G-Morphic Rings and G-regular Rings

Abstract

A ring R is called left G-morphic if l(a) is a principal left ideal for each a ∈ R. A ring R is called left G-regular if R is left G-morphic and left P-injective. Several properties of the two classes of rings are investigated, conditions under which left G-regular rings are regular rings as well as semisimple artinian rings are given, respectively.

1 Introduction

Throughout this paper, R denotes an associative ring with identity and all modules considered are unitary, m, n are positive integers unless otherwise specified. We call a ring regular if it is von Neumann regular. For any module M, M ^∗ denotes Hom _R (M, R), and M ⁺ denotes $\text {Hom}_{\mathbb {Z}}(M, \mathbb {Q}/\mathbb {Z})$, where $\mathbb {Q}$ is the set of rational numbers, and $\mathbb {Z}$ is the set of integers. For an element a of the ring R, the right and left annihilators of a are denoted by l(a) and r(a), respectively.

First, we recall some concepts.

• (1) R is called left coherent if every finitely generated left ideal of R is finitely presented.

• (2) R is called left n -coherent [9] if every n-generated left ideal of R is finitely presented. Clearly, R is left 1-coherent if and only if l(a) is finite generated for each a ∈ R. Left 1-coherent rings are also called left (1, 1)-coherent in [11].

• (3) An element a in R is called left morphic if l(a) ≅ R/R a; the ring R is called a left morphic ring if every element in R is left morphic [6].

• (4) An element a in R is called left generalized morphic if l(a) ≅ R/R b for some b ∈ R; the ring R is called a left generalized morphic ring if every element in R is left generalized morphic [12]. By [12, Corollary 2.3], a ring R is a left generalized morphic ring if and only if l(a) is a principal left ideal for each a ∈ R.

• (5) A ring R is called left PP [5] if every principal left ideal of R is projective. Clearly, left PP rings are left generalized morphic, but left generalized morphic rings need not be left PP. For example, the ring $R=\mathbb {Z}_{4}$ is a commutative generalized morphic ring, but it is not PP (see Example 2).

• (6) A left R-module M is called P-injective [7] if every R-homomorphism from a principal left ideal of R to M extends to a homomorphism of R to M; the ring R is called left P-injective if _R R is P-injective. The P-injective dimension P-i d( _R M) of a module _R M is defined to be the smallest integer n ≥ 0 such that $\text {Ext}^{n+1}_{R}(R/Ra, M)=0$ for all a ∈ R. If no such n exists, set P- i d( _R M) = ∞. l.P-i dim(R) is defined as sup{P-i d(M)∣M ∈ R- Mod} [12].

• (7) A right R-module M is called n -flat [3] if the canonical map M _R ⊗I → M is monic for every n-generated left ideal I of R. 1-flat modules are also called P-flat in some literatures such as [12]. It is easy to see that M _R is P-flat if and only if $\text {Tor}^{R}_{1}(M, R/Ra)=0$ for all a ∈ R. The P-flat dimension P- f d(M _R ) of a module M _R is defined to be the smallest integer n ≥ 0 such that $\text {Tor}^{R}_{n+1}(M, R/Ra)=0$ for all a ∈ R. If no such n exists, set P- f d( _R M) = ∞. r.P- fdim(R) is defined to be sup{P-f d(M)∣M ∈ Mod-R} [12].

• (8) A left R-module M is called (m, n)-presented [11, 13] if there exists an exact sequence of left R-modules 0 → K → R ^m → M → 0, where K is n-generated.

• (9) A left R-module M is called (m, n)-injective [2] if every R-homomorphism from an n-generated submodule of R ^m to M extends to the one from R ^m to M. Clearly, a module is P-injective if and only if it is (1, 1)-injective.

• (10) A left R-module M is called (m, n)-flat [11, 13] if the canonical map I⊗ _R M → R ^m⊗ _R M is monic for every n-generated submodule I of the right R-module R ^m. Clearly, a module is P-flat if and only if it is (1, 1)-flat.

