Free Covers and Minimal Sets of Generators

Abstract

We continue our study of the relation between existence of projective covers and existence of minimal sets of generators for right modules over an arbitrary ring.

1 Introduction

In our previous papers [3, 4], we studied the relations between the existence of projective covers and the existence of minimal sets of generators for right modules over a local ring. Right perfect rings appear to have a particularly good behaviour for the existence of minimal sets of generators. In this paper, we extend part of the results obtained in those two papers to the case of right modules over an arbitrary ring, not necessarily local.

To this end, it is necessary to replace the notion of projective cover with that of free cover, that is, a projective cover that is a free module (Proposition 2.1(1)). In fact, we prove that over an arbitrary ring, the relation is between the existence of free covers and the existence of minimal sets of generators. This explains why right perfect rings appear so naturally in the study of existence of minimal sets of generators in the local case, in which projective modules are free, and free covers always exist. As in our previous papers [3, 4], our results are heavily based on results by Hrbek and Růžička [7, 8] and on the article [9].

Similarly to [3, 4], we consider three classes of right R-modules: the class ${\mathcal {A}}$ of all right R-modules with a free cover; the class ${\mathcal {B}}$ of right R-modules for which every set of generators contains a minimal set of generators; and the class ${\mathcal {C}}$ of all right R-modules with a minimal set of generators. One has that ${\mathcal {A}}\cup {\mathcal {B}}\subseteq {\mathcal {C}}$. The class ${\mathcal {A}}\cap {\mathcal {B}}$ consists of all finitely generated right R-modules $M_R$ with M/MJ(R) a free right R/J(R)-module if the ring R is not right perfect (Theorem 5.4). If R is local and right perfect, then ${\mathcal {A}}={\mathcal {B}}={\mathcal {C}}={\text {Mod-\!}}R$ [3, Theorem 4.3]. In the case of R right perfect but not local, little is known [8, Problem 5.1]. We still even don’t know whether every set of generators of a free right module over a semisimple artinian ring always contains a minimal set of generators.

The rings we consider are associative rings R with identity $1_R$. The Jacobson radical of R is denoted by J(R).

2 Free Covers

2.1 Elementary Facts

A submodule N of a module $M_R$ is superfluous (or small, or inessential) in $M_R$ if, for every submodule L of $M_R$, $N+L=M_R$ implies $L=M_R$. An epimorphism $g:M_R \rightarrow N_R$ is superfluous if $\ker g$ is a superfluous submodule of $M_R$. It is easy to see that an epimorphism $g:M \rightarrow N$ is superfluous if and only if for every module L and every homomorphism $h:L \rightarrow M$, if gh is onto, then h is onto. A projective cover of a module $M_R$ is a pair $(P_R,p)$ where $P_R$ is a projective right R-module and $p:P \rightarrow M$ is a superfluous epimorphism.

Let (P, p) be a projective cover of a right R-module M. If Q is a projective module and $q:Q \rightarrow M$ is an epimorphism, then Q has a direct-sum decomposition $Q=P' \oplus P''$ where $P' \cong P$, $P'' \subseteq \ker (q)$ and $(P',q|_{P'}:P' \rightarrow M)$ is a projective cover. Projective covers don’t exist in general, but when they exist, they are unique up to isomorphism in the following sense. If (P, p), (Q, q) are any two projective covers of a right R-module M, there is an isomorphism $h:Q \rightarrow P$ such that $p \circ h=q$.

The notion of projective cover has been extended to that of ${\mathcal {C}}$-cover, replacing the class of projective right R-modules with another class ${\mathcal {C}}$ of right R-modules. More precisely, in the standard definition of ${\mathcal {C}}$-precover and ${\mathcal {C}}$-cover ( [2, 6, Section 5.1], [10, Section 1.2]), it is usually required that the class ${\mathcal {C}}$ of right R-modules is closed under isomorphism and direct summands. In this paper, we need free covers and precovers, in which ${\mathcal {C}}$ is the class of all free right R-modules, and therefore some care is needed.

Hence, for us, a free precover of a right module $M_R$ is any module epimorphism $\varphi :F_R\rightarrow M_R$ with $F_R$ a free right R-module. Equivalently, a module morphism $\varphi :F_R\rightarrow M_R$ with $F_R$ a free right R-module is a free precover if and only if for every free R-module $F'_R$ the group morphism ${\text {Hom}}(F'_R,\varphi ):{\text {Hom}}(F'_R,F_R)\rightarrow {\text {Hom}}(F'_R,M_R)$ is a surjective mapping.

A free cover of a right module $M_R$ is any free precover $\varphi :F_R\rightarrow M_R$ such that, for every endomorphism f of $F_R$, the equality $\varphi f=\varphi$ implies that f is an automorphism of $F_R$.

