Regularity of Idempotent Reflexive GP-V’-Rings

Abstract

This paper discusses the regularity of the GP-V’-rings in conjunction with idempotent reflexivity for the first time. We mainly discuss the weak and strong regularity of the GP-V’-rings using generalized weak ideals, weakly right ideals, and quasi-ideals. We show the following: (1) If $R$ is an idempotent reflexive semi-abelian left GP-V’-ring whose every maximal essential left ideal is a generalized weak ideal, a weakly right ideal, or a quasi-ideal, then $R$ is a reduced left weakly regular ring. (2) $R$ is a strongly regular ring if and only if $R$ is an idempotent reflexive semi-commutative left GP-V’-ring whose every maximal essential left ideal is a generalized weak ideal, a weakly right ideal, or a quasi-ideal. (3) If $R$ is a semi-primitive idempotent reflexive ring whose every simple singular left $R$-module is flat, and every maximal left ideal is a generalized weak ideal, then, for any nonzero element $a\in R$, there exists a positive integer $n$ such that $a^n\ne 0$, and $RaR+l\left(a^n\right)=R$.

1. Introduction

In 1936, von Neumann first proposed the concept of (von Neumann) regular rings in the journal of Proc. Nat. Acad. Sci. USA (if there is always an element $b$ in the ring $R$ such that $a=aba$ for any element $a\in R$, then the ring $R$ is called a regular ring). Regular rings play an important role in the study of operator algebras. Since von Neumann’s regular rings are a kind of ring with good properties, it is very important to study the von Neumann regularity of rings. In order to further research the semi-simplicity of the rings, Faith C. proposed the concept of V-rings (i.e., every simple $R$-module over $R$ is an injective module) in reference [1] in 1967. Kaplansky I. proved a very interesting result: a commutative ring $R$ is a von Neumann regular if and only if every simple $R$-module is injective. This result combines regular rings and V-rings closely. However, the rings we generally discuss are noncommutative associative rings. For noncommutative rings, what is the relationship between them? In 1973, Michler G.O. and Villamayor O.E. proved that they are neither sufficient nor necessary as discussed in reference [2], so the relationship between V-rings and regular rings is very interesting for noncommutative rings.

In 1974, Roger, Y.C.M. first proposed the concept of P-injective modules in reference [3]. In 1985, Roger, Y.C.M. weakened the condition of P-injective modules and proposed the concept of YJ-injective modules in reference [4] (which is consistent with the GP-injective modules in this paper cf. [5]). A left $R$-module M is called a $YJ$-injective module. This means that for any element $a$ in ring $R$, there is a positive integer $n$ meaning that any left $R$-module homomorphism from ${Ra}^n$ to $M$, can be extended to the homomorphism from $R$ to $M$. Therefore, the concept of V-rings is naturally extended to GP-V-rings (that is, every simple $R$-module over a ring $R$ is GP-injective). Since regular rings must be PV-rings, they must also be GP-V-rings. Since the PV-ring is not necessarily a regular ring, the GP-V-ring is also not necessarily a regular ring. In recent years, the topic of studying the regularity of rings whose every singular $R$-module is GP-injective has attracted the interest of many mathematicians (such as [6,7,8,9,10]). For instance, reference [7] analyzed the relationship between the von Neumann regularity of endomorphisms and the perspectivity of direct summands in modules in 2021. Reference [10] investigated the regularity of rings whose essential maximal right ideals are GP-injective in 2022. Reference [9] researched the structure of C-regular rings that satisfy the von Neumann regularity of commutators in 2024. Kim, J.Y. first introduced the concepts of reflexive rings and idempotent reflexive rings in reference [11]. However, until now, no researcher has studied idempotent reflexive GP-V’-rings and their regularity properties. In light of this, we systematically discuss the regularity of idempotent reflexive GP-V’-rings in this paper. We first give a brief introduction of the relevant definitions and lemmas below.

2. Preliminaries

In this section, some definitions and lemmas are provided to contribute to expounding the corresponding results more distinctly.

Definition 1 

([11]). Let $R$ be a ring and $L$ be the left ideal of $R$. If $aRb\subseteq L\Rightarrow bRa\subseteq L$ for $\forall a,b\in R$ , the left ideal $L$ is said to be reflexive. The ring $R$ is said to be reflexive if 0 is a reflexive ideal of the ring $R$.

