Some remarks on $u$-$S$-Noetherian and $u$-$S$-coherent rings

TL;DR

This paper provides new characterizations of $u$-$S$-Noetherian and $u$-$S$-coherent rings using uniform $S$-versions of modules and addresses an open question in the field.

Contribution

It introduces novel characterizations of these rings via uniform $S$-modules and resolves a previously posed question by Bouziri.

Findings

New characterizations of $u$-$S$-Noetherian rings.

New characterizations of $u$-$S$-coherent rings.

Negative answer to Bouziri's question.

Abstract

In this paper, we give some new characterizations of $u$-$S$-Noetherian rings and $u$-$S$-coherent rings in terms of uniform $S$-version of injective precovers, flat preenvelopes and absolutely pure modules, respectively. Moreover, we give a negative answer to a question proposed by Bouziri [3].

Equations

$$0\to T_1\to F\stackrel{f}{\to }M\to T_2\to 0$$

$$T^{\prime }{\otimes }_R{\prod }_{i\in \Lambda }{F}_i\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$K^{\prime }{\otimes }_R{\prod }_{i\in \Lambda }{F}_i\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$0\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$I{\otimes }_R{\prod }_{i\in \Lambda }R\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$L_I{\otimes }_R{\prod }_{i\in \Lambda }R\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$L_I{\otimes }_R{R}^{\Lambda }\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$sx=h(\sum _{j=1}^n{m}_j\otimes (r_{j,i}{)}_{i\in L_I})=(\sum _{j=1}^n{m}_j{r}_{j,i}{)}_{i\in L_I}.$$

$${\prod }_{i\in \Lambda }{F}_i\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$0\to \prod _{i\in \Lambda }{F}_i\stackrel{f}{\to }F\to \mathrm{Coker}(f)\to 0$$

$$M\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$0\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$0\to T_1\to K^{\prime }\stackrel{i_{K^{\prime }}}{\to }L\stackrel{{\pi }_L}{\to }T\to 0$$

$$0\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$K\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$0\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$I\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$0\to T_1\to F\to I\to I/\mathrm{Im}({\delta }_i)\to 0.$$

$$0\to {\mathrm{Hom}}_R(C,E)\to {\mathrm{Hom}}_R(B,E)\to {\mathrm{Hom}}_R(A,E)\to 0$$

$$E_i\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$0\to \mathrm{Ker}(f)\to E\to \underset{i\in \Lambda }{⨁}{E}_i\to 0$$

$$E^{\prime }\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

Full text

Some remarks on $u$-$S$-Noetherian and $u$-$S$-coherent rings

Xiaolei Zhang, Wei Qi

School of Mathematics and Statistics, Tianshui Normal University, 741001 Tianshui, China

Abstract

In this paper, we give some new characterizations of $u$-$S$-Noetherian rings and $u$-$S$-coherent rings in terms of uniform $S$-versions of injective precovers, flat preenvelopes and absolutely pure modules. Moreover, we give a negative answer to a question proposed by Bouziri.

Key Words: $u$-$S$-Noetherian ring, $u$-$S$-coherent ring, $u$-$S$-absolutely pure module, $u$-$S$-precover, $u$-$S$-preenvelope.

2020 Mathematics Subject Classification: 13C11.

1. Introduction

Throughout this paper, $R$ is always a commutative ring with an identity and $S$ is always a multiplicative subset of $R$, that is, $1\in S$ and $s_{1}s_{2}\in S$ for any $s_{1}\in S$ and any $s_{2}\in S$. Let $M$ be an $R$-module. Denote by $M^{+}={\rm Hom}_{\mathbb{Z}}(M,\mathbb{Q}/\mathbb{Z})$.

In 2002, Anderson [1] introduced the notions of $S$-finite modules and $S$-Noetherian rings which can be seen as $S$-versions of finitely generated modules and Noetherian rings, respectively. An $R$-module $M$ is said to be $S$-finite (with respect to $s$) provided that there is a finitely generated submodule $F$ of $M$ such that $sM\subseteq F$ for some $s\in S$; and a ring $R$ is called an $S$-Noetherian ring if every ideal of $R$ is $S$-finite. In 2018, Bennis and Hajoui [2] introduced the notions of $S$-finitely presented modules and $S$-coherent rings which can be seen as $S$-versions of finitely presented modules and coherent rings, respectively. An $R$-module $M$ is called $S$-finitely presented provided that there exists an exact sequence of $R$-modules $0\rightarrow K\rightarrow F\rightarrow M\rightarrow 0$ with $K$ $S$-finite and $F$ finitely generated free. A ring $R$ is called an $S$-coherent ring if $R$ itself is an $S$-coherent $R$-module, that is, every finitely generated ideal of $R$ is $S$-finitely presented.

However, the selected $s$ in the definitions of $S$-Noetherian rings and $S$-coherent rings is not uniform on ideals. So Chen et al. [4] introduced the notion of $u$-$S$-Noetherian rings. A ring $R$ is called a $u$-$S$-Noetherian ring (abbreviates uniformly $S$-Noetherian ring) provided that there exists $s\in S$ such that any ideal $I$ of $R$ is $S$-finite with respect to $s$. Subsequently, Zhang [18] introduced the notion of uniformly $S$-coherent rings. An $R$-module $M$ is said to be $u$-$S$-finitely presented (abbreviates uniformly $S$-finitely presented) (with respect to $s$) provided that there is an exact sequence

[TABLE]

with $F$ finitely presented and $sT_{1}=sT_{2}=0$. A ring $R$ is called a $u$-$S$-coherent ring (abbreviates uniformly $S$-coherent ring) (with respect to $s$) provided that there exists $s\in S$ such that any finitely generated ideal of $R$ is $u$-$S$-finitely presented with respect to $s$. It was proved in [18, Proposition 3.10] that every $u$-$S$-Noetherian ring is $u$-$S$-coherent.

It is well-known that a ring $R$ is Noetherian if and only if any direct sum of injective $R$-modules is injective, if and only if every $R$-module has a injective precover. A ring $R$ is coherent if and only if any pure quotient module of an absolutely pure module is absolutely pure, if and only if any direct limit of absolutely pure modules is absolutely pure, if and only if the class of flat modules is preenveloping. The main aim of this paper is to give uniform $S$-versions of several classical results.

