#### Abstract

Assume that is an Abelian ring. In this paper, we characterize the structure of whenever is -regular. It is also proved that an Abelian -regular ring is isomorphic to the subdirect sum of some metadivision rings.

#### 1. Introduction

Throughout this paper, we only consider the associative rings with identity . By , we denote the full set of idempotent elements of the ring . Write to denote the center of the ring , and write , the full set of nilpotent elements of . The ring is called Abelian if . The ring is said to be regular if, for any , there exists an such that . And is called -regular if, for any , there exist and a positive integer such that . We also say a ring is strongly -regular if, for any , there are and positive integer such that . It follows that strongly -regular rings are -regular and Abelian -regular rings are strongly -regular. The ring is said to be a metadivision if any element in is either a unit or a nilpotent element.

In ring theory, there has been much interest in finding connections between the -regularity and the condition that every prime ideal is maximal, which has been studied, for example, in [16]. A pretty result due to Storrer in [6] is if is a commutative ring with identity then is -regular if and only if every prime ideal of is maximal. A commutative ring is obviously an Abelian ring; thus, the following result is a generalization of Storrer’s result.

Theorem A. Let be an Abelian ring. Then is -regular if and only if the following statements hold:(1) is an ideal of ;(2)every one-sided ideal containing is an ideal in ;(3)every completely prime ideal is a maximal ideal in .

A classical result due to McCoy in [7] states that a commutative -regular ring is isomorphic to the subdirect sum of some metadivision rings. We generalize this result as follows.

Theorem B. Assume that is an Abelian -regular ring and denotes the collection of all completely prime ideals of . Then is isomorphic to the subdirect sum of the ’s for all .

#### 2. Abelian -Regular Rings

Recall that an ideal of is primary if for any ; we always obtain or for some positive integer .

Definition 1. The ring is said to be a metadivision ring if any element in is either a unit or a nilpotent element.

In fact, there is an intimate connection between the primary ideas and the metadivision rings as follows.

Lemma 2. Let be an Abelian -regular ring. Then is a primary ideal of if and only if is a metadivision ring.

Proof. The “only if” part: for any element , there exists the equality for a positive integer and . Set ; then is an idempotent. If is not nilpotent, then and so ; thus, . Since lies in the primary ideal , it follows that for some . Also is an idempotent; we have , and so ; that is, .
In the preceding proof, if we set , in a similar manner, then we can obtain another element such that . Hence we obtain that is unit and thus is a metadivision ring.
The “if” part: assume that with . If neither nor lies in for any positive integer , then both of and are unit in (which is a metadivision ring). Thus there exist such that , and so . Also ; it follows that , a contradiction. Therefore we conclude that is a primary ideal of .

Proposition 3. Suppose that is an Abelian -regular ring and for . Then, for any , for some .

Proof. Let ; then is an idempotent and so lies in . Thus it is immediate that . Because where , the application of the inductive argument on yields the desired result.

Lemma 4. Let be an Abelian -regular ring and a prime ideal of . Then is a minimal primary ideal of .

Proof. It is no loss to assume that is a proper ideal of . Because , is an ideal. For and , the application of the above proposition yields that there exists a positive integer such that and . Given and , then are idempotent for , and thus . Considering is prime, either or lies in and so in . We further get that or , and hence is primary in . Suppose now that is a primary ideal of . Then there exists ; we therefore have that and for some . It follows that and thus , a contradiction. Therefore is a minimal primary ideal of .

Lemma 5. Let be an Abelian ring. Then is Abelian -regular if and only if is an ideal and is strongly regular.

Proof. This is Theorem 3 of [1].

Lemma 6. Assume that is an Abelian -regular ring, and is a one-sided ideal of . If , then is an ideal of and for any . In particular, a maximal one-sided ideal of is a maximal two-sided ideal of .

Proof. First we assert that if is Abelian and regular, then a left ideal of is an ideal. For , with some . Let ; then is idempotent, and thus for any . We get that is an ideal.
Now assume that is a left ideal of and . Then is a left ideal of . By Lemma 5, is strongly regular, so is an ideal of . For any , , , and so ; hence, is a right ideal. By the same argument, we may also deduce that if is a right ideal, then is a left ideal. Therefore, the one-sided ideal is an ideal in when contains .
For , both and are ideals of . Since , for any , . Likewise, , and thus we get that , as desired.
If is a maximal one-sided ideal of , then , and thus is an ideal by the next to last paragraph; it is automatically a maximal ideal.

Lemma 7. Assume that is an Abelian -regular ring; then the following statements are equivalent.(1) is a completely prime ideal.(2) is a prime ideal and .(3) is a division ring.(4) is a maximal ideal of .

Proof. (2) (3). First we claim that if is a prime ideal of an Abelian regular , then is a division ring. Let and for and . Then, and . Also since , we have and so ; that is, . Thus is a division ring. Now assume that is a prime ideal of the Abelian -regular ring and . Applying the above lemma, is a prime ideal of the Abelian regular ring , and then is a division ring, as wanted. The proofs of the others are obvious.

