Abstract

Let be a ring with involution having a nontrivial symmetric idempotent element . If is any appropriate multiplicative generalized reverse CE-derivation of with involution , then under some suitable restrictions on , is centrally-extended additive.

1. Introduction

In [1], Bell and Daif introduced the notion of centrally-extended derivations as follows. Let be a ring with center , a map of is called a centrally-extended derivation (CE-derivation) if for each and . They discussed the existence of such map which is not a derivation and gave some commutativity results. In [2], the authors generalized this notion to other kinds of maps and extended some results due to Bell and Daif. Recently, in [3], the authors gave the notion of Jordan CE-derivations and, under some conditions, they proved that every Jordan CE-derivation of a prime ring is a CE-derivation.

Martindale [4] has asked the following question: When is a multiplicative mapping additive? He answered his question for a multiplicative isomorphism of a ring . In [5], Daif has given an answer to that question when the mapping is a multiplicative derivation on . Also, in [6–8], a generalization of this question can be found for the case of multiplicative generalized derivations, multiplicative generalized reverse derivations, and multiplicative left centralizers.

In this article, we generalized the idea of Martindale [4] and Daif [5] for the notion of the multiplicative generalized reverse CE-derivation.

2. Preliminaries

In this note, we introduce the notion of the multiplicative generalized reverse CE-derivation of a ring with involution to be a mapping of into such that , for all , where is a reverse CE-derivation from into ; i.e., for all and . In other words, we can write the maps and by and , where and are central elements depend on the choice of and and related to the mappings and , respectively.

Here, we ask the following question: When is a multiplicative generalized reverse CE-derivation a CE-additive? Under suitable conditions, we give an answer for this question.

As in [9], let be a nontrivial symmetric idempotent element so that (S need not have an identity). We will formally set and The two-sided Peirce decomposition of relative to the idempotents and takes the form . So letting : , we may write . An element of the subring will be denoted by . If since , then , so we conclude that Also, we formally use the symbol for referring to the subring

By the definition of , we note that . However, since is bijection and, for all , So that Then, we have which is a two sided central ideal in . Since if , then and this gives also Similarly, is a two sided central ideal in and and

Moreover, . If we express and use the two expressions of , we get and . Consequently, we have the following equation:

By the same manner, if is a multiplicative generalized reverse CE-derivation associated with a reverse CE-derivation ,then , where and we can write and using the values of and . we conclude that and so,

In our work we will need the following facts.

Proposition 1 [7]. Let . Then, where Moreover,

Lemma 2. and where .

Proof. For any element by expanding both sides of we get the following equation:Now using equation (1) in equation (3), we get and since this meansNow, Since is bijection, there exist such that , so we can rewrite and using that we get and which implies and And again equation (4) gives and , which gives and this means that is a left and right annihilator of the two subrings and Now, for any which gives Since , then Also, since , we get and
To achieve our main result, we assume that the ring endowed with an involution contains a nontrivial symmetric idempotent and satisfies the following conditions:  implies And implies   implies And is any multiplicative generalized reverse CE-derivation of associated with a reverse CE-derivation of
The following lemma is fruitful in our proofs:

Lemma 3. The ideals , and are central ideals in S, where , , and .

Proof. First, using Lemma 2, for any , we get and using condition , we get Secondly, assume that and since so using equation (2), we have and this gives and , where Now, using Lemma 2, for any , we get and this gives Also, if then By a similar method one can prove the other cases.

Remark 4. An example of a reverse CE-derivation, if is any fixed element in , the map which satisfies where is a central ideal, we can call it an inner reverse CE-derivation. Now, using Lemma 3 we can show that the map given by is a reverse CE-derivation and with equation (1), we get the following equation:

Remark 5. An example of a generalized reverse CE-derivation is if and are any two fixed elements in , the map which satisfies where is a central ideal, we can call it an inner generalized reverse CE-derivation associated with the inner reverse CE-derivation which is given by .
Again, using Lemma 3, we can show that the map given by is a generalized reverse CE-derivation associated with the inner reverse CE-derivation and with equation (2) we get the following equation:

Remark 6. For simplification, we will replace, without loss of generality, the reverse CE-derivation by the reverse CE-derivation which by using equation (5) bring us to and the multiplicative generalized reverse CE-derivation by the multiplicative generalized reverse CE-derivation with by equation (6). Also, and . One can easily show that both of and generates a two sided central ideal in .
To prove our main theorem we need the following lemmas.

Lemma 7. For any element there exists and such that(1)(2)

Proof. For , we have to prove two separable cases:(a)Let be an arbitrary element of and let . Then, which gives and , so we get Similarly, which means and we get (b)Assume that write so , so which means and Likewise, , so , so that and thus , where Also, For , we have to prove two separable cases:(a)Assume that , so that Also, we have which gives Comparing between the two values of , we get and having and we get Now, , hence and this gives which means and So, we arrive to (b)Assume that , so that Also, we have which gives Comparing the two expressions of , we get , and we get Now, , hence which means and we have

Lemma 8. For any element , we have for some and

Proof. Since , for every and it follows that for every , we have because and by Lemma 7 and , so we have that . Now, assume that , then which gives We conclude that and with and as required.