Since the paper involves uniformly torsion theory, we give a quick review (see [16] for more details). An $R$-module $T$ is called $u$-$S$-torsion (with respect to $s$) provided that there exists $s\in S$ such that $sT=0$. An $R$-sequence $A\xrightarrow{f}B\xrightarrow{g}C$ is said to be $u$-$S$-exact at $B$ (with respect to $s$) if $s{\rm Ker}(g)\subseteq{\rm Im}(f)$ and $s{\rm Im}(f)\subseteq{\rm Ker}(g)$ for some $s\in S$. An $R$-sequence $\cdots\rightarrow A_{n-1}\xrightarrow{f_{n}}A_{n}\xrightarrow{f_{n+1}}A_{n+1}\rightarrow\cdots$ is called a $u$-$S$-exact sequence if it is $u$-$S$-exact at each $A_{n}$. An $R$-homomorphism $f:M\rightarrow N$ is a $u$-$S$-monomorphism $($resp., $u$-$S$-epimorphism, $u$-$S$-isomorphism$)$ (with respect to $s$) provided $0\rightarrow M\xrightarrow{f}N$ $($resp., $M\xrightarrow{f}N\rightarrow 0$, $0\rightarrow M\xrightarrow{f}N\rightarrow 0$ $)$ is $u$-$S$-exact (with respect to $s$). Let $M$ and $N$ be $R$-modules. We say that $M$ is $u$-$S$-isomorphic to $N$ if there exists a $u$-$S$-isomorphism $f:M\rightarrow N$. If so, it was proved in [19, Lemma 1.2] that there is $s\in S$ and a $u$-$S$-isomorphism $f^{\prime}:N\rightarrow M$ such that $f\circ f^{\prime}=s{\rm Id}_{N}$ and $f^{\prime}\circ f=s{\rm Id}_{M}.$

2. Characterizing $u$-$S$-coherent rings by $u$-$S$-flat modules

Let $R$ be a ring, $S$ be a multiplicative subset of $R$ and $s\in S$. Recall from [16] that an $R$-module $F$ is said to be $u$-$S$-flat (abbreviates uniformly $S$-flat) (with respect to $s$) provided that for any exact sequence $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$, the induced sequence $0\rightarrow A\otimes_{R}F\rightarrow B\otimes_{R}F\rightarrow C\otimes_{R}F\rightarrow 0$ is $u$-$S$-exact (with respect to $s$). It follows by [18, Proposition 4.3] that an $R$-module $F$ is $u$-$S$-flat if and only if there exists $s\in S$ such that ${\rm Tor}_{1}^{R}(R/I,F)$ is $u$-$S$-torsion with respect to $s$ for any ideal $I$ of $R$ provided that $S$ is regular, i.e., every element in $S$ is a nonzero-divisor.

It follows from [18, Theorem 4.4] that a ring $R$ is a $u$-$S$-coherent ring if and only if there is $s\in S$ such that any direct product of flat modules (projective modules, copies of $R$) is $u$-$S$-flat with respect to $s$. It is well-known that a ring $R$ is coherent if and only if every $R$-module has a flat preenvelope. Now, we will try to give the uniform $S$-version of this result.

First, we introduce a uniform $S$-version of the (pre)enveloping properties for a class of modules.

Definition 2.1**.**

Let $M$ be an $R$-module and $\mathcal{C}$ a class of $R$-modules.

1. (1)

An $R$-homomorphism $f\in{\rm Hom}_{R}(M,C)$ with $C\in\mathcal{C}$ is called a $\mathcal{C}$ $u$-$S$*-preenvelope of $M$ $($with respect to $s)$ provided that, for any $C^{\prime}\in\mathcal{C}$, ${\rm Hom}_{R}(f,C^{\prime}):{\rm Hom}_{R}(C,C^{\prime})\rightarrow{\rm Hom}_{R}(M,C^{\prime})$ is a $u$-$S$-epimorphism *$($with respect to $s)$. 2. (2)

A $\mathcal{C}$ $u$-$S$-preenvelope $f\in{\rm Hom}_{R}(M,C)$ of $M$ is called a $\mathcal{C}$ $u$-$S$*-envelope $($with respect to $s)$ of $M$, provided that, for any $\alpha\in{\rm End}_{R}(C)$, $f=\alpha\circ f$ implies that $\alpha$ is a $u$-$S$-isomorphism *$($with respect to $s)$. 3. (3)

If every $R$-module has a $\mathcal{C}$ $u$-$S$-preenvelope (resp., $u$-$S$-envelope), then $\mathcal{C}$ is called a $u$-$S$-preenveloping (resp., $u$-$S$-enveloping) class.

Recall from [19] that a short $u$-$S$-exact sequence $\xi:0\rightarrow A\xrightarrow{f}B\xrightarrow{g}C\rightarrow 0$ is said to be * $u$-$S$-split* (with respect to $s$) provided that there is $s\in S$ and $R$-homomorphism $f^{\prime}:B\rightarrow A$ such that $f^{\prime}(f(a))=sa$ for any $a\in A$, that is, $f^{\prime}\circ f=s{\rm Id}_{A}$.

Lemma 2.2**.**

Let $\xi:0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ be a $u$-$S$-split short $u$-$S$-exact sequence. Suppose $B$ is $u$-$S$-flat. Then so is $A$ and $C$.

Proof.

Let $\xi:0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ be a $u$-$S$-split short $u$-$S$-exact sequence with $B$ $u$-$S$-flat. It follows by [17, Proposition 2.3] that $\xi$ is $u$-$S$-pure. Hence $C$ is $u$-$S$-flat by [17, Proposition 2.5]. By the proof of [19, Lemma 2.4], there is a $u$-$S$-split short $u$-$S$-exact sequence $\xi^{\prime}:0\rightarrow C\rightarrow B\rightarrow A\rightarrow 0$ induced by $\xi$. Hence $A$ is also $u$-$S$-flat. ∎

For any $s\in S$, let $s\mathcal{F}$ denote the class of all $R$-modules that are $u$-$S$-flat with respect to $s$. It was proved in [3] that if there exists $s\in S$ such that every $R$-module has an $s\mathcal{F}$-preenvelope, then $R$ is $u$-$S$-coherent. However, the converse is still open (see [3, Question 2.12]). Next, we will give a new characterization of $u$-$S$-coherent rings in terms of uniform $S$-version of preenvelopes under a mild condition.

Theorem 2.3**.**

Let $R$ be a ring, $S$ be a regular multiplicative subset of $R$. Consider the following statements:

1. (1)

$R$* is a $u$-$S$-coherent ring;* 2. (2)

for all $s\in S$, any direct product of $R$-modules in $s\mathcal{F}$ is $u$-$S$-flat; 3. (3)

any direct product of flat $R$-modules $($projective $R$-modules, copies of $R)$ is $u$-$S$-flat; 4. (4)

any $R$-module has a $u$-$S$-flat $u$-$S$-preenvelope.

