Red-injective modules

TL;DR

This paper introduces the concept of Red-injective modules, explores their properties, and uses them to characterize important classes of rings such as quasi-Frobenius and V-rings.

Contribution

It defines Red-injective modules inspired by Soc-injective modules and applies this concept to characterize specific classes of rings.

Findings

Red-injective modules are characterized and their properties are studied.

Red-injective modules are used to characterize quasi-Frobenius rings.

Red-injective modules help in understanding V-rings.

Abstract

Let $\text{Red}(M)$ be the sum of all reduced submodules of a module $M$. For modules over commutative rings, $\text{Soc}(M)\subseteq \text{Red}(M)$. By drawing motivation from how $\text{Soc}$-injective modules were defined by Amin et. al. in \cite{amin2005}, we introduce $\text{Red}$-injective modules, study their properties and use them to characterize quasi-Frobenius rings and $V$-rings.

Equations

$$\text{Red}(M):=\sum _{i\in I}\{N_i|N_i\text{is a reduced submodule of}M\}.$$

$$\text{simple}\Rightarrow \text{semi-simple}\Rightarrow \text{reduced}\Rightarrow \text{semi-reduced}.$$

$$f_{M_i}:\text{Red}(N)\to M_i\text{such that}{\pi }_{{}_{M_i}}\circ f=f_{M_i}$$

$$f_{M_j}:\text{Red}(N)\to M_j\text{such that}{\pi }_{{}_{M_j}}\circ f=f_{M_j},$$

$$f_{M_i}=f_{M_i}^{\prime }\circ h\text{and}{f}_{M_j}=f_{M_j}^{\prime }\circ h.$$

$$k:\text{Red}(S)\to \text{Red}(N);g:\text{Red}(S)\to S;h:\text{Red}(N)\to N\text{and}\iota :S\to N.$$

Full text

Department of Mathematics

Makerere University, P.O BOX 7062, Kampala Uganda

E-mail addresses: kasozi@cns.mak.ac.ug, ssevviiri@cns.mak.ac.ug, umutabazivincent@yahoo.fr

Keywords: Injective modules; Red-injective modules; Soc-injective modules; Quasi-Frobenius rings; $V$-rings

MSC 2010 Mathematics Subject Classification: 16D50, 16D60, 16L60

1 Introduction

For a not necessarily commutative ring $R$, Lee and Zhou in [7] defined an $R$-module $M$ to be reduced if for all $r\in R$ and $m\in M$, $mr=0$ implies that $Mr\cap mR=\{0\}$. This definition is equivalent to saying that for all $r\in R$ and $m\in M$, $mr^{2}=0$ implies that $mRr=\{0\}$, see [10] for the proof. However, for modules over commutative rings we get Definition 1 below.

Definition 1**.**

An $R$-module $M$ is reduced if for all $r\in R$ and $m\in M$, $mr^{2}=0$ implies that $mr=0$.

Except in Example 3.1, all rings are unital, commutative and associative. Modules are right unital defined over rings. A submodule is reduced if it is reduced as a module. A submodule of a reduced module is reduced but a factor module of a reduced module need not be reduced. The $\mathbb{Z}$-module $\mathbb{Z}$ is reduced but its factor module $\mathbb{Z}/n\mathbb{Z}$ is not reduced for a non-square free integer $n$. The socle of an $R$-module $M$, denoted by $\text{Soc}(M)$ is the sum of simple submodules of $M$. Let $\text{Red}(M)$ denote the sum of reduced submodules of $M$, i.e.,

[TABLE]

Definition 2**.**

An $R$-module $M$ is semi-reduced if $\text{Red}(M)=M$.

Proposition 1**.**

For any $R$-module $M$, the following implications hold:

[TABLE]

Proof: We prove that a semi-simple module is reduced. The other implications follow from the definition of semi-simple and semi-reduced modules respectively. Since a simple module is prime222An $R$-module $M$ for which $RM\neq\{0\}$ is prime if for all $a\in R$ and every $m\in M$, $am=0$ implies that $m=0$ or $aM=\{0\}$. and every prime module is reduced, a simple module is reduced. Suppose that $M$ is a semi-simple module and $mr^{2}=0$ where $m\in M$ and $r\in R$. Then, $(m_{1},m_{2},\cdot\cdot\cdot,m_{i},\cdot\cdot\cdot)r^{2}=0$ where $(m_{1},m_{2},\cdot\cdot\cdot,m_{i},\cdot\cdot\cdot)=m\in M=\bigoplus_{i\in I}M_{i}$ for some simple modules $M_{i}$. Since every simple module is reduced, $m_{i}r^{2}=0\Rightarrow m_{i}r=0~{}\forall~{}i\in I$. Hence, $mr=0$, and $M$ is reduced. $\Box$

Corollary 1**.**

For any $R$-module $M$, $\text{Soc}(M)\subseteq\text{Red}(M)$.