• (11) If U is a right R-module and U ^′ is a submodule of U, then U ^′ is called an (m, n)-pure submodule of U if the canonical map U ^′⊗ _R V → U⊗ _R V is monic for every (m, n)-presented left R-module V. In this case, the exact sequence 0 → U ^′ → U → U/U ^′ → 0 is called (m, n)-pure [13].

In this paper, we shall further investigate left generalized morphic rings; several of properties of this class of rings will be given. Especially, left generalized morphic left P-injective rings will be studied; relations between this class of rings and regular rings, strongly regular rings, as well as semisimple artinian rings will be given.

Using the standard techniques, one can prove the following propositions.

Proposition 1

[13, Theorem 1.5] Let U ^′ be a submodule of the right R-module U, then the following statements are equivalent:

• (1) U ^′ is (m, n)-pure in U.

• (2) For every (n, m)-presented left R-module V, the canonical map Hom _R (V, U) → Hom _R (V,U/U ^′) is epic.

• (3) (U ^′)^m ∩ U ⁿ C = (U ^′)ⁿ C for all C ∈ R ^{n × m}.

Proposition 2

[13, Theorem 2.4] Suppose that A _R ≤ B _R and B _R is (m, n)-injective, then A is (m, n)-injective if and only if A is (n, m)-pure in B.

Proposition 3

[13, Theorem 3.6] Let $U^{\prime }_{R}\leq U_{R}$. Then,

• (1) If U/U ^′ is (m, n)-flat, then U ^′ is (m,n)-pure in U.

• (2) If U ^′ is (m, n)-pure in U and U is (m, n)-flat, then also U/U ^′ is (m, n)-flat.

Proposition 4

Let M be a left R-module. Then, the following statements are equivalent:

• (1) M is P-injective.

• (2) $\text {Ext}^{1}_{R}(R/Ra, M)=0$ for all a ∈ R.

• (3) Every exact sequence of left R-modules 0 → M → M ^′ → M ^′′ → 0 is (1, 1)-pure.

• (4) There exists a (1, 1)-pure exact sequence of left R-modules 0 → M → M ^′ →M ^′′ →0, where M ^′ is P-injective.

• (5) r _M l(a) = aM for all a ∈ R.

Proof

(1) ⇒ (3) by [13, Theorem 2.2]. (3) ⇒ (4) is clear. (4) ⇒ (1) by Proposition 2. The other implications are easy. □

2 G-Morphic Rings

In order to facilitate, we call a ring R left G-morphic if it is left generalized morphic. Clearly, left G-morphic rings are left 1-coherent. However, the converse is false. For example, Zhu and Ding give a left and right artinian ring which is not right G-morphic [12, Example 2.7]. In fact, even if R is a commutative artinian local ring, it need not be G-morphic.

Example 1

Let F be a field, and let

$$R=\left\{\left.\left[ \begin{array}{lll} a & b & c \\ &a & 0 \\ &&a \end{array} \right] ~\right|~ a, b, c\in F\right\}. $$

Then, R is a commutative artinian local ring, but R is not G-morphic.

Proof

It is obvious that R is a commutative local ring with unit element

$$1_{R}=\left[ \begin{array}{lll} 1_{F} & 0 & 0\\ & 1_{F} & 0 \\ && 1_{F} \end{array} \right], $$

the set of zero divisors of R is

$$D=\left\{\left.\left[ \begin{array}{lll} 0 & b & c \\ & 0 & 0 \\ && 0 \end{array} \right]~\right|~ b, c\in F\right\},\quad D^{2}=0, $$

D is the unique maximal ideal of R. Noting that

$${{~}_{R}}D=R\left[ \begin{array}{lll} 0 & 1_{F} & 0\\ & 0 & 0 \\ && 0 \end{array} \right] \oplus R\left[ \begin{array}{lll} 0& 0& 1_{F}\\ & 0& 0 \\ && 0 \end{array} \right] $$

and that $R\left [ \begin {array}{lll}0&1_{F}&0\\&0&0 \\&&0 \end {array} \right ]$ and $R\left [ \begin {array}{lll} 0& 0 & 1_{F}\\&0&0 \\&&0 \end {array} \right ]$ are simple R-modules, we have that R is artinian. We assert that

$$\mathbf{l}\left[ \begin{array}{lll} 0 & 1_{F} & 1_{F}\\ & 0 & 0 \\ && 0 \end{array} \right]=D\neq Rd\quad\text{ for all }~d\in D. $$

