# On the Structures of Abelianπ-Regular Rings

Authors: Jianwei Lu; Liguo He
Journal: International Journal of Mathematics and Mathematical Sciences
 (2014)

---

## Abstract

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


---

## Body

## 1. Introduction

Throughout this paper, we only consider the associative rings with identity1. 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 calledAbelian if E
(
R
)
⊆
C
(
R
). The ring R is said to beregular if, for any a
∈
R, there exists an x
∈
R such that a
=
a
x
a. And R is called π-regular if, for any a
∈
R, there exist x
∈
R and a positive integer n such that a
n
=
a
n
x
a
n. We also say a ring R isstrongly
π-regular if, for any a
∈
R, there are x
∈
R and positive integer n such that a
n
=
x
a
n
+
1. It follows that strongly π-regular rings are π-regular and Abelian π-regular rings are strongly π-regular. The ring R is said to be ametadivision if any element in R is either a unit or a nilpotent element.In ring theory, there has been much interest in finding connections between theπ-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 R is a commutative ring with identity then R is π-regular if and only if every prime ideal of R is maximal. A commutative ring is obviously an Abelian ring; thus, the following result is a generalization of Storrer’s result.Theorem A.
LetR be an Abelian ring. Then R is π-regular if and only if the following statements hold:(1)
N
i
l
(
R
) is an ideal of R;
(2)
every one-sided ideal containingN
i
l
(
R
) is an ideal in R;
(3)
every completely prime ideal is a maximal ideal inR.A classical result due to McCoy in [7] states that a commutative π-regular ring is isomorphic to the subdirect sum of some metadivision rings. We generalize this result as follows.Theorem B.
Assume thatR is an Abelian π-regular ring and ℳ denotes the collection of all completely prime ideals of R. Then R is isomorphic to the subdirect sum of the R
/
E
(
P
α
)
P
α’s for all P
α
∈
ℳ.