It is easily seen ([1, Section 27], [10, Theorem 1.2.12]) that:

Proposition 2.1

(1) A free cover of a module $M_R$ is exactly a projective cover $\varphi :F_R\rightarrow M_R$ of $M_R$ with $F_R$ a free R-module.

(2) A right R-module has a free cover if and only if it is isomorphic to a module $F_R/S$, where $F_R$ is a free R-module and S is a superfluous submodule of $F_R$.

(3) Every right R-module has a free cover if and only if R is a right perfect ring over which every projective right R-module is free, if and only if $R=0$ or R is a local right perfect ring.

Proof

The statement is trivial, except perhaps for the fact that if $R\ne 0$ is a right perfect ring over which every projective right R-module is free, then R is local. Now if $R\ne 0$ is a right perfect ring over which every projective right R-module is free, R has at least a maximal right ideal M, and the simple module $R_R/M$ has a free cover $R^{(X)}\rightarrow R/M$. This is in particular a projective cover, and the canonical projection $R_R \rightarrow R/M$ is an epimorphism. Hence $R^{(X)}$ is a direct summand of $R_R$, so that $R^{(X)}$ is isomorphic to eR for some non-zero idempotent $e\in R$. The non-zero R-module eR has a maximal superfluous submodule because it is a projective cover of the simple module R/M. Since it is a finitely generated projective module, and it must be free, it must be isomorphic to $R^n_R$ for some $n\ge 0$. But if $R^n_R$ has a maximal superfluous submodule, then $n=1$, and R is a local ring.$\square$

A finitely generated right R-module has a free cover if and only if it is isomorphic to a module $R^n_R/S$, where S is a submodule of $J(R)^n$.

2.2 Superfluous Submodules of Free Modules

We now recall a result due to Sexauer and Warnock [9]. Let R be a ring with identity and X be a set. Consider the endomorphism ring of the free right R-module $F_R=R_R^{(X)}=\bigoplus _{x\in X}xR$. The endomorphism ring $E:={\text {End}}(F_R)$ of the module $F_R$ is isomorphic to the ring of all column-finite $X\times X$ matrices with entries in R. For any such matrix A, let

${\mathcal {A}}(A,x)$ denote the right ideal of R generated by all entries on the x-th row of A. Let $\pi _x:R_R^{(X)}\rightarrow R_R$ denote the canonical projection of $F_R$ onto the x-th direct summand of $R_R^{(X)}$, and, for any column-finite matrix $A\in E$, let $\varphi _A$ indicate the endomorphism of $F_R$ corresponding to the matrix A. Then ${\mathcal {A}}(A,x)=\pi _x(\varphi _A(F_R))$.

An indexed family of right ideals ${\mathcal {I}}_x,\ x\in X$, of R is left vanishing [9] if for every sequence $x_1,x_2,x_3,\dots$ of distinct elements of X and every sequence $a_{i}$ of elements of ${\mathcal {I}}_{x_i}$ there exists a positive integer m such that $a_ma_{m-1}\cdots a_2a_1=0$. When the set X is finite, all families of right ideals indexed in X are left vanishing.

Theorem 2.2

(Sexauer and Warnock [9]) Let R be a ring, X be a set, and $F_R$ the free right R-module $R_R^{(X)}=\bigoplus _{x\in X}xR$. Then an endomorphism $\varphi$ of $F_R$ belongs to the Jacobson radical $J({\text {End}}(F_R))$ if and only if the indexed family of right ideals $\pi _x(\varphi (F_R))$, $x\in X$, is a left vanishing family of right ideals of R and $\pi _x(\varphi (F_R))\subseteq J(R)$ for every $x\in X$.

We thus obtain a characterization of superfluous submodules of $F_R$:

Proposition 2.3

A submodule of the free right module $F_R=R_R^{(X)}=\bigoplus _{x\in X}xR$ over a ring R is a superfluous submodule of $F_R$ if and only if it is contained in a submodule of $F_R$ of the form $\bigoplus _{x\in X}x{\mathcal {I}}_x$ for some left vanishing indexed family $\{\,{\mathcal {I}}_x\mid x\in X\,\}$ of right ideals of R with ${\mathcal {I}}_x\subseteq J(R)$ for every $x\in X$.

The proof of Proposition 2.3 is the same as the proof of [3, Proposition 2.3].

Proposition 2.4

If R is an integral domain, an infinitely generated right R-module has a free cover if and only if it is the direct sum of a free module and a finitely generated module with a free cover. More generally, if R is a ring, P is a completely prime ideal of R and $\{\,{\mathcal {I}}_x\mid x\in X\,\}$ is a left vanishing indexed family of right ideals of R, then ${\mathcal {I}}_x\subseteq P$ for almost all $x\in X$.

Proof

If R is a ring and ${\mathcal {I}}_x\nsubseteq P$ for infinitely many indices $x\in X$, it is easy to construct a sequence $a_n\in {\mathcal {I}}_x\setminus P$ with $a_n\dots a_1\notin P$ for every $n\ge 1$.