Proof: The proof follows from the fact that a semi-simple module is semi-reduced which is proved in Proposition 1. $\Box$

Note that for semi-simple modules and for modules without nonzero reduced submodules, $\text{Soc}(M)=\text{Red}(M)$.

Example 1.1**.**

A reduced module need not be semi-simple. $\mathbb{Z}$ and $\mathbb{Q}$ are reduced $\mathbb{Z}$-modules but they are not semi-simple.

1.1 Other basic definitions

Definition 3**.**

[1, Definition 1.1] Let $M$ and $N$ be $R$-modules. $M$ is socle-$N$-injective (Soc-$N$-injective) if any $R$-homomorphism $f:\text{Soc}(N)\rightarrow M$ extends to $N$. Equivalently, for any semi-simple submodule $K$ of $N$, any $R$-homomorphism $f:K\rightarrow M$ extends to $N$. An $R$-module $M$ is Soc*-quasi-injective* if $M$ is Soc-$M$-injective. $M$ is Soc*-injective* if $M$ is Soc-$R$-injective. $R$ is right (self-) Soc*-injective*, if the module $R_{R}$ is Soc-injective (equivalently, if $R_{R}$ is Soc-quasi-injective).

Definition 4**.**

[1, Definition 1.2] An $R$-module $M$ is called strongly Soc-injective, if $M$ is Soc-$N$-injective for all $R$-modules $N$. A ring $R$ is called strongly Soc-injective, if the module $R_{R}$ is strongly Soc-injective.

Definitions 3 and 4 together with Corollary 1 motivate us to have Definitions 5 and 6 respectively.

Definition 5**.**

An $R$-module $M$ is called Red-$N$-injective if any $R$-homomorphism $f:K\rightarrow M$ extends to $N$ for any semi-reduced submodule $K$ of $N$. $M$ is called Red*-quasi-injective* if it is Red-$M$-injective. $M$ is called Red*-injective* if it is Red-$R$-injective.

Definition 6**.**

An $R$-module $M$ is called strongly-Red-injective, if $M$ is Red-$N$-injective for all $R$-modules $N$.

In Definition 7, we recall different generalizations of injective modules that we later use in the sequel. As with Soc-injective and Red-injective modules defined above, these generalizations of injective modules were defined by relaxing conditions on the lifting property of homomorphisms.

Definition 7**.**

If $M$ and $N$ are $R$-modules, then

$M$ is $N$-injective if every $R$-homomorphism from a submodule of $N$ into $M$ can be extended to an $R$-homomorphism from $N$ into $M$. 2. 2.

$M$ is quasi-injective if it is $M$-injective. 3. 3.

$M$ is $N$-simple-injective if for any submodule $L$ of $N$, any homomorphism $\theta:L\rightarrow M$ with $\theta(L)$ simple, can be extended to a homomorphism $\beta:N\rightarrow M$. 4. 4.

$M$ is simple-injective if it is simple $R$-injective. 5. 5.

$M$ is strongly simple-injective, if $M$ is simple-$N$-injective for all right $R$-modules $N$. 6. 6.

$M$ is min-$N$-injective if, for every simple submodule $L$ of $N$, every homomorphism $\gamma:L\rightarrow M$ extends to $N$. 7. 7.

$M$ is min-injective if it is min-$R$-injective. 8. 8.

$M$ is strongly min-injective, if it is min-$N$-injective for all $R$-modules $N$. 9. 9.

$M$ is pseudo-injective if any monomorphism from a submodule of $M$ to $M$ extends to an endomorphism of $M$.

1.2 Notation

Throughout this paper, $N\subseteq^{e}M$, $N\oplus M$, $N\subseteq^{\oplus}M$, and $N\leq M$, mean that $N$ is an essential submodule of $M$, a direct sum of $N$ and $M$, $N$ is a direct summand of $M$, $N$ is a submodule of $M$ respectively.

1.3 Paper roadmap

In Section 1, we have given the introduction, defined key terms, given the notation used and the roadmap for the paper.

Section 2 is devoted to obtaining properties of Red-injective modules and their generalizations. An equivalent definition of a Red-injective module is obtained. It is shown that any injective module is strongly Red-injective and a Red-injective module is Soc-injective. Other implications with known generalizations of injective modules are given. The class of (strongly) Red-injective $R$-modules is closed under isomorphisms, direct products and summands. If $M$ is a Noetherian module, then a direct sum of Red-$M$-injective is Red-$M$-injective. For a family of $R$-modules $\{M_{i}~{}:~{}i\in I\}$, an $R$-module $N$ is Red-$(\oplus_{i\in I}M_{i})$-injective if and only if it is Red-$M_{i}$-injective for each $i$. For a projective $R$-module $M$, every quotient of a Red-$M$-injective $R$-module is Red-$M$-injective if and only if $\text{Red}(M)$ is projective if and only if every quotient of an injective $R$-module is Red-$M$-injective. Over a principal ideal domain a free module is Red-injective if each of its submodule is Red-injective. Red$(N)$-lifting modules are introduced. It is shown that if a module $N$ is Red($N$)-lifting, then any $R$-module $K$ is Red-$N$-lifting if and only if $K$ is $N$-injective. It is shown that Red-quasi-injective modules inherit a weaker version of C2-condition and C3-condition.