Then $(1)\Leftrightarrow(2)\Leftrightarrow(3)\Leftarrow(4)$. If, moreover, every $u$-$S$-flat module is $u$-$S$-isomorphic to a flat module, then $(3)\Rightarrow(4)$.

Proof.

$(2)\Rightarrow(3)$ Trivial.

$(1)\Rightarrow(2)$ Suppose $R$ is $u$-$S$-coherent with respect to $s^{\prime}\in S$. Let $\{F_{i}\mid i\in\Lambda\}$ be a family of $u$-$S$-flat $R$-modules with respect to $s$ and $I$ a finitely generated ideal of $R$. Then $I$ is $u$-$S$-finitely presented with respect to $s^{\prime}$. So we have an exact sequence $0\rightarrow T^{\prime}\xrightarrow{t^{\prime}}K\xrightarrow{f}I\xrightarrow{t}T\rightarrow 0$ with $K$ finitely presented and $s^{\prime}T=s^{\prime}T^{\prime}=0$. Set ${\rm Im}(f)=K^{\prime}$, $f_{1}:K\rightarrow K^{\prime}$ and $f_{2}:K^{\prime}\rightarrow I$ to be the natural homomorphisms.

Consider the following commutative diagram with exact rows:

[TABLE]

By [15, Theorem 2.6.11], $\gamma$ is an isomorphism. Since each $T^{\prime}\otimes_{R}F_{i}$ is $u$-$S$-torsion with respect to $s^{\prime}$, $\prod_{i\in\Lambda}(T^{\prime}\otimes_{R}F_{i})$ and hence $X$ is also $u$-$S$-torsion with respect to $s^{\prime}$. We claim that ${\rm Ker}(\beta)$ is a $u$-$S$-torsion with respect to $s^{\prime}$. Indeed, let $x\in{\rm Ker}(\beta)$. Then there is $y\in K\otimes_{R}\prod_{i\in\Lambda}F_{i}$ such that $(\prod f_{1}\otimes{\rm Id}_{F_{i}})\circ\gamma(y)=0,$ and $f_{1}\otimes{\rm Id}_{\prod F_{i}}(y)=x.$ Hence, there is $z\in{\rm Im}(\prod t^{\prime}\otimes{\rm Id}_{F_{i}})$ such that $\gamma(y)=z$. So $s^{\prime}x=s^{\prime}f_{1}\otimes{\rm Id}_{\prod F_{i}}(y)=f_{1}\otimes{\rm Id}_{\prod F_{i}}(s^{\prime}y)=$ $f_{1}\otimes{\rm Id}_{\prod F_{i}}\gamma^{-1}\gamma(s^{\prime}y)=f_{1}\otimes{\rm Id}_{\prod F_{i}}\gamma^{-1}s^{\prime}\gamma(y)=f_{1}\otimes{\rm Id}_{\prod F_{i}}\gamma^{-1}(s^{\prime}z)=f_{1}\otimes{\rm Id}_{\prod F_{i}}\gamma^{-1}(0)=0$. The claim holds.

Consider the following commutative diagram:

[TABLE]

where $X_{i}:={\rm Ker}(f_{2}\otimes{\rm Id}_{F_{i}}).$ We claim that ${\rm Ker}(\theta)$ is also $u$-$S$-torsion with respect to $s(s^{\prime})^{2}$. Indeed, let $a\in{\rm Ker}(\theta)$. Since $T\otimes_{R}\prod_{i\in\Lambda}F_{i}$ is $u$-$S$-torsion with respect to $s^{\prime}$, we have $t\otimes{\rm Id}_{\prod F_{i}}(s^{\prime}a)=s^{\prime}(t\otimes{\rm Id}_{\prod F_{i}}(a))=0$. So there $b\in K^{\prime}\otimes_{R}\prod_{i\in\Lambda}F_{i}$ such that $f_{2}\otimes{\rm Id}_{\prod F_{i}}(b)=s^{\prime}a.$ Then $(\prod f_{2}\otimes{\rm Id}_{F_{i}})\circ\beta(b)=\theta\circ(f_{2}\otimes{\rm Id}_{\prod F_{i}})(b)=\theta(s^{\prime}a)=s^{\prime}\theta(a)=0$. Since each $F_{i}$ is $u$-$S$-flat with respect to $s$, we have $\prod_{i\in\Lambda}X_{i}$ is $u$-$S$-torsion with respect to $s$. Hence $\beta(sb)=s\beta(b)=0$. Since ${\rm Ker}(\beta)$ is $u$-$S$-torsion with respect to $s^{\prime}$, we have $ss^{\prime}b=0$. Consequently, $s(s^{\prime})^{2}a=ss^{\prime}f_{2}\otimes{\rm Id}_{\prod F_{i}}(b)=f_{2}\otimes{\rm Id}_{\prod F_{i}}(ss^{\prime}b)=0.$

Now we consider the following commutative diagram with exact rows:

[TABLE]

Since each $F_{i}$ is $u$-$S$-flat with respect to $s$, $\prod_{i\in\Lambda}{\rm Tor}_{1}^{R}(R/I,F_{i})$ is $u$-$S$-torsion with respect to $s$. Hence $s{\rm Tor}_{1}^{R}(R/I,\prod_{i\in\Lambda}F_{i})\subseteq{\rm Ker}(\theta)$. So ${\rm Tor}_{1}^{R}(R/I,\prod_{i\in\Lambda}F_{i})$ is $u$-$S$-torsion. Hence $\prod_{i\in\Lambda}F_{i}$ is $u$-$S$-flat by Proposition [18, Theorem 4.3].

$(4)\Rightarrow(1)$ Let $\mathscr{I}=\{I_{i}\mid i\in\Gamma\}$ be the set of all finitely generated ideals of $R$, and let $0\rightarrow L_{I_{i}}\rightarrow F_{i}\rightarrow I_{i}\rightarrow 0$ be an exact sequence with $F_{i}$ finitely generated free for each $i\in\Gamma$. Let $\Lambda$ be cardinal greater than the sum of all ordinals of $L_{I_{i}}$s.