Definition 2 

([11]). Let $R$ be a ring and $L$ be a left ideal of the $ringR$. If $aRe\subseteq L\Rightarrow eRa\subseteq L$ for $\forall a,e\in R$ and $e^2=e$ , the left ideal $L$ is said to be idempotent reflexive. If $aRe=0\Rightarrow eRa=0$, then $R$ is said to be an idempotent reflexive ring.

Note that the prime ideal is reflexive. Since the intersection of reflexive left ideals is reflexive, so is the semiprime ideal. It is clear that both abelian rings and semiprime rings are idempotent reflexive. Birkenmeier G.F., Kim J.Y. and Park J.K. have given an example in the literature [12] (Example 8) to show that an idempotent reflexive ring is not necessarily reflexive.

Definition 3. 

An element $a$ in the ring $R$ is said to be a left weakly regular element if it satisfies $a\in RaRa$.

Definition 4. 

A ring $R$ is called semi-commutative if $aRb=0$ for $\forall a,b\in R$ and $ab=0$.

Definition 5. 

A ring $R$ is semi-abelian if and only if either $eR(1-e)=0$ or $(1-e)Re=0$ for any idempotent $e$ of ring $R$.

Definition 6. 

An additive semigroup $L$ of ring $R$ is a weakly left (right) ideal if there exists a positive integer $n$ such that ${\left(Rx\right)}^n\subseteq L({\left(xR\right)}^n\subseteq L)$ for every $x\in R$.

Definition 7. 

A left (right) ideal $M$ of ring  $R$ is a generalized weak ideal if there exists a positive integer $n$ such that $a^{n}R\subseteq M(R{a}^n\subseteq M)$ for every $a\in R$.

Definition 8. 

An additive semigroup $I$ of ring $R$ is a quasi-ideal if $rar\in I, ara\in I$ for every $r\in I, a\in R$.

Definition 9 

([13,14]). A ring R is called left GP-V-ring (GP-V’-ring) if every simple (singular) left $R$-module is YJ-injective.

In the following, we discuss the regularity of GP-V’-rings with idempotent reflexivity. In order to obtain the corresponding theorems, we first give the following several lemmas. Among them, Lemma 1, Lemma 2 and Lemma 3 have been proven in the literature [11,15,16], respectively. Therefore, we only give the statements of these three lemmas and omit their detailed proofs.

Lemma 1 

([11]). Let $R$ be a semi-primitive ring. If the maximal left (right) ideal of $R$ is a generalized weak ideal, then $R$ is a reduced ring.

Lemma 2 

([15]). Let $R$ be an idempotent reflexive $ring$. If $a\in R$ is not a left weakly regular element, then every maximal left ideal $M$ of $R$ containing $RaR+l\left(a\right)$ must be essential.

Lemma 3 

([16]). If every maximal left ideal of the $ringR$ is a weakly right ideal, then $R/J$ is reduced.

Lemma 4. 

If $R$ is an idempotent reflexive semi-abelian left GP-V’-ring whose every maximal essential left ideal is a quasi-ideal, then $R$ is reduced.

Proof. 

Suppose there exists $0\ne a\in R$ such that $a^2=0$, then $a\in l\left(a\right)$. Therefore, there exists a maximal left ideal $M$ of $R$ which contains $l\left(a\right)$, and it is easy to see that $M$ is essential. In fact, if $M$ is not essential, then $M$ is a direct sum term of $R$ and $M=l\left(e\right)$, where $0\ne e^2=e\in R,a\in l\left(a\right)\subseteq M=l\left(e\right),$ then $ae=0$. Since $R$ is semi-abelian, then $ea=0$, i.e., $e\in l\left(a\right)\subseteq M=l\left(e\right).\mathrm{T}$ hus $e=e^2=0$, contradicting the hypothesis. Therefore, $M$ is essential. Noting that $l\left(a\right)\subseteq M$, we can define a left $R$-module homomorphism as

$$f:Ra\to R/M,f\left(ra\right)=r+M.$$

Since $R$ is a left GP-V’-ring, the simple singular left $R$-module $R/M$ is GP-injective. Thus, there exists $b\in R$ such that

$$1+M=f\left(a\right)=ab+M,1-ab\in M,$$

where $a\in l\left(a\right)\subseteq M$. In view of the fact that $M$ is a left ideal and a quasi-ideal,

$$\left(1+b\right)a\left(1+b\right)\in M,bab\in M,ba\in M.$$

Thus, we obtain

$$ab=\left(1+b\right)a\left(1+b\right)-a-ba-bab\in M.$$

Hence, $1=(1-ab)+ab\in M$, which contradicts the maximality of $M$. This implies that the supposition is not true, that is to say, ring $R$ is reduced. □

If the condition “a quasi-ideal” in Lemma 4 is replaced by “a weakly right ideal” or “a generalized weak ideal”, the lemma still holds. Therefore, we can obtain the following two lemmas.