## 2. Abelianπ-Regular Rings

Recall that an idealQ of R isprimary if a
b
∈
Q for any a
,
b
∈
R; we always obtain a
n
∈
Q or b
n
∈
Q for some positive integer n.Definition 1.
The ringR is said to be a metadivision ring if any element in R 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.
LetR be an Abelian π-regular ring. Then Q is a primary ideal of R if and only if R
/
Q is a metadivision ring.Proof.
The “only if” part: for any elementa
∈
R, there exists the equality a
n
=
a
n
r
a
n for a positive integer n and r
∈
R. Set e
=
r
a
n; then e is an idempotent. If a
¯
∈
R
/
Q is not nilpotent, then a
¯
n
≠
0
¯ and so e
¯
≠
0
¯; thus, e
∉
Q. Since e
(
e
-
1
)
=
0 lies in the primary ideal Q, it follows that (
e
-
1
)
m
∈
Q for some m
>
0. Also 1
-
e is an idempotent; we have 1
-
e
=
(
1
-
e
)
m
∈
Q, and so e
¯
=
1
¯; that is, r
¯
a
¯
n
=
1
¯.
In the preceding proof, if we sete
=
a
n
r, in a similar manner, then we can obtain another element s
¯ such that a
¯
s
¯
=
1
¯. Hence we obtain that a
¯ is unit and thus R
/
Q is a metadivision ring.
The “if” part: assume thata
,
b
∈
R with a
b
∈
Q. If neither a
n nor b
n lies in Q for any positive integer n, then both of a
¯ and b
¯ are unit in R
/
Q (which is a metadivision ring). Thus there exist r
1
,
r
2
∈
R such that r
1
¯
a
¯
=
1
¯
=
b
¯
r
¯
2, and so r
¯
1
a
¯
b
¯
r
¯
2
=
1
¯. Also a
b
∈
Q; it follows that 0
¯
=
r
¯
1
0
¯
r
¯
2
=
r
¯
1
a
¯
b
¯
r
¯
2
=
1
¯, a contradiction. Therefore we conclude that Q is a primary ideal of R.Proposition 3.
Suppose thatR is an Abelian π-regular ring and a
n
=
a
n
x
a
n for a
,
x
∈
R. Then, for any m
>
n, a
m
=
a
m
y
a
m for some y
∈
R.Proof.
Lete
=
a
n
x; then e is an idempotent and so lies in C
(
R
). Thus it is immediate that a
n
=
e
a
n
=
a
n
e. Because
(1)
a
n
+
1
=
a
n
a
=
a
n
e
a
=
a
2
n
x
2
a
n
+
1
=
a
n
+
1
(
a
n
-
1
x
2
)
a
n
+
1
=
a
n
+
1
y
a
n
+
1
,
where y
=
a
n
-
1
x
2, the application of the inductive argument on m yields the desired result.Lemma 4.
LetR be an Abelian π-regular ring and P a prime ideal of R. Then E
(
P
)
P is a minimal primary ideal of R.Proof.
It is no loss to assume thatP is a proper ideal of R. Because E
(
P
)
P
=
P
E
(
P
), E
(
P
)
P is an ideal. For a
,
b
∈
R and a
b
∈
E
(
P
)
P, the application of the above proposition yields that there exists a positive integer n such that a
n
=
a
n
r
1
a
n and b
n
=
b
n
r
2
b
n. Given e
1
=
r
1
a
n and e
2
=
r
2
b
n, then e
i are idempotent for i
=
1,2, and thus e
1
e
2
=
e
1
r
2
b
n
=
r
2
b
n
-
1
e
1
b
=
r
2
b
n
-
1
r
1
a
n
-
1
a
b
∈
E
(
P
)
P
⊆
P. Considering P is prime, either e
1 or e
2 lies in P and so in E
(
P
)
P. We further get that a
n
=
a
n
e
1
∈
E
(
P
)
P or b
n
=
b
n
e
2
∈
E
(
P
)
P, and hence E
(
P
)
P is primary in R. Suppose now that Q
⊊
E
(
P
)
P is a primary ideal of R. Then there exists e
∈
E
(
P
)
-
Q; we therefore have that e