Otherwise, if D = R d for some $d=\left [ \begin {array}{lll}0&b&c\\&0&0 \\&&0 \end {array} \right ]$, let

$$\left[ \begin{array}{lll} 0 & 1_{F} & 1_{F}\\ & 0 & 0 \\ && 0 \end{array} \right]= \left[ \begin{array}{lll} a_{1} & b_{1} & c_{1}\\ & a_{1} & 0 \\ && a_{1} \end{array} \right] \left[ \begin{array}{lll} 0 & b & c\\ & 0 & 0 \\ && 0 \end{array} \right], $$

then b = c. Observing that

$$\left[ \begin{array}{lll} 0 & 0 & 1_{F}\\ & 0 & 0 \\ && 0 \end{array} \right]\in D= Rd, $$

we have 0=1 _F , a contradiction. Therefore, R is not G-morphic. □

Theorem 1

Let R be a left G-morphic ring, and let M be a (1, 1)-presented left R-module. Then, fd(M) = pd(M).

Proof

Clearly, f d(M) ≤ p d(M). Conversely, suppose that f d(M) = n < ∞. Since M is (1, 1)-presented, there exists an exact sequence 0→ R a ₁→ R → M → 0. But R is left G-morphic, we have an exact sequence 0→ R a ₂→ R → R a ₁→0. Continue in this way, we obtain an exact sequence 0→ R a _n → R → ⋯→ R → M → 0. Since f d(M) = n, R a _n is flat. Note that R a _n is finite presented, it is projective. Therefore, p d(M) ≤ n. □

Theorem 2

Let R be a left G-morphic ring, a ∈ R, n ≥ 0. Then, pd(R/Ra)≤n if and only if $\text {Ext}^{n+1}_{R}(R/Ra, R/Rb)=0$ for all b∈R.

Proof

“ ⇒” It is clear.

“ ⇐” We use induction on n. If n = 0, by the exact sequence

$$ 0\rightarrow Ra\rightarrow R\rightarrow R/Ra\rightarrow 0 $$

(1)

we have an exact sequence $\text {Hom}_{R}(R, Ra)\rightarrow \text {Hom}_{R}(Ra,Ra)\rightarrow \text {Ext}^{1}_{R}(R/Ra, Ra)$. Since R is left G-morphic, R a is (1, 1)-presented, and then $\text {Ext}^{1}_{R}(R/Ra, Ra)=0$ by hypothesis. Hence, the homomorphism Hom _R (R, R a)→Hom _R (R a, R a) is epic. Thus, the exact sequence (1) is split, this follows that R/R a is projective, that is, p d(R/R a)=0. If n ≥ 1, for every (1, 1)-presented left R-module B, we have an exact sequence $0=\text {Ext}^{n}_{R}(R, B)\rightarrow \text {Ext}^{n}_{R}(Ra, B)\rightarrow \text {Ext}^{n+1}_{R}(R/Ra, B)\rightarrow \text {Ext}^{n+1}_{R}(R, B)=0$, this follows that $\text {Ext}^{n}_{R}(Ra, B)\cong \text {Ext}^{n+1}_{R}(R/Ra, B)=0$. Since R is left G-morphic, R a is (1, 1)-presented, hence p d(R a) ≤ n−1 by induction hypothesis, and whence p d(R/R a) ≤ n. □

Corollary 1

Let R be a left G-morphic ring, a ∈ R, n ≥ 0. If pd(R/Ra) = n, then $\text {Ext}^{n}_{R}(R/Ra, R)\neq 0$.