Lemma 5. 

If $R$ is an idempotent reflexive semi-abelian left GP-V’-ring whose every maximal essential left ideal is a weakly right ideal, then $R$ is reduced.

Proof. 

If $R$ is not reduced, then there exists $0\ne a\in R$ such that $a^2=0$. Thus, $l\left(a\right)$ is contained in a maximal left ideal $M$ of $R$, and it is easy to see that $M$ is essential. Indeed, if $M$ is not essential, then $M$ is a direct sum term of $R$ and $M=l\left(e\right)$, where $0\ne e^2=e\in R,a\in l\left(a\right)\subseteq M=l\left(e\right)$, then $ae=0$. Since the ring $R$ is semi-abelian, then $ea=0$, i.e., $e\in l\left(a\right)\subseteq M=l\left(e\right).$ Therefore, $e=e^2=0$, contradicting the hypothesis. Hence, $M$ is essential. Noting that $l\left(a\right)\subseteq M$, we can define a left $R$-module homomorphism

$$f:Ra\to R/M,$$

such that $f\left(ra\right)=r+M$. Since $R$ is a left GP-V’-ring, the simple singular left $R$-module $R/M$ is GP-injective. Thus, there exists $b\in R$ such that

$$1+M=f\left(a\right)=ab+M,1-ab\in M.$$

where $a\in l\left(a\right)\subseteq M$. By the condition that $M$ is a left ideal and a weak right ideal, there exists a positive integer $n$ such that $(ab{)}^n\in M$. Since

$$1-(ab{)}^n=\left[1+ab+(ab{)}^2+\cdots (ab{)}^{n-1}\right](1-ab)\in M.$$

and $1=\left[1-(ab{)}^n\right]+(ab{)}^n\in M$, which contradicts the maximality of $M$. Thus, $R$ is reduced. □

Lemma 6. 

If $R$ is an idempotent reflexive semi-abelian left GP-V’-ring whose every maximal essential left ideal is a generalized weak ideal, then $R$ is reduced.

Proof. 

If $R$ is not reduced, then there exists $0\ne a\in R$ such that $a^2=0$. Thus, $l\left(a\right)$ is contained in a maximal left ideal $M$ of $R$, and it is easy to see that $M$ is essential. Indeed, if $M$ is not essential, then $M$ is a direct sum term of $R$ and $M=l\left(e\right)$ where $0\ne e^2=e\in R,a\in l\left(a\right)\subseteq M=l\left(e\right),$ then $ae=0$. Since $R$ is semi-abelian, then $ea=0$, i.e., $e\in l\left(a\right)\subseteq M=l\left(e\right)$, so we have

$$e=e^2=0,$$

contradicting the hypothesis. Therefore, $M$ is essential. Noting that $l\left(a\right)\subseteq M$, we can define a left $R$-module homomorphism

$$f:Ra\to R/M,$$

such that $f\left(ra\right)=r+M$. Since $R$ is a left GP-V’-ring, the simple singular left $R$-module $R/M$ is GP-injective. Thus, there exists $b\in R$ such that

$$1+M=f\left(a\right)=ab+M,1-ab\in M,$$

where $a\in l\left(a\right)\subseteq M$. By the condition that $M$ is a left ideal, we can obtain $ba\in M,b(1-ab)=b-bab\in M$. Since $M$ is a generalized weak ideal, there exists a positive integer $n$ such that $(ba{)}^{n}b\in M$. Note that

$$\begin{array}{c}(ba{)}^{n-1}b=(ba{)}^{n-1}(b-bab)+(ba{)}^{n}b\in M,\\ (ba{)}^{n-2}b=(ba{)}^{n-2}(b-bab)+(ba{)}^{n-1}b\in M,\\ .....................\\ bab=ba(b-bab)+(ba{)}^{2}b\in M,\\ b=(b-bab)+bab\in M.\end{array}$$

Since $M$ is a left ideal, $ab\in M$. Since $1=(1-ab)+ab\in M$, it contradicts the maximality of $M$. Therefore, $R$ is reduced. □

Lemma 7. 

If $R$ is an idempotent reflexive semi-commutative left GP-V’-ring whose every maximal essential left ideal is a quasi-ideal, then $R$ is reduced.

Proof. 

