max-projective modules

TL;DR

This paper characterizes right almost-$QF$ and max-$QF$ rings, showing their equivalence under certain conditions and providing structural descriptions for these classes of rings.

Contribution

It introduces the concepts of max-projective modules and right max-$QF$ rings, and characterizes these rings in terms of their structural decomposition and properties.

Findings

Right Hereditary right Noetherian rings are right max-$QF$ iff they are right almost-$QF$.

A right Hereditary ring is max-$QF$ iff every injective simple module is projective.

A commutative Noetherian ring is almost-$QF$ iff it decomposes into a product of a $QF$ ring and a small ring.

Abstract

A right $R$-module $M$ is called max-projective provided that each homomorphism $f:M \to R/I$ where $I$ is any maximal right ideal, factors through the canonical projection $\pi : R \to R/I$. We call a ring $R$ right almost-$QF$ (resp. right max-$QF$) if every injective right $R$-module is $R$-projective (resp. max-projective). This paper attempts to understand the class of right almost-$QF$ (resp. right max-$QF$) rings. Among other results, we prove that a right Hereditary right Noetherian ring $R$ is right almost-$QF$ if and only if $R$ is right max-$QF$ if and only if $R=S\times T$ , where $S$ is semisimple Artinian and $T$ is right small. A right Hereditary ring is max-$QF$ if and only if every injective simple right $R$-module is projective. Furthermore, a commutative Noetherian ring $R$ is almost-$QF$ if and only if $R$ is max-$QF$ if and only if $R=A \times B$, where $A$ is $QF$ and $B$ is a small ring.

Equations

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Full text

1. Introduction and Preliminaries

Throughout, $R$ will denote an associative ring with identity, and modules will be unital right $R$-modules, unless otherwise stated. Let $M$ and $N$ be $R$-modules. $M$ is called $N$-projective (projective relative to $N$) if every $R$-homomorphism from $M$ into an image of $N$ can be lifted to an $R$-homomorphism from $M$ into $N$. $M$ is called $R$-projective if it is projective relative to the right $R$-module $R_{R}$. The module $M$ is called projective if $M$ is $N$-projective, for every $R$-module $N$. A right $R$-module $M$ is called max-projective provided that each homomorphism $f:M\to R/I$ where $I$ is any maximal right ideal, factors through the canonical projection $\pi:R\to R/I$. This notion properly generalizes the notions $R$-projective and rad-projective modules studied in [3].

Characterizing rings by projectivity of some classes of their modules is a classical problem in ring and module theory. A result of Bass [6, Theorem 28.4] states that a ring $R$ is right perfect if and only if each flat right $R$-module is projective. On the other hand, the ring $R$ is $QF$ if and only if each injective right $R$-module is projective, [14]. Recently, the notion of $R$-projectivity and its generalizations are considered in [1, 2, 3, 5, 24]. The rings whose flat right $R$-modules are $R$-projective and max-projective are characterized in [5, 4] and [8], respectively.

We call a ring $R$ right almost-$QF$ (resp. right max-$QF$) in case all injective right $R$-modules are $R$-projective (resp. max-projective). Right almost $QF$-rings are max-$QF$. The ring of integers is almost-$QF$, since $\operatorname{Hom}(E,\mathbb{Z}/n\mathbb{Z})=0$ for each injective $\mathbb{Z}$-module $E$.

In this paper, we investigate some properties of max-projective $R$-modules, and give some characterizations of almost-$QF$ and max-$QF$ rings.

We organize the paper as follows. In Section 2, some properties of max-projective $R$-modules are investigated. We obtain that $R$-projectivity and max-projectivity coincide over the ring of integers and over right perfect rings. Characterizations of semiperfect, perfect and $QF$ rings in terms of max-projectivity are given. As an application, we show that a ring $R$ is right (semi)perfect if and only if every (finitely generated) right R-module has a max-projective cover if and only if every (simple) semisimple right $R$-module has a max-projective cover. By [1, Lemma 2.1] any finitely generated $R$-projective right $R$-module is projective. This result is not true when $R$-projectivity is replaced with max-projectivity. We prove that if $R$ is either a semiperfect or nonsingular self-injective ring, then finitely generated max-projective right $R$-modules are projective. We show that any max-projective right $R$-module of finite length is projective.

In Section 3, we give some characterizations of almost-$QF$ and max-$QF$ rings. Every right small ring is right max-$QF$, while a right small ring is right almost-$QF$ provided $R$ is right Hereditary or right Noetherian. A right Hereditary right Noetherian ring $R$ is right almost-$QF$ if and only if $R$ is right max-$QF$ if and only if $R=S\times T$ , where $S$ is a semisimple Artinian and $T$ is a right small ring. A right Hereditary ring $R$ is right max-$QF$ if and only if every simple injective right $R$-module is projective. A commutative Noetherian ring $R$ is almost-$QF$ if and only if $R$ is max-$QF$ if and only if $R=A\times B$, where $A$ is $QF$ and $B$ is a small ring. A right Noetherian local ring is almost-$QF$ if and only if $R$ is $QF$ or right small.

As usual, we denote by Mod$-R$ the category of right $R$-modules. For a module $M$, $E(M)$, $Z(M)$, $\operatorname{Rad}(M)$ and $Soc(M)$ denote the the injective hull, singular submodule, Jacobson radical and socle of $M$, respectively. The notation $K\ll M$ means that $K$ is a superfluous submodule of $M$ in the sense that $K+L\neq M$ for any proper submodule $L$ of $M$.

2. Max-projective modules

Definition 1**.**

A right $R$-module $M$ is said to be max-projective if for every epimorphism $f:R\rightarrow R/I$ with $I$ is a maximal right ideal of $R$, and every homomorphism $g:M\rightarrow R/I$, there exists a homomorphism $h:M\rightarrow R$ such that $fh=g$.

Example 1*.*

Every projective $R$-module is max-projective.

The $\mathbb{Z}$-module $\mathbb{Q}$ is max-projective, since $\operatorname{Hom}(\mathbb{Q},\,\mathbb{Z}_{p})=0$ for each simple $\mathbb{Z}$-module $\mathbb{Z}_{p}$.

Every simple max-projective $R$-module is projective. For if $S$ is a simple right $R$-module and $1_{S}:S\to S$ is the identity map, then by max-projectivity of $S$ there is a homomorphism $f:S\to R$ such that $\pi f=1_{S}$, where $\pi:R\rightarrow S$ is the natural epimorphism. Then $R\cong K\oplus S$, and so $S$ is projective.

Any $R$-module $M$ with $\operatorname{Rad}(M)=M$ is max-projective, since $M$ has no simple factors.

Given modules $M$ and $N$, $M$ is said to be $N$-subprojective if for every homomorphism $f:M\rightarrow N$ and for every epimorphism $g:B\rightarrow N$, there exists a homomorphism $h:M\rightarrow B$ such that $gh=f$ (see [16]).

Lemma 1**.**

For an $R$-module $M$, the following are equivalent.

1. (1)

$M$* is max-projective.* 2. (2)

$M$* is $S$-subprojective for each simple $R$-module $S$.* 3. (3)

For every epimorphism $f:N\rightarrow S$ with $S$ simple, and homomorphism $g:M\rightarrow S$, there exists a homomorphism $h:M\rightarrow N$ such that $fh=g$.

Proof.

$(2)\Leftrightarrow(3)$ By definition. $(3)\Rightarrow(1)$ is clear.

$(1)\Rightarrow(3)$ Let $f:N\rightarrow S$ be an epimorphism with $S$ is simple $R$-module and $g:M\rightarrow S$ a homomorphism. Since $S$ is simple, there exists an epimorphism $\pi:R\rightarrow S$. By the hypothesis there exists a homomorphism $h:M\rightarrow R$ such that $\pi h=g$. Since $R$ is projective, there exists a homomorphism $h^{\prime}:R\rightarrow N$ such that $fh^{\prime}=\pi$. Then $f(h^{\prime}h)=\pi h=g$, and so $M$ is max-projective. ∎

We need the following result in the sequel.

Lemma 2**.**

The following conditions are true.

1. (1)

A direct sum $\oplus_{i\in I}A_{i}$ of modules is max-projective (resp. $R$-projective) if and only if each $A_{i}$ is max-projective (resp. $R$-projective). 2. (2)

If $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ is an exact sequence and $M$ is $B$-projective, then $M$ is projective relative to both $A$ and $C$.

Proof.

$(1)$ Since it is similar to the one provided in [6, Proposition 16.10] for $R$-projective modules, the proof is omitted for max-projective modules.

$(2)$ is clear by [6, Proposition 16.12]. ∎

Corollary 1**.**

For a ring $R$, the following are equivalent.

1. (1)

$R$* is semisimple.* 2. (2)

Every right $R$-module is max-projective. 3. (3)

Every finitely generated right $R$-module is max-projective. 4. (4)

Every cyclic right $R$-module is max-projective. 5. (5)

Every simple right $R$-module is max-projective.