For each $I\in\mathscr{I}$, consider the following commutative diagram with exact rows:

[TABLE]

By assumption, $\prod_{i\in\Lambda}R$ is a $u$-$S$-flat module with respect to some fixed element $s\in S$, it follows that $f$ is a $u$-$S$-monomorphism with respect to $s$. So $g$ is also a $u$-$S$-monomorphism with respect to $s$. Consider the following commutative diagram with exact rows:

[TABLE]

Since $g$ is a $u$-$S$-monomorphism with respect to $s$, $h$ is a $u$-$S$-epimorphism with respect to $s$. We will show $L_{I}$ is $S$-finite with respect to $s$. Indeed, consider the following exact sequence

[TABLE]

with $T$ a $u$-$S$-torsion module with respect to $s$. Consider an element $x\in\prod_{i\in\Lambda}L_{I_{i}}$ such that every element of $L_{I}$ appears at least once among the components of $x$. Then $sx\subseteq{\rm Im}h$. Subsequently, there exist $m_{j}\in L_{I},r_{j,i}\in R,i\in\Lambda,j=1,\dots,n$ such that

[TABLE]

Set $U=\langle m_{j}\mid j=1,\dots,n\rangle$ be the finitely generated submodule of $L_{I}$. Now, for any $m\in L_{I}$, $sm\in\langle\sum_{j=1}^{n}m_{j}r_{j,m}\rangle\subseteq U$, thus the embedding map $U\hookrightarrow L_{I}$ is a $u$-$S$-isomorphism with respect to $s$ and so $L_{I}$ is $S$-finite with respect to $s$. Consequently, $I$ is $u$-$S$-finitely presented with respect to $s$. Hence, $R$ is $u$-$S$-coherent with respect to $s$.

(4)$\Rightarrow$ (3): Let $\{F_{i}\mid i\in\Lambda\}$ be a family of flat modules. Then, by assumption, $\prod_{i\in\Lambda}F_{i}$ has a $u$-$S$-flat $u$-$S$-preenvelope $f:\prod_{i\in\Lambda}F_{i}\rightarrow F$ with $F$ $u$-$S$-flat. Hence there exists $s\in S$ such that for any $i\in\Lambda$ we have the following commutative diagram:

[TABLE]

Consequently, $s{\rm Id}_{\prod_{i\in\Lambda}F_{i}}=(\prod_{i\in\Lambda}g_{i})\circ f$. So

[TABLE]

is a $u$-$S$-split short $u$-$S$-exact sequence. It follows by Lemma 2.2 that $\prod_{i\in\Lambda}F_{i}$ is $u$-$S$-flat.

Now, we assume that every $u$-$S$-flat module is $u$-$S$-isomorphic to a flat module.

(3)$\Rightarrow$ (4): Let $M$ be an $R$-module and Card$M\leq\aleph_{\beta}$. Then by [7, Lemma 5.3.12] there is an infinite cardinal $\aleph_{\alpha}$ such that if $F$ is a flat module and $S$ is a submodule of $F$ with Card$S\leq\aleph_{\alpha}$, there is a pure, hence flat, submodule $G$ of $F$ with $S\subseteq G$ and Card$G\leq\aleph_{\alpha}$. So for any flat $R$-module $F_{i}$ and any homomorphism $\phi_{i}^{\prime}:M\rightarrow F_{i}$, there is a flat $R$-submodule $G_{i}$ of $F_{i}$ satisfying that $G_{i}$ contains $f(M)$ and Card$G_{i}\leq\aleph_{\alpha}$. So, we have a natural factorization $M\xrightarrow{\phi_{i}}G_{i}\rightarrow F_{i}$.

Note Card$G_{i}\leq\aleph_{\alpha}$ for each $i$, so all such $G_{i}$ (up to isomorphism) constitute a set, which is denoted by $\{G_{i}\mid i\in\Lambda\}$. We claim that the natural $R$-homomorphism $\phi:=\prod_{i\in\Lambda}\phi_{i}:M\rightarrow\prod_{i\in\Lambda}G_{i}$ is $u$-$S$-flat $u$-$S$-preenvelope of $M$. Indeed, by assumption, $\prod_{i\in\Lambda}G_{i}$ is $u$-$S$-flat. Let $f:M\rightarrow F$ be an $R$-homomorphism with $F$ $u$-$S$-flat. Then $F$ is $u$-$S$-isomorphic to a flat module $F^{\prime}$. Hence there is $s\in S$ and $u$-$S$-isomorphisms $t:F^{\prime}\rightarrow F$ and $u$-$S$-isomorphism $t^{\prime}:F\rightarrow F^{\prime}$ such that $t\circ t^{\prime}=s{\rm Id}_{F}$ and $t^{\prime}\circ t=s{\rm Id}_{F^{\prime}}$ by [19, Lemma 1.2]. By the property of each $G_{i},$ there exists an $R$-homomorphism $g:\prod_{i\in\Lambda}G_{i}\rightarrow F^{\prime}$ such that $g\circ\phi=t^{\prime}\circ f$:

[TABLE]

Hence, $(t\circ g)\circ\phi=t\circ t^{\prime}\circ f=sf$. Consequently, $\phi$ is a $u$-$S$-flat $u$-$S$-preenvelope. ∎

Remark 2.4*.*

The equivalence of $(1)$ and $(3)$ in Theorem 2.3 is slightly different with that in [18, Theorem 4.4]. As to the equivalence of $(1)$ and $(4)$, we don’t know whether every $u$-$S$-flat module is $u$-$S$-isomorphic to a flat module in general.

3. Characterizing $u$-$S$-coherent rings by $u$-$S$-absolutely pure modules

It is well-known that a ring $R$ is coherent if and only if any pure quotient of an absolutely pure $R$-module is absolutely pure, if and only if any direct limit of absolutely pure $R$-modules is absolutely pure (see [13]). In this section, we will give a uniform $S$-versions of this result.

Recall the notion of uniformly $S$-absolutely pure modules from [17]. An $R$-module $E$ is said to be $u$-$S$-absolutely pure (abbreviates uniformly $S$-absolutely pure) (with respect to $s$) if there exists an element $s\in S$ satisfying that for any finitely presented $R$-module $N$, ${\rm Ext}_{R}^{1}(N,E)$ is $u$-$S$-torsion with respect to $s$, or equivalently, there exists an element $s\in S$ satisfying that if $P$ is finitely generated projective, $K$ is a finitely generated submodule of $P$ and $f:K\rightarrow E$ is an $R$-homomorphism, then there is an $R$-homomorphism $g:P\rightarrow E$ such that $sf=gi$.

Lemma 3.1**.**

Let $R$ be a ring, $S$ be a multiplicative subset of $R$, and $M$ an $R$-module. Suppose $M$ is $u$-$S$-coherent with respect to some $s\in S$. Then $M$ is $u$-$S$-finitely presented with respect to $s^{2}$.

Proof.