If $R$ is not reduced, then there exists $0\ne a\in R$ such that $a^2=0$. Thus, $l\left(a\right)$ is contained in a maximal left ideal $M$ of $R$, and it is easy to see that $M$ is essential. In fact, if $M$ is not essential, then $M$ is a direct sum term of $R$ and $M=l\left(e\right)$, where $0\ne e^2=e\in R,a\in l\left(a\right)\subseteq M=l\left(e\right),$ then $ae=0$. Since $R$ is semi-commutative, then $aRe=0$. Since $R$ is idempotent reflexive, then $eRa=0$. Thus, $ea=0$, i.e., $e\in l\left(a\right)\subseteq M=l\left(e\right).\mathrm{T}$ hus, $e=e^2=0$, contradicting the hypothesis. Therefore, $M$ is essential. Noting that $l\left(a\right)\subseteq M$, we can define a left $R$-module homomorphism

$$f:Ra\to R/M,f\left(ra\right)=r+M.$$

Since $R$ is a left GP-V’-ring, the simple singular left $R$-module $R/M$ is GP-injective. Thus, there exists $b\in R$ such that

$$1+M=f\left(a\right)=ab+M,1-ab\in M,$$

where $a\in l\left(a\right)\subseteq M$. In view of the fact that $M$ is a left ideal and a quasi-ideal,

$$\left(1+b\right)a\left(1+b\right)\in M,bab\in M,ba\in M.$$

So, we can obtain

$$ab=\left(1+b\right)a\left(1+b\right)-a-ba-bab\in M.$$

Hence, $1=(1-ab)+ab\in M$, which contradicts the maximality of $M$. This implies that the ring $R$ is reduced. □

Lemma 8. 

If $R$ is an idempotent reflexive semi-commutative left GP-V’-ring whose every maximal essential left ideal is a weakly right ideal, then $R$ is reduced.

Proof. 

If $R$ is not reduced, then there exists $0\ne a\in R$ such that $a^2=0$. Thus, $l\left(a\right)$ is contained in a maximal left ideal $M$ of $R$, and it is easy to see that $M$ is essential. Indeed, if $M$ is not essential, then $M$ is a direct sum term of $R$ and $M=l\left(e\right)$, where $0\ne e^2=e\in R,a\in l\left(a\right)\subseteq M=l\left(e\right)$, then $ae=0$. Since $R$ is semi-commutative, then $aRe=0$. Since $R$ is idempotent reflexive, then $eRa=0$. Thus, $ea=0$, i.e., $e\in l\left(a\right)\subseteq M=l\left(e\right)$. Therefore, $e=e^2=0$, contradicting the hypothesis. Hence, $M$ is essential. Noting that $l\left(a\right)\subseteq M$, we can define a left $R$-module homomorphism

$$f:Ra\to R/M,$$

such that $f\left(ra\right)=r+M$. Since $R$ is a left GP-V’-ring, the simple singular left $R$-module $R/M$ is GP-injective. Thus, there exists $b\in R$ such that

$$1+M=f\left(a\right)=ab+M,1-ab\in M.$$

where $a\in l\left(a\right)\subseteq M$. By the condition that $M$ is a left ideal and a weak right ideal, there exists a positive integer $n$ such that $(ab{)}^n\in M$. Since

$$1-(ab{)}^n=\left[1+ab+(ab{)}^2+\cdots (ab{)}^{n-1}\right](1-ab)\in M,$$

and $1=\left[1-(ab{)}^n\right]+(ab{)}^n\in M$, which contradicts the maximality of $M$. Thus, $R$ is reduced. □

Lemma 9. 

If $R$ is an idempotent reflexive semi-commutative left GP-V’-ring whose every maximal essential left ideal is a generalized weak ideal, then $R$ is reduced.

Proof. 

If $R$ is not reduced, then there exists $0\ne a\in R$ such that $a^2=0$. Thus, $l\left(a\right)$ is contained in a maximal left ideal $M$ of $R$, and it is easy to see that $M$ is essential. Indeed, if $M$ is not essential, then $M$ is a direct sum term of $R$ and $M=l\left(e\right)$, where $0\ne e^2=e\in R,a\in l\left(a\right)\subseteq M=l\left(e\right),$ then $ae=0$. Since $R$ is semi-commutative, then $aRe=0$. Since $R$ is idempotent reflexive, then $eRa=0$. Thus, $ea=0$, i.e., $e\in l\left(a\right)\subseteq M=l\left(e\right)$, so we have

$$e=e^2=0,$$

contradicting the hypothesis. Therefore, $M$ is essential. Noting that $l\left(a\right)\subseteq M$, we can define a left $R$-module homomorphism