In Section 3, we characterize quasi-Frobenius rings and right $V$-rings in terms of strongly Red-injective modules. A ring $R$ is quasi-Frobenius if and only if every strongly Red-injective $R$-module is projective. A ring $R$ is a right $V$-ring if and only if every simple $R$-module is strongly Red-injective. A question is raised as to whether Red-quasi-injective modules and Soc-quasi-injective modules are clean and or satisfy the exchange property.

2 Red-injective modules

Proposition 2**.**

For $R$-modules $K,~{}M$ and $N$, the following statements are equivalent:

Any $R$-homomorphism $f:K\rightarrow M$ extends to $N$ for any semi-reduced submodule $K$ of $N$. 2. 2.

Any $R$-homomorphism $f:\text{Red}(N)\rightarrow M$ extends to $N$.

Proof:

$1\Rightarrow 2$

since $\text{Red}(N)$ is semi-reduced.

$2\Rightarrow 1.$

Suppose $f:K\rightarrow M$ is an $R$-homomorphism and $K$ is a semi-reduced submodule of $N$. Since $K\leq\text{Red}(N)$, then $f$ extends to $N$.

$\Box$

Proposition 3**.**

If $N$ is an $R$-module, then

any injective module is strongly Red-injective, 2. 2.

a $\text{Red}(N)$-injective module is $\text{Soc}(N)$-injective.

Proof:

Let $M$ be an injective module. Then $M$ is $N$-injective for every $R$-module $N$. For every submodule $K$ of $N$, any $R$-homomorphism $f:K\rightarrow M$ extends to $N$. For every module $N$, any $R$-homomorphism $f:\text{Red}(N)\rightarrow M$ extends to $N$. Hence, $M$ is strongly Red-injective. 2. 2.

Suppose $f:\text{Soc}(N)\rightarrow M$ is an $R$-homomorphism and $M$ is $\text{Red}(N)$-injective. By Proposition 1, $\text{Soc}(N)$ is a semi-reduced submodule of $N$. Hence by Definition 5, $f$ extends to $N$. Thus, $M$ is Soc-$N$-injective.

$\Box$

Every projective module over a right Noetherian right self-injective ring is strongly Red-injective. Let $R$ be a ring for which each module $M$ has $\text{Red}(M)=\{0\}$. Then, $M$ is strongly Red-injective.

A Red-injective module need not be injective. The module $\mathbb{Z}_{\mathbb{Z}}$ is Red-injective but not injective.

Theorem 1**.**

Let $\{M_{i}:i\in I\}$ be a family of $R$-modules and $N$, $M$, $A$, $C$, $S$ and $K$ be $R$-modules. Then the following conditions hold:

A direct product $\prod_{i\in I}M_{i}$ is Red-$N$-injective if and only if each $M_{i}$ is Red-$N$-injective. 2. 2.

For $S\leq N$, if $M$ is Red-$N$-injective, then $M$ is Red-$S$-injective. 3. 3.

For $M\cong N$; $M$ is Red-$S$-injective if and only if $N$ is Red-$S$-injective. 4. 4.

For $A\cong B$; $C$ is Red-$A$-injective if and only if it is Red-$B$-injective. 5. 5.

For $N\subseteq^{\bigoplus}M$, if $M$ is Red-$K$-injective, then $N$ is Red-$K$-injective.

Proof:

We prove only for $M=M_{i}\times M_{j}$ where $i,~{}j\in I$. The proof for the general case is analogous. Let $M_{i}$ and $M_{j}$ be Red-$N$-injective $R$-modules, $h:\text{Red}(N)\rightarrow N$ and $f:\text{Red}(N)\rightarrow M_{i}\times M_{j}$ be any $R$-homomorphisms.

Define

[TABLE]

and

[TABLE]

where $\pi_{{}_{M_{i}}}:M_{i}\times M_{j}\rightarrow M_{i}$ and $\pi_{{}_{M_{j}}}:M_{i}\times M_{j}\rightarrow M_{j}$ are $R$-homomorphisms. Since $M_{i}$ and $M_{j}$ are Red-$N$-injective there exists $f^{\prime}_{M_{i}}:N\rightarrow M_{i}$ and $f^{\prime}_{M_{j}}:N\rightarrow M_{j}$ such that

[TABLE]