Proof.

$(1)\Rightarrow(2)\Rightarrow(3)\Rightarrow(4)\Rightarrow(5)$ are clear.

$(5)\Rightarrow(1)$ Example 1$(c)$ and the hypothesis implies that each simple right $R$-module is projective. Thus $R$ is semisimple. ∎

In [3], the module $M$ is called rad-projective if, for any epimorphism $\sigma:R\rightarrow K$ where $K$ is an image of $R/J(R)$ and any homomorphism $f:M\rightarrow K$, there exists a homomorphism $g:M\rightarrow R$ such that $f=\sigma g$. We have the following implications:

projective $\Rightarrow$ R-projective $\Rightarrow$ rad-projective $\Rightarrow$ max-projective

Proposition 1**.**

Let $R$ be a semilocal ring and $M$ an $R$-module. Then the following are equivalent.

1. (1)

$M$* is rad-projective.* 2. (2)

$M$* is max-projective.* 3. (3)

Every homomorphism $f:M\rightarrow R/J(R)$ can be lifted to a homomorphism $g:M\rightarrow R$.

Proof.

$(1)\Rightarrow(2)$ Clear. $(3)\Rightarrow(1)$ By [2, Proposition 3.14].

$(2)\Rightarrow(3)$ Since $R/J(R)$ is semisimple, $R/J(R)=\oplus^{n}_{i=1}K_{i}$, with each $K_{i}$ simple as an $R$-module. Let $\pi_{i}:\oplus^{n}_{i=1}K_{i}\rightarrow K_{i}$, and $\pi:R\rightarrow\oplus^{n}_{i=1}K_{i}$. Set $h:=\pi_{i}\pi$. By the hypothesis, there exists a homomorphism $g:M\rightarrow R$ such that $hg=\pi_{i}f$. Since $R/J(R)$ is semisimple, $\pi_{i}$ splits and there exists a homomorphism $\varepsilon_{i}:K_{i}\rightarrow\oplus^{n}_{i=1}K_{i}$ such that $\varepsilon_{i}\pi_{i}=1_{R/J(R)}$. Then, $\pi g=\varepsilon_{i}hg=\varepsilon_{i}\pi_{i}f=f$. ∎

In the next Proposition we provide a sufficient condition for an $R$-module to be max-projective. We establish a converse in the case of self-injective rings.

Proposition 2**.**

If $M$ is a right $R$-module such that $Ext^{1}_{R}(M,I)=0$ for every maximal right ideal $I$ of $R$, then $M$ is max-projective. The converse is true when $R$ is a right self-injective ring.

Proof.

By applying $\operatorname{Hom}(M,-)$ to the short exact sequence $0\rightarrow I\rightarrow R\rightarrow R/I\rightarrow 0$, with $I$ being a maximal right ideal of $R$, we obtain the following exact sequence:

$0\rightarrow\operatorname{Hom}(M,I)\rightarrow\operatorname{Hom}(M,R)\rightarrow\operatorname{Hom}(M,R/I)\rightarrow Ext^{1}_{R}(M,I)\rightarrow Ext^{1}_{R}(M,R)\rightarrow...$. If $Ext^{1}_{R}(M,I)=0$ for every maximal right ideal $I$ of $R$, it follows that $M$ is max-projective. Conversely, since $R$ is right self injective, $Ext^{1}_{R}(M,R)=0$. If $M$ is a max-projective right $R$-module, then the map $\operatorname{Hom}(M,R)\rightarrow\operatorname{Hom}(M,R/I)$ is onto, and so $Ext^{1}_{R}(M,I)=0$ for any maximal right ideal $I$ of $R$. ∎

Proposition 3**.**

Let $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ be a short exact sequence. If $M$ is $A$-subprojective and $C$-subprojective, then $M$ is $B$-subprojective.

Proof.

Let $\gamma:F\rightarrow B$ be an epimorphism with $F$ projective. Then using the pullback diagram of $\gamma:F\rightarrow B$ and $\beta:A\rightarrow B$, and applying $\operatorname{Hom}(M,-)$, we get a commutative diagram with exact rows and columns:

[TABLE]

Since $M$ is $A$-subprojective and $C$-subprojective, ${\theta}$ and ${\phi}$ are epic. Hence, ${\gamma^{*}}$ is epic by [6, Five Lemma 3.15].

∎

Proposition 4**.**

Let $M$ be an $R$-module. $M$ is max-projective if and only if $M$ is $N$-subprojective for any $R$-module $N$ with composition length $cl(N)<\infty$.

Proof.

Let $M$ be a max-projective $R$-module and $N$ be an $R$-module with $cl(N)=n$. Then there exists a composition series $0=S_{0}\subset S_{1}...\subset S_{n}=N$ with each composition factor $S_{i+1}/S_{i}$ simple. Consider the short exact sequence $0\rightarrow S_{1}\rightarrow S_{2}\rightarrow S_{2}/S_{1}\rightarrow 0$. Since $M$ is max-projective, by Lemma 1, $M$ is $S_{1}$-subprojective and $S_{2}/S_{1}$-subprojective. So, by Proposition 3, $M$ is $S_{2}$-subprojective. Continuing in this way, $M$ is $S_{i}$-subprojective for each $0\leq i\leq n$. Hence, $M$ is $N$-subprojective. Conversely, since each simple right $R$-module has finite length, $M$ is max-projective by Lemma 1. ∎

Corollary 2**.**

A $\mathbb{Z}$-module $M$ is max-projective if and only if $M$ is $\mathbb{Z}$-projective.

Proof.

By the Fundamental Theorem of Abelian Groups, a cyclic $\mathbb{Z}$-module $M$ is isomorphic either to $\mathbb{Z}$ or to a finite direct sum of $\mathbb{Z}$-modules of finite length. Now the proof is clear by Proposition 4. ∎

Corollary 3**.**

Let $M$ be an $R$-module with finite composition length. If $M$ is max-projective, then it is projective.

Proof.

Let $f:R^{n}\to M$ be an epimorphism. The module $M$ is $M$-subprojective by Proposition 4. That is, there is a homomorphism $g:M\to R^{n}$ such that $1_{M}=fg$. Thus the map $f$ splits, and so $M$ is projective. ∎

Submodules of max-projective $R$-modules need not be max-projective. Consider the ring $R=\mathbb{Z}/p^{2}\mathbb{Z}$, for some prime integer $p$. $R$ is max-projective, whereas the simple ideal $pR$ is not max-projective, since the epimorphism $R\to pR\to 0$ does not split.

Recall that a ring $R$ is called right $V$-ring (resp. right $GV$-ring) if all simple (resp. all singular simple) right $R$-modules are injective.

Proposition 5**.**

Consider the following conditions for a ring $R$:

1. (1)

$R$* is a right $GV$-ring.* 2. (2)

Submodules of max-projective right $R$-modules are max-projective. 3. (3)

Submodules of projective right $R$-modules are max-projective. 4. (4)

Every right ideal of $R$ is max-projective.

Then, $(1)\Rightarrow(2)\Rightarrow(3)\Rightarrow(4)$. Also, if $R$ is a right self injective ring, then $(4)\Rightarrow(1)$.

Proof.

$(1)\Rightarrow(2)$ Let $N$ be a submodule of a max-projective right $R$-module $M$. Consider the following diagram:

[TABLE]

where $S$ is a simple right $R$-module, $i:N\rightarrow M$ is the inclusion map and $\pi:R\rightarrow S$ is the canonical quotient map. Since the simple module $S$ is either projective or singular, the former implies $\pi:R\rightarrow S$ splits and there exists a homomorphism $\varepsilon:S\rightarrow R$ such that $\varepsilon\pi=1_{R}$. In the latter one, $S$ is singular, so it is injective by the hypothesis. Thus, there is a homomorphism $g:M\rightarrow S$ such that $gi=f$. Since $M$ is max-projective, there is a homomorphism $h:M\rightarrow R$ such that $\pi h=g$. Hence, $\pi(hi)=gi=f$. In either case, there exists a homomorphism from $N$ to $R$ that makes the diagram commute. This implies that $N$ is max-projective.

$(2)\Rightarrow(3)\Rightarrow(4)$ Clear. $(4)\Rightarrow(1)$ Let $I$ be a right ideal of $R$ and $J$ a maximal right ideal of $R$. Consider the following diagram:

[TABLE]

where $R/J$ is a simple right $R$-module, $i:I\rightarrow R$ is the inclusion map and $\pi:R\rightarrow R/J$ is the canonical quotient map. Since $I$ is max-projective, there is a homomorphism $h:I\rightarrow R$ such that $\pi h=f$. Since $R$ is injective, there exists a homomorphism $\lambda:R\rightarrow R$ such that $\lambda i=h$. Now the map $\beta=\pi\lambda:R\rightarrow R/J$ satisfies $\beta i=\pi\lambda i=\pi h=f$, as required. ∎

Proposition 6**.**

Let $R$ be a commutative or semilocal ring. Then pure submodules of max-projective $R$-modules are max-projective.