Suppose $M$ is $u$-$S$-coherent with respect to some $s\in S$. Then $M$ is $S$-finite with respect to $s$, and hence there is a finitely generated $R$-submodule $K$ of $M$ such that $sM\subseteq K\stackrel{{\scriptstyle i}}{{\hookrightarrow}}M$. Note that $K$ is $u$-$S$-finitely presented with respect to $s$. So there is an exact sequence $0\rightarrow T_{1}\rightarrow F\xrightarrow{f}K\rightarrow T_{2}\rightarrow 0$ with $F$ finitely presented and $sT_{1}=sT_{2}=0$. Then one can construct an exact sequence $0\rightarrow T_{1}\rightarrow F\xrightarrow{f\circ i}M\rightarrow T_{3}\rightarrow 0$ with $s^{2}T_{3}=0$. Hence, $M$ is $u$-$S$-finitely presented with respect to $s^{2}$. ∎

Let $\xi:0\rightarrow A\xrightarrow{f}B\xrightarrow{g}C\rightarrow 0$ be a $u$-$S$-short exact sequence. Then $\xi$ is said to be $u$-$S$-split (with respect to $s$) provided that there is $s\in S$ and $R$-homomorphism $f^{\prime}:B\rightarrow A$ such that $f^{\prime}(f(a))=sa$ for any $a\in A$, that is, $f^{\prime}\circ f=s{\rm Id}_{A}$.

Lemma 3.2**.**

Let $\xi:0\rightarrow A\xrightarrow{f}B\xrightarrow{g}C\rightarrow 0$ be a $u$-$S$-split short exact sequence of $R$-modules with respect to $s$. Suppose $B$ is finitely presented, then $C$ is $u$-$S$-finitely presented with respect to $s$.

Proof.

First, we claim that $A$ is $S$-finite with respect to $s$. Indeed, let $f^{\prime}:B\rightarrow A$ be $R$-homomorphisms such that $f^{\prime}\circ f=s{\rm Id}_{A}$. Then for any $a\in A$, we have $sa=f^{\prime}(f(a))$. Hence, $sA\subseteq{\rm Im}(f^{\prime})\subseteq A$. Since $B$ is finitely presented, we have ${\rm Im}(f^{\prime})$ is finitely generated. Consequently, $A$ is $S$-finite with respect to $s$.

Next, we claim that $C$ is $u$-$S$-finitely presented with respect to $s$. Indeed, Consider the following commutative:

[TABLE]

Then $sT=0.$ Since ${\rm Im}(f^{\prime})$ is finitely generated and $B$ is finitely presented, $L$ is also finitely presented. It follows by the snake Lemma that ${\rm Ker}(\pi)\cong T$, and hence $s{\rm Ker}(\pi)=0$. Considering the exact sequence $0\rightarrow{\rm Ker}(\pi)\rightarrow L\rightarrow C\rightarrow 0$, we have $C$ is $u$-$S$-finitely presented with respect to $s$. ∎

Recently, Bouziri [3] characterized $u$-$S$-coherent rings by dualising $u$-$S$-absolutely pure modules. Now, we give a new characterization of $u$-$S$-coherent rings by considering the pure quotients or direct limits of $u$-$S$-absolutely pure modules. Recall that a multiplicative subset $S$ of $R$ is said to be anti-Archimedean if $\bigcap\limits_{n\geq 1}s^{n}R\bigcap S\not=\emptyset.$

Theorem 3.3**.**

Let $R$ be a ring, $S$ be an anti-Archimedean multiplicative subset of $R$. Then the following statements are equivalent:

1. (1)

$R$* is $u$-$S$-coherent;* 2. (2)

there is $s\in S$ such that any pure quotient of an absolutely pure $R$-module is $u$-$S$-absolutely pure with respect to $s$; 3. (3)

there is $s\in S$ such that any direct limit of absolutely pure $R$-modules is $u$-$S$-absolutely pure with respect to $s$.

Proof.

$(1)\Rightarrow(2)$: Suppose $R$ is a $u$-$S$-coherent ring with respect to some $s^{\prime}\in S$. Let $B$ be an absolutely pure $R$-module. Let $A$ be a pure submodule of $B$, and let $C=B/A$. Let $s\in\bigcap\limits_{n\geq 1}(s^{\prime})^{n}R\bigcap S$. We need to show that $C$ is $u$-$S$-absolutely pure with respect to $s$. Let $P$ be a finitely generated projective $R$-module and $K$ a finitely generated submodule of $P$. Let $i:K\rightarrow P$ be the natural embedding map, and $f_{C}:K\rightarrow C$ an $R$-homomorphism. We claim that there is an $R$-homomorphism $g:P\rightarrow C$ such that $sf_{C}=gi$. Consider the following exact sequence $0\rightarrow L\xrightarrow{i_{L}}P^{\prime}\xrightarrow{\pi_{P^{\prime}}}K\rightarrow 0$ with $P^{\prime}$ finitely generated projective. Consider the following commutative diagram with exact sequences.

[TABLE]

Since $R$ is $u$-$S$-coherent with respect to $s^{\prime}$, the finitely generated projective $R$-module $P$ is a $u$-$S$-coherent $R$-module with respect to some $t^{\prime}\in S$, a power of $s^{\prime}$, and so is its finitely generated $R$-submodule $K$. It follows by [18, Theorem 3.2(2)] that $L$ is also $u$-$S$-coherent with respect to $t^{\prime}$, and so is $u$-$S$-finitely presented with respect to $t=t^{\prime 2}$ by Lemma 3.1. Therefore there is an exact sequence

[TABLE]

with $tT=tT_{1}=0$ and $K^{\prime}$ finitely presented. Consider the following push-out:

[TABLE]

It follows by [11, Lemma 4] that $A$ is absolutely pure. Consider the following commutative diagram:

[TABLE]

Since ${\rm Im}(i_{K^{\prime}})$ is finitely generated, then there is an $R$-homomorphism $g_{A}:P^{\prime}\rightarrow A$ such that $f_{A}\circ e_{L}=g_{A}\circ e_{P^{\prime}}$ (see [12, Corollary 2]). And it follows by [14, Exercise 1.60] that there is an $R$-homomorphism $g_{B}:X\rightarrow B$ such that $\pi_{B}\circ g_{B}=f_{C}\circ\pi_{K}$. Since $K$ is $u$-$S$-isomorphic to $X$ with respect to $t$, by the proof of [17, Proposition 1.1] there exist $\pi^{\prime}_{K}:K\rightarrow X$ and $t_{1}\in S$, which is also a power of $s^{\prime}$ say $t_{1}=s^{\prime n}$, such that $\pi_{K}\circ\pi^{\prime}_{K}=t_{1}{\rm Id}_{K}$ and $\pi^{\prime}_{K}\circ\pi_{K}=t_{1}{\rm Id}_{X}$.