By the uniqueness part of the universal property of direct product there exists an $R$-homomorphism $f^{\prime}:N\rightarrow M_{i}\times M_{j}$ such that $f=f^{\prime}\circ h$. It follows that $\pi_{{}_{M_{i}}}\circ(f^{\prime}\circ h)=f_{M_{i}}$ and $\pi_{{}_{M_{j}}}\circ(f^{\prime}\circ h)=f_{M_{j}}$. By the uniqueness of the universal property we conclude that $f=f^{\prime}\circ h$. Hence, $f:\text{Red}(N)\rightarrow M_{i}\times M_{j}$ extends to $N$. Thus $M_{i}\times M_{j}$ is Red-$N$-injective. Conversely, assume that $M_{i}\times M_{j}$ is Red-$N$-injective. Let $h:\text{Red}(N)\rightarrow N$ and $f_{M_{i}}:\text{Red}(N)\rightarrow M_{i}$ be any $R$-homomorphisms. Choose $f_{M_{j}}:\text{Red}(N)\rightarrow M_{j}$ to be the zero $R$-homomorphism. We obtain $f^{\prime}:N\rightarrow M_{i}\times M_{j}$ such that $f=f^{\prime}\circ h$. Finally we obtain $f_{M_{i}}=\pi_{M_{i}}\circ f=(\pi_{M_{i}}\circ f^{\prime})\circ h$. Hence $\pi_{M_{i}}\circ f^{\prime}:N\rightarrow M_{i}$ is an extension of $f_{M_{i}}$. Thus, $M_{i}$ is Red-$N$-injective. Similarly, $M_{j}$ is Red-$N$-injective. 2. 2.

Consider the diagram in Figure 1, where $M$ is Red-$N$-injective.

Since $S\leq N$, $\text{Red}(S)\leq\text{Red}(N)$. Consider inclusion maps

[TABLE]

$f^{\prime}\circ l:S\rightarrow M$ is an extension for any $R$-homomorphism $q:\text{Red}(S)\rightarrow M$. Thus $M$ is Red-$S$-injective. 3. 3.

Let $N\cong M$ where $\theta:N\rightarrow M$ is an $R$-isomorphism between them. Let $f_{N}:\text{Red}(S)\rightarrow N$ be any $R$-homomorphism. Since $M$ is Red-$S$-injective, any $R$-homomorphism $f_{M}:\text{Red}(S)\rightarrow M$ extends to $f^{\prime}_{M}:S\rightarrow M$. So for any $R$-homomorphism $h:\text{Red}(S)\rightarrow S$, $f_{M}=f^{\prime}_{M}\circ h$. Since $M$ and $N$ are isomorphic there exists an inverse homomorphism $\theta^{-1}:M\rightarrow N$ such that $\theta^{-1}\circ f^{\prime}_{M}:S\rightarrow N$ is an $R$-homomorphism. Define $f^{\prime}_{N}=\theta^{-1}\circ f^{\prime}_{M}:S\rightarrow N$. Then, $f^{\prime}_{N}$ is an extension of $f_{N}$. Thus, $N$ is Red-$S$-injective. Similarly, if $N$ is Red-$S$-injective then $M$ is Red-$S$-injective. 4. 4.

Suppose that $A\cong B$ and $C$ is Red-$A$-injective. We show that $C$ is Red-$B$-injective.

Consider the diagram in Figure 2, where $f^{\prime}_{A}:A\rightarrow C$ is the extension of $f_{A}:\text{Red}(A)\rightarrow C$. Let also $f_{B}:\text{Red}(B)\rightarrow C$ be an $R$-homomorphism. Define $f^{\prime}_{B}=f^{\prime}_{A}\circ\theta:B\rightarrow C$. Then $f^{\prime}_{B}:B\rightarrow C$ is the extension of $f_{B}$. Thus $C$ is Red-$B$-injective. A similar argument works for the converse. 5. 5.

Let $N\subseteq^{\oplus}M$ and $M$ be Red-$K$-injective. We show that $N$ is Red-$K$-injective. Since $N\subseteq^{\oplus}M$, there exists an $R$-submodule $N^{\prime}$ of $M$ such that $N\bigoplus N^{\prime}=M$. Let $\pi_{{}_{N}}:N\bigoplus N^{\prime}\rightarrow N$ be the projection $R$-homomorphism. Since $M$ is Red-$K$-injective, any $R$-homomorphism $f_{M}:\text{Red}(K)\rightarrow M$ extends to $f^{\prime}_{M}:K\rightarrow M$. Suppose $f_{N}=\pi_{{}_{N}}\circ f_{M}:\text{Red}(K)\rightarrow N$. Define $f^{\prime}_{N}=\pi_{{}_{N}}\circ f^{\prime}_{M}:K\rightarrow N$. Then $f^{\prime}_{N}:K\rightarrow N$ is the extension of $f_{N}$. Hence, $N$ is Red-$K$-injective.

$\Box$

Corollary 2**.**

Let $N$ be an $R$-module, then

a finite direct sum of Red-$N$-injective modules is again Red-$N$-injective. In particular, a finite direct sum of Red-injective (resp., strongly Red-injective) modules is again Red-injective (resp., strongly Red-injective); 2. 2.

a direct summand of Red-quasi-injective (resp., Red-injective, strongly Red-injective) module is again Red-quasi-injective (resp., Red-injective, strongly Red-injective) module.

