Quasi-dual Baer modules

Abstract

Let R be a ring and let M be an R-module with $S={\text {End}}_R(M)$. The module M is called quasi-dual Baer if for every fully invariant submodule N of M, $\{\phi \in S \mid Im\phi \subseteq N\} = eS$ for some idempotent e in S. We show that M is quasi-dual Baer if and only if $\sum _{\varphi \in \mathfrak {I}} \varphi (M)$ is a direct summand of M for every left ideal $\mathfrak {I}$ of S. The R-module $R_R$ is quasi-dual Baer if and only if R is a finite product of simple rings. Other characterizations of quasi-dual Baer modules are obtained. Examples which delineate the concepts and results are provided.

1 Introduction

Throughout this paper, R is an associative ring with identity, and all the modules are unital right R-modules. Let M be an R-module. By ${\text {End}}_R(M)$ and E(M), we denote the endomorphism ring of M and the injective hull of M, respectively. The notation $N \subseteq M$ means that N is a subset of M and $N \le M$ means that N is a submodule of M. By $\mathbb {Q}$ and $\mathbb {Z}$, we denote the ring of rational numbers and integer numbers, respectively. For any prime number p, $\mathbb {Z}(p^{\infty })$ denotes the Prüfer p-group.

A submodule K of M is called fully invariant if $f(K)\subseteq K$ for all $f \in {\text {End}}_R(M)$. In 1967, Clark introduced the concept of quasi-Baer rings. A ring R is called right quasi-Baer if the right annihilator of any right ideal of R is generated as a right ideal by an idempotent. Rizvi and Roman generalized the notion of right quasi-Baer rings to a module theoretic version. A module M is called quasi-Baer if the right annihilator in M of any two-sided ideal of $End_R(M)$ is a direct summand of M. This notion was dualized by Amouzegar–Talebi in 2013 [1]. A module M is said to be quasi-dual Baer if for every fully invariant submodule N of M, there exists an idempotent e in S such that $\{\phi \in S \mid Im\phi \subseteq N\} = eS$. In Sect. 2, we continue the study of quasi-dual Baer modules. We show that an R-module M is quasi-dual Baer if and only if $\sum _{\varphi \in \mathfrak {I}} \varphi (M)$ is a direct summand of M for every left ideal $\mathfrak {I}$ of ${\text {End}}_R(M)$. We provide examples to separate the properties of quasi-dual Baer modules and dual Baer modules. We prove that for a ring R, the R-module $R_R$ is quasi-dual Baer if and only if R is a finite product of simple rings. We obtain conditions under which an indecomposable module is quasi-dual Baer. It is also shown that if M is a quasi-dual Baer module over a commutative ring R, then $M\mathfrak {a}$ is a direct summand of M for any ideal $\mathfrak {a}$ of R. We give a characterization of when a direct sum of a module $M_1$ with ${\text {Rad}}(M_1)=M_1$ and a semisimple module $M_2$ is quasi-dual Baer. It is also shown that over a commutative perfect ring, every quasi-dual Baer module is semisimple.

2 Some results on quasi-dual Baer modules

Let M be an R-module with $S={\text {End}}_R(M)$. For a submodule N of M, the right ideal $\{\phi \in S \mid Im\phi \subseteq N\}$ of S will be denoted by $\mathfrak {D}(N)$. For a subset $\mathfrak {X}$ of S and a submodule N of M, let $\mathfrak {X}(N)$ denote the submodule $\sum _{\varphi \in \mathfrak {X}} \varphi (N)$ of M. It is easily seen that if N is a fully invariant submodule of M, then $\mathfrak {D}(N)$ is a two-sided ideal of S and if $\mathfrak {I}$ is a two-sided ideal of S, then $\mathfrak {I}(M)$ is a fully invariant submodule of M. Therefore, we have the following two maps:

$\mathfrak {D}:\{$fully invariant submodules of $M\} \longrightarrow \{$two-sided ideals of S} and

$\mathfrak {T}:\{$two-sided ideals of $S\} \longrightarrow \{$fully invariant submodules of M}

with $\mathfrak {T}(\mathfrak {I})=\mathfrak {I}(M)$ for any two-sided ideal $\mathfrak {I}$ of S.

Following [1, 4], the module M is called dual Baer (quasi-dual Baer) if for every submodule (fully invariant submodule) N of M, there exists an idempotent e in S such that $\mathfrak {D}(N)=eS$. In this section, we continue the study of quasi-dual Baer modules. We begin with the following characterization of this class of modules.

Proposition 2.1

Let M be an R-module with $S={\text {End}}_R(M)$. Then, the following statements are equivalent:

• (i) M is a quasi-dual Baer module;

• (ii) For any fully invariant submodule N of M, there exists a decomposition $M=D_1 \oplus D_2$ with $D_1 \subseteq N$ and ${\text {Hom}}_R(M, N \cap D_2)=0$.