Proof

Since p d(R/R a) = n, by Theorem 2, there exists b ∈ R such that $\text {Ext}^{n}_{R}(R/Ra,\allowbreak R/Rb)\neq 0$. But R b is (1, 1)-presented because R is left G-morphic, $\text {Ext}^{n+1}_{R}(R/Ra, Rb)\allowbreak = 0$ again by Theorem 2. Then, we get an exact sequence

$$\text{Ext}^{n}_{R}(R/Ra, R)\rightarrow \text{Ext}^{n}_{R}(R/Ra,R/Rb)\rightarrow 0, $$

it shows that $\text {Ext}^{n}_{R}(R/Ra, R)\neq 0$. □

Lemma 1

Every (m,n)-presented (n,m)-flat left R-module is projective.

Proof

Let A be an (m, n)-presented (n, m)-flat left R-module. Since A is (m, n)-presen-ted, there exists an exact sequence of left R-modules 0→ K → R ^m → A → 0, where K is n-generated. But A is (n, m)-flat, by Proposition 3(1), K is (n, m)-pure in R ^m. So by Proposition 1, the canonical map Hom _R (A, R ^m)→Hom _R (A, A) is epic, which implies that A is isomorphic to a direct summand of R ^m, and hence A is projective. □

Corollary 2

[8, Corollary 3.58] Every finitely presented flat module is projective.

Theorem 3

The following statements are equivalent for a ring R.

• (1) R is left PP.

• (2) R is left G-morphic and every principal left ideal of R is P-flat.

Proof

(1) ⇒ (2) is trivial.

(2) ⇒ (1). Assume (2). Then, for each a ∈ R, R a is (1, 1)-presented and P-flat, so R a is projective by Lemma 1. □

Corollary 3

If R is a domain, then R is left PP if and only if every principal left ideal of R is P-flat.

Proposition 5

Let R be a left G-morphic ring. If N ₁ is a P-injective submodule of a P-injective left R-module N, then N/N ₁ is P-injective.

Proof

Let a ∈ A. Since R is left G-morphic, R a≅R/R b for some b ∈ R, and so $\text {Ext}^{1}_{R}(Ra, N_{1})=0$. Noting that $\text {Ext}^{1}_{R}(Ra, N_{1})\cong \text {Ext}^{2}_{R}(R/Ra, N_{1})$, we have $\text {Ext}^{2}_{R}(R/Ra, N_{1})=0$. Thus, from the exact sequence

$$0=\text{Ext}^{1}_{R}(R/Ra, N)\rightarrow \text{Ext}^{1}_{R}(R/Ra,N/N_{1})\rightarrow \text{Ext}^{2}_{R}(R/Ra, N_{1})=0, $$

we get that $\text {Ext}^{1}_{R}(R/Ra, N/N_{1})=0$. Hence, N/N ₁ is P-injective. □

3 G-Regular Rings

In this section, we study left G-morphic left P-injective rings. Let M be a right R-module. The group M ^∗=Hom _R (M, R) becomes a left R-module which we call the dual of M. If we do it again to get M ^∗∗=Hom _R (M ^∗,R), which is a right R-module. There is a natural homomorphism of M to $M^{\ast \ast }, ~M\overset {\sigma }{\rightarrow } M^{\ast \ast }$ caused by considering the elements of M as homomorphisms of M ^∗ into R. Following the terminology of Bass [1], M is said to be torsionless if σ is a monomorphism, reflexive if σ is an isomorphism.

Theorem 4

Let R be a left G-morphic ring. Then, the following statements are equivalent:

• (1) R is left P-injective.

• (2) $\text {Ext}^{1}_{R}(R/Ra, R)=0$ for all a ∈ R.

• (3) Every injective right R-module is P-flat.

• (4) Every P-flat left R-module is P-injective.

• (5) Every projective left R-module is P-injective.

• (6) Every right R-module is a submodule of a P-flat right R-module.

• (7) The right R-module R/aR is reflexive for each a ∈ R.

• (8) The right R-module R/aR is torsionless for each a ∈ R.

Proof

(1) ⇒ (2); (4) ⇒ (5) ⇒ (1); (3) ⇒ (6); and (7) ⇒ (8) are obvious.