Proof.

Let $M$ be a max-projective (right) module and $N$ a pure submodule of $M$. Let $S$ be a simple (right) module and $f:N\to S$ be a homomorphism. Since $S$ is pure-injective and $N$ is a pure submodule of $M$, there is $g:M\to S$ such that $gi=f$, where $i:N\to M$ is the inclusion map. By max-projectivity of $M$, there is a homomorphism $h:M\to R$ such that $g=\pi h$, where $\pi:R\to S$ is the natural epimorphism. Now we have $f=gi=\pi hi$, i.e. $hi:N\to R$ lifts $f$. This proves that $N$ is max-projective. ∎

Lemma 3**.**

Let $R$ be a ring and $\tau$ be a preradical with $\tau(R)=0$. If $M$ is a max-projective $R$-module, then $M/\tau(M)$ is max-projective.

Proof.

Let $M$ be a max-projective $R$-module and $f:M/\tau(M)\rightarrow S$ a homomorphism with $S$ simple $R$-module. Consider the following diagram:

[TABLE]

Since M is max-projective, there exists a homomorphism $g:M\rightarrow R$ such that $f\pi=\eta g$. Since $g(\tau(M))\subseteq\tau(R)=0$, $\tau(M))\subseteq\operatorname{Ker}(g)$, and so there exists a homomorphism $h:M/\tau(M)\rightarrow R$ such that $h\pi=g$. Now, since $\eta h\pi=\eta g=f\pi$ and $\pi$ is an epimorphism, $\eta h=f$, and so $M/\tau(M)$ is max-projective. ∎

Remark 1*.*

Recall that any finitely generated $R$-projective module is projective, [1, Lemma 2.1]. This is not true for max-projective modules in general. Let $R$ be a right $V$-ring which is not right semihereditary. Then $R$ has a finitely generated right ideal which is not projective. By Proposition 5, each right ideal of $R$ is max-projective.

Proposition 7**.**

Let $R$ be a right nonsingular right self-injective ring. Every finitely generated max-projective right $R$-module is projective.

Proof.

Let $M$ be a finitely generated max-projective right $R$-module. As $R$ is a right nonsingular ring, by Lemma 3, $M/Z(M)$ is max-projective. Since $M/Z(M)$ is finitely generated, there exists an epimorphism $f:F\rightarrow M/Z(M)$ such that $F$ is finitely generated free. This means $\operatorname{Ker}(f)$ is closed in $F$. By the injectivity of $F$, $\operatorname{Ker}(f)$ is a direct summand of $F$, and so $M/Z(M)$ is projective. Then, $M=Z(M)\oplus K$ for some projective submodule $K$ of $M$. We claim that $Z(M)=0$. Assume to the contrary that $Z(M)\neq 0$. Since, $Z(M)$ is a finitely generated submodule of $M$, there exists a nonzero epimorphism $g:Z(M)\rightarrow S$ for some simple right $R$-module $S$. Then, by Lemma 2, $Z(M)$ is max-projective, and so there exists a nonzero homomorphism $h:Z(M)\rightarrow R$ such that $\pi h=g$, where $\pi:R\rightarrow S$ is the natural epimorphism. But then $h(Z(M))\subseteq Z(R_{R})=0$, a contradiction. Thus we must have $Z(M)=0$, whence $M$ is projective. ∎

A ring $R$ is called right max-ring if every nonzero right $R$-module $M$ has a maximal submodule i.e. $\operatorname{Rad}(M)\neq M$.

Proposition 8**.**

The following conditions are true.

1. (1)

Over a semiperfect ring $R$, every max-projective right $R$-module with small radical is projective. 2. (2)

A ring $R$ is right perfect if and only if $R$ is semilocal and every max-projective right $R$-module is projective.

Proof.

$(1)$ Let $M$ be a max-projective right $R$-module with $\operatorname{Rad}(M)\ll M$. Since $R$ is semilocal, $M$ is rad-projective by Proposition 1. Hence $M$ is projective by [2, Theorem 4.7].

$(2)$ Since over a right perfect ring $R$ every right $R$-module has small radical, it follows from $(1)$ that every max-projective right $R$-module is projective. Conversely, assume that $R$ is semilocal and every max-projective right $R$-module is projective. Let $M$ be a nonzero right $R$-module. We claim that $\operatorname{Rad}(M)\neq M$. Assume to the contrary that $M$ has no maximal submodule, i.e. $\operatorname{Rad}(M)=M$. Since $\operatorname{Hom}(M,S)=0$ for any simple right $R$-module, $M$ is max-projective. Thus $M$ is projective, by the hypothesis. Since projective modules have maximal submodules, this is a contradiction. Hence, every right $R$-module has a maximal submodule. Since $R$ is semilocal, $R$ is right perfect by [6, Theorem 28.4]. ∎

Recall that if $R$ is a right perfect ring, every $R$-projective right $R$-module is projective, [24]. Thus the following result follows from Proposition 8(2).

Corollary 4**.**

Let $R$ be a right perfect ring and $M$ be a right $R$-module. Then the following are equivalent.

1. (1)

$M$* is projective.* 2. (2)

$M$* is $R$-projective.* 3. (3)

$M$* is max-projective.*

The following Remark is an example of a right nonperfect ring $R$ such that every max-projective module is $R$-projective.

Remark 2*.*

Let $K$ be a field, and $R$ the subalgebra of $K^{\omega}$ consisting of all eventually constant sequences in $K^{\omega}$. For each $i<\omega$, we let $e_{i}$ be the idempotent in $K^{\omega}$ whose ith component is $1$ and all the other components are [math]. Notice that $\{e_{i}:i<\omega\}$ a set of pairwise orthogonal idempotents in $R$, so $R$ is not perfect, [25, Lemma 2.3]. By [25, Lemma 2.3 and Lemma 2.4], $R/Soc(R)$ is simple $R$-module and a module $M$ is $R$-projective if and only if it is projective with respect to the projection $\pi:R\rightarrow R/Soc(R)$. Thus, an $R$-module $M$ is max-projective if and only if $M$ is $R$-projective.

The following Corollary follows from [25, Theorem 3.3] and Remark 2.

Corollary 5**.**

Let $K$ be a field of cardinality $\leq 2^{\omega}$ and $R$ the subalgebra of $K^{\omega}$ consisting of all eventually constant sequences in $K^{\omega}$. Assume Gödel’s Axiom of Constructibility $(V=L)$. Then all max-projective $R$-modules are projective.

Lemma 4**.**

If $M_{R}$ is max-projective and $\bar{R}=R/J(R)$, then $(M/\operatorname{Rad}(M))_{\bar{R}}$ is max-projective.

Proof.

Let $\pi:\bar{R}_{\bar{R}}\rightarrow K_{\bar{R}}$ be an $\bar{R}$-epimorphism with $K_{\bar{R}}$ simple $\bar{R}$-module. Consider the following diagram:

[TABLE]

Since $M$ is max-projective, there exists a homomorphism $\lambda:M_{R}\rightarrow\bar{R}_{R}$ such that $\pi\lambda=f\eta$. Since $\lambda(\operatorname{Rad}(M))\subseteq\operatorname{Rad}(R/J(R))=0$, $\operatorname{Rad}(M)\subseteq\operatorname{Ker}(\lambda)$, and so there exists a homomorphism $\delta:(M/\operatorname{Rad}(M))_{R}\rightarrow\bar{R}_{R}$ such that $\delta\eta=\lambda$. Now, since $\pi\delta\eta=\pi\lambda=f\eta$ and $\eta$ is an epimorphism, $\pi\delta=f$, and so $(M/\operatorname{Rad}(M))_{\bar{R}}$ is a max-projective $\bar{R}$-module. ∎

It is well-known that a ring $R$ is semiperfect if and only if every simple $R$-module has a projective cover. In the next Proposition, we extend this result by replacing projective covers with max-projective covers. Let $R$ be a ring and $\Omega$ be a class of right $R$-modules which is closed under isomorphisms. A homomorphism $f:P\rightarrow M$ is called an $\Omega$-cover of the right $R$-module $M$, if $P\in\Omega$ and $f$ is an epimorphism with small kernel. That is to say, if $\Omega$ is the class of max-projective right $R$-modules, the homomorphism $f:P\rightarrow M$ is called max-projective cover of $M$. With the help of an argument similar to the one provided in [3, Theorem 18], we can establish the next Proposition.

Proposition 9**.**

For a ring $R$, the following are equivalent.

1. (1)

$R$* is semiperfect.* 2. (2)

Every finitely generated right $R$-module has a max-projective cover. 3. (3)

Every cyclic right $R$-module has a max-projective cover. 4. (4)

Every simple right $R$-module has a max-projective cover.

Proof.

$(1)\Rightarrow(2)\Rightarrow(3)\Rightarrow(4)$ are clear.