Since $B$ is absolutely pure, it follows by [12, Corollary 2] again that there exists $R$-homomorphism $g^{\prime}_{B}:P\rightarrow B$ such that the following diagram is commutative:

[TABLE]

We have $t_{1}f_{C}=f_{C}\circ\pi_{K}\circ\pi^{\prime}_{K}=\pi_{B}\circ g_{B}\circ\pi^{\prime}_{K}=\pi_{B}\circ g^{\prime}_{B}\circ i=g^{\prime}\circ i$, where $g^{\prime}:=\pi_{B}\circ g^{\prime}_{B}$. Since $s\in\bigcap\limits_{n\geq 1}(s^{\prime})^{n}R\bigcap S$, $s=s^{\prime n}r_{n}=t_{1}r_{n}\in S$ and $r_{n}\in R$. Let $g=r_{n}g^{\prime}$. Then $sf_{C}=g\circ i$. It follows by [17, Theorem 3.2] that $C$ is a $u$-$S$-absolutely pure $R$-module with respect to $s$.

$(2)\Rightarrow(3)$: Let $\{M_{i}\}_{i\in\Gamma}$ be a direct system of absolutely pure $R$-modules. Then there is a pure exact sequence $0\rightarrow K\rightarrow\bigoplus\limits_{i\in\Gamma}M_{i}\rightarrow{\lim\limits_{\longrightarrow_{i\in\Gamma}}}M_{i}\rightarrow 0$. Note that $\bigoplus\limits_{i\in\Gamma}M_{i}$ is absolutely pure, so is ${\lim\limits_{\longrightarrow_{i\in\Gamma}}}M_{i}$ by (2).

$(3)\Rightarrow(1)$: Suppose the ring $R$ satisfies that any direct limit of absolutely pure $R$-modules is $u$-$S$-absolutely pure with respect to $s$. Let $I$ be a finitely generated ideal of $R$, $\{M_{i}\}_{i\in\Gamma}$ a direct system of $R$-modules. Let $\alpha:I\rightarrow\lim\limits_{\longrightarrow_{i\in\Gamma}}M_{i}$ be a homomorphism. For any $i\in\Gamma$, $E(M_{i})$ is the injective envelope of $M_{i}$. Then $E(M_{i})$ is absolutely pure. By (3), we have $s{\rm Ext}_{R}^{1}(R/I,\lim\limits_{\longrightarrow_{i\in\Gamma}}M_{i})=0$. So there exists an $R$-homomorphism $\beta:R\rightarrow\lim\limits_{\longrightarrow_{i\in\Gamma}}E(M_{i})$ such that the following diagram commutes:

[TABLE]

Thus, by [9, Lemma 2.7], there exists $j\in\Gamma$, such that $\beta$ can factor through $R\xrightarrow{\beta_{j}}E(M_{j})$. Consider the following commutative diagram:

[TABLE]

Since the composition $I\rightarrow R\rightarrow E(M_{j})\rightarrow E(M_{j})/M_{j}$ becomes to be [math] in the direct limit, we can assume $I\rightarrow R\rightarrow E(M_{j})$ can factor through some $I\xrightarrow{s\alpha_{j}}M_{j}$. Thus $s\alpha$ can factor through $M_{j}$. Consequently, the natural homomorphism $\lim\limits_{\longrightarrow_{i\in\Gamma}}{\rm Hom}_{R}(I,M_{i})\xrightarrow{\phi}{\rm Hom}_{R}(I,\lim\limits_{\longrightarrow_{i\in\Gamma}}M_{i})$ is a $u$-$S$-epimorphism with respect to $s$. Now suppose $\{M_{i}\}_{i\in\Gamma}$ is a direct system of finitely presented $R$-modules such that $\lim\limits_{\longrightarrow_{i\in\Gamma}}M_{i}=I$. Then there exists $f\in{\rm Hom}_{R}(I,M_{j})$ with $j\in\Gamma$ such that the identity map $s{\rm Id}_{I}=\phi(u_{j}(f))$ where $u_{j}$ is the natural homomorphism ${\rm Hom}_{R}(I,M_{j})\rightarrow\lim\limits_{\longrightarrow_{i\in\Gamma}}{\rm Hom}_{R}(I,M_{i})$. That is, we have the following commutative diagram:

[TABLE]

So, by the proof of [19, Lemma 2.4], $0\rightarrow{\rm Ker}{(\delta_{i})}\rightarrow M_{i}\xrightarrow{\delta_{i}}{\rm Im}(\delta_{i})\rightarrow 0$ is a $u$-$S$-split short exact sequence with respect to $s^{3}$. It follows by the proof of Lemma 3.2 that there is a finitely presented $R$-module $F$ and a $u$-$S$-torsion module $T_{1}$ (with respect to $s^{3}$) such that the sequence $0\rightarrow T_{1}\rightarrow F\rightarrow{\rm Im}(\delta_{i})\rightarrow 0$ exact. Combining the exact sequence $0\rightarrow{\rm Im}(\delta_{i})\rightarrow I\rightarrow I/{\rm Im}(\delta_{i})\rightarrow 0$, we obtain a long exact sequence

[TABLE]

Note that $s(I/{\rm Im}(\delta_{i}))=0.$ Consequently, $I$ is $u$-$S$-finitely presented with respect to $s^{3}$. Hence $R$ is a $u$-$S$-coherent ring. ∎

Remark 3.4*.*

It is well-known that a ring $R$ is coherent if and only if the class of absolutely pure $R$-modules is (pre)covering (see [5, 6]). It is a natural question that does the following statement hold?

A ring $R$ is $u$-$S$-coherent if and only if any $R$-module has a $u$-$S$-absolutely pure $u$-$S$-(pre)cover (see Definition 4.5).

4. Characterizing $u$-$S$-Noetherian rings by $u$-$S$-injective modules

Let $R$ be a ring and $S$ a multiplicative subset of $R$. Recall from [4] that an $R$-module $E$ is said to be $u$-$S$-injective (with respect to $s$) provided that the induced sequence

[TABLE]

is $u$-$S$-exact (with respect to $s$) for any exact sequence $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ of $R$-modules, or equivalently, ${\rm Ext}_{R}^{1}(M,E)$ is $u$-$S$-torsion with respect to $s$ for any $R$-module $M$ (see [3, Lemma 2.4]).