Now suppose R an integral domain. Let $M_R$ be an infinitely generated right R-module with a free cover. Then $M_R\cong R^{(X)}_R/S$ for some infinite set X and a superfluous submodule S of $R^{(X)}_R$. Therefore $S\subseteq \bigoplus _{x\in X}x{\mathcal {I}}_x$ for some left vanishing indexed family $\{\,{\mathcal {I}}_x\mid x\in X\,\}$ of right ideals of R with ${\mathcal {I}}_x\subseteq J(R)$ for every $x\in X$. From the previous paragraph (with $P=0$), ${\mathcal {I}}_x=0$ for almost all $x\in X$. So $F:=\{\,x\in X\mid {\mathcal {I}}_x\ne 0\,\}$ is a finite set. Then $M_R$ is isomorphic to the direct sum of the free right R-module $R^{(X\setminus F)}_R$ and the finitely generated module with a free cover $R^{(F)}_R/S$. $\square$

3 Minimal Sets of Generators

A minimal set of generators for a module $M_R$ is a set of generators X for $M_R$ with the property that, for every $x\in X$, the subset $X\setminus \{x\}$ generates a proper submodule of $M_R$.

Let us show that superfluous epimorphisms are exactly the module morphisms that behave well as far as sets of generators and minimal sets of generators are concerned. Of course, free covers are exactly the superfluous epimorphisms $f:A_R\rightarrow B_R$ with $A_R$ free.

Proposition 3.1

[4, Proposition 6.6]. The following conditions are equivalent for a morphism $f:A_R\rightarrow B_R$ between right modules $A_R,B_R$ over an arbitrary ring R:

(a) $f:A_R\rightarrow B_R$ is a superfluous epimorphism.

(b) For every subset X of $A_R$, X generates $A_R$ if and only if f(X) generates $B_R$.

By ${{\,\mathrm{rad}\,}}(A_R)$ we denote the radical of a module $A_R$, that is, the intersection of all its maximal submodules.

Corollary 3.2

Let $f:A_R\rightarrow B_R$ be a superfluous epimorphism between two right modules $A_R,B_R$ over an arbitrary ring R. Then:

(a) The sets of generators of $A_R$ are the subset X of $A_R$ such that f(X) generates $B_R$.

(b) The minimal sets of generators of $A_R$ are the subset X of $A_R$ such that f(X) is a minimal sets of generators of $B_R$ and $f(x)\ne f(x')$ for every pair $x,x'$ of distinct elements of X.

(c) There is a one-to-one correspondence between the set of all maximal submodules M of $A_R$ and the set of all maximal submodules N of $B_R$ given by $M\mapsto f(M)$.

(d) $f({{\,\mathrm{rad}\,}}(A_R))={{\,\mathrm{rad}\,}}(B_R)$ and $f^{-1}({{\,\mathrm{rad}\,}}(B_R))={{\,\mathrm{rad}\,}}(A_R)$.

Proof

(a) and (b) appear in [4, Corollary 6.7]. (c) and (d) follow from the fact that any maximal submodule contains all superfluous submodules [5, proof of Lemma 2.15].$\square$

By (b), if $f:A_R\rightarrow B_R$ is a superfluous epimorphism, the minimal sets of generators of $A_R$ are the subset g(Y), where Y is a minimal set of generators of $B_R$ and $g:Y\rightarrow A_R$ is any mapping such that $fg:Y\rightarrow B_R$ is the inclusion of Y into $B_R$.

The inverse of the one-to-one correspondence in (c) is $N\mapsto f^{-1}(N)$.

Let R be any ring. There are three classes of right R-modules interconnected with each other:

(a) The class ${\mathcal {A}}$ of right R-modules with a free cover.

(b) The class ${\mathcal {B}}$ of right R-modules for which every set of generators contains a minimal set of generators.

(c) The class ${\mathcal {C}}$ of right R-modules with a minimal set of generators.

Trivially, ${\mathcal {B}}\subseteq {\mathcal {C}}$. Also, ${\mathcal {A}}\subseteq {\mathcal {C}}$, as the next Proposition shows. Its proof is easy.

Proposition 3.3

A right R-module $M_R$ has a free cover if and only if it is has a minimal set X of generators and the kernel of the canonical mapping $\varphi :R_R^{(X)}\rightarrow M_R$ is a superfluous submodule of $R_R^{(X)}$.

As far as the modules in the class ${\mathcal {B}}$ are concerned, we have the following Proposition:

Proposition 3.4

Let $M_R$ be a module over an arbitrary ring R, and S be a superfluous submodule of $M_R$. If the R-module $M_R$ has the property that every set of generators contains a minimal set of generators, then the R-module $M_R/S$ has the same property.