A regular ring is obviously a -regular ring. The following lemma is its converse under some conditions.

Lemma 8. Let be an Abelian ring and . If is -regular, then is regular.

Proof. Let . Then where for some and a positive integer ; then . Since , it follows that , and so ; thus, is a regular ring.

The next result is Theorem B.

Theorem 9. Let be an Abelian ring. Then is -regular if and only if the following statements hold. (1) is an ideal of .(2)Every one-sided ideal containing is an ideal in .(3)Every completely prime ideal is a maximal ideal in .

Proof. The “if part” immediately follows from Lemmas 57.
Now we prove the “only if” part. Suppose to the contrary that is not a -regular ring. Lemmas 5 and 8 imply that is not -regular, and so not a strong -regular ring; then there exists such that , which is obviously an infinite descending chain of left ideals of . Let be a collection of ideals of such that the left ideal chains , , are infinitely descending. And is not empty in view of . For any ascending chain in , , it is clear ; thus, has a maximal element . We claim that is not a completely prime ideal. Write to denote the preimage of the natural homomorphism from onto and let with ; then, . If is a prime ideal of , then or , and so or . Hence is a completely prime ideal of . Part (3) implies that is a maximal ideal of , and thus is a maximal ideal of . Part (2) implies that or for a positive integer , and thus or . This is a contradiction, however, because . Hence is not a completely prime ideal, and so there exist , such that , whereas neither nor lies in . Set and ; then, both and are ideals of , which properly contain . Thus neither nor lies in ; consequently, there exists some positive integer such that and . And so there are , such that and ; hence, . Since for any (by Part (2)), it follows that there exist , such that and . Thus we get that Also and we have that and so . This is a contradiction, however, because is a maximal element of . Therefore is -regular, and the proof is finished.

Corollary 10. Assume that is an Abelian ring such that is an ideal of and the one-sided ideals containing are all ideals. Then the following statements are equivalent.(1) is a -regular ring.(2) is a metadivision ring for any completely prime ideal .(3)For any completely prime ideals , if , then .(4)For any completely prime ideal , R/P is -regular.

Proof. (1) (2). is primary ideal of R by Lemma 4 thus is ametadivision ring by Lemma 2.
(2) (1). If is a completely prime ideal, then, under the natural homomorphism , is a completely prime ideal of . Because is a metadivision ring, it follows that is a unique maximal ideal, and so ; thus, . We get that is a division ring. Hence is also a division ring and is a maximal ideal of . Theorem 9 shows that is an Abelian -regular ring.
(1) (3). Let , be completely prime ideals of and . If , then it is no loss to assume that but . Set where for some and positive integer . Since is a completely prime ideal, it follows that and so . Also and ; we have and so . This contradiction shows that .
(3) (1). Let be a completely prime ideal of ; then there exists a maximal ideal . Note that since . If were not completely prime, we may pick , with such that and . If and , then and . Thus, we get
Since , , we have . This contradiction implies that is a completely prime ideal. Fix , because , . Also ; it follows that , and then ; thus, ; hence ; that is, is a maximal ideal. Theorem 9 shows that is an Abelian -regular ring.
(1) (4). Every division ring is -regular and R/P is a division ring by Lemma 7, thus R/P is -regular.
(4) (1). By Theorem 9, it suffices to prove that every completely prime ideal is maximal. If was not maximal, then there exists a maximal ideal . Thus, there exists an but , and so for all . Let for . Since , we get and so . Since , we have and so . Also and ; it follows that and so . This is a contradiction, which shows is a maximal ideal, as wanted.

The following result is Theorem A.

Theorem 11. Assume that is an Abelian -regular ring and denotes the collection of all completely prime ideals of . Then is isomorphic to the subdirect sum of the ’s for all .

Proof. Let . It suffices to prove that in order to prove the desired result. Suppose to the contrary that . Then there exists . We first claim that there exist with for any . Since , for all . Let where , , , and then . Set ; then and . If , then ; thus, (since ); hence . We get that for some , and so , a contradiction. Therefore we have .
Set is an ideal of , , and for any . Since , it follows that and so is not empty. For any ascending series of , , it is clear that , and so has a maximal element . We assert that is a completely prime ideal. If not, then there exist , such that but neither nor lies in . Applying Lemma 6, both and are ideals of . Because either or does not lie in , it follows that none of or lies in , and so there exist idempotent elements and such that and ; thus, . Also ; it follows that Moreover . This is a contradiction since is a maximal element of . Hence the assertion is true. By the last paragraph, there is an element such that , again a contradiction since . Therefore and the desired result follows. The proof is complete.

#### Conflict of Interests

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

#### Acknowledgments

This project is supported by NSF of China (no. 61074005) and by NSF of Liaoning Education Department (no. 2008516).