On φ-u-S-flat modules and nonnil-u-S-injective modules

Hwankoo Kim; Najib Mahdou; El Houssaine Oubouhou

doi:10.1515/gmj-2023-2117

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On φ-u-S-flat modules and nonnil-u-S-injective modules

Hwankoo Kim , Najib Mahdou and El Houssaine Oubouhou

Published/Copyright: January 2, 2024

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From the journal Georgian Mathematical Journal Volume 31 Issue 5

Abstract

This paper introduces and studies the ϕ-u-S-flat (resp., nonnil-u-S-injective) modules, which are a generalization of both ϕ-flat modules and u-S-flat modules (resp., both nonnil-injective modules and u-S-injective modules). We give the Cartan–Eilenberg–Bass theorem for nonnil-u-S-Noetherian rings. Finally, we offer some new characterizations of the ϕ-von Neumann regular ring.

Keywords: nonnil-injective

MSC 2020: 13C05; 13C11; 13C12; 13D07; 13D30

1 Introduction

Throughout this paper, it is assumed that all rings are commutative with non-zero identity. If R is a ring, we denote by Nil ⁢ ( R ) the ideal of all nilpotent elements of R. Recall that an ideal I of R is said to be nonnil if I ⊈ Nil ⁢ ( R ) . Denote by NN ⁡ ( R ) the set of all nonnil ideals of R. A non-empty subset S of R is said to be multiplicative if 1 ∈ S and for every a , b ∈ S , we have a ⁢ b ∈ S .

Let R be a ring and let M be an R-module. Set

ϕ ⁢ - ⁢ tor ⁡ ( M ) := { x ∈ M ∣ s ⁢ x = 0 ⁢ for some ⁢ s ∈ R ∖ Nil ⁢ ( R ) } .

If ϕ ⁢ - ⁢ tor ⁡ ( M ) = M , then M is called a ϕ- torsion module, and if ϕ ⁢ - ⁢ tor ⁡ ( M ) = 0 , then M is called a ϕ- torsion-free module. Recall from [17] that an R-module F is said to be ϕ-flat if for every R-monomorphism f : A → B with Coker ⁡ ( f ) a ϕ-torsion R-module, 1 F ⊗ R f : F ⊗ R A → F ⊗ R B is an R-monomorphism.

In [2], Badawi introduced and studied the concept of nonnil-Noetherian rings. Recall that a commutative ring R is said to be nonnil-Noetherian if every nonnil ideal of R is finitely generated. Many of the properties of Noetherian rings are proved analogously for the nonnil-Noetherian rings. In [2], the trivial extension construction is provided to give examples of nonnil-Noetherian rings that are not Noetherian rings. Then, in 2006, Yang [16] introduced nonnil-injective modules by replacing the ideals in Baer’s criterion for injective modules with nonnil ideals: An R-module E is called nonnil-injective if the induced sequence 0 → Hom R ⁢ ( C , E ) → Hom R ⁢ ( B , E ) → Hom R ⁢ ( A , E ) → 0 is exact for any exact sequence 0 → A → B → C → 0 with C ϕ-torsion; equivalently, Ext 1 R ⁢ ( R / I , E ) = 0 for any nonnil ideal I of R, and they gave the Cartan–Eilenberg–Bass theorem for nonnil-Noetherian rings. See, for example, [10, 16].

It is known that the notion of flat modules (resp., injective modules) has hereditary property; that is, if the second and third (resp., first and second) terms of a short exact sequence are flat (resp., injective), then the first (resp., third) term is also flat (resp., injective). Interestingly, neither ϕ-flat modules nor nonnil-injective modules have the hereditary property [15]. So they introduced strongly ϕ-flat modules and strongly nonnil-injective modules to obtain the hereditary property.

Let R be a ring with Nil ⁢ ( R ) a prime ideal of R and let M be an R-module. Then:

M is called strongly ϕ-flat if Tor n R ⁡ ( T , M ) = 0 for any ϕ-torsion module T and any n ≥ 1 .
M is called strongly nonnil-injective if Ext R n ⁡ ( T , M ) = 0 for any ϕ-torsion module T and any n ≥ 1 .

On the other hand, in [1], Anderson and Dumitrescu introduced the notion of S-Noetherian rings as a generalization of Noetherian rings; i.e., every ideal of R is S-finite. Recall from [1] that an R-module M is said to be S - finite if there exist a finitely generated submodule F of M and s ∈ S such that s ⁢ M ⊆ F . In [4], Kwon and Lim introduced the notion of nonnil-S-Noetherian rings as a generalization of both nonnil-Noetherian rings and S-Noetherian rings. Here R is said to be a nonnil-S-Noetherian ring if every nonnil ideal of R is S-finite.

Recall from [11] that an R-module M is called a u-S-torsion module if there exists an element s ∈ S such that s ⁢ M = 0 . An R-sequence M ⁢ → 𝑓 ⁢ N ⁢ → 𝑔 ⁢ L is called u-S-exact (at N) if there is an element s ∈ S such that s ⁢ Ker ⁡ ( g ) ⊆ Im ⁡ ( f ) and s ⁢ Im ⁡ ( f ) ⊆ Ker ⁡ ( g ) . An R-homomorphism f : M → N is an S - monomorphism (resp., S - epimorphism, S - isomorphism) if Ker ⁡ ( f ) is u-S-torsion (resp., Coker ⁡ ( f ) is u-S-torsion, both Ker ⁡ ( f ) and Coker ⁡ ( f ) are u-S-torsion). It is easy to verify that an R-homomorphism f : M → N is an S-monomorphism (resp., S-epimorphism, S-isomorphism) if 0 → M ⁢ → 𝑓 ⁢ N (resp., M ⁢ → 𝑓 ⁢ N → 0 , 0 → M ⁢ → 𝑓 ⁢ N → 0 ) is u-S-exact. In [11], Zhang introduced the class of u-S-flat modules F for which the functor F ⊗ R preserves u-S-exact sequences; equivalently, Tor 1 R ⁢ ( M , F ) is u-S-torsion for any R-module M by [11, Theorem 3.2]. In [12], Zhang introduced and studied the S-flat dimension S- fd R ⁡ ( M ) of an R-module M as the length of the shortest S-flat S-resolution of M, and the S-weak global dimension S- w . gl . ⁢ dim ⁡ ( R ) of a commutative ring R is the supremum of S-flat dimensions of all R-modules. The authors in [6] introduced the uniformly S-Noetherian (u-S-Noetherian for short) ring. A ring is said to be u-S-Noetherian if there exists an element s ∈ S such that for any ideal I of R, s ⁢ I ⊂ K for some finitely generated subideal K of I. Trivially, Noetherian rings are u-S-Noetherian; if S consists of units of R, then the notion of u-S-Noetherian rings is identical to that of the nonnil-Noetherian ring. If S consists of finite elements, then a ring R is a u-S-Noetherian ring if and only if R is an S-Noetherian ring. The u-S-Noetherian ring has been studied in [6] by using the trivial ring extension and the amalgamation of rings along an ideal, in addition to the Eakin–Nagata–Formanek theorem for u-S-Noetherian rings and the Cartan–Eilenberg–Bass theorem for u-S-Noetherian rings.

Our paper consists of four sections, including an introduction. In Section 2 we introduce and study the notion of ϕ-u-S-flat modules, which is a generalization of ϕ-flat and u-S-flat. We say that an R-module F is ϕ- u-S-flat (abbreviates ϕ-uniformly S-flat) provided that for any u-S-exact sequence 0 → A → B → C → 0 with C ϕ-torsion, the induced sequence 0 → A ⊗ R F → B ⊗ R F → C ⊗ R F → 0 is u-S-exact. It is well known that an R-module F is ϕ-flat if and only if Tor 1 R ⁢ ( M , F ) = 0 for any ϕ-torsion R-module M [17, Theorem 3.2]. We obtain the S-analog of this result (see Theorem 2.2). If an R-module F is ϕ-u-S-flat, then S - 1 ⁢ F is ϕ-flat over S - 1 ⁢ R (see Corollary 2.8). However, the converse does not hold. Also, a new local characterization of ϕ-flat modules is given in Proposition 2.11.

In Section 3, we introduce and study the notion of nonnil-u-S-injective modules. We say that an R-module E is nonnil-u-S-injective if the induced sequence 0 → Hom R ⁢ ( C , E ) → Hom R ⁢ ( B , E ) → Hom R ⁢ ( A , E ) → 0 is u-S-exact for any u-S-exact sequence 0 → A → B → C → 0 with C ϕ-torsion. Baer’s criterion for nonnil-u-S-injective modules is given in Proposition 3.5. Finally, we obtain the Cartan–Eilenberg–Bass theorem for nonnil-u-S-Noetherian rings (see Theorem 3.9).

Section 4 shows that a ϕ-ring R is a ϕ-von Neumann regular ring if and only if every R-module is ϕ-u-S-flat, and we will give some other characterizations of the ϕ-von Neumann regular ring.

The reader is referred to [3, 8, 9] for any undefined terminology and notation.

2 On ϕ-u-S-flat modules

First, we define an S-analog of ϕ-flat modules.

Definition 2.1.

Let R be a ring and S be a multiplicative subset of R. An R-module F is called ϕ- uniformly S-flat (for short ϕ- u-S-flat) if for any u-S-exact sequence 0 → A → B → C → 0 with C ϕ-torsion, the induced sequence 0 → A ⊗ R F → B ⊗ R F → C ⊗ R F → 0 is u-S-exact.

For brevity, we say that a multiplicative subset S of a ring R is regular if S consists entirely of regular elements of R. Recall from [17] that an R-module F is ϕ-flat if and only if Tor 1 R ⁢ ( M , F ) = 0 for any ϕ-torsion R-module M. We give an S-analog of this result.

Theorem 2.2.

Let R be a ring and let S be a multiplicative subset of R such that Nil ⁢ ( R ) is a prime ideal of R or S is a regular multiplicative subset of R, and let F be an R-module. Then the following statements are equivalent:

F is ϕ - u - S - flat.
For any short exact sequence 0 → A ⁢ → 𝑓 ⁢ B ⁢ → 𝑔 ⁢ C → 0 with C ϕ - torsion, the induced sequence

0 → A ⊗ R F → f ⊗ R 1 F B ⊗ R F → g ⊗ R 1 F C ⊗ R F → 0

is u - S - exact.
Tor 1 R ⁢ ( M , F ) is u - S - torsion for any ϕ - torsion R - module M.

