On the Structures of Abelian π-Regular Rings

Throughout this paper, we only consider the associative rings with identity 1. By E(R), we denote the full set of idempotent elements of the ring R. Write C(R) to denote the center of the ring R, and write Nil(R), the full set of nilpotent elements of R. The ring R is called Abelian if E(R) ⊆ C(R). The ring R is said to be regular if, for any a ∈ R, there exists an x ∈ R such that a = axa. And R is called π-regular if, for any a ∈ R, there exist x ∈ R and a positive integer n such that an =


Introduction
Throughout this paper, we only consider the associative rings with identity 1.By (), we denote the full set of idempotent elements of the ring .Write () to denote the center of the ring , and write Nil(), 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   =  +1 .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][2][3][4][5][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) Nil() is an ideal of ; (2) every one-sided ideal containing Nil() is an ideal in ; (3) every completely prime ideal is a maximal ideal in .[7] states that a commutative -regular ring is isomorphic to the subdirect sum of some metadivision rings.We generalize this result as follows.

A classical result due to McCoy in
Theorem B. Assume that  is an Abelian -regular ring and M denotes the collection of all completely prime ideals of .Then  is isomorphic to the subdirect sum of the /(  )  's for all   ∈ M.

Abelian 𝜋-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.

International Journal of Mathematics and Mathematical Sciences
In the preceding proof, if we set  =   , in a similar manner, then we can obtain another element  such that   = 1.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  1 ,  2 ∈  such that  1  = 1 =  2 , and so  1  2 = 1.Also  ∈ ; it follows that 0 =  1 0 2 =  1  2 = 1, 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  =  −1  2 , 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   =    1   and   =    2   .Given  1 =  1   and  2 =  2   , then   are idempotent for  = 1, 2, and thus Considering  is prime, either  1 or  2 lies in  and so in ().We further get that   =    1 ∈ () or   =    2 ∈ (), and hence () is primary in .Suppose now that  ⊊ () is a primary ideal of .Then there exists  ∈ () − ; we therefore have that (1 − ) = 0 ∈  and 1 −  = (1 − )  ∈  for some  > 0. It follows that 1 −  ∈ () and thus 1 = 1 −  +  ∈ (), a contradiction.Therefore () is a minimal primary ideal of .Lemma 5. Let  be an Abelian ring.Then  is Abelian regular if and only if Nil() is an ideal and /Nil() 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  ⊇ Nil(), 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  ⊇ Nil().Then /Nil() is a left ideal of /Nil().By Lemma 5, /Nil() is strongly regular, so /Nil() is an ideal of /Nil().For any  ∈ ,  ∈ ,   ∈  = /Nil(), and so  ∈ ; hence,  is a right ideal.By the same argument, we may also deduce that if  ⊇ Nil() is a right ideal, then  is a left ideal.Therefore, the one-sided ideal  is an ideal in  when  contains Nil().
If  is a maximal one-sided ideal of , then  ⊇ () = Nil(), 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.
(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 0 ̸ =  ∈ / and  =  =  for  ∈  and  = .Then,  ̸ = 0 and  ∉ .Also since ( − 1) = 0 ∈ , we have  − 1 ∈  and so  = 1; that is,   = 1.Thus / is a division ring.Now assume that  is a prime ideal of the Abelian -regular ring  and  ⊇ Nil().Applying the above lemma,  = /Nil() is a prime ideal of the Abelian regular ring  = /Nil(), 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 Nil() = 0.If  is -regular, then  is regular.
The next result is Theorem B. Theorem 9. Let  be an Abelian ring.Then  is -regular if and only if the following statements hold.
(2) Every one-sided ideal containing Nil() is an ideal in .
(3) Every completely prime ideal is a maximal ideal in .