Proof

Apply Proposition 3.1 and Corollary 3.2 to the canonical projection $M_R\rightarrow M_R/S$, which is a superfluous epimorphism. $\square$

Similarly:

Theorem 3.5

Let $M_R$ be a module over an arbitrary ring R. If the R-module $M_R$ has the property that every set of generators contains a minimal set of generators, then the R/J(R)-module $M_R/M_RJ(R)$ has the same property.

Proof

Let $M_R$ be a right R-module with the property that every set of generators of $M_R$ contains a minimal set of generators. Let ${\mathcal {F}}\subseteq M_R/M_RJ(R)$ be a set of generators of the R/J(R)-module $M_R/M_RJ(R)$ (equivalently, of the R/J(R)-module $M_R/M_RJ(R)$). Let $g:M_R/M_RJ(R)\rightarrow M_R$ be a mapping with $\pi g=1_{M/MJ(R)}$, where $\pi :M_R\rightarrow M_R/M_RJ(R)$ denotes the canonical projection. Then the disjoint union $g({\mathcal {F}}\setminus \{0_{M/MJ(R)})\cup MJ(R)$ generates $M_R$, so that there exist subsets ${\mathcal {G}}\subseteq ({\mathcal {F}}\setminus \{0_{M/MJ(R)})$ and $Y\subseteq MJ(R)$ such that ${\mathcal {G}}\cup Y$ is a minimal set of generators of $M_R$. Then $\pi ({\mathcal {G}})$ generates $M_R/M_RJ(R)$. Let us show that $\pi ({\mathcal {G}})$ is a minimal set of generators for $M_R/M_RJ(R)$. Fix $y_0\in \pi ({\mathcal {G}})$. Then $y_0=\pi (g_0)$ for some $g_0\in {\mathcal {G}}$. It remains to show that $\pi ({\mathcal {G}})\setminus \{y_0\}$ generates a proper subset of $M_R/M_RJ(R)$. Now $({\mathcal {G}}\setminus \{y_0\})\cup Y$ generates a proper submodule of $M_R$, because ${\mathcal {G}}\cup Y$ is a minimal set of generators of $M_R$. Thus $T:= \sum _{g\in {\mathcal {G}}\setminus \{y_0\}}gR+\sum _{y\in Y}yR$ is a proper submodule of $M_R$, and $M_R/T$ is a non-zero cyclic R-module. By Nakayama’s Lemma, $(M_R/T)J(R)$ is properly contained in $M_R/T$. Hence $M_RJ(R)+T$ is properly contained in $M_R$. But $M_RJ(R)+T=M_RJ(R)+\sum _{g\in {\mathcal {G}}\setminus \{y_0\}}gR$. Applying the canonical projection $\pi :M_R\rightarrow M_R/M_RJ(R)$ to the inclusions $M_RJ(R)\subseteq M_RJ(R)+\sum _{g\in {\mathcal {G}}\setminus \{y_0\}}gR\subset M_R$, we get that $\pi (M_RJ(R)+\sum _{g\in {\mathcal {G}}\setminus \{y_0\}}gR)$ is properly contained in $M_R/M_RJ(R)$, that is, $\sum _{g\in {\mathcal {G}}\setminus \{y_0\}}\pi (g)R)$ is properly contained in $M_R/M_RJ(R)$. This proves that $\pi ({\mathcal {G}})$ is a minimal set of generators for $M_R/M_RJ(R)$. $\square$

Propositions 2.1(2) and 3.3, which describe modules with a free cover, can be weakened to describe modules with a minimal set of generators, as follows.

Proposition 3.6

A right R-module has a minimal set of generators if and only if it is isomorphic to a module $F_R/S$, where $F_R=R_R^{(X)}$ is a free R-module and S is a submodule of $F_R$ such that $S+R_R^{(X\setminus \{x\})}\ne R_R^{(X)}$ for every $x\in X$.

Notice that $R_R^{(X\setminus \{x\})}=\ker (\pi _x)$, where $\pi _x:R_R^{(X)}\rightarrow R_R$ is the canonical projection onto the x-th direct summand of $R_R^{(X)}$. Thus $S+R_R^{(X\setminus \{x\})}\ne R_R^{(X)}$ if and only if $\pi _x(S)\ne R_R$. That is:

Corollary 3.7

A right R-module has a minimal set of generators if and only if it is isomorphic to a module $F_R/S$, where $F_R=R_R^{(X)}$ is a free R-module and S is a submodule of $\bigoplus _{x\in X}x{\mathcal {I}}_x$ for some indexed family $\{\,{\mathcal {I}}_x\mid x\in X\,\}$ of proper right ideals of R.

In the following, we need two well known lemmas:

Lemma 3.8

Let $A_R\le B_R\le C_R$ be R-modules. Then $B_R$ is superfluous in $C_R$ if and only if $A_R$ is superfluous in $C_R$ and $B_R/A_R$ is superfluous in $C_R/A_R$.