$$f:Ra\to R/M,$$

such that $f\left(ra\right)=r+M$. Since $R$ is a left GP-V’-ring, the simple singular left $R$-module $R/M$ is GP-injective. Thus, there exists $b\in R$ such that

$$1+M=f\left(a\right)=ab+M,1-ab\in M,$$

where $a\in l\left(a\right)\subseteq M$. By the condition that $M$ is a left ideal, we can obtain $ba\in M,b(1-ab)=b-bab\in M$. Since $M$ is a generalized weak ideal, there exists a positive integer $n$ such that $(ba{)}^{n}b\in M$. Note that

$$\begin{array}{c}(ba{)}^{n-1}b=(ba{)}^{n-1}(b-bab)+(ba{)}^{n}b\in M,\\ (ba{)}^{n-2}b=(ba{)}^{n-2}(b-bab)+(ba{)}^{n-1}b\in M,\\ .....................\\ bab=ba(b-bab)+(ba{)}^{2}b\in M,\\ b=(b-bab)+bab\in M.\end{array}$$

Since $M$ is a left ideal, $ab\in M$. Since $1=(1-ab)+ab\in M$, it contradicts the maximality of $M$. Therefore, $R$ is reduced. □

3. Main Results

In 2005, Kim, J.Y. proved in the literature [11] that if $R$ is an idempotent reflexive left GP-V’-ring, then for any nonzero element $a$ in $R$, there exists a positive integer $n=n\left(a\right)$ such that $a^n\ne 0,RaR+l\left(a^n\right)=R$ and $J\left(R\right)=0$. Inspired by the result, we obtained the following theorem.

Theorem 1. 

Let $R$ is a semi-primitive idempotent reflexive ring. If every simple singular left $R$-module is flat, and every maximal left ideal is a generalized weak ideal, then for any nonzero element $a$ in $R$, there exists a positive integer $n=n\left(a\right)$ such that $a^n\ne 0$, and $RaR+l\left(a^n\right)=R$.

Proof. 

If $a\in R$ is a left weak regular element, then the conclusion clearly holds. In the following, we consider that if $a$ is not a left weak regular element, then $RaR+l\left(a\right)\ne R$. Firstly, we assume that $a^n\ne 0$, and $a^{n+1}=0$. We claim that $RaR+l\left(a^m\right)=R$. Otherwise, there exists a maximal left ideal $M$ containing $RaR+l\left(a^m\right)$, and by Lemma 2, $M$ is an essential left ideal of $R$. Thus, $R/M$ is flat and ${\left(a^n\right)}^2=0$. Since $a^n\in M$, there exists $u\in M$ such that $a^n=a^{n}u$, and so $(1-u)\in r\left(a^n\right)$. By Lemma 1, $R$ is a reduced ring. Therefore,

$$r\left(a^n\right)=l\left(a^n\right)\subseteq M,1\in M.$$

This contradicts the maximality of $M$. So $RaR+l\left(a^n\right)=R$.

Moreover, we considered the case where $a$ is not a nilpotent element. Consider the following chain

$$RaR+l\left(a\right)\subseteq RaR+l\left(a^2\right)\subseteq \cdots .$$

Set ${\bigcup }_{i-1}^{\mathrm{\infty }}\left[RaR+l\left(a^i\right)\right]=I$. If $I\ne R$, then $I$ is contained in the maximal left ideal $M$ of $R$. Similarly, by Lemma 2, $M$ is essential. Thus, $R/M$ is flat. Since $a^n\in M$, there exists $u\in M$ such that $a^n=a^{n}u$. Hence,

$$\left(1-u\right)\in r\left(a^n\right)=l\left(a^n\right)\subseteq M,1\in M.$$

This contradicts the maximality of $M$. Thus, $RaR+l\left(a^n\right)=R$. □

In the following, we give a sufficient condition for the idempotent reflexive GP-V’-ring to be left weakly regular by using the condition “every maximal essential left ideal is a quasi-ideal”.

Theorem 2. 

If $R$ is an idempotent reflexive semi-abelian left GP-V’-ring whose every maximal essential left ideal is a quasi-ideal, then $R$ is a reduced left weakly regular ring.

Proof. 