Proof.

(1) ⇒ (2) and (3) ⇒ (2) These are straightforward.

(2) ⇒ (3) Let 0 → L → P → M → 0 be a short exact sequence, where P is projective and M is ϕ-torsion. Then there exists a long exact sequence

0 → Tor 1 R ⁡ ( M , F ) → L ⊗ F → P ⊗ F → M ⊗ F → 0 .

Thus Tor 1 R ⁡ ( M , F ) is u-S-torsion by (2).

(2) ⇒ (1) Let F be an R-module satisfying (2). Assume that 0 → A ⁢ → 𝑓 ⁢ B ⁢ → 𝑔 ⁢ C → 0 is a u-S-exact sequence with C ϕ-torsion. Then there is an exact sequence B ⁢ → 𝑔 ⁢ C → T → 0 , where T := Coker ⁡ ( g ) is u-S-torsion. Taking the tensor product of F over R with the exact sequence above, we have an exact sequence

B ⊗ R F → g ⊗ R 1 F C ⊗ R F → T ⊗ R F → 0 .

Then T ⊗ R F is u-S-torsion. Thus 0 → A ⊗ R F → f ⊗ R 1 F B ⊗ R F → g ⊗ R 1 F C ⊗ R F → 0 is u-S-exact at C ⊗ R F .

Of course, there are two short exact sequences

0 → Ker ⁡ ( f ) → A → Im ⁡ ( f ) → 0 and 0 → Im ⁡ ( f ) → B → Coker ⁡ ( f ) → 0 .

Consider the induced exact sequence

Ker ⁡ ( i Ker ⁡ ( f ) ⊗ R 1 F ) → Ker ⁡ ( f ) ⊗ R F → i Ker ⁡ ( f ) ⊗ R 1 F A ⊗ R F → Im ⁡ ( f ) ⊗ R F → 0 .

Since Ker ⁡ ( f ) is u-S-torsion, Ker ⁡ ( i Ker ⁡ ( f ) ⊗ R 1 F ) is also u-S-torsion. On the other hand, let x ∈ B . Since C is ϕ-torsion, there exists t ∈ R ∖ Nil ⁢ ( R ) such that t ⁢ g ⁢ ( x ) = 0 , and hence s ⁢ t ⁢ x ∈ s ⁢ Ker ⁡ ( g ) ⊆ Im ⁡ ( f ) for some s ∈ S . Note that if Nil ⁢ ( R ) is a prime ideal of R or S is a regular multiplicative subset of R, we have s ⁢ t ∈ R ∖ Nil ⁢ ( R ) . Therefore, Coker ⁡ ( f ) is a ϕ-torsion R-module. Consider the induced exact sequence

Ker ⁡ ( i Im ⁡ ( f ) ⊗ R 1 F ) → Im ⁡ ( f ) ⊗ R F → i Im ⁡ ( f ) ⊗ 1 F B ⊗ R F → Coker ⁡ ( f ) ⊗ R F → 0 .

Then Ker ⁡ ( i Ker ⁡ ( f ) ⊗ R 1 F ) is u-S-torsion by (2). Thus we have the following pullback diagram:

Note that Ker ⁡ ( f ) ⊗ R F is u-S-torsion, since Ker ⁡ ( f ) is so, and hence Im ⁡ ( i Ker ⁡ ( f ) ⊗ R 1 F ) is u-S-torsion. Therefore, Y is also u-S-torsion by [11, Proposition 2.8], which gives that f ⊗ R 1 F : A ⊗ R F → B ⊗ R F is a u-S-monomorphism. Consequently,

0 → A ⊗ R F → f ⊗ R 1 F B ⊗ R F → g ⊗ R 1 F C ⊗ R F → 0

is u-S-exact at A ⊗ R F .

Since the sequence 0 → A ⁢ → 𝑓 ⁢ B ⁢ → 𝑔 ⁢ C → 0 is u-S-exact at B, there exists s 1 ∈ S such that s 1 ⁢ Ker ⁡ ( g ) ⊆ Im ⁡ ( f ) and s ⁢ Im ⁡ ( f ) ⊆ Ker ⁡ ( g ) . So there are two short exact sequences

0 → s 1 ⁢ Ker ⁡ ( f ) → Im ⁡ ( f ) → Im ⁡ ( f ) / s 1 ⁢ Ker ⁡ ( f ) → 0

and

0 → s 1 ⁢ Im ⁡ ( f ) → Ker ⁡ ( f ) → Ker ⁡ ( f ) / s 1 ⁢ Im ⁡ ( f ) → 0 .

Note that s 1 2 ⁢ Im ⁡ ( f ) / s 1 ⁢ Ker ⁡ ( f ) = s 1 2 ⁢ Ker ⁡ ( f ) / s 1 ⁢ Im ⁡ ( f ) = 0 and s 1 2 ∈ R ∖ Nil ⁢ ( R ) . Hence Ker ⁡ ( f ) / s 1 ⁢ Im ⁡ ( f ) and Im ⁡ ( f ) / s 1 ⁢ Ker ⁡ ( f ) are ϕ-torsion R-modules. Thus, by (2), there are two induced exact sequences

0 → T 1 → s 1 ⁢ Ker ⁡ ( g ) ⊗ R F → Im ⁡ ( f ) ⊗ R F → ⋯

with s 2 ⁢ T 1 = 0 for some s 2 ∈ S , and

0 → T 2 → s 1 ⁢ Im ⁡ ( f ) ⊗ R F → Ker ⁡ ( g ) ⊗ R F → ⋯

with s 3 ⁢ T 2 = 0 for some s 3 ∈ S . Since Coker ⁡ ( g ) is a ϕ-torsion R-module, the natural exact sequence

0 → Ker ⁡ ( g ) → B → Coker ⁡ ( g ) → 0

gives the exact sequence

0 → T → Ker ⁡ ( g ) ⊗ R F → B ⊗ R F → Coker ⁡ ( g ) ⊗ R F → 0

with s 4 ⁢ T = 0 for some s 4 ∈ S . Set s := s 1 ⁢ s 2 ⁢ s 3 ⁢ s 4 . Then it is easy to show that s ⁢ Ker ⁡ ( g ⊗ R F ) ⊆ Im ⁡ ( f ⊗ R F ) and s ⁢ Im ⁡ ( f ⊗ R F ) ⊆ Ker ⁡ ( g ⊗ R F ) . Thus 0 → A ⊗ R F → B ⊗ R F → C ⊗ R F → 0 is u-S-exact at B ⊗ R F . ∎

The following example shows that the condition “ Tor 1 R ⁢ ( M , F ) is u-S-torsion for any ϕ-torsion R-module M” in Theorem 2.2 cannot be replaced by “ Tor 1 R ⁢ ( R / I , F ) is u-S-torsion for any nonnil ideal I of R”.

Example 2.3 ([11, Example 3.3]).

Let ℤ be the ring of integers, p a prime number in ℤ , and S := { p n ∣ n ≥ 0 } . Set M := ℤ ( p ) / ℤ . Then Tor 1 R ⁢ ( R / I , M ) is u-S-torsion for any ideal I of R. However, M is not ϕ-u-S-flat, since it is not u-S-flat.

However, if Nil ⁢ ( R ) is a prime ideal, we have the following characterization of ϕ-u-S-flat modules using the nonnil ideals.

Theorem 2.4.

Let R be a ring with Nil ⁢ ( R ) a prime ideal, S a multiplicative subset of R, and F an R-module. Then the following statements are equivalent:

F is ϕ - u - S - flat.
There exists s ∈ S such that for any nonnil ideal I of R, s ⁢ Tor 1 R ⁡ ( R / I , F ) = 0 .
There exists s ∈ S such that for any nonnil ideal I of R , the natural homomorphism σ I : I ⊗ R F → I ⁢ F is a u - S - isomorphism with respect to s.
There exists s ∈ S such that s ⁢ Tor 1 R ⁡ ( R / K , F ) = 0 for every finitely generated nonnil ideal K of R.
There exists s ∈ S such that for any finitely generated nonnil ideal K of R , the natural homomorphism σ K : K ⊗ R F → K ⁢ F is a u - S - isomorphism with respect to s.

Proof.

(2) ⇔ (3), (4) ⇔ (5), and (2) ⇒ (4) These are simple.

(1) ⇒ (2) Set N := ⊕ I ∈ NN ⁡ ( R ) R / I . Then N is a ϕ-torsion R-module by [17, Theorem 2.5]. Thus there exists an element s ∈ S such that s ⁢ Tor 1 R ⁡ ( F , N ) = 0 by Theorem 2.2, and so s ⁢ ⊕ I ∈ NN ⁡ ( R ) Tor 1 R ⁡ ( F , R / I ) = 0 . Therefore, s ⁢ Tor 1 R ⁡ ( F , R / I ) = 0 for every nonnil ideal I of R.

(2) ⇒ (1) Let N be a ϕ-torsion R-module and assume that N is generated by { n i ∣ i ∈ Γ } . Set N 0 := 0 and N α = 〈 n i ∣ ⁢ i ⁢ < α 〉 for each α ∈ Γ . Then N has a continuous filtration { N α ∣ α ∈ Γ } with N α + 1 / N α ≅ R / I α + 1 and I α = Ann R ⁡ ( n α + N α ∩ R ⁢ n α ) is a nonnil ideal of R. Since s ⁢ Tor 1 R ⁡ ( F , R / I α ) = 0 for every α ∈ Γ , it is easy to verify that s ⁢ Tor 1 R ⁡ ( F , N α ) = 0 by transfinite induction on α. So s ⁢ Tor 1 R ⁡ ( F , N ) = 0 .