Lemma 3.9

[1, Lemma 28.3]. The following conditions are equivalent for a right ideal I of a ring R:

(a) I is right T-nilpotent.

(b) $M_RI$ is superfluous in $M_R$ for every right R-module $M_R$.

(c) There exists an infinite set X for which $R_R^{(X)}I=I^{(X)}$ is a superfluous submodule of $R_R^{(X)}$.

Proposition 3.10

Let R be a ring, I a two-sided ideal of R contained in the Jacobson radical J(R) of R, $M_R$ an R-module that is not finitely generated, and suppose $MI=0$. Then the following conditions are equivalent:

(1) The module $M_R$ has a free cover.

(2) The module $M_{R/I}$ has a free cover and I is a right T-nilpotent ideal of R.

Proof

Suppose that (1) holds. Let $\pi :R_R^{(X)}\rightarrow M_R$ be a free cover for $M_R$. The set X is infinite because $M_R$ is not finitely generated. Then $\pi (I^{(X)})=\pi (R^{(X)}I)=M_RI=0$. Hence we have an induced mapping ${\overline{\pi }}:(R/I)^{(X)}\rightarrow M_R$ and three right R-modules $I^{(X)}\le \ker (\pi )\le R_R^{(X)}$. As $\ker (\pi )$ is superfluous in $R_R^{(X)}$, by Lemma 3.8 we have that $I^{(X)}$ is superfluous in $R_R^{(X)}$ and $\ker (\pi )/I^{(X)}=\ker ({\overline{\pi }})$ is superfluous in $R_R^{(X)}/I^{(X)}$. In particular, ${\overline{\pi }}$ is a free cover of the module $M_{R/I}$. Since $I^{(X)}=R^{(X)}I$ is superfluous in $R_R^{(X)}$, we get that I is right T-nilpotent by Lemma 3.9.

Conversely, suppose (2) holds. Let ${\pi '}:(R/I)^{(X)}\rightarrow M_R$ be a free cover of $M_{R/I}$. Let ${\pi ''}:R_R^{(X)}\rightarrow (R/I)^{(X)}$ be the canonical projection. The kernel of the composite mapping $\pi '\pi ''$ contains $I^{(X)}$ and is such that $\ker (\pi '\pi '')/I^{(X)}=\ker (\pi ')$, which is a superfluous submodule of $(R/I)^{(X)}$. By Lemma 3.9, from I right T-nilpotent we get that $I^{(X)}$ is superfluous in $R_R^{(X)}$. From Lemma 3.8, we obtain that $\ker (\pi '\pi '')$ is superfluous in $R_R^{(X)}$. Hence $\pi '\pi ''$ is the required free cover for $M_R$.$\square$

4 Modules for Which Every Set of Generators Contains a Minimal Set of Generators

Lemma 4.1

Let R be a ring with Jacobson radical J(R) and $M_R$ a right R-module. Assume that every set of generators of $M_R$ contains a minimal set of generators. Let ${\mathcal {F}}:=\{\,x_\lambda \mid \lambda \in \Lambda \,\}$ be an indexed family of elements of $M_R$. If $\{\,x_\lambda +MJ(R)\mid \lambda \in \Lambda \,\}$ is a minimal set of generators for M/MJ(R), then ${\mathcal {F}}$ is a minimal set of generators for $M_R$.

Proof

Set $N_R:=\sum _{\lambda \in \Lambda } x_\lambda R$, so that $N_R+MJ(R)=M_R$. The disjoint union${\mathcal {F}}\cup MJ(R)$ generates $M_R$, so that there exist subsets ${\mathcal {G}}\subseteq {\mathcal {F}}$ and $Y\subseteq MJ(R)$ such that ${\mathcal {G}}\cup Y$ is a minimal set of generators of $M_R$. Then ${\mathcal {G}}={\mathcal {F}}$, because if ${\mathcal {G}}\subset {\mathcal {F}}$, then ${\mathcal {G}}\cup Y$ generates $M_R$ implies ${\mathcal {G}}\cup MJ(R)$ generates $M_R$, so $\overline{{\mathcal {G}}}:=\{\,x_\lambda +MJ(R)\mid \lambda \in \Lambda ,\ x_\lambda \in {\mathcal {G}}\,\}$ generates M/MJ(R), which is not. Thus ${\mathcal {F}}\cup Y$ is a minimal set of generators for $M_R$. Let us show that $\overline{{\mathcal {Y}}}:=\{\,y+N_R\mid y\in Y\,\}$ is a minimal set of generators for $M_R/N$. For every $y_0\in Y$, we know that $\sum _{y\in Y\setminus \{y_0\}}yR+N_R\ne M_R$ because ${\mathcal {F}}\cup Y$ is a minimal set of generators for $M_R$. It follows that $\overline{{\mathcal {Y}}}$ is a minimal set of generators for $M_R/N$. Moreover $(M_R/N)J(R)=M_R/N_R$ because $N_R+MJ(R)=M_R$.