(
1
-
e
)
=
0
∈
Q and 1
-
e
=
(
1
-
e
)
m
∈
Q for some m
>
0. It follows that 1
-
e
∈
E
(
P
)
P and thus 1
=
1
-
e
+
e
∈
E
(
P
)
P, a contradiction. Therefore E
(
P
)
P is a minimal primary ideal of R.Lemma 5.
LetR be an Abelian ring. Then R is Abelian π-regular if and only if N
i
l
(
R
) is an ideal and R
/
N
i
l
(
R
) is strongly regular.Proof.
This is Theorem 3 of [1].Lemma 6.
Assume thatR is an Abelian π-regular ring, and I is a one-sided ideal of R. If I
⊇
N
i
l
(
R
), then I is an ideal of R and R
a
+
I
=
a
R
+
I for any a
∈
R. In particular, a maximal one-sided ideal of R is a maximal two-sided ideal of R.Proof.
First we assert that ifR is Abelian and regular, then a left ideal A of R is an ideal. For r
∈
A, r
=
r
y
r with some y
∈
R. Let e
=
y
r; then e is idempotent, and thus r
x
=
r
e
x
=
r
x
e
∈
A for any x
∈
R. We get that A is an ideal.
Now assume thatI is a left ideal of R and I
⊇
Nil
(
R
). Then I
/
Nil
(
R
) is a left ideal of R
/
Nil
(
R
). By Lemma 5, R
/
Nil
(
R
) is strongly regular, so I
/
Nil
(
R
) is an ideal of R
/
Nil
(
R
). For any r
∈
I, x
∈
R, r
¯
x
¯
∈
I
¯
=
I
/
Nil
(
R
), and so r
x
∈
I; hence, I is a right ideal. By the same argument, we may also deduce that if I
⊇
Nil
(
R
) is a right ideal, then I is a left ideal. Therefore, the one-sided ideal I is an ideal in R when I contains Nil
(
R
).
Fora
∈
R, both R
a
+
I and a
R
+
I are ideals of R. Since a
x
∈
R
a
+
I, for any x
∈
R, a
R
+
I
⊆
R
a
+
I. Likewise, R
a
+
I
⊆
a
R
+
I, and thus we get that R
a
+
I
=
a
R
+
I, as desired.
IfP is a maximal one-sided ideal of R, then P
⊇
J
(
R
)
=
Nil
(
R
), and thus P is an ideal by the next to last paragraph; it is automatically a maximal ideal.Lemma 7.
Assume thatR is an Abelian π-regular ring; then the following statements are equivalent.(1)
P is a completely prime ideal.
(2)
P is a prime ideal and P
⊇
Nil
(
R
).
(3)
R
/
P is a division ring.
(4)
P is a maximal ideal of R.Proof.
(2)⇒ (3). First we claim that if A is a prime ideal of an Abelian regular R, then R
/
A is a division ring. Let 0
¯
≠
a
¯
∈
R
/
A and a
=
a
r
a
=
a
e for r
∈
R and e
=
r
a. Then, e
¯
≠
0
¯ and e
∉
A. Also since e
(
e
-
1
)
=
0
∈
A, we have e
-
1
∈
A and so e
¯
=
1
¯; that is, r
¯
a
¯
=
1
¯. Thus R
/
A is a division ring. Now assume that P is a prime ideal of the Abelian π-regular ring R and P
⊇
Nil
(
R
). Applying the above lemma, P
¯
=
P
/
Nil
(
R
) is a prime ideal of the Abelian regular ring R
¯
=
R
/
Nil
(
R
), and then R
/
P
≅
R
¯
/
P
¯ is a division ring, as wanted. The proofs of the others are obvious.A regular ring is obviously aπ-regular ring. The following lemma is its converse under some conditions.Lemma 8.
LetR be an Abelian ring and N
i
l
(
R
)
=
0. If R is π-regular, then R is regular.Proof.
Leta
∈
R. Then a
n
=
a
n
x
a
n
=
a
n
e where e
=
r
a
n for some r
∈