$(4)\Rightarrow(1)$ We first show that $\bar{R}=R/J(R)$ is a semisimple ring. Let $S$ be a simple right $\bar{R}$-module. By the hypothesis $S_{R}$ has a max-projective cover $P_{R}$, say $f:P\rightarrow S$ with $Rad(P)=\operatorname{Ker}(f)\ll P$. Since $S$ is simple and $P/\operatorname{Rad}(P)\cong S$, $P/\operatorname{Rad}(P)$ is a simple $\bar{R}$-module. So, $(P/\operatorname{Rad}(P))_{\bar{R}}$ is max-projective by Lemma 4, whence $(P/\operatorname{Rad}(P))_{\bar{R}}$ is projective. Consider the map $\tilde{f}:P/\operatorname{Rad}(P)\rightarrow S$. This map induces an isomorphism. Since $P/\operatorname{Rad}(P)$ is projective $\bar{R}$-module, $P/\operatorname{Rad}(P)$ is the projective cover of $S_{\bar{R}}$. Hence, $\bar{R}$ is a semiperfect ring. Therefore, $\bar{R}$ is semisimple as an $\bar{R}$-module, and hence semisimple as an $R$-module. Write $\bar{R}=R/J(R)=\oplus^{n}_{i=1}K_{i}$, with each $K_{i}$ simple as a right $R$-module, and let $L_{i}$ be a max-projective cover of $K_{i}$, $1\leq i\leq n$, as right $R$-modules. Now, in order to prove that R is a semiperfect ring, it is enough to show that each $L_{i}$, $1\leq i\leq n$, is projective as a right $R$-module. Clearly, $L=\oplus^{n}_{i=1}L_{i}$, as a right R-module, is a max-projective cover of $\bar{R}_{R}$. Consider the diagram

[TABLE]

with $f$ being the max-projective cover of $\bar{R}_{R}$, and $\pi$ the canonical $R$-epimorphism. By the max-projectivity of $L_{R}$, $f$ can be lifted to a map $g:L_{R}\rightarrow R_{R}$ such that $\pi g=f$. Since $R=\operatorname{Im}(g)+J(R)$ and $J(R)\ll R$ we infer that $R=\operatorname{Im}(g)$ and $g$ is onto. By the projectivity of $R$, the map $g$ splits and $L_{R}=\operatorname{Ker}(g)\oplus A$ for a submodule $A$ of $L_{R}$. Since $\operatorname{Ker}(g)\subseteq\operatorname{Ker}(f)\ll L_{R}$, $\operatorname{Ker}(g)=0$ and $L_{R}\cong R_{R}$ is projective. Therefore, each $L_{i}$, $1\leq i\leq n$, is projective as a right $R$-module, and $R$ is semiperfect. ∎

Proposition 10**.**

For a ring $R$, the following conditions are equivalent.

1. (1)

$R$* is right perfect.* 2. (2)

Every right $R$-module has a max-projective cover. 3. (3)

Every semisimple right $R$-module has a max-projective cover.

Proof.

$(1)\Rightarrow(2)\Rightarrow(3)$ are clear.

$(3)\Rightarrow(1)$ By Proposition 9, $R$ is a semiperfect ring. Let $M$ be a semisimple right $R$-module and $f:P\rightarrow M$ be a max-projective cover of $M$. Since $Rad(P)=\operatorname{Ker}(f)\ll P$, $P$ is projective by Proposition 8(1). Thus every semisimple right $R$-module has a projective cover, and so $R$ is right perfect. ∎

Let $R$ be any ring and $M$ be an $R$-module. A submodule $N$ of $M$ is called radical submodule if $N$ has no maximal submodules, i.e. $N=\operatorname{Rad}(N)$. By $P(M)$ we denote the sum of all radical submodules of a module $M$. For any module $M$, $P(M)$ is the largest radical submodule of $M$, and so $\operatorname{Rad}(P(M))=P(M)$. Moreover, $P$ is an idempotent radical with $P(M)\subseteq\operatorname{Rad}(M)$ and $P(M/P(M))=0$, (see [7]).

In [11, Lemma 1], the authors prove that over a right nonsingular right $V$-ring, max-projective right $R$-modules are nonsingular. Regarding the converse of this fact, we have the following.

Proposition 11**.**

If every max-projective right $R$-module is nonsingular, then $R$ is right nonsingular and right max-ring.

Proof.

Clearly the ring $R$ is right nonsingular. If $R$ is right $V$-ring, then $\operatorname{Rad}(M)=0$ for any right $R$-module $M$. Thus $R$ is a max-ring. Suppose $R$ is not right $V$-ring and let $S$ be a noninjective simple right $R$-module. We shall first see that $P(E(S))=0$. Suppose $\operatorname{Rad}(P(E(S)))=P(E(S))\neq 0$. Then $P(E(S))/S$ is singular. Furthermore, since $\operatorname{Rad}(P(E(S))/S)=P(E(S))/S$, $P(E(S))/S$ is max-projective. This contradicts with the hypothesis. Therefore, for every simple right $R$-module $S$, $P(E(S))=0$. Let $M$ be a nonzero right $R$-module. We claim that $\operatorname{Rad}(M)\neq M$. Assume to the contrary that $\operatorname{Rad}(M)=M$. Let $0\neq x\in M$ and $K$ be a maximal submodule of $xR$. Then the simple right $R$-module $S=xR/K$ is noninjective, because $S$ small. Now, the obvious map $xR\to E(S)$ extends to a nonzero map $f:M\rightarrow E(S)$. Since $P(\operatorname{Im}(f))\subseteq P(E(S))=0$, $P(M/\operatorname{Ker}(f))=0$. This contradicts with $P(M)=M$. Hence $\operatorname{Rad}(M)\neq M$ for every right $R$-module $M$, and so $R$ is a right max-ring. ∎

Corollary 6**.**

For a ring $R$, the following are equivalent.

1. (1)

$R$* is semilocal and every max-projective right $R$-module is nonsingular.* 2. (2)

$R$* is right perfect and right nonsingular.*

3. Almost-$QF$ and max-$QF$ rings

Recall that a ring $R$ is $QF$ if and only if every injective (right) $R$-module is projective (see, [14]). We slightly weaken this condition and obtain the following definition.

Definition 2**.**

A ring $R$ is called right almost-$QF$ if every injective right $R$-module is $R$-projective. We call $R$ right max-$QF$, if every injective right $R$-module is max-projective. Left almost-$QF$ and left max-$QF$ rings are defined similarly.

Clearly, we have the following inclusion relationship:

{$QF$ rings} $\subseteq$ {right almost-$QF$ rings} $\subseteq$ { right max-$QF$ rings}.

Example 2*.*

The ring of integers $\mathbb{Z}$, is a right almost-$QF$ but not $QF$: For every injective $\mathbb{Z}$-module $E$, we have $\operatorname{Rad}(E)=E$. Thus $\operatorname{Hom}(E,\mathbb{Z}/n\mathbb{Z})=0$, for each cyclic $\mathbb{Z}$-module $\mathbb{Z}/n\mathbb{Z}$. This means that each injective $\mathbb{Z}$-module is $\mathbb{Z}$-projective,and so $\mathbb{Z}$ is almost-$QF$.

Remark 3*.*

Sandomierski [24] proved that if $R$ is a right perfect ring, then every $R$-projective right module is projective. Thus a ring $R$ is right perfect and right almost-$QF$ if and only if $R$ is $QF$.

Proposition 12**.**

Let $R$ and $S$ be Morita equivalent rings. Then, $R$ is right almost-$QF$ if and only if $S$ is right almost-$QF$.

Proof.

An R-module $M$ is $R$-projective if and only if $M$ is $N$-projective for any finitely generated projective $R$-module $N$. Now, by [6, propositions 21.6 and 21.8 ], since injectivity, relative projectivity and being finitely generated are preserved by Morita equivalence, the proof is clear. ∎

Lemma 5**.**

Let $R_{1}$ and $R_{2}$ be rings. Then $R=R_{1}\times R_{2}$ is right almost-$QF$ (resp. right max-$QF$) if and only if $R_{1}$ and $R_{2}$ are both right almost-$QF$ (resp. right max-$QF$).

Proof.

Let $M$ be an injective right $R_{1}$-module. Then $M$ is an injective right $R$-module, as well as an $R$-projective module by the hypothesis. Hence, by Lemma 2, $M$ is $R_{1}$-projective, and so $R_{1}$ is right almost-$QF$. Similarly, $R_{2}$ is right almost-$QF$. Conversely, let $M$ be an injective right $R$-module. Since we have the decomposition $M=MR_{1}\oplus MR_{2}$, $MR_{1}$ is an injective right $R$-module, whence it is an injective right $R_{1}$-module. On the other hand, since $(MR_{2})R_{1}=0$, $MR_{2}$ is an $R_{1}$-module, and so it is an injective $R_{1}$-module. This means that $MR_{1}$ and $MR_{2}$ are $R_{1}$-projective by the hypothesis. Then, by Lemma 2, $M=MR_{1}\oplus MR_{2}$ is $R_{1}$-projective. Similarly, $M$ is $R_{2}$-projective. Therefore, $M$ is $R$-projective. Since it is similar to the one provided for almost-$QF$ rings, the proof is omitted for max-$QF$ rings. ∎

Proposition 13**.**

Let $R$ be a right Hereditary ring and $E$ be an indecomposable injective right $R$-module. Then the following are equivalent.