Suppose $M_R/N_R\ne 0$. Then $M_R/N_R$ has a maximal submodule, $A_R/N_R$ say [7, Lemma 3.1]. Hence J(R) annihilates the simple module $M_R/A_R$, so $MJ(R)\subseteq A_R$. which contradicts $(M_R/N)J(R)=M_R/N_R$. The contradiction proves that $M_R/N_R=0$, so $Y=\emptyset$. Thus ${\mathcal {F}}$ is a minimal set of generators for $M_R$.$\square$

Theorem 4.2

The following conditions are equivalent for a right module $M_R$ over a ring R:

(1) Every set of generators of $M_R$ contains a minimal set of generators.

(2) Every set of generators of $M_R/MJ(R)$ contains a minimal set of generators, and the submodule MJ(R) of $M_R$ is superfluous.

Proof

(1)${}\Rightarrow {}$(2) Suppose that (1) holds. Let $X'$ be a set of generators of$M_R/MJ(R)$. Let $g:M_R/MJ(R)\rightarrow M_R$ be a mapping such that $\pi g=1_{M/MJ(R)}$, where $\pi :M_R\rightarrow M_R/MJ(R)$ is the canonical projection. Then $X'\setminus \{0_{M/MJ(R)}\}$ is a set of generators of $M_R/MJ(R)$ and the disjoint union $g(X'\setminus \{0_{M/MJ(R)}\})\cup MJ(R)$ is a set of generators of $M_R$. By (1), there exist subsets ${\mathcal {G}}\subseteq g(X'\setminus \{0_{M/MJ(R)}\})$ and $Y\subseteq MJ(R)$ such that ${\mathcal {G}}\cup Y$ is a minimal set of generators of $M_R$. Then $\pi ({\mathcal {G}})$ is a minimal set of generators of M/MJ(R) contained in $X'$.

Let us prove that MJ(R) is superfluous in $M_R$. Let A be a submodule of $M_R$ such that $A+MJ(R)=M_R$. By what we have proved in the previous paragraph, $\pi (A)$ contains a minimal set ${\mathcal {G}}'$ of generators of M/MP. The restriction $\pi ':A_R\rightarrow M/MJ(R)$ of the canonical projection $\pi :M_R\rightarrow M_R/MJ(R)$ is an epimorphism. Let $g':M_R/MJ(R)\rightarrow A_R$ be a mapping such that $\pi ' g'=1_{M/MJ(R)}$. By Lemma 4.1 applied to ${\mathcal {F}}:=g'({\mathcal {G}}')$, we know that ${\mathcal {F}}$ is a minimal set of generators for $M_R$ contained in $A_R$.

(2)${}\Rightarrow {}$(1) Assume that (2) holds. Let X be any set of generators for $M_R$. Then $\{\,x+MJ(R)\mid x\in X\,\}$ is a set of generators for $M_R/MJ(R)$, hence contains a minimal set of generators for $M_R/MJ(R)$. That is, there exists a subset Y of X such that $\{\,y+MJ(R)\mid y\in Y\,\}$ is a minimal set of generators for $M_R/MJ(R)$. Therefore $\sum _{y\in Y}yR+MJ(R)=M_R$. By (2), $\sum _{y\in Y}yR=M_R$. It follows that Y is a minimal set of generators for $M_R$.

$\square$

The next proposition generalizes and completes Lemma 3.9 for the Jacobson radical.

Proposition 4.3

The following conditions are equivalent for a ring R:

(a) J(R) is right T-nilpotent.

(b) There exists an infinitely generated module $M_R$ with a free cover and with $M_RJ(R)$ superfluous in $M_R$.

Proof

(a)${}\Rightarrow {}$(b) follows from Lemma 3.9((a)${}\Rightarrow {}$(c)).

(b)${}\Rightarrow {}$(a). Let $M_R$ be an infinitely generated module with a free cover and with $M_RJ(R)$ superfluous in $M_R$. Then $M_R\cong R_R^{(X)}/S$ for some infinite set X, S a superfluous submodule of $R_R^{(X)}$, $S\subseteq J(R)_R^{(X)}$ (Proposition 2.3). From $M_RJ(R)$ superfluous in $M_R$ we get that $J(R)_R^{(X)}/S$ is superfluous in $R_R^{(X)}/S$. Thus $J(R)_R^{(X)}$ is superfluous in $R_R^{(X)}$ (Lemma 3.8), so J(R) is right T-nilpotent by Lemma 3.9. $\square$

Now let R be a fixed ring. Let us go back to the previous three classes ${\mathcal {A}},{\mathcal {B}},{\mathcal {C}}$ of right R-modules, with ${\mathcal {A}}\cup {\mathcal {B}}\subseteq {\mathcal {C}}$.