Proof

(i) $\Rightarrow$ (ii) Let N be a fully invariant submodule of M. Then $\mathfrak {D}(N)=eS$ for some idempotent e of S. Set $D_1=e(M)$ and $D_2=(1-e)(M)$. Therefore, $\mathfrak {T}(\mathfrak {D}(N))=D_1 \subseteq N$. Since $M=D_1 \oplus D_2$, we have $N=D_1 \oplus (N \cap D_2)$. Now, let h be an endomorphism of M such that $h(M) \subseteq N \cap D_2$. Hence, $h \in \mathfrak {D}(N)$ and so $h=eg$ for some $g \in S$. Thus, $h(M) \subseteq e(M)$. But $h(M) \subseteq (1-e)(M)$. Then, $h=0$. It follows that ${\text {Hom}}_R(M, N \cap D_2)=0$.

(ii) $\Rightarrow$ (i) Let N be a fully invariant submodule of M. So there exists a decomposition $M=D_1 \oplus D_2$ with $D_1 \subseteq N$ and ${\text {Hom}}_R(M, N \cap D_2)=0$. Let e be an idempotent of S with $D_1=e(M)$. Let us prove that $\mathfrak {D}(N)=eS$. It is clear that $eS \subseteq \mathfrak {D}(N)$. Now, let $f \in \mathfrak {D}(N)$. Let $\pi :N=D_1 \oplus (N \cap D_2) \rightarrow N \cap D_2$ be the projection map. Then, $\pi f \in Hom_R(M, N \cap D_2)=0$. This implies that $f(M) \subseteq e(M)$. Thus, $(1-e)f=0$. Since $f=ef+(1-e)f$, we obtain $f=ef \in eS$. Therefore, $\mathfrak {D}(N)=eS$ and so M is quasi-dual Baer. $\square$

The following corollary follows directly from Proposition 2.1.

Corollary 2.2

Let M be an R-module. Assume that every fully invariant submodule of M is a direct summand of M. Then, M is quasi-dual Baer.

Example 2.3

• (i) Clearly, every dual Baer module is quasi-dual Baer. From [9, Proposition 4.16], it follows that every torsion-free injective module over a commutative domain is quasi-dual Baer.

• (ii) Let M be an R-module such that ${\text {Hom}}_R(M, N)=0$ for every fully invariant proper submodule N of M. Then, clearly M is quasi-dual Baer (see Proposition 2.1). For example, the $\mathbb {Z}$-modules $\mathbb {Q}$ and $\mathbb {Z}(p^{\infty })$ (where p is a prime number) are quasi-dual Baer modules.

• (iii) Recall that a nonzero ring R is called a simple ring if (0) and R are the only two-sided ideals in R. Let M be a module and assume that M has no nonzero proper fully invariant submodules. Then, M is quasi-dual Baer. For example, for any simple ring S, the right S-module $S_S$ is quasi-dual Baer.

Next, we present another characterization of quasi-dual Baer modules.

Proposition 2.4

Let M be an R-module with $S={\text {End}}_R(M)$. Then, the following are equivalent:

• (i) M is a quasi-dual Baer R-module;

• (ii) For every left ideal $\mathfrak {I}$ of S, $\mathfrak {I}(M)$ is a direct summand of M;

• (iii) For every two-sided ideal $\mathfrak {I}$ of S, $\mathfrak {I}(M)$ is a direct summand of M.

Proof

(i) $\Rightarrow$ (ii) Let $\mathfrak {I}$ be a left ideal of S. Put $N=\mathfrak {I}(M)$. Clearly, N is a fully invariant submodule of M. By hypothesis, $\mathfrak {D}(N)=eS$ for some idempotent $e \in S$. Therefore, $e(M) \subseteq N$. Now, let $\varphi \in \mathfrak {I}$. Then $Im \varphi \subseteq \mathfrak {I}(M)$. Hence, $\varphi \in \mathfrak {D}(N)$. This implies that $\varphi = e\psi$ for some $\psi \in S$. Thus $\varphi (M) \subseteq e(M)$. It follows that $N=\mathfrak {I}(M) \subseteq e(M)$. Consequently, $N=e(M)$ is a direct summand of M.

(ii) $\Rightarrow$ (iii) This is obvious.

(iii) $\Rightarrow$ (i) Let L be a fully invariant submodule of M. Note that $\mathfrak {I}= \mathfrak {D}(L)$ is a two-sided ideal of S. By (iii), $\mathfrak {I}(M)=e(M)$ for some $e^2=e \in S$. Let $\phi \in \mathfrak {I}$. Then, $Im \phi \subseteq e(M)$. Hence, $(1_S-e)\phi = 0$. Since $1_S=e+(1_S-e)$, we have $\phi = e\phi \in eS$. Therefore, $\mathfrak {I} \subseteq eS$. Now let $\psi \in S$. Then, $e\psi (M) \subseteq e(M) \subseteq \mathfrak {I}(M) \subseteq L$. So $e\psi \in \mathfrak {I}$. This yields $eS \subseteq \mathfrak {I}$. It follows that $\mathfrak {D}(L)=eS$. This completes the proof. $\square$

As an application of Proposition 2.4, we provide another way of proving [1, Theorem 2.1].