R and a positive integer n; then (
a
(
1
-
e
)
)
n
=
a
n
(
1
-
e
)
n
=
a
n
(
1
-
e
)
=
a
n
-
a
n
e
=
0. Since Nil
(
R
)
=
0, it follows that a
(
1
-
e
)
=
0, and so a
=
a
e
=
a
r
a
n
=
a
(
r
a
n
-
1
)
a; thus, R is a regular ring.The next result is Theorem B.Theorem 9.
LetR be an Abelian ring. Then R is π-regular if and only if the following statements hold. (1)
N
i
l
(
R
) is an ideal of R.
(2)
Every one-sided ideal containingN
i
l
(
R
) is an ideal in R.
(3)
Every completely prime ideal is a maximal ideal inR.Proof.
The “if part” immediately follows from Lemmas5–7.
Now we prove the “only if” part. Suppose to the contrary thatR is not a π-regular ring. Lemmas 5 and 8 imply that R
¯
=
R
/
Nil
(
R
) is not π-regular, and so not a strong π-regular ring; then there exists a
¯
∈
R
¯ such that R
¯
a
¯
⊋
R
¯
a
¯
2
⊋
⋯
⊋
R
¯
a
¯
n
⊋
⋯, which is obviously an infinite descending chain of left ideals of R. Let 𝒮 be a collection of ideals I
¯ of R
¯ such that the left ideal chains R
¯
a
¯
i
+
I
¯, i
=
1,2
,
…, are infinitely descending. And 𝒮 is not empty in view of 0
¯
∈
𝒮. For any ascending chain in R, I
¯
1
⊂
I
¯
2
⊂
⋯, it is clear ∪
I
¯
i
∈
𝒮; thus, 𝒮 has a maximal element H
¯. We claim that H
¯ is not a completely prime ideal. Write σ
-
1
(
H
¯
) to denote the preimage of the natural homomorphism σ from R onto R
¯ and let a
′
,
b
′
∈
R with a
′
b
′
∈
σ
-
1
(
H
¯
); then, a
′
¯
b
′
¯
∈
H
¯. If H
¯ is a prime ideal of R
¯, then a
′
¯
∈
H
¯ or b
′
¯
∈
H
¯, and so a
′
∈
σ
-
1
(
H
¯
) or b
′
∈
σ
-
1
(
H
¯
). Hence σ
-
1
(
H
¯
) is a completely prime ideal of R. Part (3) implies that σ
-
1
(
H
¯
) is a maximal ideal of R, and thus H
¯ is a maximal ideal of R
¯. Part (2) implies that H
¯
+
R
¯
a
¯
k
+
1
=
H
¯ or H
¯
+
R
¯
a
¯
k
+
1
=
R
¯ for a positive integer k, and thus H
¯
+
R
¯
a
¯
k
+
1
=
H
¯
+
R
¯
a
¯
k
+
2 or H
¯
+
R
¯
a
¯
k
+
1
=
H
¯
+
R
¯
a
¯
k. This is a contradiction, however, because H
¯
∈
𝒮. Hence H
¯ is not a completely prime ideal, and so there exist a
¯
1, b
¯
1
∈
R
¯ such that a
¯
1
b
¯
1
∈
H
¯, whereas neither a
¯
1 nor b
¯
1 lies in H
¯. Set H
¯
1
=
R
¯
a
¯
1
+
H
¯ and H
¯
2
=
b
¯
1
R
¯
+
H
¯; then, both H
¯
1 and H
¯
2 are ideals of R
¯, which properly contain H
¯. Thus neither H
¯
1 nor H
¯
2 lies in 𝒮; consequently, there exists some positive integer n such that R
¯
a
¯
n
+
H
¯
1
=
R
¯
a
¯
n
+
1
+
H
¯
1 and R
¯
a
¯
n
+
H
¯
2
=
R
¯
a
¯
n
+
1
+
H
¯
2. And so there are x
¯, y
¯
∈
R
¯ such that a
¯
n
-
x
¯
a
¯
n
+
1
∈
H
¯
1 and a
¯
n
-
y
¯
a
¯
n
+
1
∈
H
¯
2; hence, (
a
¯
n
-
x
¯
a
¯
n
+
1
)
(
a
¯
n
-
y
¯
a
¯
n
+
1
)
∈
H
¯
1
H
¯
2. Since r
¯
R
¯
=
R
¯
r
¯ for any r
¯
∈
R
¯ (by Part (2)), it follows that there exist y
¯
1, y
¯
2
∈
R
¯ such that a
¯
n
y
¯
=
y
¯
1
a
¯
n and a
¯
n
+
1
y
¯
=
y
¯
2
a
¯
n
+
1. Thus we get that
(2)
(
a
¯
n
-
x
¯
a
¯
n
+
1
)
(
a
¯
n
-
y
¯
a
¯
n
+
1
)
=
a
¯
2
n
-