(4) ⇒ (2) Let I be a nonnil ideal of R. Then I is the direct limit of all finitely generated nonnil subideals I i of I, i.e., I = lim → ⁡ I i . Hence s ⁢ Tor 2 R ⁡ ( I i , F ) ≅ s ⁢ Tor 1 R ⁡ ( R / I i , F ) = 0 , and so

s ⁢ Tor 2 R ⁡ ( lim → ⁡ I i , F ) ≅ s ⁢ lim → ⁡ Tor 2 R ⁡ ( I i , F ) ≅ lim → ⁡ s ⁢ Tor 2 R ⁡ ( I i , F ) = 0

by [14, Lemma 4.1]. Therefore,

s ⁢ Tor 1 R ⁡ ( R / I , F ) ≅ s ⁢ Tor 2 R ⁡ ( I , F ) ≅ s ⁢ Tor 2 R ⁡ ( lim → ⁡ I i , F ) = 0 . ∎

Corollary 2.5.

Let R be a ring with Nil ⁢ ( R ) a prime ideal and S be a multiplicative subset of R. Then:

Every u - S - flat module is ϕ - u - S - flat, and if R is a domain, then the two notions are identical.
Every ϕ - flat module is ϕ - u - S - flat, and if S consists of units of R , then the two notions are identical.

Proof.

This is a straightforward consequence of Theorem 2.4, [12, Proposition 2.3], and [17, Theorem 3.2]. ∎

The following result is an answer to the question: when are ϕ-u-S-flat modules u-S-flat?

Theorem 2.6.

Let R be a ϕ-ring and S be a regular multiplicative subset of R. Then the following assertions are equivalent:

R is an integral domain.
Every ϕ - u - S - flat module is u - S - flat.
Every ϕ - flat module is u - S - flat.

Proof.

(1) ⇒ (2) ⇒ (3) These are trivial.

(3) ⇒ (1) Let t ∈ Nil ⁢ ( R ) . Since R / Nil ⁢ ( R ) is ϕ-flat by [13, Proposition 1.7], R / Nil ⁢ ( R ) is u-S-flat by (3). Then

Tor 1 R ⁢ ( R / t ⁢ R , R / Nil ⁢ ( R ) ) ≅ ( t ⁢ R ∩ Nil ⁢ ( R ) ) / t ⁢ Nil ⁡ ( R ) = t ⁢ R / t ⁢ Nil ⁡ ( R )

is u-S-torsion by Theorem 2.4, and so there exists s ∈ S such that s ⁢ t ∈ t ⁢ Nil ⁢ ( R ) . Now, consider J := s ⁢ R . Since S ∩ Nil ⁢ ( R ) = ∅ , we get that J is a nonnil ideal of R, and so Nil ⁢ ( R ) = J ⁢ Nil ⁢ ( R ) by [13, Lemma 1.6]. Thus

t ⁢ J ⊆ t ⁢ Nil ⁢ ( R ) = t ⁢ J ⁢ Nil ⁢ ( R ) ⊆ t ⁢ J .

So t ⁢ J = t ⁢ J ⁢ Nil ⁢ ( R ) . Since tJ is finitely generated (in fact, it is principal), we have t ⁢ J = 0 by Nakayama’s lemma. Hence s ⁢ t = 0 , and so t = 0 , since s is a regular element. Thus Nil ⁢ ( R ) = 0 . Therefore, R is an integral domain. ∎

Let R be a ring and M an R-module. Then M is said to be strongly ϕ-S-flat if Tor n R ⁡ ( T , M ) is u-S-torsion for any ϕ-torsion module T and any n ≥ 1 .

Proposition 2.7.

Let R be a ring and S a multiplicative subset of R such that Nil ⁢ ( R ) is a prime ideal of R or S is a regular multiplicative subset of R. Then the following statements hold:

Every pure quotient of a ϕ - u - S - flat module is ϕ - u - S - flat.
Every finite direct sum of ϕ - u - S - flat modules is ϕ - u - S - flat.
Let 0 → A ⁢ → 𝑓 ⁢ B ⁢ → 𝑔 ⁢ C → 0 be a u - S - exact sequence, where C is a ϕ - u - S - flat module. If A is ϕ - u - S - flat, then so is B , and if B is strongly ϕ - u - S - flat, then A is ϕ - u - S - flat.
Let A → B be a u - S - isomorphism. Then A is ϕ - u - S - flat if and only if B is ϕ - u - S - flat.

Proof.

(1) Let 0 → A → B → C → 0 be a pure exact sequence with B a ϕ-u-S-flat module. Let M be a ϕ-torsion R-module. Then there exists an exact sequence Tor 1 R ⁡ ( M , B ) → Tor 1 R ⁡ ( M , C ) → 0 . Since Tor 1 R ⁡ ( M , B ) is u-S-torsion, there exists s ∈ S such that s ⁢ Tor 1 R ⁡ ( M , B ) = 0 , and so s ⁢ Tor 1 R ⁡ ( M , C ) = 0 . Hence C is u-S-flat.

(2) Let F 1 , … , F n be ϕ-u-S-flat modules and let M be a ϕ-torsion R-module. Then for each i = 1 , … , n , there exists s i ∈ S such that s i ⁢ Tor 1 R ⁡ ( M , F i ) = 0 . Set s := s 1 ⁢ ⋯ ⁢ s n . Then s ⁢ Tor 1 R ⁡ ( M , ⊕ i = 1 n F i ) ≅ ⊕ i = 1 n s ⁢ Tor 1 R ⁡ ( M , F i ) = 0 . So ⊕ i = 1 n F i is ϕ-u-S-flat.

(3) Let 0 → A ⁢ → 𝑓 ⁢ B ⁢ → 𝑔 ⁢ C → 0 be a u-S-exact sequence. Then Ker ⁡ ( f ) and Coker ⁡ ( g ) are all u-S-torsion and s ⁢ Ker ⁡ ( g ) ⊆ Im ⁡ ( f ) and s ⁢ Im ⁡ ( f ) ⊆ Ker ⁡ ( g ) for some s ∈ S . On the other hand, we have three short exact sequences

0 → Ker ⁡ ( f ) → A → Im ⁡ ( f ) → 0 , 0 → Ker ⁡ ( g ) → B → Im ⁡ ( g ) → 0 ,

and

0 → Im ⁡ ( g ) → C → Coker ⁡ ( g ) → 0 .

Let M be a ϕ-torsion R-module. Suppose A and C are ϕ-u-S-flat. Then the sequence

Tor 1 R ⁡ ( M , A ) → Tor 1 R ⁡ ( M , Im ⁡ ( f ) ) → M ⊗ R Ker ⁡ ( f )

is exact. Since Ker ⁡ ( f ) is u-S-torsion, it follows that M ⊗ R Ker ⁡ ( f ) is u-S-torsion. Note that Tor 1 R ⁡ ( M , A ) is u-S-torsion, so Tor 1 R ⁡ ( M , Im ⁡ ( f ) ) is u-S-torsion. Note that the sequence

Tor 2 R ⁡ ( M , Coker ⁡ ( g ) ) → Tor 1 R ⁡ ( M , Im ⁡ ( g ) ) → Tor 1 R ⁡ ( M , C )

is exact. Since Coker ⁡ ( g ) is u-S-torsion, it follows that Tor 2 R ⁡ ( M , Coker ⁡ ( g ) ) is u-S-torsion. Hence Tor 1 R ⁡ ( M , Im ⁡ ( g ) ) is u-S-torsion, since Tor 1 R ⁡ ( M , C ) is u-S-torsion. We also note that

Tor 1 R ⁡ ( M , Ker ⁡ ( g ) ) → Tor 1 R ⁡ ( M , B ) → Tor 1 R ⁡ ( M , Im ⁡ ( g ) )

is exact. Thus, to verify that Tor 1 R ⁡ ( M , B ) is u-S-torsion, we only need to show that Tor 1 R ⁡ ( M , Ker ⁡ ( g ) ) is u-S-torsion. Set N := Ker ⁡ ( g ) + Im ⁡ ( f ) . Then s ⁢ N ⊆ Ker ⁡ ( g ) and s ⁢ N ⊆ Im ⁡ ( f ) . Thus N / Ker ⁢ ( g ) and N / Im ⁢ ( f ) are all u-S-torsion. So the two exact sequences

0 → Ker ⁡ ( g ) → N → N / Ker ⁢ ( g ) → 0 ⁢ and ⁢ 0 → Im ⁡ ( f ) → N → N / Im ⁢ ( f ) → 0

give the following two induced exact sequences:

Tor 2 R ⁡ ( M , N / Im ⁢ ( f ) ) → Tor 1 R ⁡ ( M , Ker ⁡ ( g ) ) → Tor 1 R ⁡ ( M , N )

and

Tor 1 R ⁡ ( M , Im ⁡ ( f ) ) → Tor 1 R ⁡ ( M , N ) → Tor 1 R ⁡ ( M , N / Ker ⁡ ( g ) ) .

Since N / Ker ⁡ ( g ) is u-S-torsion, it follows that Tor 1 R ⁡ ( M , N / Ker ⁢ ( g ) ) is u-S-torsion. So Tor 1 R ⁡ ( M , N ) is u-S-torsion, since Tor 1 R ⁡ ( M , Im ⁡ ( f ) ) is u-S-torsion. Therefore, Tor 1 R ⁡ ( M , Ker ⁡ ( g ) ) is u-S-torsion since Tor 1 R ⁡ ( M , Im ⁡ ( f ) ) is u-S-torsion. Consequently, B is ϕ-u-S-flat.

Conversely, suppose B is strongly ϕ-u-S-flat. As above, there are three short exact sequences

0 → Ker ⁡ ( f ) → A → Im ⁡ ( f ) → 0 ,

0 → Ker ⁡ ( g ) → B → Im ⁡ ( g ) → 0 ,

0 → Im ⁡ ( g ) → C → Coker ⁡ ( g ) → 0 .