Proposition 4**.**

If $M$ is a Noetherian $R$-module, then a direct sum of Red-$M$-injective modules is Red-$M$-injective.

Proof: For $D=\bigoplus_{i\in I}D_{i}$, a direct sum of Red-$M$-injective modules, let $f:K\rightarrow D$ be an $R$-homomorphism, where $K$ is any semi-reduced submodule of $M$. Since $K$ is finitely generated, $f(K)\leq\bigoplus_{i=1}^{n}D_{i}$ for some positive integer $n$. Since $\bigoplus_{i=1}^{n}D_{i}$ is Red-$M$-injective, then $f$ can be extended to an $R$-homomorphism $\hat{f}:M\rightarrow D$. $\Box$

Corollary 3**.**

Let $R_{R}$ be a Noetherian module. Then, a direct sum of Red-injective modules is Red-injective.

Proposition 5**.**

Let $\{M_{i}:i\in I\}$ be a family of $R$-modules and $N$ be an $R$-module. Then, $N$ is Red-$(\bigoplus_{i\in I}M_{i})$-injective if and only if it is Red-$M_{i}$-injective for each $i$.

Proof:

$(\Rightarrow).$

Suppose that $N$ is Red-$(\bigoplus_{i\in I}M_{i})$-injective. Let $f:\text{Red}(M_{i})\rightarrow N$ be any $R$-homomorphism. By hypothesis, any $R$-homomorphism $g:\text{Red}(\bigoplus_{i\in I}M_{i})\rightarrow N$ extends to $\bar{g}:\bigoplus_{i\in I}M_{i}\rightarrow N$. The required extension of $f$ is $\bar{g}\circ\iota$ where $\iota$ is the injection $\iota:M_{i}\rightarrow\bigoplus_{i\in I}M_{i}$.

$(\Leftarrow).$

Suppose that $N$ is Red-$M_{i}$-injective for each $i\in I$. Since $N$ is Red-$M_{i}$-injective for each $i\in I$; let $\theta_{i}:M_{i}\rightarrow N$ be the extension of $f_{i}:\text{Red}(M_{i})\rightarrow N$ for each $i\in I$. Let also $g:\text{Red}(\bigoplus_{i\in I}M_{i})\rightarrow N$ be any $R$-homomorphism. By the fundamental property of direct sum of modules, there exists an $R$-homomorphism $\theta=\langle\theta_{i}\rangle:\bigoplus_{i\in I}M_{i}\rightarrow N$ such that $\theta\circ\iota_{i}=\theta_{i}$ for all $i\in I$; where $\iota_{i}:M_{i}\rightarrow\bigoplus_{i\in I}M_{i}$ is the injection $R$-homomorphism for each $i\in I$. Then $\theta$ is an extension of $g:\text{Red}(\bigoplus_{i\in I}M_{i})\rightarrow N$. Hence, $N$ is Red-$(\bigoplus_{i\in I}M_{i})$-injective.

$\Box$

Corollary 4**.**

If $A$, $B$, $C$, and $Q$ are $R$-modules and the short exact sequence $\{0\}\rightarrow A\xrightarrow{\mu}B\xrightarrow{\varepsilon}C\rightarrow\{0\}$ splits, then the following conditions hold:

$Q$* is Red-$C$-injective if and only if it is Red-$(B/\mu(A))$-injective.* 2. 2.

$Q$* is Red-$B$-injective if and only if it is Red-$A$-injective and Red-$C$-injective.*

Proof:

This follows from the fact that $B/\mu(A)\cong C$. 2. 2.

Follows from Proposition 4, Theorem 1 and the fact that $B\cong A\oplus C$.

$\Box$

Proposition 6**.**

Let $N$ and $M$ be $R$-modules. Then the following conditions hold:

$M$* is injective $\Rightarrow$ $M$ is $N$-injective $\Rightarrow$ $M$ is Red-$N$-injective $\Rightarrow$ $M$ is Soc-$N$-injective $\Rightarrow$ $M$ is min-$N$-injective.* 2. 2.

$M$* is injective $\Rightarrow$ $M$ is strongly Red-injective $\Rightarrow$ $M$ is strongly Soc-injective $\Rightarrow$ $M$ is strongly $min$-injective $\Leftrightarrow$ $M$ is strongly simple-injective.*

Proof: Elementary. $\Box$

Proposition 7**.**

For an $R$-module $M$, if $\text{Red}(M)$ is a direct summand of $M$, then every $R$-module is Red-$M$-injective.

