called a semigroup right ideal (resp., semigroup left ideal) if (). Finally, is called a semigroup ideal if it is a right as well as a left semigroup ideal. A nearring is called prime, if or for all . We refer to Pilz [21] for the basics definitions and properties of nearring.

As noted in [22], an additive mapping is called a derivation of if holds for all . An additive mapping is said to be a right generalized derivation associated with if and is said to be a left generalized derivation associated with if is said to be a generalized derivation associated with if it is a right as well as a left generalized derivation associated with .

The purpose of this note is to prove some results which are of independent interest and related to generalized derivations on nearrings.

2. Ideals and Generalized Derivations in Nearrings

Over the last several years, many authors [7, 19, 20, 23] studied the commutativity in prime and semiprime rings admitting derivations and generalized derivations. On other hand, there are several results asserting that prime nearrings with certain constrained derivations have ring-like behavior. It is natural to look for comparable results on nearrings, and this has been done [22, 24–26]. In this section, we investigate some results of nearrings satisfying certain identities involving generalized derivation.

In order to prove our theorems, we will make extensive use of the following lemma.

Lemma A. If is prime and a generalized derivation on associated with of , then

Proof. Clearly , and, also, we obtain . Comparing these two expressions for gives the desired conclusion.

Lemma B (see [25, Lemma 1.5]). Let be a prime nearring and let be a semigroup ideal of . If , then is commutative.

Lemma C (see [25, Lemma 1.4]). Let be a prime nearring and let be a semigroup ideal of . If is an element of such that or , then .

Lemma D. Let be a prime nearring and a semigroup ideal of . If is a derivation on such that , then .

Proof. From hypothesis, we get That is, Thus, we then concluded the required result by Lemma  C.

Theorem 1. Let be a noncommutative prime nearring, a nonzero semigroup ideal of , and a generalized derivation associated with of such that , for all . Then is trivial.

Proof. From hypothesis, we have Replacing with in (6) and using it, we get Again replacing with in (7) and using it, we obtain that is, It follows from Lemma  C that either or , for all , . Therefore, in view of hypothesis and from Lemma  B, we are forced to consider later case and so by Lemma . Hence, our hypothesis becomes Let , for all , and so , for all . Then, the last equality can be rewritten as Taking instead of in (11) and using Lemma  A, we find Replacing with in the last equality, we obtain It implies that Thus, we then concluded, by the primeness of , that either or for all , , that is,  . , then we conclude that is commutative by Lemma  B, which is a contradiction. Hence, this completes the proof.

A slight modification in the proof of Theorem yields the following.

Theorem 2. Let be a noncommutative prime nearring, a nonzero semigroup ideal of , and admit a generalized derivation associated with such that , for all . Then is trivial.

As a consequence of above theorems, we obtain the following remarks.

Remark 3. Suppose is a prime nearring and a nonzero semigroup ideal of . If admits a generalized derivation associated with such that , for all or , for all , then is commutative or is trivial.

Remark 4. Suppose is a prime nearring. If admits a generalized derivation associated with of such that , for all or , for all , then is commutative.

Proof. If , then we have the desired conclusion. Now, we consider , and following the same technique as in the proof of Theorem 1, we reach , for all . Taking instead of in the last relation, we obtain , for all . Since is prime, we obtain the required result by hypothesis.Similar argument can be adapted in the case for all , and we can omit the similar proof.

Here, we try to construct an example to demonstrates that the above result do not hold for arbitrary rings.

Example 5. Let , where is a commutative ring and . We define a map by ; then it is easy to check that is a generalized derivation associated with , where define as on . However, satisfies the properties of Theorems 1 and 2 and Remarks 3 and 4, but neither is commutative nor is trivial.

Remark 6. In Remark 4, the hypothesis of primeness may be weakened by assuming that is not a right as well left zero divisor of , where is a nearring. Then the same proof will lead to the conclusion that is commutative.

Conclusion of Theorems 1 and 2 is still hold if we replace the product by . In fact, we obtain the following results.

Theorem 7. Let be a noncommutative prime nearring, a nonzero semigroup ideal of , and admit a generalized derivation associated with such that , for all . Then is trivial.

Proof. From hypothesis, we have Replacing by in (15) and using it, we obtain that Replacing by in the last expression and using it, we reach for all , , that is, Equation (17) is the same as (9) in Theorem 1. Thus, by same argument of Theorem 1, we can conclude the result here.

Proceeding the same line as above with necessary variation, we can prove the following.

Theorem 8. Let be a noncommutative prime nearring, a nonzero semigroup ideal of , and admit a generalized derivation associated with such that , for all . Then is trivial.

As a consequence of above theorems, we obtain the following remarks.

Remark 9. Suppose is a prime nearring and a nonzero semigroup ideal of . If admits a generalized derivation associated with such that , for all or , for all , then is commutative or is trivial.

Remark 10. Suppose is a prime nearring. If admits a generalized derivation associated with such that , for all or , for all , then is commutative.

Proof. For any , we have , and following the same technique as in proof of Theorem 7, we reach , for all . Taking instead of in last relation and using it, we obtain , for all . This implies that , for all . Due to hypothesis and primeness of , we get the required result.Similar results hold in case , for all .

The following example demonstrates that the above results do not hold for arbitrary rings.

Example 11. Let and a nonzero ideal of , where is a noncommutative ring with condition , for all . We define a map by . Then, it is easy to check that is a generalized derivation associated with , where is defined as on . However, satisfies the properties of Theorems 7 and 8 and Remarks 9 and 10, but neither is commutative nor is trivial.

Remark 12. In Remark 10, the hypothesis of primeness may be weakened by assuming that is not a right as well left zero divisor of , where is a nearring. Then, the same proof will lead to the conclusion that is commutative.

In conclusion of our paper, it would be interesting to prove or disprove the following problem.

Problem 13. Let be a fixed positive number. Suppose is a prime nearring. Let be a nonzero ideal of and let admit a generalized derivation associated with a derivation . (i)Does the condition that , for all or , for all imply that is commutative?(ii)Does the condition that , for all or , for all imply that is commutative?

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