Then Ker ⁡ ( f ) and Coker ⁡ ( g ) are all u-S-torsion and s ⁢ Ker ⁡ ( g ) ⊆ Im ⁡ ( f ) and s ⁢ Im ⁡ ( f ) ⊆ Ker ⁡ ( g ) for some s ∈ S . Let M be a ϕ-torsion R-module. Note that the following sequence

Tor 1 R ⁡ ( M , Ker ⁡ ( f ) ) → Tor 1 R ⁡ ( M , A ) → Tor 1 R ⁡ ( M , Im ⁡ ( f ) ) → M ⊗ R Ker ⁡ ( f )

is exact. Since Ker ⁡ ( f ) is u-S-torsion, it follows that Tor 1 R ⁡ ( M , Ker ⁡ ( f ) ) and M ⊗ R Ker ⁡ ( f ) are u-S-torsion. It is only necessary to prove that Tor 1 R ⁡ ( M , Im ⁡ ( f ) ) is u-S-torsion. So we are only need to show that Tor 1 R ⁡ ( M , Ker ⁡ ( g ) ) is u-S-torsion. Since the sequence

Tor 2 R ⁡ ( M , Im ⁡ ( g ) ) → Tor 1 R ⁡ ( M , Ker ⁡ ( g ) ) → Tor 1 R ⁡ ( M , B )

is exact and Tor 1 R ⁡ ( M , B ) is u-S-torsion, we only need to show that Tor 2 R ⁡ ( M , Im ⁡ ( g ) ) is u-S-torsion. Note that the sequence:

Tor 3 R ⁡ ( M , Coker ⁡ ( g ) ) → Tor 2 R ⁡ ( M , Im ⁡ ( g ) ) → Tor 2 R ⁡ ( M , C )

is exact. Since Coker ⁡ ( g ) is u-S-torsion and C is strongly ϕ-u-S-flat, we have Tor 3 R ( M , Coker ( g ) ) and Tor 2 R ⁡ ( M , C ) are u-S-torsion. So Tor 2 R ⁡ ( M , Im ⁡ ( g ) ) is u-S-torsion.

(4) This is easily deduced from (3). ∎

Corollary 2.8.

Let R be a ring and S be a multiplicative subset of R. If F is ϕ-u-S-flat over R, then S - 1 ⁢ F is ϕ-flat over S - 1 ⁢ R .

Proof.

Let J be a finitely generated nonnil ideal of S - 1 ⁢ R . Then J = S - 1 ⁢ I for some finitely generated nonnil ideal I of R. So R / I is a ϕ-torsion R-module. Then there exists s ∈ S such that s ⁢ Tor 1 R ⁢ ( R / I , F ) = 0 . Thus

Tor 1 S - 1 ⁢ R ⁢ ( S - 1 ⁢ R / J , S - 1 ⁢ F ) ≅ S - 1 ⁢ Tor 1 R ⁢ ( R / I , F ) = 0 .

Hence S - 1 ⁢ F is a ϕ-flat S - 1 ⁢ R -module. ∎

Note that the converse of Corollary 2.8 is not generally true, as the following example shows.

Example 2.9.

Let the ℤ -module M := ℤ ( p ) / ℤ be as in Example 2.3, and let S = { p n ∣ n ≥ 0 } . Then S - 1 ⁢ M = 0 , and thus is flat over S - 1 ⁢ R . However, M is not ϕ-u-S-flat over ℤ .

Proposition 2.10.

Let R be a ring, S a multiplicative subset of R consisting of finite elements such that Nil ⁢ ( R ) is a prime ideal of R or S is a regular multiplicative subset of R, and let F be an R-module. Then F is ϕ-u-S-flat over R if and only if S - 1 ⁢ F is ϕ-flat over S - 1 ⁢ R .

Proof.

We just need to show that if S - 1 ⁢ F is ϕ-flat over S - 1 ⁢ R , then F is ϕ-u-S-flat over R. Set S := { s 1 , … , s n } . Let 0 → A ⁢ → 𝑓 ⁢ B → C → 0 be a short exact sequence over R, where C is ϕ-torsion. Then

0 → T → A ⊗ R F → f ⊗ R 1 F B ⊗ R F → C ⊗ R F → 0

is exact with T := Ker ⁡ ( f ⊗ R 1 F ) . Taking the tensor product of S - 1 ⁢ R with this exact sequence, we have an exact sequence:

0 → S - 1 ⁢ T → S - 1 ⁢ A ⊗ S - 1 ⁢ R S - 1 ⁢ F → S - 1 ⁢ B ⊗ S - 1 ⁢ R F S → S - 1 ⁢ C ⊗ S - 1 ⁢ R S - 1 ⁢ F → 0

of S - 1 ⁢ R -modules. Since S - 1 ⁢ F is ϕ-flat over S - 1 ⁢ R and S - 1 ⁢ C is a ϕ-torsion S - 1 ⁢ R -module, S - 1 ⁢ T = 0 . Thus for every x ∈ T , s x ⁢ x = 0 for some s x ∈ S . Set s := s 1 ⁢ ⋯ ⁢ s n . Then s ⁢ T = 0 . Hence

0 → T → A ⊗ R F → f ⊗ R 1 F B ⊗ R F → C ⊗ R F → 0

is a u-S-exact sequence. So F is ϕ-u-S-flat over R. ∎

Let P be a prime ideal of R. We say that an R-module F is ϕ- u-P-flat if F is ϕ-u- ( R ∖ P ) -flat.

Proposition 2.11.

Let R be a ring with Nil ⁢ ( R ) a prime ideal and M be an R-module. Then the following conditions are equivalent:

M is ϕ - flat.
M is ϕ - u - P - flat for every prime ideal P of R.
M is ϕ - u - 𝔪 - flat for every maximal ideal 𝔪 of R.

Proof.

(1) ⇒ (2) ⇒ (3) These are straightforward.

(3) ⇒ (1) Let F be a ϕ-torsion R-module. Then for every maximal ideal 𝔪 of R, Tor 1 R ⁢ ( M , F ) is u- ( R ∖ 𝔪 ) -torsion. Thus there exists s 𝔪 ∈ R ∖ 𝔪 such that s 𝔪 ⁢ Tor 1 R ⁢ ( M , F ) = 0 . Since the ideal generated by all s 𝔪 is R, there exist a 𝔪 1 , … , a 𝔪 n ∈ R such that 1 = a 𝔪 1 ⁢ s 𝔪 1 + ⋯ + a 𝔪 n ⁢ s 𝔪 n . Therefore,

Tor 1 R ⁢ ( M , F ) = a 𝔪 1 ⁢ s 𝔪 1 ⁢ Tor 1 R ⁢ ( M , F ) + ⋯ + a 𝔪 n ⁢ s 𝔪 n ⁢ Tor 1 R ⁢ ( M , F ) = 0 .

So F is ϕ-flat. ∎

3 On nonnil-u-S-injective modules

The well-known Cartan–Eilenberg–Bass theorem for the nonnil-Noetherian ring states that a ring R is nonnil-Noetherian if and only if any direct sum of nonnil-injective modules is nonnil-injective (see [2, Theorem 2.5]). To obtain the Cartan–Eilenberg–Bass theorem for nonnil uniformly S-Noetherian rings, we first introduce the S-analog of nonnil-injective modules.

Definition 3.1.

Let R be a ring and let S be a multiplicative subset of R. An R-module E is said to be nonnil-u-S-injective if the induced sequence 0 → Hom R ⁢ ( C , E ) → Hom R ⁢ ( B , E ) → Hom R ⁢ ( A , E ) → 0 is u-S-exact for any u-S-exact sequence 0 → A → B → C → 0 with C ϕ-torsion.

Theorem 3.2.

Let R be a ring and let S be a multiplicative subset of R such that Nil ⁢ ( R ) is a prime ideal of R or S is a regular multiplicative subset of R, and E be an R-module. Then the following assertions are equivalent:

E is a nonnil-u - S - injective module.
For any short exact sequence 0 → A ⁢ → 𝑓 ⁢ B ⁢ → 𝑔 ⁢ C → 0 with C ϕ - torsion, the induced sequence

0 → Hom R ⁡ ( C , E ) → g * Hom R ⁡ ( B , E ) → f * Hom R ⁡ ( A , E ) → 0

is u - S - exact.
Ext R 1 ⁡ ( M , E ) is u - S - torsion for any ϕ - torsion R - module M.

Proof.

(1) ⇒ (2) This is simple.

(2) ⇒ (3) Let M be a ϕ-torsion R-module and let 0 → L → P → M → 0 be a short exact sequence with P projective. Then there exists an exact sequence

0 → Hom R ⁡ ( M , E ) → Hom R ⁡ ( P , E ) → Hom R ⁡ ( L , E ) → Ext R 1 ⁡ ( M , E ) → 0 .

Thus Ext R 1 ⁡ ( M , E ) is u-S-torsion by (2).

(3) ⇒ (2) Let 0 → A ⁢ → 𝑓 ⁢ B ⁢ → 𝑔 ⁢ C → 0 be a short exact sequence with C a ϕ-torsion R-module. Then we have the following long exact sequence:

0 → Hom R ⁡ ( C , E ) ⁢ → g * ⁢ Hom R ⁡ ( B , E ) ⁢ → f * ⁢ Hom R ⁡ ( A , E ) ⁢ → 𝛿 ⁢ Ext R 1 ⁢ ( C , E ) → 0 .

By (3), Ext R 1 ⁡ ( C , E ) is u-S-torsion, and so the sequence

0 → Hom R ⁡ ( C , E ) ⁢ → 9 * ⁢ Hom R ⁡ ( B , E ) ⁢ → f * ⁢ Hom R ⁡ ( A , E ) → 0

is u-S-exact.

(2) ⇒ (1) Let E be an R-module satisfying (2). Suppose that 0 → A ⁢ → 𝑓 ⁢ B ⁢ → 𝑔 ⁢ C → 0 is a u-S-exact sequence with C ϕ-torsion. Set T := Coker ⁡ ( g ) . Then there exists an exact sequence B ⁢ → 𝑔 ⁢ C → T → 0 , where T is u-S-torsion. Hence we have an exact sequence

0 → Hom R ⁡ ( T , E ) → Hom R ⁡ ( C , E ) → Hom R ⁡ ( B , E ) .

Therefore, Hom R ⁡ ( T , E ) is u-S-torsion. So

0 → Hom R ⁡ ( C , E ) ⁢ → g * ⁢ Hom R ⁡ ( B , E ) ⁢ → f * ⁢ Hom R ⁡ ( A , E ) → 0

is u-S-exact at Hom R ⁡ ( C , E ) .

There are also two short exact sequences

0 → Ker ⁡ ( f ) → i A A → π Im ⁡ ( f ) Im ⁡ ( f ) → 0

and

0 → Im ⁡ ( f ) → i B B → Coker ⁡ ( f ) → 0 .