Proposition 4.4

Let $f:M_R\rightarrow N_R$ be a superfluous epimorphism between right modules $M_R,N_R$ over an arbitrary ring R. Then:

(a) $M_R\in {\mathcal {A}}$ if and only if $N_R\in {\mathcal {A}}$.

(b) $M_R\in {\mathcal {B}}$ if and only if $N_R\in {\mathcal {B}}$.

(c) $M_R\in {\mathcal {C}}$ if and only if $N_R\in {\mathcal {C}}$.

Proof

(a) is an easy exercise that follows from the characterization of superfluous epimorphisms $\alpha :X_R\rightarrow Y_R$ as the epimorphisms $\alpha :X_R\rightarrow Y_R$ such that $\alpha g$ epi implies g epi for every morphism $g:Z_R\rightarrow X_R$.

(b) and (c) are essentially a restatement of Corollary 3.2.$\square$

5 The Special Case of Rings R with J(R) T-Nilpotent

Lemma 5.1

Let $M_R$ be a right module over a ring R.

(a) If $M_R$ has a free cover, then $M_R/M_RJ(R)$ is a free right R/J(R)-module.

(b) If $M_R$ has a projective cover, then $M_R/M_RJ(R)$ is a projective right R/J(R)-module and there exists a projective right R-module $P_R$ with

$$\begin{aligned} P_R/P_RJ(R)\cong M_R/M_RJ(R). \end{aligned}$$

Proof

(a) If $M_R$ has a free cover, then $M_R\cong F_R/S$ for some free R-module $F_R$ and some superfluous submodule S of $F_R$ (Proposition 2.1(2)). Hence $M_R/M_RJ(R)\cong (F_R/S)/(F_R/S)J(R)\cong F_R/(F_RJ(R)+S)=F_R/F_RJ(R)$, because $F_RJ(R)\supseteq S$ (Proposition 2.3). Thus $M_R/M_RJ(R)$ is isomorphic to the free right R/J(R)-module $F_R/F_RJ(R)$.

(b) Let $P_R\rightarrow M_R$ be a projective cover. There exists a right R-module $Q_R$ such that $P_R\oplus Q_R$ is a free right R-module. Thus there is a free cover $P_R\oplus Q_R\rightarrow M_R\oplus Q_R$. Since $M_R\oplus Q_R$ has a free cover, by (a) $M_R/M_RJ(R)\oplus Q_R/Q_RJ(R)$ is a free right R/J(R)-module. Therefore its direct summand $M_R/M_RJ(R)$ is a projective right R/J(R)-module.

Moreover, like in (a), $M_R\cong P_R/S$ for some superfluous submodule S of $P_R$. Hence $M_R/M_RJ(R)\cong (P_R/S)/(P_R/S)J(R)\cong P_R/(P_RJ(R)+S)=P_R/P_RJ(R)$, because $P_RJ(R)\supseteq S$ ( [5, Lemma 2.15 and Proposition 2.160]).$\square$

Theorem 5.2

Let R be a ring with J(R) T-nilpotent, and let $M_R$ be a right R-module. Then:

(a) The R-module $M_R$ has a free cover if and only if $M_R/M_RJ(R)$ is a free right R/J(R)-module.

(b) The R-module $M_R$ has a projective cover if and only if $M_R/M_RJ(R)$ is a projective right R/J(R)-module and there exists a projective right R-module $P_R$ with $P_R/P_RJ(R)\cong M_R/M_RJ(R)$.

Proof

In view of the previous lemma, it suffices to prove one of the two implications, both in (a) and in (b).

(a) Assume that $M_R/M_RJ(R)$ is a free right R/J(R)-module, $M_R/M_R J(R)\cong (R/J(R))^{(X)}$ say. Then there is an epimorphism $\pi :R_R^{(X)}\rightarrow M_R/M_R J(R)$ with kernel $(J(R))^{(X)}$. By the projectivity of $R_R^{(X)}$, there is a morphism $f:R_R^{(X)}\rightarrow M_R$ such that the composite mapping of f with the canonical projection $M_R\rightarrow M_R/M_RJ(R)$ is $\pi$. Then $f(R_R^{(X)})+M_RJ(R)=M_R$ and $\ker (f)\subseteq \ker (\pi )=(J(R))^{(X)}=R_R^{(X)}J(R)$. By Lemma 3.9((a)${}\Rightarrow {}$(b)), it follows that f is an epimorphism, and its kernel is superfluous in $R_R^{(X)}$ by Lemma 3.8. Hence f is a free cover for $M_R$.