(2) ⇒ (3). Let E be an injective right R-module. Then, for any a ∈ R, since R is left G-morphic, R/R a has a projective resolution each of whose terms is R, by a remark of [8, Theorem 9.51], we have

$$\text{Tor}^{R}_{1}(E, R/Ra)\cong \text{Tor}^{R}_{1}(\text{Hom}_{R}(R, E), R/Ra)\cong \text{Hom}_{R}(\text{Ext}^{1}_{R}(R/Ra, R), E). $$

By (2), $\text {Ext}^{1}_{R}(R/Ra, R)=0$, so $\text {Tor}^{R}_{1}(E, R/Ra)=0$, i.e., E is P-flat.

(3)⇒(1). Let C _R be an injective cogenerator and a ∈ R. Since R is left G-morphic, by a remark of [8, Theorem 9.51] again, we have

$$\text{Tor}^{R}_{1}(C, R/Ra)\cong \text{Tor}^{R}_{1}(\text{Hom}_{R}(R, C), R/Ra)\cong \text{Hom}_{R}(\text{Ext}^{1}_{R}(R/Ra, R), C). $$

By (3), $\text {Tor}^{R}_{1}(C, R/Ra)=0$. Thus, $\text {Ext}^{1}_{R}(R/Ra, R)=0$, and then R is left P-injective.

(1) ⇒ (4). Let U be a P-flat left R-module. Take an exact sequence of left R-modules 0→ K → F → U → 0, where F is a free module. Since _R R is P-injective, F is also P-injective. But U is P-flat, by Proposition 3(1), K is (1, 1)-pure in F. And hence K is also P-injective by Proposition 2. Since R is left G-morphic, by Proposition 5, U is P-injective.

(6) ⇒ (3). Let E be an injective right R-module. By (6), we have an exact sequence of right R-modules 0→ E → U → U/E → 0 with U P-flat. Hence, U≅E⊕U/E, and then E is P-flat.

(1) ⇒ (7). Since R is left G-morphic and left P-injective, by [12, Lemma 3.6], $\text {Ext}^{n}_{R}(R/Ra, R)=0$ for all a ∈ R and all positive integers n. Let $R\overset {d_{1}}{\rightarrow } R \overset {d_{0}}{\rightarrow } R/aR\rightarrow 0$ be exact, by [4, Lemma 2.2], we have an exact sequence

$$0\rightarrow \text{Ext}^{1}_{R}(N, R)\rightarrow R/aR\rightarrow (R/aR)^{\ast\ast}\rightarrow \text{Ext}^{2}_{R}(N, R)\rightarrow 0, $$

where $N=R^{\ast }/\text {Im}(d^{\ast }_{1})$ is a (1, 1)-presented left R-module. Thus

$$\text{Ext}^{1}_{R}(N, R)= \text{Ext}^{2}_{R}(N, R)=0, $$

and then R/R a → (R/R a)^∗∗ is an isomorphism, as required.

(8) ⇒ (1). To prove (1), we need only to prove that if a, b ∈ R with l(a) ⊆ l(b) then b = a c for some c ∈ R by [7, Lemma 1.1]. Indeed, if b∉a R, then 0≠b+a R ∈ R/a R. Since R/a R is torsionless, it embeds in a direct product of R and so there exists a right R-homomorphism g:R/a R → R such that g(b+a R)≠0. Then, g(1+a R)∉l(b) but g(1+a R) ∈ l(a), a contradiction. □

Recall that a ring R is regular if and only if R is left PP and left P-injective.

Definition 1

A ring R is called left G-regular, if R is left G-morphic and left P-injective.

Clearly, regular rings are left G-regular, but the inverse implication is not true.

Example 2

The ring $R=\mathbb {Z}_{4}=\{\overline { 0}, \overline { 1}, \overline { 2}, \overline { 3}\}$ is a commutative G-regular ring, but it is not regular.

Proof

It is obvious that R is a commutative G-regular ring. Since $\mathbf {l}(\overline { 2})=\{\overline { 0}, \overline {2}\}$ contains no nonzero idempotent elements, it is not a direct summand of _R R, so that $R\overline {2}$ is not projective, and thus R is not PP. Therefore, R is not regular. □

Example 3

• (1) Let R be the ring of 2 × 2 upper triangular matrices over the field $\mathbb {Z}_{2}$. Then, R is left G-morphic, but it is not left G-regular.