Consider the induced exact sequences

0 → Hom R ⁡ ( Im ⁡ ( f ) , E ) → π Im ⁡ ( f ) * Hom R ⁡ ( A , E ) → i A * Hom R ⁡ ( Ker ⁡ ( f ) , E )

and

0 → Hom R ⁡ ( Coker ⁡ ( f ) , E ) → Hom R ⁡ ( B , E ) → i B * Hom R ⁡ ( Im ⁡ ( f ) , E ) .

Since Ker ⁡ ( f ) is u-S-torsion, so is Im ⁡ ( i A * ) . On the other hand, as in Theorem 2.2, Coker ⁡ ( f ) is a ϕ-torsion R-module, and hence by (2) Coker ⁡ ( i B * ) is u-S-torsion. We have the following pushout diagram:

Since Im ⁡ ( i A * ) and Coker ⁡ ( i B * ) are all u-S-torsion, it follows that Y is also u-S-torsion [11, Proposition 2.8]. Thus f * : Hom R ⁡ ( B , E ) → Hom R ⁡ ( A , E ) is a u-S-epimorphism. So

0 → Hom R ⁡ ( C , E ) ⁢ → a * ⁢ Hom R ⁡ ( B , E ) ⁢ → f * ⁢ Hom R ⁡ ( A , E ) → 0

is u-S-exact at Hom R ⁡ ( A , E ) .

As 0 → A ⁢ → 𝑓 ⁢ B ⁢ → 𝑎 ⁢ C → 0 is u-S-exact at B and C, there exists s ∈ S such that s ⁢ Ker ⁡ ( g ) ⊆ Im ⁡ ( f ) , s ⁢ Im ⁡ ( f ) ⊆ Ker ⁡ ( g ) , and s ⁢ Coker ⁡ ( g ) = 0 . We claim that s 2 ⁢ Im ⁡ ( g * ) ⊆ Ker ⁡ ( f * ) and s 2 ⁢ Ker ⁡ ( f * ) ⊆ Im ⁡ ( g * ) . Indeed, consider the following diagram:

Let h ∈ Im ⁡ ( g * ) . Then there exists u ∈ Hom R ⁡ ( C , E ) such that h = u ∘ g . Hence for any a ∈ A ,

s ⁢ h ∘ f ⁢ ( a ) = s ⁢ u ∘ g ∘ f ⁢ ( a ) = u ∘ g ∘ s ⁢ f ⁢ ( a ) = 0

as s ⁢ Im ⁡ ( f ) ⊆ Ker ⁡ ( g ) . Thus s ⁢ h ∘ f = 0 , and so s ⁢ Im ⁡ ( g * ) ⊆ Ker ⁡ ( f * ) . Hence we have s 2 ⁢ Im ⁡ ( g * ) ⊆ Ker ⁡ ( f * ) . Now, let h ∈ Ker ⁡ ( f * ) . Consider the natural exact sequence

0 → Ker ⁡ ( g ) → i Ker ⁡ ( g ) B → π B Im ⁡ ( g ) → 0 .

Since h ∘ f = 0 , it follows that Ker ⁡ ( h ) ⊇ Im ⁡ ( f ) ⊇ s ⁢ Ker ⁡ ( g ) . So s ⁢ h ∘ i Ker ⁡ ( g ) = 0 . Consequently, there is a well-defined R-homomorphism v : Im ⁡ ( g ) → E such that v ∘ π B = s ⁢ h by (2) as Im ⁡ ( g ) is a ϕ-torsion module. Consider the following exact sequence:

Hom R ⁡ ( Coker ⁡ ( g ) , E ) → Hom R ⁡ ( C , E ) → Hom R ⁡ ( Im ⁡ ( g ) , E ) → Ext R 1 ⁡ ( Coker ⁡ ( g ) , E )

induced by 0 → Im ⁡ ( g ) → C → Coker ⁡ ( g ) → 0 , where Coker ⁡ ( g ) is a ϕ-torsion module. Since

s ⁢ Hom R ⁡ ( Coker ⁡ ( g ) , E ) = s ⁢ Ext R 1 ⁡ ( Coker ⁡ ( g ) , E ) = 0 ,

it follows that s ⁢ Hom R ⁡ ( Im ⁡ ( g ) , E ) ⊆ i Im ⁡ ( g ) * ⁢ ( Hom R ⁡ ( C , E ) ) . Thus there is a homomorphism u : C → E such that s 2 ⁢ h = v ∘ g . Then s 2 ⁢ Ker ⁡ ( f * ) ⊆ Im ⁡ ( g * ) . Hence

0 → Hom R ⁡ ( C , E ) → g * Hom R ⁡ ( B , E ) → f * Hom R ⁡ ( A , E ) → 0

is u-S-exact at Hom R ⁡ ( B , E ) . ∎

Let R be a ring and let M be an R-module. Then M is called strongly nonnil-u-S-flat if Ext R n ⁡ ( T , M ) is u-S-torsion for any ϕ-torsion module T and any n ≥ 1 .

Proposition 3.3.

Let R be a ring, S a multiplicative subset of R such that Nil ⁢ ( R ) is a prime ideal of R or S is a regular multiplicative subset of R, and let E be an R-module. Then the following statements hold:

Every finite direct sum of nonnil-u - S - injective modules is nonnil-u - S - injective.
Let 0 → A ⁢ → 𝑓 ⁢ B ⁢ → 𝑔 ⁢ C → 0 be a u - S - exact sequence such that A is a nonnil-u - S - injective module. If C is nonnil-u - S - injective, so is C , with the equivalence if A is a strongly nonnil-u - S - injective module.
Let A → B be a u - S - isomorphism. If one of A and B is nonnil-u - S - injective, so is the other.

Proof.

(1) Suppose E 1 , … , E n are nonnil-u-S-injective modules. Let M be a ϕ-torsion R-module. Then there exists s i ∈ S such that s i ⁢ Ext R 1 ⁡ ( M , E i ) = 0 for each i = 1 , … , n . Set s := s 1 ⁢ … ⁢ s n . Then

s ⁢ Ext R 1 ⁡ ( M , ⊕ i = 1 n E i ) ≅ ⊕ i = 1 n s ⁢ Ext R 1 ⁡ ( M , E i ) = 0 .

Thus ⊕ i = 1 n E i is nonnil-u-S-injective.

(2) Suppose A and C are nonnil-u-S-injective modules and 0 → A ⁢ → 𝑓 ⁢ B ⁢ → 𝑔 ⁢ C → 0 is a u-S-exact sequence. Then there are three short exact sequences

0 → Ker ⁡ ( f ) → A → Im ⁡ ( f ) → 0 ,

0 → Ker ⁡ ( g ) → B → Im ⁡ ( g ) → 0 ,

0 → Im ⁡ ( g ) → C → Coker ⁡ ( g ) → 0 .

Then Ker ⁡ ( f ) and Coker ⁡ ( g ) are all u-S-torsion and s ⁢ Ker ⁡ ( g ) ⊆ Im ⁡ ( f ) and s ⁢ Im ⁡ ( f ) ⊆ Ker ⁡ ( g ) for some s ∈ S . Let M be a ϕ-torsion R-module. Then

Ext R 1 ⁡ ( M , A ) → Ext R 1 ⁡ ( M , Im ⁡ ( f ) ) → Ext R 2 ⁡ ( M , Ker ⁡ ( f ) )

is exact. Since Ker ⁡ ( f ) is u-S-torsion and A is nonnil-u-S-injective, it follows that Ext R 1 ⁡ ( M , Im ⁡ ( f ) ) is u-S-torsion. Note that

Hom R ⁡ ( M , Coker ⁡ ( g ) ) → Ext R 1 ⁡ ( M , Im ⁡ ( g ) ) → Ext R 1 ⁡ ( M , C )

is exact. Since Coker ⁡ ( g ) is u-S-torsion, it follows that Hom R ⁡ ( M , Coker ⁡ ( g ) ) is u-S-torsion. Thus Ext R 1 ⁡ ( M , Im ⁡ ( g ) ) is u-S-torsion as Ext R 1 ⁡ ( M , C ) is u-S-torsion. We also note that

Ext R 1 ⁡ ( M , Ker ⁡ ( g ) ) → Ext R 1 ⁡ ( M , B ) → Ext R 1 ⁡ ( M , Im ⁡ ( g ) )

is exact. Thus, to verify that Ext R 1 ⁡ ( M , B ) is u-S-torsion, we just need to show that Ext R 1 ⁡ ( M , Ker ⁡ ( g ) ) is u-S-torsion. Set N := Ker ⁡ ( g ) + Im ⁡ ( f ) . Consider the following two exact sequences:

0 → Ker ⁡ ( g ) → N → N / Ker ⁢ ( g ) → 0 and 0 → Im ⁡ ( f ) → N → N / Im ⁢ ( f ) → 0 .

Then it is easy to verify that N / Ker ⁢ ( g ) and N / Im ⁢ ( f ) are all u-S-torsion. Consider the following two induced exact sequences:

Hom R ⁡ ( M , N / Im ⁢ ( f ) ) → Ext R 1 ⁡ ( M , Ker ⁡ ( g ) ) → Ext R 1 ⁡ ( M , N ) → Ext R 1 ⁡ ( M , N / Im ⁢ ( f ) )

and

Hom R ⁡ ( M , N / Ker ⁢ ( g ) ) → Ext R 1 ⁡ ( M , Im ⁡ ( f ) ) → Ext R 1 ⁡ ( M , N ) → Ext R 1 ⁡ ( M , N / Ker ⁢ ( g ) ) .

It follows that Ext R 1 ⁡ ( M , Ker ⁡ ( g ) ) is u-S-torsion if and only if Ext R 1 ⁡ ( M , Im ⁡ ( f ) ) is u-S-torsion. Consequently, B is nonnil-u-S-injective, since Ext R 1 ⁡ ( M , Im ⁡ ( f ) ) is u-S-torsion.

Conversely, as above, there are three short exact sequences

0 → Ker ⁡ ( f ) → A → Im ⁡ ( f ) → 0 ,

0 → Ker ⁡ ( g ) → B → Im ⁡ ( g ) → 0 ,

0 → Im ⁡ ( g ) → C → Coker ⁡ ( g ) → 0 .