1. (1)

$E$* is $R$-projective.* 2. (2)

$E$* is max-projective.* 3. (3)

Either $\operatorname{Rad}(E)=E$ or $E$ is projective.

Proof.

$(1)\Rightarrow(2)$ Clear.

$(2)\Rightarrow(3)$ Assume that $\operatorname{Rad}(E)\neq E$. Then $E$ has a simple factor module isomorphic to $R/I$. Let $f:E\rightarrow R/I$ be a nonzero homomorphism. Since $E$ is max-projective, there exists a homomorphism $g:E\rightarrow R$ such that $\operatorname{Im}(g)\neq 0$. By the fact that $R$ is right Hereditary, $\operatorname{Im}(g)$ is projective, whence $E\cong\operatorname{Im}(g)\oplus K$ for some right $R$-module $K$. Since $E$ is indecomposable, either $K=0$ or $\operatorname{Im}(g)=0$, where the latter case implies that $g=0$ which is a contradiction. In the former case $K=0$, implying that $E$ is projective.

$(3)\Rightarrow(1)$ Conversely, if $E$ is projective then $E$ is clearly $R$-projective. Now suppose $\operatorname{Rad}(E)=E$ and let $f:E\rightarrow R/I$ be a homomorphism. Then $f(E)=f(\operatorname{Rad}(E))\subseteq\operatorname{Rad}(R/I)\ll R/I$. Moreover $f(E)$ is a direct summand of $R/I$ since $R$ is right Hereditary. Therefore $f(E)=0$, and so $f$ can be lifted to $R$. ∎

Lemma 6**.**

(See [22, 3.3]) For a ring $R$ the following are equivalent.

1. (1)

$R$* is a right small ring.* 2. (2)

$\operatorname{Rad}(E)=E$* for every injective right $R$-module $E$.* 3. (3)

$\operatorname{Rad}(E(R))=E(R)$.

Corollary 7**.**

If $R$ is a right Hereditary right small ring, then $R$ is right almost-$QF$.

Proposition 14**.**

If $R$ is a right semihereditary right small ring, then $\operatorname{Hom}(E,R)=0$, for any injective right $R$-module $E$. In particular, $R$ is right almost-$QF$ if and only if $\operatorname{Hom}(E,R/I)=0$ for any right ideal $I$ of $R$.

Proof.

Let $E$ be an injective right $R$-module and $f\in\operatorname{Hom}(E,R)$. Then $f(E)=f(\operatorname{Rad}(E))\subseteq J(R)$. Since $R$ is right semihereditary, $f(E)$ is absolutely pure. This means that $R/f(E)$ is flat by [18, Corollary 4.86]. Then, by [18, §4 Exercise 20], $f(E)=0$, i.e. $\operatorname{Hom}(E,R)=0$. Hence, the rest is clear. ∎

Recall that by Example 1$(d)$, any right small ring $R$ is right max-$QF$. Moreover, if $R$ is right Noetherian, we have the following.

Proposition 15**.**

If $R$ is a right Noetherian and right small ring, then $R$ is right almost-$QF$.

Proof.

Let $E$ be an injective right $R$-module. Then, by Lemma 6, $\operatorname{Rad}(E)=E$. Now let $f:E\rightarrow R/I$ be a homomorphism for any right ideal $I$ of $R$. This implies that $f(E)\subseteq R/I$ and since $\operatorname{Rad}(E)=E$, we have $\operatorname{Rad}(f(E))=f(E)$. By the right Noetherian assumption, $R/I$ is a Noetherian right $R$-module and its submodule $f(E)$ is finitely generated, i.e. $\operatorname{Rad}(f(E))\neq f(E)$. Also since $\operatorname{Rad}(f(E))=f(E)$, this means that $f(E)=0$, whence $f:E\rightarrow R/I$ can be lifted to $R$. Consequently, $E$ is $R$-projective. ∎

Theorem 1**.**

Let $R$ be a right Hereditary and right Noetherian ring. Then the following are equivalent.

1. (1)

$R$* is right almost-$QF$.* 2. (2)

$R$* is right max-$QF$.* 3. (3)

Every injective right $R$-module $E$ has a decomposition $E=A\oplus B$ where $\operatorname{Rad}(A)=A$ and $B$ is projective and semisimple. 4. (4)

$R=S\times T$, where $S$ is a semisimple Artinian ring and $T$ is a right small ring.

Proof.

$(1)\Rightarrow(2)$ Clear.

$(2)\Rightarrow(3)$ Let $E$ be an injective right $R$-module. Then $E$ has an indecomposable decomposition $E=\oplus_{i\in\Gamma}A_{i}$ where $A_{i}$’s are either projective or $\operatorname{Rad}(A_{i})=A_{i}$ by proposition 13. Let $\Lambda=\{j\in\Gamma:A_{j}$ is projective}. So the decomposition of $E$ can be written as $E=(\oplus_{j\in\Lambda}A_{j})\oplus(\oplus_{i\in\Gamma-\Lambda}A_{i})$. We claim that each $A_{j}$ is simple for $j\in\Lambda$. Since $A_{j}$ is projective for $j\in\Lambda$, $\operatorname{Rad}(A_{j})\neq A_{j}$. So there exists a simple factor $B_{j}$ of $A_{j}$ i.e. $B_{j}\cong A_{j}/N\cong R/I$ for some maximal submodule $N$ of $A_{j}$ and for some maximal right ideal $I$ of $R$. Since $B_{j}$ is injective, by $(2)$, the following diagram commutes.

[TABLE]

With the Hereditary assumption on $R$, $\operatorname{Im}(g)\cong B_{j}$ is projective and so $A_{j}\cong N\oplus B_{j}$. However $A_{j}$ is indecomposable, whence $N=0$. Consequently, each $A_{j}$ is simple for $j\in\Lambda$.

$(3)\Rightarrow(1)$ Let $E$ be an injective right $R$-module. By the assumption, $E=A\oplus B$ where $\operatorname{Rad}(A)=A$ and $B$ is semisimple and projective. Since $B$ is $R$-projective, we only need to show that $A$ is $R$-projective. By the Noetherian assumption, the injective $R$-module $A$ has a decomposition $A=\oplus_{i\in\Gamma}A_{i}$ where each $A_{i}$ is indecomposable injective with $\operatorname{Rad}(A_{i})=A_{i}$. Proposition 13 implies that each $A_{i}$ is $R$-projective, whence $A$ is $R$-projective by Lemma 2. Therefore, $M=A\oplus B$ is $R$-projective by Lemma 2.

$(2)\Rightarrow(4)$ Let $S$ be the sum of minimal injective right ideals of $R$. Then $S$ is injective since $R$ is right Noetherian. Thus we have the decomposition $R=S\oplus T$ for some right ideal $T$ of $R$ such that $\operatorname{Soc}(S)=S$ and $T$ has no simple injective submodule. If $f:S\rightarrow T$ is a nonzero homomorphism, then $f(\operatorname{Soc}(S))=f(S)\subseteq\operatorname{Soc}(T)$, where $f(S)$ is injective by the Hereditary assumption, and so $\operatorname{Soc}(T)$ contains a semisimple injective direct summand $f(S)$. This means that $f(S)=0$, a contradiction. Thus, we have $\operatorname{Hom}(S,T)=0$ , and so $S$ is a two sided ideal. On the other hand, if $g:T\rightarrow S$ is a nonzero homomorphism, then $T/\operatorname{Ker}(g)\cong\operatorname{Im}(g)\subseteq S$, and so $T/\operatorname{Ker}(g)$ is projective by Hereditary assumption. Also since $S$ is a semisimple injective $R$-module, $T/\operatorname{Ker}(g)$ is semisimple injective, whence $K/\operatorname{Ker}(g)$ is semisimple injective for any maximal submodule $K/\operatorname{Ker}(g)$ of $T/\operatorname{Ker}(g)$. This implies that $T/\operatorname{Ker}(g)\cong K/\operatorname{Ker}(g)\oplus T/K$. Then the simple $R$-module $T/K$ is injective and projective, and so $T$ contains an isomorphic copy of a simple injective $R$-module $T/K$, yielding a contradiction. Therefore, $\operatorname{Hom}(T,S)=0$, and so $T$ is a two sided ideal. Consequently, $R=S\oplus T$ is a ring decomposition. Now let $E(T)$ be the injective hull of $T$ as an $R$-module. The injective hull $E(T)$ is also a $T$-module by the fact that $E(T)S=0$. We claim that $\operatorname{Rad}(E(T))=E(T)$. Suppose the contrary and let $K$ be a maximal submodule of $E(T)$. Then $E(T)/K$ is injective by the Hereditary assumption and it is max-projective by (2). Since $E(T)/K$ is a simple right $R$-module, it is isomorphic to $R/I$ for some maximal right ideal $I$ of $R$, and so $R/I$ is injective. Then, the isomorphism $\alpha:E(T)/K\rightarrow R/I$ lifts to $\beta:E(T)/K\rightarrow R$ i.e. the following diagram commutes.