• (2) The ring $\mathbb {Z}$ of integers is G-morphic but not G-regular.

Proof

(1). Let e _{i j} be the 2 × 2 matrices over the field $\mathbb {Z}_{2}$ having a lone 1 as its (i, j)-entry and all other entries 0, i, j = 1, 2. Then, by a routine computation, we get l(0) = R, l(e ₁₁ + e ₂₂) = l(e ₁₁ + e ₁₂ + e ₂₂)=0,l(e ₁₁) = R e ₂₂, l(e ₁₂) = R e ₂₂, l(e ₂₂) = R e ₁₁, l(e ₁₁ + e ₁₂) = R e ₂₂, l(e ₁₂ + e ₂₂) = R(e ₁₁ + e ₁₂); hence, R is left G-morphic. But since

$$\mathbf{r}\mathbf{l}(e_{12})= \left[\begin{array}{cc} {\mathbb Z}_{2} &~ {\mathbb Z}_{2} \\ 0 &~ 0 \end{array}\right]\neq \left[\begin{array}{cc} 0 &~ {\mathbb Z}_{2} \\ 0 &~ 0 \end{array}\right]=e_{12}R, $$

R is not left P-injective by Proposition 4, and so R is not left G-regular.

(2). It is obvious. □

Theorem 5

For a ring R, the following statements are equivalent:

• (1) R is left G-regular.

• (2) For any a ∈ R, there exists b ∈ R such that l(a) = Rb and r(b) = aR.

Proof

(1) ⇒ (2). Let a ∈ R. Since R is left G-morphic, by [12, Corollary 2.3], there exists b ∈ R such that l(a) = R b, which implies that r l(a) = r(b). But R is left P-injective, by Proposition 4, we have r l(a) = a R, and hence r(b) = a R.

(2) ⇒ (1). Assume (2), then it is clear that R is left G-morphic. For any a ∈ R, by hypothesis, there exists b ∈ R such that l(a) = R b and r(b) = a R, so r l(a) = r(b) = a R, and thus R is left P-injective by Proposition 4. Hence, R is left G-regular. □

Theorem 6

For a left G-regular ring R, the following statements are equivalent:

• (1) R is regular.

• (2) M ^∗ is P-flat for every left R-module M.

• (3) R is left PP.

• (4) R is right PP.

• (5) Every principal right ideal of R is P-flat.

• (6) Every principal left ideal of R is P-flat.

Proof

(1) ⇒ (2) through (6) are clear. (6) ⇒ (3) by Theorem 3.

(2) ⇒ (1). For any a ∈ R, since R is left G-regular, by Theorem 4, R/a R≅(R/a R)^∗∗=((R/a R)^∗)^∗. But ((R/a R)^∗)^∗ is P-flat by hypothesis, so R/a R is P-flat and hence projective by Lemma 1. It follows that a R is a direct summand of R _R . Consequently, R is regular.

(3) ⇒ (1). By [10, Theorem 3], left P-injective left PP ring is regular.

(4) ⇒ (3). Since R is right PP, every principal right ideal of R is projective and hence flat. By [5, Theorem 2.2], every principal left ideal of R is flat. Note that R is left G-morphic , by Theorem 3, R is left PP.

(5) ⇒ (1). For any a ∈ R, since R is left P-injective, we have r l(a) = a R. Since R is left G-morphic, l(a) = R b for some b ∈ R. Thus r(b) = a R, and hence R/a R = R/r(b) ≅ b R is P-flat by hypothesis. By Lemma 1, R/a R is projective, so that a R is a direct summand of R _R and (1) follows. □

Theorem 7

The following statements are equivalent for a ring R:

• (1) R is a strongly regular ring.

• (2) R is a reduced left G-regular ring.

Proof

(1) ⇒ (2). It is obvious.

(2) ⇒ (1). Let a ∈ R. Since R is reduced, we have l(a ²) = l(a). But R is left P-injective, by Proposition 4, we have a ² R = a R, and so a = a ² b for some b ∈ R, as required. □

Lemma 2

If R is a left G-regular ring with ACC on principal left ideals, then R is left perfect.