Proof: Suppose that $K$ is an $R$-module and $\text{Red}(M)\subseteq^{\oplus}M$. We show that $K$ is Red-$M$-injective. Let $f:\text{Red}(M)\rightarrow K$ be any $R$-homomorphism. Since $\text{Red}(M)$ is a direct summand of $M$, there exists a proper $R$-submodule $P$ of $M$ such that $M=\text{Red}(M)\oplus P$. There exists an $R$-homomorphism $f^{\prime}:M\rightarrow\text{Red}(M)$ such that $f^{\prime}(n+p)=n$, for all $n\in\text{Red}(M)$ and $p\in P$. Then, the $R$-homomorphism $f\circ f^{\prime}:M\rightarrow K$ is an extension of $f$ because $(f\circ f^{\prime})(n+p)=f(f^{\prime}(n+p))=f(n)$ for all $n+p~{}\in M$. Hence, $K$ is Red-$M$-injective. $\Box$

Theorem 2**.**

For a projective $R$-module $M$, the following conditions are equivalent:

Every quotient of a Red-$M$-injective $R$-module is Red-$M$-injective. 2. 2.

Every quotient of an injective $R$-module is Red-$M$-injective. 3. 3.

$\text{Red}(M)$* is a projective $R$-module.*

Proof:

$(1\Rightarrow 2).$

This is due to the fact that every injective $R$-module is Red-$M$-injective.

$(2\Rightarrow 3).$

Consider the diagram in Figure 3 below:

where $E$ and $N$ are $R$-modules, $\varepsilon$ an $R$-epimorphism, and $f$ an $R$-homomorphism. By [4, Proposition 5.1], assume that $E$ is injective. Since $N$ is Red-$M$-injective $f$ can be extended to an $R$-homomorphism $g:M\rightarrow N$. Since $M$ is projective, $g$ can be lifted to an $R$-homomorphism $\widetilde{g}:M\rightarrow E$ such that $\varepsilon\circ\widetilde{g}=g$. Define $\widetilde{f}:\text{Red}(M)\rightarrow E$ by $\widetilde{f}=\widetilde{g}|_{{}_{\text{Red}(M)}}$. Then $\varepsilon\circ\widetilde{f}=\varepsilon\circ\widetilde{g}|_{{}_{\text{Red}(M)}}=f$. Hence, $\text{Red}(M)$ is projective.

$(3\Rightarrow 1).$

Let $N$ and $L$ be $R$-modules with $\varepsilon:N\rightarrow L$ an $R$-epimorphism and $N$ is Red-$M$-injective. Consider the diagram in Figure 4.

Since $\text{Red}(M)$ is projective, $f$ can be lifted to an $R$-homomorphism $g:\text{Red}(M)\rightarrow N$ such that $\varepsilon\circ g(m)=f(m)$, for all $m\in\text{Red}(M)$. Since $N$ is Red-$M$-injective, $g$ can be extended to an $R$-homomorphism $\widetilde{g}:M\rightarrow N$. Hence, $\varepsilon\circ\widetilde{g}:M\rightarrow L$ extends $f$.

$\Box$

Corollary 5**.**

The following conditions are equivalent for a reduced projective $R$-module:

Every quotient of a Red-injective $R$-module is Red-injective. 2. 2.

Every quotient of an injective $R$-module is Red-injective. 3. 3.

$\text{Red}(R_{R})$* is a projective module.*

In addition, if every semi-reduced submodule of a projective $R$-module is projective, then $\text{Red}(R_{R})$ is a projective module.

Proof: $1\Leftrightarrow 1\Leftrightarrow 4$ follows from Theorem 2. The additional case follows from the fact that $\text{Red}(R_{R})$ is a semi-reduced submodule of a projective module $R_{R}$. $\Box$

Proposition 8**.**

Let $R$ be a Principal Ideal Domain (PID) and $N$ be an $R$-module. Then, the following statements hold:

If every free $R$-module is Red-$N$-injective then each of its submodules is Red-$N$-injective. 2. 2.

If every projective $R$-module is Red-$N$-injective then each of its submodules is Red-$N$-injective. 3. 3.

Every projective $R$-module is Red-$N$-injective if and only if every free $R$-module is Red-$N$-injective.

Proof:

Suppose that every free $R$-module $M$ is Red-$N$-injective, and $L\leq M$. Since over a PID a submodule of a free module is free, $L$ is free. By hypothesis, $L$ is Red-$N$-injective. 2. 2.

Suppose that every projective $R$-module $P$ is Red-$N$-injective, and $K\leq P$. Since over a PID a submodule of a projective $R$-module is projective, $K$ is projective. By hypothesis, $K$ is Red-$N$-injective. 3. 3.

Over a PID every projective module is free. The converse holds since any free module is projective.

$\Box$

Definition 8**.**

Let $X$ be a submodule of a module $M$. We say that $\text{Red}(M)$ respects $X$ if there exists a direct summand $A$ of $M$ contained in $X$ such that $X=A\oplus B$ and $B\leq\text{Red}(M)$. $M$ is called $\text{Red}(M)$-lifting if $\text{Red}(M)$ respects every submodule of $M$.

Proposition 9**.**

Let $N$ be an $R$-module. If $N$ is $\text{Red}(N)$-lifting, then any $R$-module $K$ is Red-$N$-injective if and only if $K$ is $N$-injective.