[TABLE]

Since $\beta$ is monic and $E(T)/K$ injective, $U=\beta(E(T)/K)$ is a direct summand of $R$. It is easy to see that $U$ is also a right $T$-module and so $U\subseteq T$. On the other hand, since $U$ is minimal and injective, $U$ is also contained in $S$, a contradiction. So we must have $\operatorname{Rad}(E(T))=E(T)$, whence $T\ll E(T)$ by Lemma 6. This proves (4).

$(4)\Rightarrow(1)$ Clear, by Lemma 5 and Proposition 15.

∎

Theorem 2**.**

Let $R$ be a right Hereditary ring. Then the following are equivalent.

1. (1)

$R$* is right max-$QF$.* 2. (2)

Every simple injective right $R$-module is projective. 3. (3)

Every singular injective right $R$-modules is $R$-projective. 4. (4)

Every singular injective right $R$-modules is max-projective. 5. (5)

$\operatorname{Rad}(E)=E$* for every singular injective right $R$-module $E$.* 6. (6)

Every injective right $R$-module $E$ can be decomposed as $E=Z(E)\oplus F$ with $\operatorname{Rad}(Z(E))=Z(E)$.

Proof.

$(1)\Rightarrow(4)$, $(3)\Rightarrow(4)$ and $(6)\Rightarrow(5)$ are clear.

$(4)\Rightarrow(2)$ Let $S$ be a simple injective right $R$-module. We claim that $S$ is projective. Assume that $S$ is not projective. Then it is singular and injective. This implies, by our hypothesis that $S$ is max-projective, hence $S$ is projective, this is a contradiction. The conclusion now follows.

$(2)\Rightarrow(1)$ Let $E$ be an injective right $R$-module and $f:E\rightarrow S$ with $S$ is a simple right $R$-module. If $f=0$, there is nothing to prove. We may assume that $f$ is a nonzero homomorphism, and so $f$ is an epimorphism. Since $R$ is right Hereditary, $S$ is injective, and so by (2), $S$ is projective. Hence, the natural epimorphism $\pi:R\rightarrow S$ splits, i.e. there exists a homomorphism $\eta:S\rightarrow R$ such that $\pi\eta=1_{S}$. Then, $\pi\eta f=f$, and so $E$ is max-projective.

$(4)\Rightarrow(5)$ Let $E$ be a singular injective right $R$-module. Assume to the contrary that $E$ has a maximal submodule $K$ such that $E/K\cong R/I$ for some maximal right ideal $I$ of $R$. So, there is a nonzero homomorphism $f:E\rightarrow R/I$, and by (4), there exists a nonzero homomorphism $g:E\rightarrow R$ such that $\pi g=f$, where $\pi:R\rightarrow R/I$ is the canonical epimorphism. Since $E$ is singular, $\operatorname{Im}(g)$ is singular. Moreover, $\operatorname{Im}(g)\subseteq R$, and so $\operatorname{Im}(g)$ is nonsingular. This implies that $g(E)=0$, yielding a contradiction.

$(5)\Rightarrow(6)$ Let $E$ be an injective right $R$-module. Since $R$ is a right nonsingular ring, $Z(E)$ is a closed submodule of $E$, and so $E=Z(E)\oplus F$ for some submodule $F$ of $E$. Then, by $(5)$, $\operatorname{Rad}(Z(E))=Z(E)$.

$(5)\Rightarrow(3)$ Let $E$ be a singular injective right $R$-module. This implies, by our hypothesis, that $\operatorname{Rad}(E)=E$. Let $f:E\rightarrow R/I$ be homomorphism for some right ideal $I$ of $R$. Since $\operatorname{Rad}(E)=E$ and $\operatorname{Rad}(R/I)\neq R/I$, $f:E\rightarrow R/I$ is not an epimorphism. By the right Hereditary assumption, $f(E)$ is injective, and so $f(E)$ is a direct summand of $R/I$. But since $f(E)\subseteq\operatorname{Rad}(R/I)$, we must have $f(E)\ll R/I$. This means, $f(E)=0$, whence $\operatorname{Hom}(E,R/I)=0$ for each right ideal $I$ of $R$. Therefore, $E$ is $R$-projective. ∎

Proposition 16**.**

Let $R$ be a local right max-$QF$ ring. Then $R$ is either right self-injective or right small.

Proof.

Let $J$ be the unique maximal right ideal of $R$ and $E$ be the injective hull of the ring $R$. Assume first that $R$ is not a small ring i.e. $\operatorname{Rad}(E)\neq E$. Then $E$ has a maximal submodule $K$ such that $\frac{E}{K}$ is isomorphic to $\frac{R}{J}$ and denote this isomorphism by $f$. Consider the composition $f\pi$ where $\pi:E\rightarrow\frac{E}{K}$ is the canonical projection. Since $R$ is right max-$QF$, there is a nonzero homomorphism $g:E\rightarrow R$ such that

[TABLE]

commutes. Furthermore, $h$ is a small epimorphism and $f\pi$ is an epimorphism, which means $g:E\rightarrow R$ is also an epimorphism and splits. Thus, $E\cong R\oplus T$ for some $T$. Hence, $R$ is a right self injective ring. ∎

Corollary 8**.**

Let $R$ be a commutative semiperfect ring. If $R$ is max-$QF$, then $R=S\times T$ where $S$ is self-injective and $T$ is small.

Proof.

Let $R$ be a commutative semiperfect ring, then by [19, Theorem 23.11], $R=R_{1}\times...\times R_{n}$, where $R_{i}$ is a local ring $(1\leq i\leq n)$. Hence, by Lemma 5 and Proposition 16, $R$ can be written as a direct product of local max-$QF$ rings and every local max-$QF$ ring either self-injective or small. ∎

Corollary 9**.**

Let $R$ be a right Noetherian local ring. Then the following are equivalent.

1. (1)

$R$* is right almost-$QF$.* 2. (2)

$R$* is right max-$QF$.* 3. (3)

$R$* is $QF$ or right small.*

Proof.

$(1)\Rightarrow(2)$ Clear. $(3)\Rightarrow(1)$ Follows from Proposition 15.

$(2)\Rightarrow(3)$ Clear by Proposition 16, since right Noetherian right self-injective rings are $QF$. ∎

We do not know whether every right chain ring is almost-$QF$. But the following result will imply that each right chain ring with $P(R)=0$ is right almost $QF$.

Proposition 17**.**

Let $R$ be a right chain ring and $J=J(R)$. Then $P(R)=\bigcap_{n\geq 1}J^{n}$.

Proof.

Assume first that $J^{n}=0$ for some $n\in\mathbb{Z}^{+}$. Then $\bigcap_{n\geq 1}J^{n}=0$, and so, by [13, Proposition 5.3(b)], $R$ is a right Noetherian ring with $P(R)=0$. On the other hand if we suppose that $J^{n}\neq 0$ for all $n\in\mathbb{Z}^{+}$, then, by [13, Proposition 5.2(d)], $A=\bigcap_{n\geq 1}J^{n}$ is a completely prime ideal. Let us now look at the case $A\neq AJ$. Then $\frac{A}{AJ}$ simple right $R$-module and $AJ\ll A$. Let $a\in A\setminus AJ$. If we have $A=aR+AJ$, then $A=aR$, whence either $A=J(R)$ or $A=0$, by [13, Proposition 5.2(f)]. If $A=\bigcap_{n\geq 1}J^{n}=0$, then $R$ is a right Noetherian ring with $P(R)=\bigcap_{n\geq 1}J^{n}=0$. Otherwise, if $A=J(R)=\bigcap_{n\geq 1}J^{n}$, then $J=J^{2}$, but since $A\neq AJ$, this is not the case. If we look at the case $A=AJ$, then $A\subseteq P(R)$. Since $P(R)=P^{2}(R)$, $P(R)$ is a completely prime ideal of $R$, and so, by [13, Lemma 5.1], $P(R)\subseteq A$. Hence, $P(R)=A=\bigcap_{n\geq 1}J^{n}$. ∎

Corollary 10**.**

Let $R$ be a right chain ring. Then $R/P(R)$ is a right almost-$QF$ ring.

Proof.