Proof

Suppose that a ₁ R ⊇ a ₂ R ⊇ ⋯. Then, l(a ₁) ⊆ l(a ₂) ⊆ ⋯. Since R is a left G-regular ring with ACC on principal left ideals, by Theorem 5, there exists a positive integer n such that l(a _n+1) = l(a _n+2) = ⋯. Noting that R is left P-injective, by [7, Lemma 1.1], we have a _n+1 R = a _n+2 R = ⋯. Therefore, R is left perfect. □

Lemma 3

If R is a left perfect semiprime ring, then any nonzero right ideal of R is not nil.

Proof

Let I be any nonzero right ideal of R. We claim that there exists 0≠b ∈ I such that b R is simple. If not, let 0 ≠ a ₁ ∈ I, then there is a ₂ ∈ R such that a ₁ a ₂ ≠ 0 and a ₁ R ≠ a ₁ a ₂ R. Since 0 ≠ a ₁ a ₂ ∈ I, there exists a ₃ ∈ R such that a ₁ a ₂ a ₃ ≠ 0 and a ₁ a ₂ R ≠ a ₁ a ₂ a ₃ R. Continuing in this way, we get a strictly descending chain of right ideals of R

$$a_{1}R\supsetneq a_{1}a_{2}R\supsetneq a_{1}a_{2}a_{3}R\supsetneq\cdots, $$

a contradiction. This proves the claim.

Since R is semiprime, there exists c ∈ R such that b c b ≠ 0, and so b R = b c b R because b R is simple. Since c b ≠ 0, c∉l(b) = l(b c b); hence, (c b)² ≠ 0. Continuing this process, we have that (c b)ⁿ ≠ 0 for every positive integer n. Therefore, I is not nil. □

Theorem 8

The following statements are equivalent for a ring R:

• (1) R is a semisimple artinian ring.

• (2) R is a left G-regular semiprime ring with ACC on principal left ideals.

Proof

(1) ⇒ (2). It is clear.

(2) ⇒ (1). Since R is a left G-regular ring with ACC on principal left ideals, by Lemma 2, R is left perfect. This follows that R is semilocal and J(R) is nil. But R is semiprime, by Lemma 3, J(R)=0. Therefore, R is semisimple artinian. □

Proposition 6

Let R be a left and right G-morphic ring. Then, the following statements are equivalent:

• (1) R is left P-injective.

• (2) Every P-injective right R-module is P-flat.

• (3) Every injective right R-module is P-flat.

Proof

(2) ⇒ (3) is obvious. (3) ⇒ (1) by Theorem 4.

(1) ⇒ (2). Let M be a P-injective right R-module. Since R is right G-morphic and $\mathbb {Q}/\mathbb {Z}$ is an injective $\mathbb {Z}$-module, by a remark of [8, Theorem 9.51], we have

$$\text{Tor}^{R}_{1}\left(\text{Hom}_{\mathbb{Z}}(M, \mathbb{Q}/\mathbb{Z}), R/aR\right)\cong \text{Hom}_{\mathbb{Z}}\left(\text{Ext}^{1}_{R}(R/aR, M), \mathbb{Q}/\mathbb{Z}\right). $$

Since M is P-injective, $\text {Ext}^{1}_{R}(R/aR,M)=0$, so $\mathrm{Tor}^{R}_{1}(\mathrm{Hom}_{\mathbb{Z}}(M, \mathbb{Q}/\mathbb{Z}),R/aR)=0$, and hence $M^{+}=\text {Hom}_{\mathbb {Z}}(M, \mathbb {Q}/\mathbb {Z})$ is a P-flat left R-module. But R is left G-morphic and left P-injective, by Theorem 4, M ⁺ is P-injective. Thus, for any b ∈ R, we have

$$0=\text{Ext}^{1}_{R}(R/Rb,\text{Hom}_{\mathbb{Z}}(M, \mathbb{Q}/\mathbb{Z}))\cong \text{Hom}_{\mathbb{Z}}\left(\text{Tor}^{R}_{1}(M, R/Rb), \mathbb{Q}/\mathbb{Z}\right). $$

Hence, $\text {Tor}^{R}_{1}(M, R/Rb)=0$ because $\mathbb {Q}/\mathbb {Z}$ is a cogenerator, i.e., M is P-flat. □

Proposition 7

Let 0 → A → B → C → 0 be an exact sequence of left R-modules. Then,

• (1) If A, C are P-flat, then B is also P-flat.