Proof:

$(\Rightarrow).$

Suppose that $K$ is Red-$N$-injective. Let $L$ be any submodule of $N$, $\iota:L\rightarrow N$ the inclusion map and $f:L\rightarrow K$ any $R$-homomorphism. Since $\text{Red}(N)$ respects $L$, $L$ has a decomposition $L=A\oplus B$ such that $A\subseteq^{\oplus}N$ and $B\leq\text{Red}(N)$. $N=A\oplus A^{\prime}$ for some submodule $A^{\prime}$ of $N$. Then, $L=A\oplus(L\cap A^{\prime})$ and $L\cap A^{\prime}$ is semi-reduced. Let $i:L\cap A^{\prime}\rightarrow L$ be the inclusion map and $f|_{L\cap A^{\prime}}:L\cap A^{\prime}\rightarrow K$. Since $K$ is Red-$N$-injective, there exists an $R$-homomorphism $g:N\rightarrow K$ such that $g\circ\iota\circ i=f|_{L\cap A^{\prime}}$. Now, define $h:N\rightarrow K$ by $h(a+a^{\prime})=f(a)+g(a^{\prime})$   $(a\in A,a^{\prime}\in A^{\prime})$. Then $h\circ\iota=f$, and hence $K$ is $N$-injective.

$(\Leftarrow).$

Every $N$-injective module is Red-$N$-injective. This is due to the fact that for every $N$-injective module $K$, any $R$-homomorphism from any submodule of $N$ to $K$ extends to $N$.

$\Box$

Note that a semi-simple module as well as a module $M$ with no reduced submodule (i.e., one for which $\text{Red}(M)=\{0\}$) is $\text{Red}(M)$-lifting.

Let $K$ and $L$ be submodules of $M$. $M$ is said to satisfy:

C1-condition if every submodule of $M$ is essential in a summand of $M$. 2. 2.

C2-condition if $K\cong L$ and $K\subseteq^{\oplus}M$, then $L\subseteq^{\oplus}M$. 3. 3.

C3-condition if $K\cap L=\{0\}$, $K\subseteq^{\oplus}M$ and $L\subseteq^{\oplus}M$, then $K\oplus L\subseteq^{\oplus}M$.

A module is quasi-continuous if it satisfies C1 and C3 conditions.

Proposition 9 shows that Red-quasi-injective modules inherit a weaker version of C2-condition and C3-conditions.

Proposition 10**.**

Suppose that an $R$-module $N$ is Red-quasi-injective.

(Red-C2) If $P$ and $Q$ are semi-reduced submodules of $N$, $P\cong Q$ and $P\subseteq^{\oplus}N$, then $Q\subseteq^{\oplus}N$. 2. 2.

(Red-C3) Let $P$ and $Q$ be semi-reduced submodules of $N$ with $P\cap Q=\{0\}$. If $P\subseteq^{\oplus}N$ and $Q\subseteq^{\oplus}N$; then $P\bigoplus Q\subseteq^{\oplus}N$.

Proof:

Since $P\cong Q$, and $P$ is Red-$N$-injective, being a direct summand of a Red-quasi-injective module $N$, $Q$ is Red-$N$-injective by Corollary 2(2). If $i:Q\rightarrow N$ is the inclusion map, the identity $id_{Q}:Q\rightarrow Q$ has an extension $\eta:N\rightarrow Q$ such that $\eta\circ i=id_{Q}$, and hence $Q\subseteq^{\oplus}N$. 2. 2.

Since both $P$ and $Q$ are direct summands of $N$; then both $P$ and $Q$ are Red-$N$-injective. Then the semi-reduced module $P\oplus Q$ is Red-$N$-injective, and so a direct summand of $N$ by an argument similar to the one given in 1.

$\Box$

3 Strongly Red-injective modules

In this section, we characterize quasi-Frobenius rings and right $V$-rings in terms of strongly Red- injective modules. A ring $R$ is called right semi-Artinian if every non-zero $R$-module has nonzero socle. A submodule $S\leq M$ is small if, for any submodule $N\leq M$, $S+N=M$ implies that $N=M$. The projective cover of an $R$-module $M$ is a projective module $P$ for which there is an epimorphism $P\rightarrow M$ whose kernel is small. A ring $R$ is left perfect if every $R$-module has a projective cover.

Proposition 11**.**

The following implications hold:

$R$* is right semi-Artinian $\Rightarrow$ every strongly Red-injective $R$-module is injective $\Rightarrow$ every strongly Red-injective $R$-module is quasi-continuous.*

In particular, over a left perfect ring $R$, every strongly Red-injective right $R$-module is injective.