Lemma 4.1**.**

Let $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ be a $u$-$S$-split short $u$-$S$-exact sequence. Suppose $B$ is $u$-$S$-injective. Then so are $A$ and $C$.

Proof.

Let $\xi:0\rightarrow A\xrightarrow{f}B\xrightarrow{g}C\rightarrow 0$ be a $u$-$S$-split short $u$-$S$-exact sequence with $B$ $u$-$S$-injective. Let $X$ be a module. We will prove that ${\rm Ext}_{R}^{1}(X,A)$ is annihilated by some element in $S$. Let $f^{\prime}:B\rightarrow A$ such that $f^{\prime}\circ f=s{\rm Id}_{A}$ with $s\in S$. Then $f,f^{\prime}$ induce the maps ${\rm Ext}_{R}^{1}(X,f):{\rm Ext}_{R}^{1}(X,A)\rightarrow{\rm Ext}_{R}^{1}(X,B)$ and ${\rm Ext}_{R}^{1}(X,f^{\prime}):{\rm Ext}_{R}^{1}(X,B)\rightarrow{\rm Ext}_{R}^{1}(X,A)$ with ${\rm Ext}_{R}^{1}(X,f^{\prime})\circ{\rm Ext}_{R}^{1}(X,f)=s{\rm Id}_{{\rm Ext}_{R}^{1}(X,A)}$. As $t{\rm Ext}_{R}^{1}(X,B)=0$ for some $t\in S$, we get $st{\rm Ext}_{R}^{1}(X,A)=0$. Consequently, $A$ is $u$-$S$-injective. It follows by [4, Proposition 4.7] that $C$ is also $u$-$S$-injective. ∎

An $R$-module $E$ is said to be $u$-$S$-cofree if $F$ is $u$-$S$-isomorphic to a cofree module. The latter is defined to be a direct product of copies of $R^{+}$.

Proposition 4.2**.**

Let $R$ be a ring and $S$ a multiplicative subset of $R$. Then an $R$-module $E$ is $u$-$S$-injective if and only if $E$ is a direct summand of a $u$-$S$-cofree module.

Proof.

Let $E$ be a $u$-$S$-injective. Then $E$ can be embeded into a cofree $R$-module $G$. Consider the $u$-$S$-split exact sequence $0\rightarrow E\rightarrow G\rightarrow G/E\rightarrow 0$. Then $G/E$ is $u$-$S$-injective, and $G$ is $u$-$S$-isomorphic to $E\oplus G/E$ by [10, Lemma 2.8]. So $E\oplus G/E$ is $u$-$S$-cofree. The converse holds obviously. ∎

It is well-known that a ring $R$ is Noetherian if and only if the class of injective modules is precovering. The main aim of this section is to give the uniform $S$-versions of this result. Denote by $u$-$S\mathcal{I}$ the class of all $u$-$S$-injective modules. The author in [3] characterized when $u$-$S\mathcal{I}$ is (pre)covering.

Theorem 4.3**.**

[3, Theorem 2.16]** Let $R$ be a ring, $S$ be a regular multiplicative subset of $R$. Consider the following statements:

1. (1)

$u$-$S\mathcal{I}$* is closed under direct sums;* 2. (2)

$u$-$S\mathcal{I}$* is closed under direct limits;* 3. (3)

$u$-$S\mathcal{I}$* is a precovering class;* 4. (4)

$u$-$S\mathcal{I}$* is a covering class;* 5. (5)

$R$* is $u$-$S$-Noetherian;*

Then $(1)\Leftrightarrow(2)\Leftrightarrow(3)\Leftrightarrow(4)\Rightarrow(5)$.

Subsequently, the author [3] proposed the following question:

Question [3, Question 2.17] Does the implication $(5)\Rightarrow(1)$ in the above theorem hold? If not, is there a way to define a uniform $S$-version of the notion of covers so that we can obtain a uniform $S$-version of Enochs-Teply’s characterization of Noetherian rings.

The following example shows that the implication $(5)\Rightarrow(1)$ in [3, Theorem 2.16] may be wrong even for Noetherian rings.

Example 4.4**.**

Let $R=\mathbb{Z}$ be the ring of integers, $p$ a prime in $\mathbb{Z}$ and $S=\{p^{n}\mid n\geq 0\}$. Then each $\mathbb{Z}/\langle p^{k}\rangle$ is $u$-$S$-torsion, and thus is $u$-$S$-injective. However, $\bigoplus\limits_{k=1}^{\infty}\mathbb{Z}/\langle p^{k}\rangle$ is not $u$-$S$-injective. Indeed, since each $\mathbb{Z}/\langle p^{k}\rangle$ is a reduced group, so is $\bigoplus\limits_{k=1}^{\infty}\mathbb{Z}/\langle p^{k}\rangle$. It follows by [8, Chapter 9, Corollary 6.7] that $\bigoplus\limits_{k=1}^{\infty}\mathbb{Z}/\langle p^{k}\rangle$ is a subgroup of ${\rm Ext}^{1}_{\mathbb{Z}}(\mathbb{Q}/\mathbb{Z},\bigoplus\limits_{k=1}^{\infty}\mathbb{Z}/\langle p^{k}\rangle)$. Since $\bigoplus\limits_{k=1}^{\infty}\mathbb{Z}/\langle p^{k}\rangle$ is not $u$-$S$-torsion, so is ${\rm Ext}^{1}_{\mathbb{Z}}(\mathbb{Q}/\mathbb{Z},\bigoplus\limits_{k=1}^{\infty}\mathbb{Z}/\langle p^{k}\rangle)$. Hence the direct sum $\bigoplus\limits_{k=1}^{\infty}\mathbb{Z}/\langle p^{k}\rangle$ is not $u$-$S$-injective.

Next, we will try to give a uniform $S$-version of Enochs-Teply’s characterization of Noetherian rings. We recall some notions on uniform $S$-versions of precovering and covering from [10].

Definition 4.5**.**

[10]** Let $M$ be an $R$-module and $\mathcal{A}$ a class of $R$-modules.