• (2) If R is right G-morphic and B, C are P-flat, then A is also P-flat.

• (3) If R is left G-regular and A, B are P-flat, then C is also P-flat.

Proof

(1). It is easy to prove by using the exact sequence

$$\text{Tor}^{R}_{1}(R/aR,A)\rightarrow \text{Tor}^{R}_{1}(R/aR, B)\rightarrow \text{Tor}^{R}_{1}(R/aR, C), $$

where a ∈ R.

(2). For any a ∈ R, the exact sequence

$$0=\text{Tor}^{R}_{2}(R,C)\rightarrow \text{Tor}^{R}_{2}(R/aR, C)\rightarrow \text{Tor}^{R}_{1}(aR, C)\rightarrow \text{Tor}^{R}_{1}(R, C)=0 $$

implies that $\text {Tor}^{R}_{2}(R/aR, C)\cong \text {Tor}^{R}_{1}(aR,C)$. Since R is right G-morphic, a R is (1, 1)-presented. Note that C is P-flat, we have $\text {Tor}^{R}_{1}(aR, C)=0$, and then $\text {Tor}^{R}_{2}(R/aR, C)=0$. This shows that $\text {Tor}^{R}_{1}(R/aR, A)=0$ from the exact sequence $0=\text {Tor}^{R}_{2}(R/aR,C)\rightarrow \text {Tor}^{R}_{1}(R/aR, A)\rightarrow \text {Tor}^{R}_{1}(R/aR, B)=0$. Therefore, A is P-flat.

(3). Since R is left G-regular and A is a P-flat left R-module, by Theorem 4, A is P-injective, and so A is a (1, 1)-pure submodule of B by Proposition 4. But B is P-flat, by Proposition 3(2), C is P-flat. □

Proposition 8

Let R be a right G-regular left G-morphic ring, and let M be a right R-module. Then, P- fd(M _R ) = 0 or ∞.

Proof

If P- f d(M _R ) ≤ 1, since R is left G-morphic, by [12, Lemma 3.8], there exists an exact sequence of right R-modules 0 → F ₁ → F ₀ → M → 0, where F ₁, F ₀ are P-flat. Since R is right G-regular, by Proposition 7(3), M is P-flat, i.e., P-f d(M _R )=0. Assume that 1< P-f d(M _R ) = n, then by [12, Lemma 3.8], there exists a P-flat resolution of $M~0\rightarrow F_{n}\overset {d_{n}}{\rightarrow } F_{n-1}\overset {d_{n-1}}{\rightarrow } \cdots \rightarrow F_{0}\overset {d_{0}}{\rightarrow } M\rightarrow 0$. The exact sequence $0\rightarrow F_{n}\overset {d_{n}}{\rightarrow } F_{n-1}\overset {d_{n-1}}{\rightarrow } \text {im}(d_{n-1})\rightarrow 0$ implies that im(d _n−1) is P-flat by Proposition 7(3). Note that the sequence $0\rightarrow \text {im}(d_{n-1})\rightarrow F_{n-2}\overset {d_{n-2}}{\rightarrow } \cdots \rightarrow F_{0}\overset {d_{0}}{\rightarrow } M\rightarrow 0$ is exact, we have that P-f d(M _R ) ≤ n−1 by [12, Lemma 3.8] because R is left G-morphic, a contradiction. So P-f d(M _R ) = 0 or ∞. □

Clearly, R is a regular ring if and only if r.P-fdim(R) = 0. By [12, Lemma 3.8], for a left G-morphic ring R, r.P-fdim(R) ≤ n if and only if $\text {Tor}^{R}_{n+1}(M, R/Ra)=0$ for all right R-modules M and all a ∈ R.

Corollary 4

Let R be a right G-regular left G-morphic ring. Then, r.P-fdim(R)=0 or ∞.