Since $R$ is a left weakly regular ring, we obtain $RaR+l\left(a\right)=R$ for any $0\ne a\in R$. Otherwise, there exists $0\ne a\in R$ such that $RaR+l\left(a\right)\ne R$. Then, there exists a left ideal $M$ containing $RaR+l\left(a\right)$, and it is easy to see that $M$ is essential. Indeed, if $M$ is not essential, then $M$ is a direct sum term of $R$. Set $M=l\left(e\right)$, where $0\ne e^2=e\in R$. Then, if $RaR\mathrm{e}=0$, we have $aRe=0$. So, we have $eRa=0$,which implies $ea=0$. Therefore,

$$e\in l\left(a\right)\subseteq M=l\left(e\right).$$

Hence, $e=e^2=0$, contradicting the hypothesis. Therefore, $M$ is essential. Since $R/M$ is GP-injective, for any $0\ne a\in R$, there exists a positive integer $n$ such that the $R$-homomorphism from $R{a}^n$ to $R/M$ can be defined as follows.

$$f:R{a}^n\to R/M,f\left(r{a}^n\right)=r+M,r\in R.$$

By Lemma 4, $R$ is reduced, so the above definition is reasonable. There exists $b\in R$ such that

$$1+M=f\left(a^n\right)=a^{n}b+M.$$

Therefore, $1-a^{n}b\in M$, where $a^n\in l\left(a\right)\subseteq M$. By the fact that $M$ is a left ideal and a quasi-ideal, then

$$(1+b)a^n(1+b)\in M,b{a}^{n}b\in M,b{a}^n\in M.$$

Hence,

$$a^{n}b=(1+b)a^n(1+b)-a^n-b{a}^n-b{a}^{n}b\in M.$$

So $1=\left(1-a^{n}b\right)+a^{n}b\in M$, which contradicts the maximality of $M$. Therefore, for any $0\ne a\in R,$ we have $RaR+l\left(a\right)=R$. Thus, $R$ is left weakly regular. □

If the condition “a quasi-ideal” is replaced by “a weakly right ideal” in Theorem 2, the conclusion of the theorem still holds. We can obtain the following theorem:

Theorem 3. 

If $R$ is an idempotent reflexive semi-abelian left GP-V’-ring whose every maximal essential left ideal is a weakly right ideal, then $R$ is a reduced left weakly regular ring.

Proof. 

By Lemma 5, the ring $R$ is reduced. Next, we prove that the ring $R$ is left weakly regular. That is, we show that there exists $RaR+l\left(a\right)=R$ for any $0\ne a\in R$. Otherwise, there exists $0\ne a\in R$ such that $RaR+l\left(a\right)\ne R$. Then, there exists a left ideal $M$ containing $RaR+l\left(a\right)$, and it is easy to see that $M$ is essential. Indeed, if $M$ is not essential, then $M$ is a direct sum term of $R$. Set $M=l\left(e\right)$, where $0\ne e^2=e\in R$. When $RaRe=0$, i.e., we have $aR\mathrm{e}=0$. Since $R$ is idempotent reflexive, we have $eRa=0$, and this implies $ea=0$. Therefore,

$$e\in l\left(a\right)\subseteq M=l\left(e\right).$$

Hence, $e=e^2=0$, contradicting the hypothesis. Therefore, $M$ is essential. Since $R/M$ is GP-injective, for any $0\ne a\in R$, there exists a positive integer $n$ such that the $R$-homomorphism from $R{a}^n$ to $R/M$ can be defined as follows.

$$f:R{a}^n\to R/M,f\left(r{a}^n\right)=r+M,r\in R.$$

Since $R$ is reduced, the above definition is reasonable. Therefore, there exists $b\in R$ such that

$$1+M=f\left(a^n\right)=a^{n}b+M.$$

Therefore, $1-a^{n}b\in M$, where $a^n\in l\left(a\right)\subseteq M$. By the fact that $M$ is a left ideal and a weakly right ideal, there exists a positive integer $m$ such that ${\left(a^{n}b\right)}^m\in M$, and

$$1-{\left(a^{n}b\right)}^m=\left[1+a^{n}b+{\left(a^{n}b\right)}^2+\cdots {\left(a^{n}b\right)}^{m-1}\right]\left(1-a^{n}b\right)\in M.$$

Since $1=\left[1-{\left(a^{n}b\right)}^m\right]+{\left(a^{n}b\right)}^m\in M$, which contradicts the maximality of $M$, we have $RaR+l\left(a\right)=R$ for any $0\ne a\in R$. Hence, $R$ is left weakly regular. □

In Theorem 3, if the condition “a weakly right ideal” is replaced by “a generalized weak ideal”, the conclusion of the theorem still holds. We can obtain the following theorem:

Theorem 4. 