x
¯
a
¯
2
n
+
1
-
a
¯
n
y
¯
a
¯
n
+
1
+
x
¯
a
¯
n
+
1
y
¯
a
¯
n
+
1
=
a
¯
2
n
-
x
¯
a
¯
2
n
+
1
-
y
¯
1
a
¯
2
n
+
1
+
x
¯
y
¯
2
a
¯
2
n
+
2
=
a
¯
2
n
-
x
¯
0
a
¯
2
n
+
1
∈
H
¯
1
H
¯
2
,
here
x
¯
0
=
x
¯
+
y
¯
1
-
x
¯
y
¯
2
a
¯
.
Also
(3)
H
¯
1
H
¯
2
=
(
R
¯
a
¯
1
+
H
¯
)
(
b
¯
1
R
¯
+
H
¯
)
=
R
¯
a
¯
1
b
¯
1
R
¯
+
H
¯
b
¯
1
R
¯
+
R
¯
a
¯
1
H
¯
+
H
¯
2
⊆
R
¯
a
¯
1
b
¯
1
R
¯
+
H
¯
⊆
H
¯
,
and we have that a
¯
2
n
-
x
¯
0
a
¯
2
n
+
1
∈
H
¯ and so R
¯
a
¯
2
n
+
H
¯
=
R
¯
a
¯
2
n
+
1
+
H
¯. This is a contradiction, however, because H
¯ is a maximal element of 𝒮. Therefore R is π-regular, and the proof is finished.Corollary 10.
Assume thatR is an Abelian ring such that N
i
l
(
R
) is an ideal of R and the one-sided ideals containing N
i
l
(
R
) are all ideals. Then the following statements are equivalent.(1)
R is a π-regular ring.
(2)
R
/
E
(
P
)
P is a metadivision ring for any completely prime ideal P.
(3)
For any completely prime idealsP
1
,
P
2, if E
(
P
1
)
=
E
(
P
2
), then P
1
=
P
2.
(4)
For any completely prime idealP,R/P is π-regular.Proof.
(1)⇒ (2). E
(
P
)
P is primary ideal ofR by Lemma 4 thus E
(
P
)
P is ametadivision ring by Lemma 2.
(2)⇒ (1). If P is a completely prime ideal, then, under the natural homomorphism R
→
R
/
E
(
P
)
P
=
R
¯, P
¯ is a completely prime ideal of R
¯. Because R
¯ is a metadivision ring, it follows that Nil
(
R
¯
) is a unique maximal ideal, and so P
¯
⊇
Nil
(
R
¯
); thus, P
¯
=
Nil
(
R
¯
). We get that R
¯
/
P
¯
=
R
¯
/
Nil
(
R
¯
) is a division ring. Hence R
/
P
≅
R
¯
/
P
¯ is also a division ring and P is a maximal ideal of R. Theorem 9 shows that R is an Abelian π-regular ring.
(1)⇒ (3). Let P
1, P
2 be completely prime ideals of R and E
(
P
1
)
=
E
(
P
2
). If P
1
≠
P
2, then it is no loss to assume that a
∈
P
2 but a
∉
P
1. Set a
n
=
a
n
r
a
n
=
a
n
e where e
=
r
a
n for some r
∈
R and positive integer n. Since P
1 is a completely prime ideal, it follows that a
n
∉
P
1 and so e
∉
P
1. Also e
=
r
a
n
∈
P
2 and e
2
=
e; we have e
∈
E
(
P
2
) and so E
(
P
1
)
≠
E
(
P
2
). This contradiction shows that P
1
=
P
2.
(3)⇒ (1). Let P be a completely prime ideal of R; then there exists a maximal ideal H
⊇
P. Note that H
⊇
Nil
(
R
) since P
⊇
Nil
(
R
). If H were not completely prime, we may pick a, b
∈
R with a
b
∈
H such that a
∉
H and b
∉
H. If a
∉
H and b
∉
H, then R
a
+
H
=
R and b
R
+
H
=
R. Thus, we get
(4)
R
2
=
(
R
a
+
H
)
(
b
R
+
H
)
=
R
a
b
R
+
R
a
H
+
H
b
R
+
H
2
⊆
R
a
b
R
+
H
.
SinceR
2
=
R, a
b
∈
H, we have R
⊆
H. This contradiction implies that H is a completely prime ideal. Fix e
∈
E
(
H
), because 1
-
e
∉
H, 1
-
e
∉
P. Also (
1
-
e
)
e
=
0
∈
P; it follows that e
∈
P, and then E
(
H
)
⊆
E
(
P
); thus, E
(
H
)
=
E
(
P
); hence H
=
P; that is, P is a maximal ideal. Theorem 9 shows that R is an Abelian π-regular ring.