Then Ker ⁡ ( f ) and Coker ⁡ ( g ) are all u-S-torsion and s ⁢ Ker ⁡ ( g ) ⊆ Im ⁡ ( f ) and s ⁢ Im ⁡ ( f ) ⊆ Ker ⁡ ( g ) for some s ∈ S . Let M be a ϕ-torsion R-module. Note that

Hom R ⁡ ( M , Coker ⁡ ( g ) ) → Ext R 1 ⁡ ( M , Im ⁡ ( g ) ) → Ext R 1 ⁡ ( M , C ) → Ext R 1 ⁡ ( M , Coker ⁡ ( g ) )

is exact. Since Coker ⁡ ( g ) is u-S-torsion, it follows that Hom R ⁡ ( M , Coker ⁡ ( g ) ) and Ext R 1 ⁡ ( M , Coker ⁡ ( g ) ) are u-S-torsion. We just need to verify that Ext R 1 ⁡ ( M , Im ⁡ ( g ) ) is u-S-torsion. Note that

Ext R 1 ⁡ ( M , B ) → Ext R 1 ⁡ ( M , Im ⁡ ( g ) ) → Ext R 2 ⁡ ( M , Ker ⁡ ( g ) )

is exact. Since Ext R 1 ⁡ ( M , B ) is u-S-torsion, we just need to verify that Ext R 2 ⁡ ( M , Ker ⁡ ( g ) ) is u-S-torsion. By the proof of (2), we just need to show that Ext R 2 ⁡ ( M , Im ⁡ ( f ) ) is u-S-torsion. Note that

Ext R 2 ⁡ ( M , A ) → Ext R 2 ⁡ ( M , Im ⁡ ( f ) ) → Ext R 3 ⁡ ( M , Ker ⁡ ( f ) )

is exact. Since Ext R 2 ⁡ ( M , A ) and Ext R 3 ⁡ ( M , Ker ⁡ ( f ) ) are u-S-torsion, it follows that Ext R 2 ⁡ ( M , Im ⁡ ( f ) ) is u-S-torsion. So C is nonnil-u-S-injective.

(3) Considering the u-S-exact sequences 0 → A → B → 0 → 0 and 0 → 0 → A → B → 0 , we have that A is nonnil-u-S-injective if and only if B is nonnil-u-S-injective by (2). ∎

Let 𝔭 be a prime ideal of R. Briefly, we say that an R module E is nonnil-u- 𝔓 -injective, provided that E is nonnil-u- U ⁢ ( R ∖ 𝔭 ) -injective. The next result gives a local characterization of nonnil-injective modules.

Proposition 3.4.

Let R be a ring with Nil ⁢ ( R ) being a prime ideal and E an R-module. Then the following statements are equivalent:

E is nonnil-injective.
E is nonnil-u - 𝔓 - injective for any 𝔓 ∈ Spec ⁡ ( R ) .
E is nonnil-u - 𝔪 - injective for any 𝔪 ∈ Max ⁡ ( R ) .

Proof.

(1) ⇒ (2) and (2) ⇒ (2) These are simple.

(3) ⇒ (1) Let M be a ϕ-torsion R-module. Then Ext R 1 ⁡ ( M , E ) is u- 𝔪 -torsion. So for every 𝔪 ∈ Max ⁡ ( R ) , there exists s 𝔪 ∈ ( R ∖ 𝔪 ) such that s 𝔪 ⁢ Ext R 1 ⁢ ( M , E ) = 0 . Since the ideal generated by all s 𝔪 is R, we get Ext R 1 ⁢ ( M , E ) = 0 . So E is nonnil-injective. ∎

We say that an R-module M is S - divisible if M = s ⁢ M for any s ∈ S . The well-known Baer’s criterion for nonnil-injectivity says that an R-module E is nonnil-injective if and only if Ext R 1 ⁢ ( R / I , E ) = 0 for any nonnil ideal I of R. The following result gives Baer’s criterion for nonnil-u-S-injective modules.

Theorem 3.5 (Baer’s criterion for nonnil-u-S-injective modules).

Let R be a ring, S a multiplicative subset of R, and E an R-module. If E is a nonnil-u-S-injective module, then there exists an element s ∈ S such that s ⁢ Ext R 1 ⁡ ( R / I , E ) = 0 for any nonnil ideal I of R. Moreover, if E = s ⁢ E , then the converse also holds.

Proof.

If E is a nonnil-u-S-injective module, then Ext R 1 ⁡ ( ⊕ I ∈ NN ⁡ ( R ) R / I , E ) is u-S-torsion by Theorem 3.2. Thus there is an element s ∈ S such that

s ⁢ Ext R 1 ⁡ ( ⊕ I ∈ NN ⁡ ( R ) R / I , E ) = s ⁢ ∏ I ∈ NN ⁡ ( R ) Ext R 1 ⁡ ( R / I , E ) = 0 .

So s ⁢ Ext R 1 ⁡ ( R / I , E ) = 0 for any nonnil ideal I of R.

Conversely, assume that there exists s ∈ S such that s ⁢ Ext R 1 ⁡ ( R / I , E ) = 0 for any nonnil ideal I of R and E is an S-divisible R-module. Let B be an R-module and A be a submodule of B such that B / A is ϕ-torsion. Let f : A → E be an R-homomorphism. Set

Γ := { ( C , d ) ∣ C is a submodule of B containing A such that B / C is ϕ - torsion and d | A = s ⁢ f } .

Since ( A , s ⁢ f ) ∈ Γ , it follows that Γ is not empty. Make a partial order on Γ by saying that ( C 1 , d 1 ) ≤ ( C 2 , d 2 ) if and only if C 1 ⊆ C 2 and d 2 | C 1 = d 1 . For any chain ( C j , d j ) , let C 0 = ⋃ C j . If c ∈ C 0 , then c ∈ C j for some j. Define d 0 : C 0 → E by d 0 ⁢ ( c ) = d j ⁢ ( c ) if c ∈ C j . Then d 0 is well-defined and extends every d j . If ϕ- tor ⁡ ( B / C 0 ) ⊈ ( B / C 0 ) , then there exists b ∈ B such that b + C 0 = b ¯ ∈ ϕ - tor ⁡ ( B / C 0 ) . Thus for all t ∈ R ∖ Nil ⁢ ( R ) we have t ⁢ b ∈ C 0 , and so for all j we have b ¯ ∈ ϕ - tor ⁡ ( B / C j ) , a contradiction. Then ( C 0 , d 0 ) is the upper bound of the chain ( C j , d j ) . By Zorn’s Lemma, there is a maximal element ( C , d ) in Γ.

We assert that C = B . Suppose, on the contrary, that x ∈ B ∖ C . Set I := { r ∈ R ∣ r ⁢ x ∈ C } . Since B / C is a ϕ-torsion R-module, it follows that I contains a non-nilpotent element of R, and thus I is a nonnil ideal of R. Since E = s ⁢ E , there exists a homomorphism h : I → E such that s ⁢ h ⁢ ( r ) = d ⁢ ( r ⁢ x ) . Then there exists an R-homomorphism g : R → E such that g ⁢ ( r ) = s ⁢ h ⁢ ( r ) = d ⁢ ( r ⁢ x ) for any r ∈ I . Set C 1 := C + R ⁢ x . Then it is easy to verify that B / C 1 is a ϕ-torsion R-module. Define d 1 : C 1 → E by d 1 ⁢ ( c + r ⁢ x ) = d ⁢ ( c ) + g ⁢ ( r ) , where c ∈ C and r ∈ R . If c + r ⁢ x = 0 , then r ∈ I , and thus d ⁢ ( c ) + g ⁢ ( r ) = d ⁢ ( c ) + s ⁢ h ⁢ ( r ) = d ⁢ ( c ) + d ⁢ ( r ⁢ x ) = d ⁢ ( c + r ⁢ x ) = 0 . So d 1 is a well-defined homomorphism such that . d 1 | A = s f . So ( C 1 , d 1 ) ∈ Γ . However, ( C 1 , d 1 ) > ( C , d ) , which contradicts the maximality of ( C , d ) . ∎

Corollary 3.6.

Let R be a ring, S a multiplicative subset of R such that Nil ⁢ ( R ) is a prime ideal of R or S is a regular multiplicative subset of R, and let E be an R-module. Then the following conditions are equivalent:

F is ϕ - u - S - flat.
Hom R ⁢ ( F , E ) is nonnil-u - S - injective for any injective module E.
Hom R ⁢ ( F , E ) is nonnil-u - S - injective for any injective cogenerator module E.

Proof.

(1) ⇒ (2) Let M be a ϕ-torsion R-module and let E be an injective R-module. Since F is ϕ-u-S-flat, it follows that s ⁢ Tor 1 R ⁢ ( M , F ) = 0 for some s ∈ S . Thus Ext R 1 ⁢ ( M , Hom R ⁢ ( F , E ) ) ≅ Hom R ⁢ ( Tor 1 R ⁢ ( M , F ) , E ) is uniformly S-torsion with respect to s ∈ S . Thus Hom R ⁢ ( F , E ) is nonnil-u-S-injective by Theorem 3.2.

(2) ⇒ (3) This is straightforward.

(3) ⇒ (1) Let M be a ϕ-torsion R-module and let E be an injective cogenerator. Since Hom R ⁢ ( F , E ) is nonnil-u-S-injective, it follows that Ext R 1 ⁢ ( M , Hom R ⁢ ( F , E ) ) ≅ Hom R ⁢ ( Tor 1 R ⁢ ( M , F ) , E ) is uniformly S-torsion by Theorem 3.2. Hence Tor 1 R ⁢ ( M , F ) is uniformly S-torsion by [14, Lemma 4.2]. Therefore, F is ϕ-u-S-flat. ∎

Corollary 3.7.

Let R be a ring with Nil ⁢ ( R ) a prime ideal and let S be a multiplicative subset of R. Then:

Every u - S - injective modules is nonnil-u - S - injective, and if R is a domain, then two notions are identical.
Every nonnil-injective modules is ϕ - u - S - flat nonnil-u - S - injective, and if S consists of units of R , then two notions are identical.

The following result is an answer to the question: when are nonnil-u-S-injective modules u-S-injective?