If $R$ is an idempotent reflexive semi-abelian left GP-V’-ring whose every maximal essential left ideal is a generalized weak ideal, then $R$ is a reduced left weakly regular ring.

Proof. 

By Lemma 6, it is clear that $R$ is reduced. Next, we prove that the ring $R$ is left weakly regular. That is, we show that for any $0\ne a\in R$, there exists $RaR+l\left(a\right)=R$. Otherwise, there exists $0\ne a\in R$ such that $RaR+l\left(a\right)\ne R$. Then, there exists a left ideal $M$ containing $RaR+l\left(a\right)$, and it is easy to see that $M$ is essential. Indeed, if $M$ is not essential, then $M$ is a direct sum term of $R.$ Set $M=l\left(e\right)$, where $0\ne e^2=e\in R$. When $RaRe=0$, we have $aR\mathrm{e}=0.$ So we have $eRa=0$, which implies $ea=0$. Therefore,

$$e\in l\left(a\right)\subseteq M=l\left(e\right).$$

Hence, $e=e^2=0$, contradicting the hypothesis. Therefore, $M$ is essential. Since $R/M$ is GP-injective, for any $0\ne a\in R$, there exists a positive integer $n$ such that the $R$-homomorphism from $R{a}^n$ to $R/M$ can be defined as follows.

$$f:R{a}^n\to R/M,f\left(r{a}^n\right)=r+M,r\in R.$$

Since $R$ is reduced, the above definition is reasonable. Therefore, there exists $b\in R$ such that

$$1+M=f\left(a^n\right)=a^{n}b+M.$$

Therefore, $1-a^{n}b\in M$, where $a^n\in l\left(a\right)\subseteq M$. By the fact that $M$ is a left ideal and a generalized weak ideal, then

$$b{a}^n\in M,b\left(1-a^{n}b\right)=b-b{a}^{n}b\in M.$$

there exists a positive integer $m$ such that ${\left(b{a}^n\right)}^{m}b\in M.$ Since $M$ is a left ideal, we have

$$\begin{array}{c}{\left(b{a}^n\right)}^{m-1}b={\left(b{a}^n\right)}^{m-1}\left(b-b{a}^{n}b\right)+{\left(b{a}^n\right)}^{m}b\in M,\\ {\left(b{a}^n\right)}^{m-2}b={\left(b{a}^n\right)}^{m-2}\left(b-b{a}^{n}b\right)+{\left(b{a}^n\right)}^{m-1}b\in M,\\ .....................\\ b{a}^{n}b=b{a}^n\left(b-b{a}^{n}b\right)+{\left(b{a}^n\right)}^{2}b\in M,\\ b=\left(b-b{a}^{n}b\right)+b{a}^{n}b\in M.\end{array}$$

Hence, $a^{n}b\in M$. Furthermore, noting that $1=\left(1-a^{n}b\right)+a^{n}b\in M$, this contradicts the maximality of $M$. Therefore, for any $0\ne a\in R$, we have $RaR+l\left(a\right)=R$. Thus, $R$ is left weakly regular. □

Strengthening the condition “semi-abelian” to “semi-commutative” in Theorem 2, Theorem 3, and Theorem 4, respectively, we obtain the following three strong regularity theorems of the rings.

Theorem 5. 

$R$ is a strongly regular ring if and only if $R$ is an idempotent reflexive semi-commutative left GP-V’-ring whose every maximal essential left ideal is a quasi-ideal.

Proof. 

The necessity of the theorem is clearly established, so we are only required to show that sufficiency holds. By Lemma 7, $R$ is reduced. Next, we prove that $Ra+l\left(a\right)=R$ for any $0\ne a\in R$. Otherwise, there exists $0\ne a\in R$ such that $Ra+l\left(a\right)\ne R$. Then, there exists a maximal left ideal $M$ of $R$ containing $Ra+l\left(a\right)$. Similar to Lemma 2, it can be shown that $M$ is essential. Thus, $R/M$ is GP-injective. Since $R$ is reduced, we can define

$$f:R{a}^n\to R/M,f\left(r{a}^n\right)=r+M.$$

Thus, there exists $b\in R$ such that

$$1+M=f\left(a^n\right)=a^{n}b+M,1-a^{n}b\in M.$$

where $a^n\in l\left(a\right)\subseteq M$. By the fact that $M$ is a left ideal and a quasi-ideal, then

$$(1+b)a^n(1+b)\in M,b{a}^{n}b\in M,b{a}^n\in M.$$

Hence,

$$a^{n}b=(1+b)a^n(1+b)-a^n-b{a}^n-b{a}^{n}b\in M.$$

Thus, $1=\left(1-a^{n}b\right)+a^{n}b\in M$, and

$$Ra+l\left(a\right)=R,\forall 0\ne a\in R.$$

Thus, $R$ is a strongly regular ring. □

Theorem 6. 