(1)⇒ (4). Every division ring is π-regular andR/P is a division ring by Lemma 7, thusR/P is π-regular.
(4)⇒ (1). By Theorem 9, it suffices to prove that every completely prime ideal is maximal. If P was not maximal, then there exists a maximal ideal Q
⊋
P. Thus, there exists an a
∈
Q but a
∉
P, and so a
n
∉
P for all n. Let a
¯
n
=
a
¯
n
r
¯
a
¯
n
=
a
¯
n
e
¯ for r
¯
∈
R
¯
=
R
/
P. Since a
¯
n
≠
0
¯, we get e
¯
≠
0
¯ and so e
∉
P. Since (
1
-
e
)
e
¯
=
(
1
¯
-
e
¯
)
e
¯
=
0
¯, we have (
1
-
e
)
e
∈
P and so 1
-
e
∈
P
⊊
Q. Also e
-
r
a
n
∈
Q and a
∈
Q; it follows that e
∈
Q and so 1
=
1
-
e
+
e
∈
Q. This is a contradiction, which shows P is a maximal ideal, as wanted.The following result is Theorem A.Theorem 11.
Assume thatR is an Abelian π-regular ring and ℳ denotes the collection of all completely prime ideals of R. Then R is isomorphic to the subdirect sum of the R
/
E
(
P
α
)
P
α’s for all P
α
∈
ℳ.Proof.
LetI
=
⋂
P
α
∈
ℳ
‍
E
(
P
α
)
P
α. It suffices to prove that I
=
{
0
} in order to prove the desired result. Suppose to the contrary that I
≠
{
0
}. Then there exists 0
≠
x
∈
I. We first claim that there exist e
α
∈
E
(
P
α
) with e
α
x
=
x for any P
α. Since x
∈
I, x
∈
E
(
P
α
)
P
α for all P
α
∈
ℳ. Let x
=
e
1
x
1
+
e
2
x
2
+
⋯
+
e
r
x
r where e
i
∈
E
(
P
α
), x
i
∈
P
α, i
=
1,2
,
…
,
r, and then (
1
-
e
1
)
⋯
(
1
-
e
r
)
x
=
0. Set e
=
1
-
(
1
-
e
1
)
⋯
(
1
-
e
r
); then e
2
=
e and e
x
=
x. If e
∉
E
(
P
α
), then e
∉
P
α; thus, 1
-
e
∈
P
α (since (
1
-
e
)
e
=
0
∈
P
α); hence (
1
-
e
1
)
(
1
-
e
2
)
⋯
(
1
-
e
r
)
∈
P
α. We get that 1
-
e
k
∈
P
α for some 1
≤
k
≤
r, and so 1
=
1
-
e
k
+
e
k
∈
P
α, a contradiction. Therefore we have e
∈
E
(
P
α
).
Set𝒮
=
{
I
∣
I is an ideal of R, I
⊇
Nil
(
R
), and e
x
≠
x for any e
∈
E
(
I
)
}. Since x
≠
0, it follows that Nil
(
R
)
∈
𝒮 and so 𝒮 is not empty. For any ascending series of 𝒮, I
1
⊂
I
2
⊂
I
3
⊂
⋯, it is clear that ∪
i
I
i
∈
𝒮, and so 𝒮 has a maximal element H. We assert that H is a completely prime ideal. If not, then there exist a, b
∈
R such that a
b
∈
H but neither a nor b lies in H. Applying Lemma 6, both R
a
+
H and b
R
+
H are ideals of R. Because either a or b does not lie in H, it follows that none of R
a
+
H or b
R
+
H lies in 𝒮, and so there exist idempotent elements e
′
∈
R
a
+
H and e
′′
∈
b
R
+
H such that e
′
x
=
x and e
′′
x
=
x; thus, e
′
e
′′
x
=
e
′
x
=
x. Also a
b
∈
H; it follows that
(5)
e
′
e
′′
∈
(
R
a
+
H
)
(
b
R
+
H
)
=
R
a
b
R
+
R
a
H
+
H
b
R
+
H
2
⊂
H
.
Moreover e
′
e
′′
∈
E
(
H
). This is a contradiction since H is a maximal element of 𝒮. Hence the assertion is true. By the last paragraph, there is an element e
∈
E
(
H
) such that e
x
=
x, again a contradiction since H
∈
𝒮. Therefore I
=
{
0
} and the desired result follows. The proof is complete.



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*Source: 842313-2014-04-01.xml*