Research Article | Open Access

# On the Structures of Abelian -Regular Rings

**Academic Editor:**A. Cigdem Ozcan

#### 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 [1–6]. 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 5–7.

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).

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#### Copyright

Copyright © 2014 Jianwei Lu and Liguo He. 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.