$R$ is a strongly regular ring if and only if $R$ is an idempotent reflexive semi-commutative left GP-V’-ring whose every maximal essential left ideal is a weakly right ideal.

Proof. 

The necessity of the theorem is clearly established, so we are only required to show that sufficiency holds. By Lemma 8, $R$ is reduced. Next, we prove that $Ra+l\left(a\right)=R$, for any $0\ne a\in R$. Otherwise, there exists $0\ne a\in R$ such that $Ra+l\left(a\right)\ne R$. Then, there exists a maximal left ideal $M$ of $R$ containing $Ra+l\left(a\right)$. Similar to the proof of Lemma 2, we can prove that $M$ is essential. Noting that $l\left(a\right)\subseteq M$, the left $R$-module homomorphism can be defined as follows.

$$f:Ra\to R/M,f\left(ra\right)=r+M.$$

Since $R$ is a left GP-V’-ring, the simple singular left $R$-module $R/M$ is GP-injective. Thus, there exists $b\in R$ such that

$$1+M=f\left(a\right)=ab+M,1-ab\in M.$$

Since $M$ is a weakly right ideal, there exists a positive integer $n$ such that $(ab{)}^n\in M$. Similar to the proof of Lemma 3, we can obtain $1\in M$. This contradicts the maximality of $M$. Therefore,

$$Ra+l\left(a\right)=R,\forall 0\ne a\in R.$$

Hence, $R$ is a strongly regular ring. □

Theorem 7. 

$R$ is a strongly regular ring if and only if $R$ is an idempotent reflexive semi-commutative left GP-V’-ring whose every maximal essential left ideal is a generalized weak ideal.

Proof. 

The necessity of the theorem is clearly established, so we are only required to show that sufficiency holds. By Lemma 9, $R$ is reduced. Next, we prove that $Ra+l\left(a\right)=R$, for any $0\ne a\in R$. Otherwise, there exists $0\ne a\in R$ such that $Ra+l\left(a\right)\ne R$. Then there exists a maximal left ideal $M$ of $R$ containing $Ra+l\left(a\right)$. Similar to Lemma 2, it can be shown that $M$ is essential. Thus, $R/M$ is GP-injective. Since $R$ is reduced, we can define

$$f:R{a}^n\to R/M,f\left(r{a}^n\right)=r+M.$$

Thus, there exists $b\in R$ such that

$$1+M=f\left(a^n\right)=a^{n}b+M,1-a^{n}b\in M.$$

Since $M$ is a left ideal, $b\left(1-a^{n}b\right)\in M$, i.e., $b-b{a}^{n}b\in M$. As $b{a}^n\in M$ and $M$ is a generalized weak ideal, there exists a positive integer $m$ such that ${\left(b{a}^n\right)}^{m}b\in M$. Note that

$$\begin{array}{c}{\left(b{a}^n\right)}^{m-1}b={\left(b{a}^n\right)}^{m-1}\left(b-b{a}^{n}b\right)+{\left(b{a}^n\right)}^{m}b\in M,\\ {\left(b{a}^n\right)}^{m-2}b={\left(b{a}^n\right)}^{m-2}\left(b-b{a}^{n}b\right)+{\left(b{a}^n\right)}^{m-1}b\in M,\\ .....................\\ b{a}^{n}b=b{a}^n\left(b-b{a}^{n}b\right)+{\left(b{a}^n\right)}^{2}b\in M.\end{array}$$

and $b=\left(b-b{a}^{n}b\right)+b{a}^{n}b\in M$. Hence,

$$a^{n}b\in M,1\in M.$$

This contradicts the maximality of $M$. Thus,

$$Ra+l\left(a\right)=R,\forall 0\ne a\in R.$$

Therefore, there exists $b\in R,c\in l\left(a\right)$, such that

$$ba+c=1$$

Thus,

$$b{a}^2=a.$$

Therefore, $R$ is a strongly regular ring. □

4. Conclusions

This paper mainly studies the regularity of idempotent reflexive GP-V’-rings. By using the concepts of generalized weak ideals, weakly right ideals, and quasi-ideals, we obtain several weakly and strongly regular theorems of idempotent reflexive GP-V’-rings.