Research Article | Open Access

Volume 2014 |Article ID 140790 | https://doi.org/10.1155/2014/140790

H. E. Bell, M. N. Daif, "On Maps of Period 2 on Prime and Semiprime Rings", International Journal of Mathematics and Mathematical Sciences, vol. 2014, Article ID 140790, 4 pages, 2014. https://doi.org/10.1155/2014/140790

# On Maps of Period 2 on Prime and Semiprime Rings

Academic Editor: Ram N. Mohapatra
Received29 Mar 2014
Accepted16 Aug 2014
Published22 Sep 2014

#### Abstract

A map f of the ring R into itself is of period 2 if for all ; involutions are much studied examples. We present some commutativity results for semiprime and prime rings with involution, and we study the existence of derivations and generalized derivations of period 2 on prime and semiprime rings.

#### 1. Introduction

Let be a ring with center , and for each , let denote the commutator . Note that and for all —facts we will use repeatedly.

Let be a nonempty subset of . A map is said to be of period 2 on if for all , and is called an -subset if . If for all , then is said to be commuting on ; if for all , then, as in , is called strong commutativity-preserving on .

We assume the reader is familiar with the definitions of derivation and involution. We define an additive map to be a right (resp. left) generalized derivation on if (resp., ) for all , where is a derivation on , called the associated derivation. If  is both a right generalized derivation and a left generalized derivation with the same associated derivation, we call a generalized derivation. (Note that this definition is different from that of Hvala in ; his generalized derivations are our right generalized derivations.)

Our purpose is to study existence and properties of involutions, derivations, and generalized derivations of period 2 on certain subsets of semiprime and prime rings.

#### 2. Two Commutativity Results for Rings with Involution

There are several known commutativity results for rings with involution (cf. [3, Chapter 3]). We now present a result showing the equivalence of two commutativity conditions on a -ideal of a semiprime ring with involution.

Theorem 1. Let be a semiprime ring with involution , and let be a -ideal of . Then is commuting on if and only if is strong commutativity-preserving on .

Proof. Assume first that is commuting on ; that is, for all . By linearizing we get It follows that and hence , and by (1) we get Since is a -ideal, (2) yields Substituting for , ,  we now get , so that . Using (3) to replace the last term in this equation by and the second term by , we see that , so by (1), Replacing by yields for all , so for all . Since an ideal of a semiprime ring is a semiprime ring, we conclude that Now (5) may be rewritten as so by replacing by we get . But by (6) ; hence for all , so that is strong commutativity-preserving on .
For the converse, we assume that is strong commutativity-preserving on , which means that for all . Substituting for , we get This is just equation (3), so we may argue as before that and for all . It follows at once that for all ; that is, is commuting on .

The proof just given yields a result for prime rings with involution. Before stating our theorem, we mention that we are using the symbols and to denote the sets of symmetric elements and skew elements, respectively, in the ring with involution .

Theorem 2. Let be a prime ring with involution , with . If is commuting on some nonzero -ideal , then .

Proof. It follows from (5) that if , then , and since in a prime ring the center of a nonzero ideal is contained in the center of , . Suppose that , and let . Then for any , so . Since is not a zero divisor, we get .
To complete the proof, we need only show that . Suppose, on the contrary, that . Then for any , ; hence . But for any , ; therefore for all , and we have contradicted a well-known result of Levitzki [4, Lemma 1.1].

#### 3. On Nonexistence of Derivations of Period 2

If is an algebra over with trivial multiplication, the map given by is a derivation of period 2. We do not know whether there exist less obvious examples.

Clearly any derivation of period 2 must be a bijection, so there exists no such that . It follows that a ring with 1 admits no derivation which is of period 2 on .

There do exist semiprime rings admitting a derivation which is a bijection, for example, the —algebra with basis , with being the usual differentiation. Obviously this example is not of period 2, and we will show that a semiprime ring admits no derivation of period 2 on .

Theorem 3. Let be a semiprime ring and a nonzero right ideal. Then admits no derivation such that is of period 2 on .

Proof. Suppose there exists a derivation on such that for all . For , and the condition that yields Since , we get and replacing by in (10), we obtain Substituting for in (11), we get ; hence But is semiprime, so for all , and by (9), Therefore which together with (10) yields for all . In particular, for all , contrary to Levitzki’s result.

Corollary 4. A semiprime ring admits no derivation of period 2 on .

Remark 5. Of course any derivation of period 2 on satisfies . It is shown in  that a noncommutative semiprime ring, though it has no derivations of period 2, may have many nonzero derivations for which ; for any noncentral idempotent , the inner derivation is an example.

#### 4. Generalized Derivations of Period 2

Any ring admits right generalized derivations of period 2, namely, the identity map and its negative. Moreover, if has 1 and with , then defines a right generalized derivation of period 2. We show that, in many prime rings, there are no other possibilities.

We will make use of several easy lemmas.

Lemma 6. Let be an arbitrary ring. If is a generalized derivation on , then .

Proof. Let and . Then , so that , where is the associated derivation of . Since , the result follows at once.

Lemma 7. Let be a prime ring with , and let be a derivation on . If the right generalized derivation given by for all is of period 2 on , then is the identity map (resp., the negative of the identity map) on .

Proof. Consider the case for all . If  is of period 2, we have ; hence Replacing by , we get ; that is, for all . In view of (15) and the assumption that , this equation gives It is well known and easy to prove that if is prime and is a nonzero derivation, then and implies . Thus, from (16) we conclude that and therefore is the identity map on . A similar argument works if for all .

Lemma 8. Let be a prime ring with , and let be a right generalized derivation on with associated derivation . If is of period 2 on , then .

Proof. For all , = ; hence Replacing by in (17) yields Letting and and replacing by in (18), we get , so by (18) we obtain If , we conclude that . But since is prime and , it is easy to show that implies ; hence as claimed.

Theorem 9. Let be a (not necessarily commutative) domain with 1, with . If is a right generalized derivation on of period 2, then is the identity map or its negative.

Proof. Note that Taking in (17) and (18) and letting , we have and for all . It follows that ; that is, ; hence If , we have or , so that for all or for all . But by Lemma 7 this would imply ; hence and Since is of period 2, and hence for all . Thus, or , so by (22), is the identity map or its negative.

Corollary 10. Let be a commutative integral domain with . If is a right generalized derivation of period 2 on , then is the identity map or its negative.

Proof. If has 1, the result is immediate from Theorem 9. If does not have 1, define on the field of fractions by . Using the fact that by Lemma 8, we can show that is well defined and is a right generalized derivation on . By Theorem 9, is the identity map or its negative on and it follows that is the identity map or its negative on .

Theorem 11. Let be a prime ring with and with . If is a generalized derivation of period 2 on with associated derivation , then is the identity map or its negative.

Proof. By Lemma 6, ; hence restricts to a generalized derivation on . Since the center of a prime ring is a commutative domain, it follows from Corollary 10 that for all or for all . If the former holds, then , together with the fact that , gives for all and . Taking gives for all , so by Lemma 7,   is the identity map on . A similar argument shows that if for all , is the negative of the identity map.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

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