Since $P(R)$ is an ideal of $R$, and every factor ring of a right chain ring is a right chain ring, without loss of generality we may assume that $P(R)=0$. Then by Proposition 17 and [13, Proposition 5.3], $R$ is a right Noetherian ring. We have two cases for $J=J(R)$: if $J$ is nilpotent, then $R$ is Artinian. This implies that $R$ is right self-injective by [13, Lemma 5.4] which then yields, $R$ is $QF$. So now assume that $J$ is not nilpotent. Then $R$ is a domain by [13, Proposition 5.2(d)], whence $R$ is right small. So, $R$ is right almost-$QF$ by Proposition 15. Thus in any case $R$ is right almost-$QF$. ∎

We shall characterize commutative Noetherian max-$QF$ rings.

Proposition 18**.**

(See [20]) Let $R$ be a commutative Noetherian ring, $P$ be a prime ideal of $R$, $E=E(R/P)$, and $A_{i}=\{x\in E:P^{i}x=0\}$. Then:

1. (1)

$A_{i}$* is a submodule of $E$, $A_{i}\subseteq A_{i+1}$, and $E=\bigcup A_{i}$.* 2. (2)

If $P$ is a maximal ideal of $R$, then $A_{i}\subseteq E(R/P)$ is a finitely generated $R$-module for every integer $i$. 3. (3)

$E(R/P)$* is Artinian.*

Lemma 7**.**

Let $R$ be a commutative Noetherian ring, and let $E=E(R/Q)$ for a maximal ideal $Q$ of $R$. The following are equivalent.

1. (1)

$E$* is $R$-projective.* 2. (2)

$E$* is max-projective.* 3. (3)

$\operatorname{Rad}(E)=E$* or $E$ is projective, local and isomorphic to an ideal of $R$.*

Proof.

$(1)\Rightarrow(2)$ is clear.

$(2)\Rightarrow(3)$ Assume that $\operatorname{Rad}(E)\neq E$. Since $R$ is commutative, $\operatorname{Rad}(E)=\bigcap_{i\in\wedge}IE$, where $\wedge$ is the set of all maximal ideals of $R$, [6, Exercises 15.(5)]. Now we will see that $IE=E$ for any maximal ideal $I$ distinct from $Q$. Let $I$ be a maximal ideal distinct from $Q$. The fact $I+Q=R$ implies $I+Q^{n}=R$ for any $n\in\mathbb{N}$. Let $x\in E$. Then $Q^{n}x=0$ for some $n\in\mathbb{N}$, by Proposition 18. We have $1=y+z$, where $y\in I$, $z\in Q^{n}$, and then $x=yx\in IE$. Hence, $\operatorname{Rad}(E)=\bigcap_{i\in\wedge}IE=QE\neq E$. Since $R$ is commutative and $(E/QE)Q=0$, $E/QE$ is a semisimple $R/Q$-module, and so $E/QE$ semisimple as an $R$-module. Then $E/QE$ is finitely generated by Artinianity of $E$, and hence $QE+K=E$ for some finitely generated submodule $K$ of $E$. Since $K$ is finitely generated, $K$ is a submodule of $A_{n}$ for some $n$, by Proposition 18. Thus $Q^{n}K=0$. Since $QE+K=E$ , $Q^{n+1}E=Q^{n}E$, implying $Q^{n}E\subseteq P(E)$. On the other hand, $Q^{2}E+QK=QE$, and so $Q^{2}E+K=E$. Continuing in this manner $Q^{n}E+K=E$, whence $E/Q^{n}E$ is finitely generated. Since $R$ is Noetherian, $P(E/Q^{n}E)=0$ and so $P(E)=Q^{n}E$. Since $E/P(E)$ is finitely generated, $E/P(E)$ has finite composition length by Proposition 18(3). By max-projectivity of $E$ and Lemma 3, $E/P(E)$ is max-projective. Thus $E/P(E)$ is projective by Corollary 3. Then, $E=P(E)\oplus L$ for some projective submodule $L$ of $E$. Since $E$ is indecomposable and $P(E)\neq E$, $E=L$. Therefore $E$ is projective. Furthermore, since $E$ is indecomposable, the endomorphism ring of $E$ is local by [13, Lemma 2.25]. By [26, Theorem 4.2], $E$ is a local module, so it is cyclic and $R\cong E\oplus I$ for some ideal $I$ of $R$. Hence $E$ is isomorphic to an ideal of $R$. This proves $(3)$.

$(3)\Rightarrow(1)$ is obvious. ∎

Lemma 8**.**

(See [17, 9.7]) Suppose $R$ commutative Noetherian or semilocal right Noetherian ring and $\{M_{i}\}_{i\in I}$ be a class of right $R$-modules. Then $\operatorname{Rad}(\prod_{i\in I}M_{i})=\prod_{i\in I}\operatorname{Rad}(M_{i})$.

Lemma 9**.**

Let $R$ be a commutative Noetherian ring. Then the following are equivalent.

1. (1)

$R$* is a small ring, i.e., $R\ll E(R)$.* 2. (2)

$\operatorname{Rad}(E(S))=E(S)$* for each simple $R$-module $S$.*

Proof.

$(1)\Rightarrow(2)$: Clear by Lemma 6.

$(2)\Rightarrow(1)$: Let $\Delta$ be a complete set of representatives of simple $R$-modules. Then $C=\oplus_{S\in\Delta}E(S)$ is an injective cogenerator. Then, for some index set $I$, the injective hull $E(R)$ of $R$ is a direct summand of $C^{I}$. By Lemma 8, $\operatorname{Rad}(C^{I})=C^{I}$. Since $E(R)$ is a direct summand of $C^{I}$, we have $\operatorname{Rad}(E(R))=E(R)$. Thus $R$ is a small ring by Lemma 6. ∎

Theorem 3**.**

Let $R$ be a commutative Noetherian ring. Then the following are equivalent.

1. (1)

$R$* is almost-$QF$.* 2. (2)

$R$* is max-$QF$.* 3. (3)

$R=A\times B$, where $A$ is $QF$ and $B$ is small.

Proof.

$(1)\Rightarrow(2)$ is clear.

$(2)\Rightarrow(3)$ First suppose that $\operatorname{Rad}(E(S))=E(S)$ for all simple $R$-module $S$. Then $R$ is a small ring by Lemma 9. On the other hand, if $\operatorname{Rad}(E(S))\neq E(S)$ for some simple $R$-module $S$, then $E(S)$ is isomorphic to a direct summand of $R$ by Lemma 7. Let $X$ be sum of minimal ideals $U$ of $R$ with $\operatorname{Rad}(E(U))\neq E(U)$. Then $E(U)$ is isomorphic to an ideal of $R$. Thus without loss of generality we can assume that $E(U)$ is an ideal of $R$. Since $R$ is Noetherian, $X$ is finitely generated, and so $A=E(X)=E(U_{1})\oplus\cdots E(U_{n})$ where each $E(U_{i})$ is an ideal of $R$. Thus $R=A\oplus B$ for some ideal $B$ of $R$. Now $A$ is injective and Noetherian, so $A$ is a $QF$ ring. On the other hand, let $V$ be a simple $B$-module, then $V$ is a simple $R$-module. Let $E(V)$ be the injective hull of $V$. As $V$ is a $B$-module, $VA=0$. If $\operatorname{Rad}(E(V))\neq E(V)$, then this would imply $V\subseteq A$, by the same arguments above. Thus $\operatorname{Rad}(E(V))=E(V)$, and so $B$ is a small ring by Lemma 9.

$(3)\Rightarrow(1)$ Clear, by Proposition 15 and Lemma 5. ∎

Proposition 19**.**

Let $R$ be a semiperfect ring. Then the following are equivalent.

1. (1)

$R$* is right almost-$QF$ and direct sum of small right $R$-modules is small.* 2. (2)

$R$* is right max-$QF$ and direct sum of small right $R$-modules is small.* 3. (3)

$R$* is right almost-$QF$ and $\operatorname{Rad}(Q)\ll Q$ for each injective right $R$-module $Q$.* 4. (4)

$R$* is right max-$QF$ and $\operatorname{Rad}(Q)\ll Q$ for each injective right $R$-module $Q$.* 5. (5)

$R$* is $QF$.*

Proof.

$(1)\Rightarrow(2)$ and $(3)\Rightarrow(4)$ Clear. $(2)\Rightarrow(3)$ and $(3)\Rightarrow(1)$ By [23, Lemma 9].

$(4)\Rightarrow(5)$ Let $M$ be an injective right $R$-module. Since $M$ is max-projective with $\operatorname{Rad}(M)\ll M$, by Proposition 8$(1)$, $M$ is projective. Hence $R$ is $QF$.

$(5)\Rightarrow(3)$ Let $M$ be an injective right $R$-module. By the hypothesis, $M$ is projective. Since $R$ is right Artinian, every right $R$-module has a small radical, whence $\operatorname{Rad}(M)\ll M$. ∎

In [10], a submodule $N$ of a right $R$-module $M$ is called coneat in $M$ if $\operatorname{Hom}(M,S)\rightarrow\operatorname{Hom}(N,S)$ is epic for every simple right $R$-module $S$. In [9], $N$ is called s-pure in $M$ if $N\otimes S\rightarrow M\otimes S$ is monic for every simple left $R$-module $S$. M is absolutely coneat (resp., absolutely s-pure) if $M$ is coneat (resp., s-pure) in every extension of it. If $R$ is commutative, then s-pure short exact sequences coincide with coneat short exact sequences, [15, Proposition 3.1].