1. (1)

An $R$-homomorphism $f\in{\rm Hom}_{R}(A,M)$ with $A\in\mathcal{A}$ is called an $\mathcal{A}$ $u$-$S$-precover of $M$ provided that ${\rm Hom}_{R}(A^{\prime},f):{\rm Hom}_{R}(A^{\prime},A)\rightarrow{\rm Hom}_{R}(A^{\prime},M)$ is a $u$-$S$-epimorphism for any $A^{\prime}\in\mathcal{A}$. 2. (2)

An $\mathcal{A}$ $u$-$S$-precover $f\in{\rm Hom}_{R}(A,M)$ of $M$ is called an $\mathcal{A}$ $u$-$S$-cover of $M$, provided that $f=f\circ\alpha$ implies that $\alpha$ is a $u$-$S$-isomorphism for any $\alpha\in{\rm End}_{R}(A)$. 3. (3)

If every $R$-module has an $\mathcal{A}$ $u$-$S$-precover $($resp., $u$-$S$-cover$)$, then $\mathcal{A}$ is called a $u$-$S$-precovering $($resp., $u$-$S$-covering$)$* class.*

For $s\in S$, let $s\mathcal{I}$ denote the class of all $R$-modules that are $u$-$S$-injective with respect to $s$. Now, we have the following new uniform $S$-version of Enochs-Teply’s characterization of Noetherian rings.

Theorem 4.6**.**

Let $R$ be a ring, $S$ be a regular multiplicative subset of $R$. Consider the following statements:

1. (1)

$R$* is $u$-$S$-Noetherian;* 2. (2)

for all $s\in S$, any direct sum of $R$-modules in $s\mathcal{I}$ is $u$-$S$-injective; 3. (3)

any direct sum of injective $R$-modules is $u$-$S$-injective; 4. (4)

every $R$-module has a $u$-$S$-injective $u$-$S$-precover.

Then $(1)\Leftrightarrow(2)\Leftrightarrow(3)\Leftarrow(4)$. If, moreover, every $u$-$S$-injective module is $u$-$S$-isomorphic to an injective module, then $(3)\Rightarrow(4)$.

Proof.

$(2)\Rightarrow(3)$ Trivial. $(3)\Rightarrow(1)$ It follows by [4, Theorem 4.10].

$(1)\Rightarrow(2)$ Let $R$ be a $u$-$S$-Noetherian ring. Then it is also a $u$-$S$-coherent ring by [18, Proposition 3.10]. Let $\{E_{i}\}_{i\in\Lambda}$ be a family of $u$-$S$-injective modules with respect to $s$. Then for each $i\in\Lambda$, $E_{i}^{+}$ is $u$-$S$-flat with respect to some fixed $s_{1}\in S$ by the proof of [18, Theorem 4.7]. Consequently, $\prod_{i\in\Lambda}E_{i}^{+}\cong(\bigoplus\limits_{i\in\Lambda}E_{i})^{+}$ is $u$-$S$-flat by Theorem 2.3. Hence, $\bigoplus\limits_{i\in\Lambda}E_{i}$ is $u$-$S$-injective by [3, Theorem 2.14].

$(4)\Rightarrow(3)$: Let $\{E_{i}\mid i\in\Lambda\}$ be a family of injective $R$-modules. Then, by assumption, $\bigoplus\limits_{i\in\Lambda}E_{i}$ has a $u$-$S$-injective $u$-$S$-precover $f:E\rightarrow\bigoplus\limits_{i\in\Lambda}E_{i}$. Hence there exists $s_{1}\in S$ such that for any $i\in\Lambda$ we have the following commutative diagram:

[TABLE]

Consequently, $s_{1}{\rm Id}_{\bigoplus\limits_{i\in\Lambda}E_{i}}=f\circ(\bigoplus\limits_{i\in\Lambda}g_{i})$. So

[TABLE]

is a $u$-$S$-split short $u$-$S$-exact sequence. It follows by Lemma 4.1 that $\bigoplus\limits_{i\in\Lambda}E_{i}$ is $u$-$S$-injective.

Now, assume that every $u$-$S$-injective module is $u$-$S$-isomorphic to an injective module.

$(3)\Rightarrow(4)$ Let $M$ be an $R$-module. Let $\{E_{i}\mid i\in\Lambda\}$ be the set of injective envelopes of all cyclic $R$-modules, $X_{i}={\rm Hom}_{R}(E_{i},M)$, and $f_{i}:E_{i}^{(X_{i})}\rightarrow M$ be the evaluation map $(\phi_{g})_{g\in X_{i}}\mapsto\sum\limits_{g\in X_{i}}g(\phi_{g})$. Then any $R$-homomorphism $E^{\prime}\rightarrow M$ with $E^{\prime}$ injective factor through the natural $R$-homomorphism $\bigoplus\limits_{i\in\Lambda}f_{i}$. We claim that $\bigoplus\limits_{i\in\Lambda}f_{i}:\bigoplus\limits_{i\in\Lambda}E_{i}^{(X_{i})}\rightarrow M$ is a $u$-$S$-injective $u$-$S$-precover. Indeed, by assumption, $\bigoplus\limits_{i\in\Lambda}E_{i}^{(X_{i})}$ is a $u$-$S$-injective $R$-module. Let $f:E\rightarrow M$ be an $R$-homomorphism with $E$ $u$-$S$-injective. Then there is a $u$-$S$-isomorphism $t^{\prime}:E\rightarrow E^{\prime}$ with $E^{\prime}$ injective by assumption. Hence there is $s^{\prime}\in S$ and a $u$-$S$-isomorphism $t:E^{\prime}\rightarrow E$ such that $t\circ t^{\prime}=s^{\prime}{\rm Id}_{E}$ and $t^{\prime}\circ t=s^{\prime}{\rm Id}_{E^{\prime}}$ by [19, Lemma 1.2]. Then there exists an $R$-homomorphism $g:E^{\prime}\rightarrow\bigoplus\limits_{i\in\Lambda}E_{i}^{(X_{i})}$ such that $\bigoplus\limits_{i\in\Lambda}f_{i}\circ g=f\circ t$:

[TABLE]

Hence, $\bigoplus\limits_{i\in\Lambda}f_{i}\circ g\circ t^{\prime}=f\circ t\circ t^{\prime}=s^{\prime}f$. Consequently, $\bigoplus\limits_{i\in\Lambda}f_{i}$ is a $u$-$S$-injecive $u$-$S$-precover. ∎

Remark 4.7*.*

The equivalence of $(1)$ and $(3)$ in Theorem 4.6 can be found in [4, Theorem 4.10]. As to the equivalence of $(1)$ and $(4)$, we don’t known whether every $u$-$S$-injective module is $u$-$S$-isomorphic to an injective module in general. It is a natural question whether the following statement holds?

A ring $R$ is $u$-$S$-Noetherian if and only if every $R$-module has a $u$-$S$-injective $u$-$S$-cover.

Acknowledgement**.**

The authors greatly appreciate the reviewer’s comments and revision suggestions on this article.