Corollary 3.8.

Let R be a ϕ-ring and let S be a regular multiplicative subset of R. Then the following assertions are equivalent:

R is an integral domain.
Every nonnil-u - S - injective R - module is u - S - injective.
Every nonnil-injective R - module is u - S - injective.

Proof.

(1) ⇒ (2) ⇒ (3) These are trivial.

(3) ⇒ (1) Let M be a ϕ-flat module. Then by [7, Proposition 1.4] Hom R ⁢ ( M , E ) is nonnil-injective for any injective R-module E. So Hom R ⁢ ( M , E ) is u-S-injective for any injective module E by (3). Thus M is u-S-flat by [14, Proposition 4.3]. Consequently, R is an integral domain by Theorem 2.6. ∎

Recall that a ring R is said to be nonnil-u-S-Noetherian if there exists s ∈ S such that every nonnil ideal of R is S-finitely generated with respect to S. This notion is explored further in [5]. Now, we give the main result of this section.

Theorem 3.9 (Cartan–Eilenberg–Bass theorem for nonnil-u-S-Noetherian rings).

Let R be a ring and let S be a regular multiplicative subset of R. Then the following assertions are equivalent:

R is nonnil-u - S - Noetherian.
Every direct union of nonnil-injective (resp., injective) submodules of a module is nonnil-u - S - injective.
Every direct sum of nonnil-injective (resp., injective) modules is nonnil-u - S - injective.
For every nonnil-injective (resp., injective) R - module E, ⊕ E is nonnil-u - S - injective.

Proof.

(1) ⇒ (2) Let { E i , f i , j } i < j be a direct system of nonnil-injective submodules of a module, where each f i , j is the embedding map. Let lim → ⁡ E i be its direct limit. Let s be an element in S such that for every nonnil ideal I of R, there exists a finitely generated subideal K of I such that s ⁢ I ⊆ K . Considering the short exact sequence 0 → I / K → R / K → R / I → 0 , we have the following long exact sequence:

Hom R ⁡ ( I / K , lim → ⁡ E i ) → Ext R 1 ⁢ ( R / I , lim → ⁡ E i ) → Ext R 1 ⁢ ( R / K , lim → ⁡ E i ) → Ext R 1 ⁢ ( I / K , lim → ⁡ E i ) .

Since R / K is finitely presented, we have

Ext R 1 ⁡ ( R / K , lim → ⁡ E i ) ≅ lim → ⁡ Ext R 1 ⁡ ( R / K , E i ) = 0

by [3, Theorem 2.1.5(3)]. Since s ⁢ I ⊆ K , we have s ⁢ Hom R ⁡ ( I / K , lim → ⁡ E i ) = 0 . Thus s ⁢ Ext R 1 ⁡ ( R / I , lim → ⁡ E i ) = 0 for any ideal I of R. Let s 0 ∈ S , x i ∈ E i and define g : R ⁢ s ⁢ s 0 → E i by g ⁢ ( s ⁢ s 0 ⁢ r ) = r ⁢ x i . Note that g is well-defined since S is composed of non-zero-divisors. Thus there is a homomorphism f : R → E i which extends sg. Hence we have s ⁢ x i = s ⁢ g ⁢ ( s 0 ⁢ s ) = f ⁢ ( s 0 ⁢ s ) = s ⁢ s 0 ⁢ f ⁢ ( 1 ) , and so x = s 0 ⁢ f ⁢ ( 1 ) . Thus every E i is S-divisible. Hence lim → ⁡ E i is also S-divisible. So lim → ⁡ E i is nonnil-u-S-injective by Theorem 3.5.

(2) ⇒ (3) ⇒ (4) These are straightforward.

(3) ⇒ (1) Set Ω := { I / J ∣ I ⊆ J ⁢ are nonnil ideals of ⁢ R } . Denote by E ⁢ ( F ) the injective envelope of F for each F ∈ Ω . Then ⊕ F ∈ Ω E ⁢ ( F ) is nonnil-u-S-injective. So there exists s ∈ S such that

s ⁢ Ext R 1 ⁡ ( R / I , ⊕ F ∈ Ω E ⁢ ( F ) ) = 0

for every nonnil ideal I of R.

Now, assume, on the contrary, that R is not a nonnil-u-S-Noetherian ring. Then there exists a nonnil ideal J of R which is not s-finite. Let a ∈ I be a non-nilpotent element in I and set I 1 := R ⁢ a . Then s ⁢ I 1 ⊈ J . Choose a 2 ∈ J ∖ I 1 . Then I 1 ⊆ I 2 and s ⁢ I 2 ⊈ I 1 . Continuing this process, we can construct a strictly ascending chain I 1 ⊂ I 2 ⊂ ⋯ of nonnil ideals of R such that for any k ≥ 1 , s ⁢ I n ⊈ I k for every n ≥ k . Set I := ⋃ i = 1 ∞ I i . Then I is a nonnil ideal of R and I / I i ≠ 0 for any i ≥ 1 . Denote by E ⁢ ( I / I i ) the injective envelope of I / I i . So

s ⁢ Ext R 1 ⁡ ( R / I , ⊕ i = 1 ∞ E ⁢ ( I / I i ) ) = 0 .

Let f i be the natural composition I → I / I i ↦ E ⁢ ( I / I i ) . Since s ⁢ I n ⊈ I i for any n ≥ i + 1 ≥ 0 , we have s ⁢ f i ≠ 0 for any i ≥ 1 . Define f : I → ⊕ i = 1 ∞ E ⁢ ( I / I i ) by f ⁢ ( a ) = ( f i ⁢ ( a ) ) . Note that for every a ∈ I , we have a ∈ I i for some i ≥ 1 , and so f is a well-defined R-homomorphism. Let π i : ⊕ i = 1 ∞ E ⁢ ( I / I i ) → E ⁢ ( I / I i ) be the natural projection. The embedding map i : I → R induces an exact sequence

Hom R ⁡ ( R , ⊕ i = 1 ∞ E ⁢ ( I / I i ) ) ⁢ → i * ⁢ Hom R ⁡ ( I , ⊕ i = 1 ∞ E ⁢ ( I / I i ) ) ⁢ → 𝛿 ⁢ Ext R 1 ⁡ ( R / I , ⊕ i = 1 ∞ E ⁢ ( I / I i ) ) → 0 .

Since s ⁢ Ext R 1 ⁡ ( R / I , ⊕ i = 1 ∞ E ⁢ ( I / I i ) ) = 0 , there exists a homomorphism g : R → ⊕ i - 1 ∞ E ⁢ ( I / I i ) such that s 1 ⁢ f = i * ⁢ ( g ) . Since i * ⁢ ( g ) ⁢ ( 1 ) ∈ ⊕ i = 1 ∞ E ⁢ ( I / I i ) , it follows that π i ⁢ i * ⁢ ( g ) ⁢ ( 1 ) = 0 for some sufficiently large i. Thus

s ⁢ π i ⁢ f ⁢ ( a ) = π i ⁢ i * ⁢ ( g ) ⁢ ( a ) = a ⁢ π i ⁢ i * ⁢ ( g ) ⁢ ( 1 ) = 0

for any a ∈ I . Therefore, s ⁢ f i = s ⁢ π i ⁢ f : I → E ⁢ ( I / I i ) is a zero homomorphism, which is a contradiction. Thus R is a nonnil-u-S-Noetherian ring.

(4) ⇒ (3) Let { E i } i ∈ Ω be a family of nonnil-injective modules. Then ∏ i ∈ Ω E i is nonnil-injective. Hence ⊕ ( ∏ i ∈ Ω E i ) is nonnil-u-S-injective by the assumption. Since ⊕ i ∈ Ω E i is a direct summand of ⊕ ( ∏ i ∈ Ω E i ) , it follows that ⊕ i ∈ Ω E i is nonnil-u-S-injective. ∎

4 On ϕ-von Neumann regular rings

Recall that a ϕ-ring R is called ϕ- von Neumann regular if for any a ∈ R ∖ Nil ⁢ ( R ) , there exists r ∈ R such that a = r ⁢ a 2 . An important topic is the study of the S-analog of ϕ-von Neumann regular rings. To study this further, we begin this section with the following result, which characterizes when S - 1 ⁢ R is ϕ-von Neumann regular.

Theorem 4.1.

Let R be a ϕ-ring and let S be a multiplicative subset of R. Then the following statements are equivalent:

S - 1 ⁢ R is ϕ - von Neumann regular.
For every a ∈ R , there exist s ∈ S and r ∈ R such that s ⁢ a = r ⁢ a 2 .
S ¯ - 1 ⁢ ( R / Nil ⁢ ( R ) ) is a field, with S ¯ = S + Nil ⁢ ( R ) .
Any S - 1 ⁢ R - module is ϕ - flat over S - 1 ⁢ R .

Proof.

(1) ⇔ (4) This is well known from [17].

(2) ⇒ (1) Let a s be a non-nilpotent element in S - 1 ⁢ R . Then a ∈ R ∖ Nil ⁢ ( R ) . Thus there exist s ′ ∈ S and r ∈ R such that s ′ ⁢ a = r ⁢ a 2 . Thus a s = s ⁢ r s ′ ⁢ ( a s ) 2 . Hence S - 1 ⁢ R is ϕ-von Neumann regular.

(1) ⇒ (2) Let a ∈ R ∖ Nil ⁢ ( R ) . If a 1 ∈ Nil ⁢ ( S - 1 ⁢ R ) , then a n 1 = 0 for some n ∈ ℕ , whence s ⁢ a n = 0 ∈ Nil ⁢ ( R ) for some s ∈ S . Since Nil ⁢ ( R ) ⁢ R is a prime ideal of R and a ∈ R ∖ Nil ⁢ ( R ) , we have s ∈ S ∩ Nil ⁢ ( R ) = ∅ , a contraction. Hence a 1 ∈ S - 1 ⁢ R ∖ Nil ⁢ ( S - 1 ⁢ R ) , and so there exists r 1 s 1 ∈ S - 1 ⁢ R such that a 1 = r 1 s 1 ⁢ a 2 1 . Thus there exists s 2 ∈ S such that s 1 ⁢ s 2 ⁢ a = s 2 ⁢ r 1 ⁢ a 2 . Set s := s 1 ⁢ s 2 and r := s 2 ⁢ r 1 . Then (2) holds naturally.