Proof: For a right semi-Artinian ring $R$, suppose that a non-zero $R$-module $M$ is strongly Red-injective. Then, $\{0\}\neq\text{Soc}(M)\subseteq^{e}M$. Amin et al., in [1, Corollary 3.2] showed that a strongly Soc-injective module with essential socle is injective. Since $M$ has essential socle, it is injective. $M$ is quasi-continuous because every injective module is quasi-continuous see [8, p.18]. The last statement follows from the fact that every left perfect ring is right semi-Artinian, see [6, Theorem 11.6.3]. $\Box$

A ring $R$ is called quasi-Frobenius if $R$ is right (or left) Artinian and right (or left) self-injective. Equivalently, $R$ is quasi-Frobenius if and only if every injective $R$-module is projective if and only if every projective $R$-module is injective. A ring whose all simple right modules are injective is called a right $V$-ring.

Theorem 3**.**

A ring $R$ is quasi-Frobenius if and only if every strongly Red-injective module is projective.

Proof: If $R$ is quasi-Frobenius, then $R$ is right semi-Artinian and so by Proposition 10 every strongly Red-injective module is injective, and hence projective since $R$ is quasi-Frobenius. Conversely, if every strongly Red-injective module is projective, then in particular every injective module is projective, and so $R$ is quasi-Frobenius. $\Box$

Theorem 4**.**

$R$* is a right $V$-ring if and only if every simple $R$-module is strongly Red-injective.*

Proof: Suppose that $M$ is a simple $R$-module where $R$ is a right $V$-ring. Then, by definition of a $V$-ring, $M$ is injective. Hence, $M$ is strongly Red-injective. Conversely, suppose that any simple module $M$ is strongly Red-injective. Since $M$ is simple, $\text{Soc}(M)=M$ and hence $\{0\}\neq\text{Soc}(M)\subseteq^{e}M$. Since $M$ has essential socle, it is injective by [1, Corollary 3.2]. Hence, $R$ is a right $V$-ring. $\Box$

Corollary 6**.**

Let $M$ be an $R$-module with essential socle. The following statements are equivalent:

$M$* is injective.* 2. 2.

$M$* is strongly Red-injective.* 3. 3.

$M$* is strongly Soc-injective.*

Proof: By Proposition 5(2), $1\Rightarrow 2\Rightarrow 3$. By [1, Corollary 3.2], $3\Rightarrow 1$ which completes the proof. $\Box$

Remark 1*.*

We make the following observations:

It is easy to check that any sort of injectivity that lies between injective and strongly Red-injective modules would lead to Theorems 3 and 4. 2. 2.

Red-injectivity is a less restricted notion than injectivity but carries most of the properties of injectivity. 3. 3.

Red-injectivity is much closer to injectivity than Soc-injectivity. 4. 4.

When the ring is not commutative, a semi-simple module need not be semi-reduced, see Example 3.1 below:

Example 3.1**.**

Let the ring $R$ be the collection of all $2\times 2$ matrices over the field of real numbers. The module $M=R_{R}$ is semi-simple but not reduced. For if m=\left(\begin{array}[]{cc}1&0\\ 0&1\\ \end{array}\right)\in M and r=\left(\begin{array}[]{cc}1&-1\\ 1&-1\\ \end{array}\right)\in R, then $mr\neq 0$ but $mr^{2}=0$. Since a direct sum of reduced modules is reduced, $M$ is a direct sum of simple modules which is not reduced. A simple module over a not necessarily commutative ring need not be reduced. $M$ is not semi-reduced.

The following implications hold:

Injective $\Rightarrow$ quasi-injective $\Rightarrow$ pseudo-injective $\Rightarrow$ Red-quasi-injective.

For the first two implications, see [9]. The last implication is trivial, it follows directly from the definitions.

Example 3.2**.**

Let $R$ be the ring of all eventually constant sequences $(x_{n})_{n\in\mathbb{N}}$ of elements in $\mathbb{F}_{2}$, the field of two elements. Then, $E(R_{R})=\prod_{n\in\mathbb{N}}\mathbb{F}_{2}$, which has only one automorphism, namely the identity automorphism. By [5, Example 9], $R_{R}$ is pseudo-injective but it is not quasi-injective. It therefore follows that $R_{R}$ is Red-quasi-injective but not injective.

An $R$-module $M$ is said to satisfy the exchange property if for every $R$-module $A$ and any two direct sum decomposition $A=M^{\prime}\bigoplus N=\bigoplus_{i\in I}A_{i}$ with $M^{\prime}=M$, there exists a submodule $B_{i}$ of $A_{i}$ such that $A=M^{\prime}\bigoplus(\bigoplus_{i\in I}B_{i})$. An $R$-module is called clean if its endomorphism ring, $\text{End}_{R}(M)$ is clean, i.e, for all $f\in\text{End}_{R}(M)$, $f=e+u$ with $e$ idempotent and $u$ a unit. Pseudo-injective modules (and hence quasi-injective and injective modules) are clean and also satisfy the exchange property, see [2] and [3]. Note that pseudo-injective modules are equivalent to automorphism-invariant modules as they are being referred to in [2] and [3]. The equivalence was proved in [5]. We now ask:

Question 1**.**

Are the Red-quasi-injective modules and Soc-quasi-injective modules clean? Do they satisfy the exchange property?