Proposition 20**.**

Consider the following conditions for a ring R:

1. (1)

$R$* is right max-$QF$.* 2. (2)

Every absolutely coneat right $R$-module is max-projective. 3. (3)

Every absolutely s-pure right $R$-module is max-projective. 4. (4)

Every absolutely pure right $R$-module is max-projective.

Then $(3)\Rightarrow(4)\Rightarrow(1)\Rightarrow(2)$. Also, if $R$ is a commutative ring, then $(2)\Rightarrow(3)$.

Proof.

$(3)\Rightarrow(4)\Rightarrow(1)$ Clear.

$(1)\Rightarrow(2)$ Let $M$ be an absolutely coneat right $R$-module. Consider the following diagram:

[TABLE]

where $S$ is a simple right $R$-module, $i:M\rightarrow E(M)$ is the inclusion map and $\pi:R\rightarrow S$ is the canonical quotient map. Since $M$ coneat in $E(M)$, there is a homomorphism $g:E(M)\rightarrow S$ such that $gi=f$. Also, by $(1)$, there exists a homomorphism $h:E(M)\rightarrow R$ such that $\pi h=g$. Hence, $(\pi h)i=gi=f$.

$(2)\Rightarrow(3)$ Let $M$ be an absolutely s-pure right $R$-module. Then $M$ is s-pure in $E(M)$. Since $R$ is commutative, $M$ is coneat in $E(M)$. Hence, $M$ is max-projective by $(2)$. ∎

In [21, Lemma 1.16], it was shown that for a projective module $M$, if $M=P+K$, where $P$ is a summand of $M$ and $K\subseteq M$, then there exists a submodule $Q\subseteq K$ with $M=P\oplus Q$. By using the same method in the proof of [5, Theorem 2.8], one can prove the following result.

Proposition 21**.**

A ring $R$ is right almost-$QF$ if and only if for every injective right $R$-module $E$, if $E=P+L$, where $P$ is a finitely generated projective summand of $E$ and $L\subseteq E$, then $E=P\oplus K$ for some $K\subseteq L$.

Let R be a ring and $\Omega$ be a class of $R$-modules which is closed under isomorphic copies. Following Enochs, a homomorphism $\varphi:G\rightarrow M$ with $G\in\Omega$ is called an $\Omega$-precover of the $R$-module $M$ if for each homomorphism $\psi:H\rightarrow M$ with $H\in\Omega$, there exists $\lambda:H\rightarrow G$ such that $\varphi\lambda=\psi$.

Lemma 10**.**

Let $R$ be a right self-injective ring. Then the following are equivalent.

1. (1)

$R$* is right almost-$QF$.* 2. (2)

Every finitely generated right $R$-module has an injective precover which is $R$-projective. 3. (3)

Every cyclic right $R$-module has an injective precover which is $R$-projective.

Proof.

$(1)\Rightarrow(2)$ Let $M$ be a finitely generated right $R$-module and $g:R^{n}\rightarrow M$ be an epimorphism. For any homomorphism $f:E\rightarrow M$ with $E$ is injective, there exists $h:E\rightarrow R^{n}$ such that $gh=f$. Since $R^{n}$ is injective, $g$ is an injective precover of $M$.

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

$(3)\Rightarrow(1)$ Let $E$ be an injective right $R$-module and $I$ be a right ideal of $R$. Suppose that $f:E\rightarrow R/I$ is a homomorphism, $\pi:R\rightarrow R/I$ is the natural epimorphism and $g:G\rightarrow R/I$ be an injective cover of $R/I$. So, there is $h:E\rightarrow G$ such that $gh=f$. By hypothesis, $G$ is $R$-projective and hence there is $k:G\rightarrow R$ such that $\pi k=g$. Let $\overline{f}=kh$. So $\pi\overline{f}=\pi kh=gh=f$. Therefore, E is $R$-projective, and so $R$ is right almost-$QF$. ∎

In [12], a module $M$ is said to be copure-injective if $\operatorname{Ext}^{1}_{R}(E,M)=0$ for any injective module $E$. Now we give the characterization of almost-$QF$ rings in terms of copure-injective modules.

Proposition 22**.**

Let $R$ be a ring. Then the followings are equivalent.

1. (1)

$R$* is right almost-$QF$ and $R_{R}$ is copure-injective.* 2. (2)

Every right ideal of $R$ is copure-injective. 3. (3)

Every submodule of a finitely generated projective right $R$-module is copure injective.

Proof.

$(1)\Rightarrow(2)$ Let $E$ be an injective right $R$-module and $I$ be a right ideal of $R$. By applying $\operatorname{Hom}(E,-)$ to the short exact sequence $0\rightarrow I\rightarrow R\rightarrow R/I\rightarrow 0$, we obtain the following exact sequence: $0\rightarrow\operatorname{Hom}(E,I)\rightarrow\operatorname{Hom}(E,R)\rightarrow\operatorname{Hom}(E,R/I)\rightarrow\operatorname{Ext}^{1}_{R}(E,I)\rightarrow\operatorname{Ext}^{1}_{R}(E,R)\rightarrow...$. Since $R_{R}$ is copure-injective, $\operatorname{Ext}^{1}_{R}(E,R)=0$. Then the map $\operatorname{Hom}(E,R)\rightarrow\operatorname{Hom}(E,R/I)$ is onto since $E$ is $R$-projective. Hence, $\operatorname{Ext}^{1}_{R}(E,I)=0$ for any injective $R$-module $E$.

$(2)\Rightarrow(3)$ Suppose that every right ideal of $R$ is copure-injective. First, by induction, we show that every submodule of $R^{n}$ is copure-injective. The case $n=1$ follows by the hypothesis. Now suppose that $n>1$ and every submodule of $R^{n-1}$ is copure-injective. Let $N$ be a submodule of $R^{n}$, and consider the exact sequence $0\rightarrow N\cap R^{n-1}\rightarrow N\rightarrow N/(N\cap R^{n-1})\rightarrow 0$. By induction hypothesis, $N\cap R^{n-1}$ is copure-injective, and $N/(N\cap R^{n-1})\cong(N+R^{n-1})/R^{n-1}\subseteq R^{n}/R^{n-1}\cong R$ is also copure-injective. Therefore, for any injective right $R$-module $E$, consider the exact sequence $\operatorname{Ext}^{1}_{R}(E,N\cap R^{n-1})\rightarrow\operatorname{Ext}^{1}_{R}(E,N)\rightarrow\operatorname{Ext}^{1}_{R}(E,N/(N\cap R^{n-1}))$. Since $\operatorname{Ext}^{1}_{R}(E,N\cap R^{n-1})=\operatorname{Ext}^{1}_{R}(E,N/(N\cap R^{n-1}))=0$, we have $\operatorname{Ext}^{1}_{R}(E,N)=0$. Therefore, $N$ is copure-injective. Now if $M$ is a submodule of a finitely generated projective right $R$-module $P$, then there is $n\geq 1$ such that $M\subseteq P\subseteq R^{n}$. By the above observation, $M$ is also copure-injective. $(3)\Rightarrow(2)$ is clear. $(2)\Rightarrow(1)$ by Proposition 2. ∎

Proposition 23**.**

Let $R$ be a ring. Then the following are equivalent.

1. (1)

$R$* is semisimple.* 2. (2)

$R$* is right almost-$QF$ right V-ring.* 3. (3)

$R$* is right almost-$QF$ and every submodule of an $R$-projective right module is $R$-projective.* 4. (4)

$R$* is right self-injective and every submodule of an $R$-projective right module is $R$-projective.*

Proof.

$(1)\Rightarrow(2)$, $(1)\Rightarrow(3)$ and $(1)\Rightarrow(4)$ are clear.

$(2)\Rightarrow(1)$ Since $R$ is a right $V$-ring, every simple right $R$-module is injective. By the hypothesis, every simple right $R$-module is $R$-projective, whence projective.

$(4)\Rightarrow(1)$ Let $M$ be a cyclic right $R$-module and $I$ a right ideal of $R$. Consider the following diagram:

[TABLE]

where $i:I\rightarrow R$ is the inclusion map and $\pi:R\rightarrow M$ is the canonical quotient map. Since $I$ is $R$-projective there exists $h:I\rightarrow R$ such that $\pi h=f$. By the injectivity of $R$, there exists $\lambda:R\rightarrow R$ such that $\lambda i=h$. Then $(\pi\lambda)i=\pi h=f$, and $\pi\lambda:R\rightarrow M$ is the required map.

$(3)\Rightarrow(1)$ Since every simple right $R$-module can be embedded in an injective $R$-module, every simple right $R$-module is $R$-projective, and so every simple right $R$-module is projective. Hence, $R$ is semisimple. ∎