(2) ⇒ (3) Let a ¯ s ¯ ∈ S ¯ - 1 ⁢ ( R / Nil ⁢ ( R ) ) ∖ 0 . Then a ∈ R ∖ Nil ⁢ ( R ) , and hence t ⁢ a = r ⁢ a 2 for some t ∈ S and r ∈ R . Then ( a ¯ s ¯ ) 2 = a ¯ s ¯ ⁢ s ⁢ r ¯ t ¯ . So S ¯ - 1 ⁢ ( R / Nil ⁢ ( R ) ) is a von Neumann regular ring. Since S ¯ - 1 ⁢ ( R / Nil ⁢ ( R ) ) is a domain, it is a field.

(3) ⇒ (2) This is easy. ∎

It is certain that for a ϕ-ring R such that S - 1 ⁢ R is von Neumann regular, the element s ∈ S such that s ⁢ a = r ⁢ a 2 for some r ∈ R depends on a ∈ R by Proposition 4.1. The natural question is whether we can define a ϕ-ring in which the element s ∈ S is uniform on any element a ∈ R ∖ Nil ⁢ ( R ) . The following theorem shows that this ring is exactly a ϕ-von Neumann regular ring.

Theorem 4.2.

Let R be a ϕ-ring and let S be a multiplicative subset of R with 0 ∉ S . Then the following statements are equivalent:

R is a ϕ - von Neumann regular ring.
There exists an element s ∈ S such that for any a ∈ R ∖ Nil ⁢ ( R ) , there exists r ∈ R such that s ⁢ a = r ⁢ a 2 .

Proof.

(1) ⇒ (2) This is straightforward.

(2) ⇒ (1) Let s ∈ S be the element in (2) and let 0 ≠ x ¯ ∈ R / Nil ⁢ ( R ) . Then x ∈ R ∖ Nil ⁢ ( R ) . Thus, by (2), s ⁢ x = a ⁢ x 2 for some a ∈ R , and hence s ¯ ⁢ x ¯ = a ¯ ⁢ x ¯ 2 . Therefore, R / Nil ⁢ ( R ) is a u- S ¯ -von Neumann regular ring. Since R / Nil ⁢ ( R ) is a domain, it follows that S ¯ is a regular multiplicative subset of R. So R / Nil ⁢ ( R ) is a von Neumann regular ring [6, Proposition 3.17]. Hence R is a ϕ-von Neumann regular ring by [17, Theorem 4.1]. ∎

Let { M j } j ∈ Γ be a family of R-modules. Let { m i , j } i ∈ Λ j ⊆ M j for each j ∈ Γ and N j = 〈 m i , j 〉 i ∈ Λ j . Recall from [6] that we say that a family of R-modules { M j } j ∈ Γ is u-S-generated by { { m i , j } i ∈ Λ j } j ∈ Γ provided that there exists an element s ∈ S such that s ⁢ M j ⊆ N j for every j ∈ Γ . It is well known that a ϕ-ring R is a ϕ-von Neumann regular ring if and only if every R-module is ϕ-flat, if and only if every nonnil principal (finitely generated) ideal is generated by an idempotent [17, Theorem 4.1]. We now give a generalization of this result.

Theorem 4.3.

Let R be a ϕ-ring and let S be a multiplicative subset of R with 0 ∉ S . Then the following statements are equivalent:

R is a ϕ - von Neumann regular ring.
For any R - modules M and N with N ϕ - torsion, there exists s ∈ S such that s ⁢ Tor 1 R ⁡ ( M , N ) = 0 .
There exists s ∈ S such that s ⁢ Tor 1 R ⁡ ( R / I , R / J ) = 0 for every ideal I and every nonnil ideal J of R.
There exists s ∈ S such that s ⁢ Tor 1 R ⁡ ( R / I , R / J ) = 0 for every S - finite ideal I and every S - finite nonnil ideal J of R.
There exists s ∈ S such that s ⁢ Tor 1 R ⁡ ( R / 〈 a 〉 , R / 〈 a 〉 ) = 0 for any element a ∈ R ∖ Nil ⁢ ( R ) .
Every R - module is ϕ - u - S - flat.
The class of all nonnil principal ideals of R is u - S - generated by idempotents.
The class of all nonnil finitely generated ideals of R is u - S - generated by idempotents.

Proof.

(2) ⇔ (6), (8) ⇒ (7), and (3) ⇒ (4) ⇒ (5) These are straightforward.

(1) ⇔ (5) This follows from the equivalences: s ⁢ Tor 1 R ⁢ ( R / R ⁢ a , R / R ⁢ a ) = 0 if and only if s ⁢ R ⁢ a / R ⁢ a 2 = 0 if and only if there exists r ∈ R such that s ⁢ a = r ⁢ a 2 for some r ∈ R .

(2) ⇒ (3) Set M = N := ⊕ I ∈ NN ⁡ ( R ) R / I . Then (3) holds naturally.

(3) ⇒ (2) Suppose that M is generated by { m i ∣ i ∈ Γ } and N is generated by { n i ∣ i ∈ 1 } . Well order Γ and Λ. Set M 0 := 0 and M α := 〈 m i ∣ ⁢ i ⁢ < α 〉 for each α ≤ Γ . Then M has a continuous filtration { M α ∣ α ≤ Γ } with M α + 1 / M α ≅ R / I α + 1 and I α = Ann R ⁡ ( m α + M α ∩ R ⁢ m α ) . Note that I α is a nonnil ideal of R since M is a ϕ-torsion R-module. Similarly, N has a continuous filtration { N β ∣ β ≤ Λ } with N β + 1 / N β ≅ R / J β + 1 and J β = Ann R ⁡ ( n β + N β ∩ R ⁢ n β ) . Since s ⁢ Tor 1 R ⁡ ( R / I α , R / J β ) = 0 for each α ≤ Γ and β ≤ Λ , it is easy to verify that s ⁢ Tor 1 R ⁡ ( M , N ) = 0 by transfinite induction on both positions of M and N.

(1) ⇒ (7) This is straightforward.

(7) ⇒ (8) Let I = R ⁢ a 1 + ⋯ + R ⁢ a n be a nonnil finitely generated ideal of R. Since R is a ϕ-ring, we can assume that all a i ∈ R ∖ Nil ⁢ ( R ) . By (3), there exists an element s ∈ S such that for every i = 1 , … , n , there is an idempotent e i ∈ R ⁢ a i such that s ⁢ R ⁢ a i ⊆ R ⁢ e i ⊆ R ⁢ a i . Set J := R ⁢ e 1 + ⋯ + R ⁢ e n . Then J is a subideal of I such that s ⁢ I ⊆ J ⊆ I . By recurrence on n, we will show that J is principally generated by an idempotent element.

The case n = 2 : Since e 1 = e 1 ⁢ ( e 1 + e 2 - e 1 ⁢ e 2 ) and e 2 = e 2 ⁢ ( e 1 + e 2 - e 1 ⁢ e 2 ) , we get R ⁢ e 1 + R ⁢ e 2 = R ⁢ ( e 1 + e 2 - e 1 ⁢ e 2 ) and it is easy to show that ( e 1 + e 2 - e 1 ⁢ e 2 ) is an idempotent element.

General case: Repeat with the case n = 2 . Hence we have J = R ⁢ e for some idempotent element e ∈ R . So I is u-S-generated by an idempotent.

(8) ⇒ (3) Let s ∈ S be the element in (8) and let J be an ideal of R. Set M = R / J and let 0 → K → F → M → 0 be an exact sequence where F is free. Let I be a finitely generated nonnil ideal of R. Then s ⁢ I ⊆ R ⁢ e ⊆ I for some idempotent e by the hypothesis. For x ∈ K ∩ F ⁢ I , we have s ⁢ x = e ⁢ y = e ⁢ y = s ⁢ e ⁢ x ∈ K ⁢ I (where y ∈ F ). Hence s ⁢ ( K ∩ F ⁢ I ) ⊆ K ⁢ I ⊆ K ∩ F ⁢ I . Then by [8, Exercise 2.7] there exists Ψ : Tor 1 ⁢ ( M , R / I ) → K ∩ F ⁢ I / K ⁢ I such that the following diagram with exact rows is commutative:

where α and β are isomorphisms. Then Ψ is an isomorphism by the Five Lemma. Since s ⁢ ( K ∩ F ⁢ I ) ⊆ K ⁢ I ⊆ K ∩ F ⁢ I , we have s ⁢ Tor 1 R ⁢ ( R / J , R / I ) = 0 . ∎

Note that a ϕ-ring R such that S - 1 ⁢ R is von Neumann regular is not necessarily ϕ-von Neumann regular.

Example 4.4.

Let D be an integral domain which is not a field and set S := D ∖ { 0 } . Then S - 1 ⁢ D is ϕ-von Neumann regular, since it is a field.

Corollary 4.5.

Let R be a ϕ-ring and let S be a multiplicative subset of R consisting of finite elements. Then R is a ϕ-von Neumann regular ring if and only if S - 1 ⁢ R is a ϕ-von Neumann regular ring.

Proof.

We just need to show that if S - 1 ⁢ R is a ϕ-von Neumann regular ring, then R is a ϕ-von Neumann regular ring. Let S = { s 1 , … , s n } . Set s := s 1 ⁢ ⋯ ⁢ s n . By Proposition 4.1, for any a ∈ R ∖ Nil ⁢ ( R ) , there exist s i ∈ S and r a ∈ R such that s i ⁢ a = r a ⁢ a 2 . So for every a ∈ R ∖ Nil ⁢ ( R ) , s ⁢ a = r ⁢ a 2 for some r ∈ R . ∎

Funding source: National Research Foundation of Korea

Award Identifier / Grant number: 2021R1I1A3047469

Funding statement: Hwankoo Kim was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2021R1I1A3047469).

Acknowledgements

The authors would like to thank the reviewers for their comments and suggestions to improve the paper, especially the two questions (Theorem 2.6 and Corollary 3.8).

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Received: 2023-07-03

Revised: 2023-09-13

Accepted: 2023-09-19

Published Online: 2024-01-02

Published in Print: 2024-10-01

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