(b) The proof is very similar to the proof of (a). Suppose that $M_R/M_RJ(R)$ is a projective right R/J(R)-module and that there exists a projective right R-module $P_R$ with $P_R/P_RJ(R)\cong M_R/M_RJ(R)$. Then there is an epimorphism $\pi :P_R\rightarrow M_R/M_RJ(R)$ with kernel $P_RJ(R)$. There is a morphism $f:P_R\rightarrow M_R$ such that the composite mapping of f with the canonical projection $M_R\rightarrow M_R/M_RJ(R)$ is $\pi$. Then $f(P_R)+M_RJ(R)=M_R$ and $\ker (f)\subseteq P_RJ(R)$. It follows that f is an epimorphism, and its kernel is superfluous in $R_R^{(X)}$. Hence f is a projective cover. $\square$

Remark 5.3

(a) and (b) in Theorem 5.2 do not necessarily hold if J(R) is not T-nilpotent. For instance, let R be a local ring with J(R) not T-nilpotent (e.g., let R be a DVR). In this case, projective modules are free, so (a) and (b) coincide. In fact, for every module $M_R$, if X is any basis of the R/J(R)-vector space $M_R/M_RJ(R)$, the free R-module $P_R:=R_R^{(X)}$ is such that $P_R/P_RJ(R)\cong M_R/M_RJ(R)$. For such a ring R, the module $M_R/M_RJ(R)$ is always a free (= projective) right R/J(R)-module. But R is not right perfect, and therefore there exist modules $M_R$ without projective covers.

We are ready for our last Theorem:

Theorem 5.4

If a ring R is not right perfect, then the class ${\mathcal {A}}\cap {\mathcal {B}}$ consists of all finitely generated right R-modules $M_R$ with M/MJ(R) a free right R/J(R)-module.

Notice that the finitely generated right R-modules $M_R$ with M/MJ(R) a free right R/J(R)-module are those isomorphic to $R^n_R/S$ for some integer $n\ge 0$ and some submodule S of $(J(R))^n$.

Proof

Let R be a ring that is not right perfect. There are two cases: either J(R) is not right T-nilpotent, or J(R) is right T-nilpotent but R/J(R) is not semisimple artinian.

Suppose J(R) is not right T-nilpotent. Assume by contradiction that the class ${\mathcal {A}}\cap {\mathcal {B}}$ properly contains the class of all finitely generated right R-modules with a free cover. Then there exists an infinitely module $M_R$ with a free cover such that every set of generators of $M_R$ contains a minimal set of generators. By Theorem 4.2, MJ(R) is superfluous in $M_R$. By Proposition 4.3, J(R) is T-nilpotent, a contradiction.

Suppose now J(R) right T-nilpotent and R/J(R) not semisimple artinian. There is the canonical functor

$$\begin{aligned} F=-\otimes _R R/J(R):{\text {Mod-\!}}R\rightarrow {\text {Mod-\!}}(R/J(R)). \end{aligned}$$

For every module $M_R$, the canonical epimorphism $M_R\rightarrow F(M_R)=M\otimes _R R/J(R)=M_R/M_RJ(R)$ is always a superfluous epimorphism (Lemma 3.9). Therefore, by Proposition 4.4, $M_R\in {\mathcal {A}}_R$ if and only if $F(M_R)\in {\mathcal {A}}_{R/J(R)}$, $M_R\in {\mathcal {B}}$ if and only if $F(M_R)\in {\mathcal {B}}_{R/J(R)}$, and $M_R\in {\mathcal {C}}$ if and only if $F(M_R)\in {\mathcal {C}}_{R/J(R)}$. In other words, F preserves the classes ${\mathcal {A}}$, ${\mathcal {B}}$ (this implies [8, Proposition 2.4(2)]) and ${\mathcal {C}}$. It also preserves the class $\mathrm{mod-}R$ of finitely generated modules. Moreover, ${\mathcal {A}}_{R/J(R)}$ coincides with the class ${\mathcal {F}}_{R/J(R)}$ of all free right R/J(R)-modules (Theorem 5.2(a)). This reduces the study of the classes ${\mathcal {A}}={\mathcal {F}},\ {\mathcal {B}}$ and ${\mathcal {C}}$ to the case of the rings R with $J(R)=0$.

Hence we suppose that $J(R)=0$ but R is not a semisimple artinian ring. In this case, ${\mathcal {A}}\cap {\mathcal {B}}={\mathcal {F}}\cap {\mathcal {B}}$ is the class of all finitely generated free right R-modules by [8, Lemma 2.2]. $\square$

If a ring R is local and right perfect, then ${\mathcal {A}}={\mathcal {B}}={\mathcal {C}}={\text {Mod-\!}}R$. See [3, Theorem 4.3]. In the case of R right perfect but not local, little is known. For example let R be a semisimple artinian ring that is not a division ring. In this case, we don’t know what ${\mathcal {A}}\cap {\mathcal {B}}={\mathcal {F}}\cap {\mathcal {B}}$ is. In this case, it is also unknown whether ${\mathcal {B}}={\text {Mod-\!}}R$ [8, Problem 5.1]. We even don’t know whether every set of generators of a free right module over a semisimple artinian ring always contains a minimal set of generators.