Corollary 2.5

Let M be a quasi-dual Baer module with $S={\text {End}}_R(M)$. Then every direct summand of M is quasi-dual Baer.

Proof

Let N be a direct summand of M. Let $\pi :M \rightarrow N$ be the projection map and $\mu :N \rightarrow M$ be the inclusion map. For every $f \in T={\text {End}}_R(N)$, let $\overline{f}=\mu f \pi \in S$. Let I be a left ideal of T. Set $\overline{I}=\{\overline{f} \mid f \in I\}$ and $\mathfrak {I}=S\overline{I}$. By Proposition 2.4, $\mathfrak {I}(M)=\mathfrak {I}(N)$ is a direct summand of M. Let $e=\mu \pi \in S$. Then, $e=e^2$ and $N=e(M)$. Since $\mathfrak {I}(M)$ is a fully invariant submodule of M, we have $\mathfrak {I}(M)=e(\mathfrak {I}(M)) \oplus (1-e)(\mathfrak {I}(M))$. Therefore $e(\mathfrak {I}(M)) = e(\mathfrak {I}(N))$ is a direct summand of N. Since $e^2=e$, we have $e(x)=x$ for any $x \in N$. Thus $\overline{I}(N) = e(\overline{I}(N)) \subseteq e(S\overline{I}(N))$. Now let $g \in S$ and $f \in I$. Then, $eg\overline{f}(N) =\mu \pi g\mu f \pi (N) = \mu (\pi g\mu f) \pi (N)$. Since I is a left ideal of T, it follows that $\pi g\mu f \in I$. This yields $eg\overline{f}(N) = \overline{\pi g\mu f}(N) \subseteq \overline{I}(N)$. Hence, $e(S\overline{I}(N)) \subseteq \overline{I}(N)$. Thus, $\overline{I}(N) = e(\mathfrak {I}(N))$. Moreover, it is easily seen that $I(N) = \overline{I}(N)$. It follows that $I(N) = e(\mathfrak {I}(N))$ is a direct summand of N. Therefore, N is a quasi-dual Baer module by Proposition 2.4. $\square$

Corollary 2.6

Let M be a quasi-dual Baer module and let N be a fully invariant submodule of M. Then, the following are equivalent:

• (i) N is a direct summand of M;

• (ii) $\mathfrak {I}(M) = N$, where $\mathfrak {I} = \mathfrak {D}(N)$.

Proof

(i) $\Rightarrow$ (ii) Let $\pi :M \rightarrow N$ be the projection map and $\mu :N \rightarrow M$ be the inclusion map. Then $\mu \pi \in \mathfrak {D}(N)$ and $\mu \pi (M)=N$. Hence $\mathfrak {I}(M) = N$.

(ii) $\Rightarrow$ (i) This follows directly from Proposition 2.4. $\square$

Next, we present more examples of quasi-dual Baer modules.

Example 2.7

• (i) By Proposition 2.4, the notion of being quasi-dual Baer only involves two-sided ideals of the endomorphism ring. It follows that any R-module M for which ${\text {End}}_R(M)$ is a simple ring is quasi-dual Baer.

• (ii) To construct an explicit example, let S be a simple ring and let $_SN_S$ be an S-S-bimodule. As in [7], consider the generalized matrix ring $R = \left( \begin{array}{rr} S &{} N \\ N&{} S \\ \end{array} \right)$ with multiplication given by

  $$\left( \begin{array}{rr} a &{} x \\ y&{} b \\ \end{array} \right) \left( \begin{array}{rr} a^{\prime } \quad &{} x^{\prime } \\ y^{\prime }\quad &{} b^{\prime } \\ \end{array} \right) = \left( \begin{array}{cc} aa^{\prime }\quad &{} ax^{\prime }+xb^{\prime } \\ ya^{\prime }+by^{\prime }\quad &{} bb^{\prime } \\ \end{array} \right) $$

  for all $a, a^{\prime }, b, b^{\prime } \in S$ and $x, x^{\prime }, y, y^{\prime } \in N$. Then, $M=N \oplus S$ is a right R-module by the following action:

  $$(n,s)\left( \begin{array}{rr} a &{} x \\ y&{} b \\ \end{array} \right) =(na+sy, sb).$$

  It is shown in [7, p. 1278] that the rings ${\text {End}}_R(M)$ and S are isomorphic. Therefore M is a quasi-dual Baer R-module.

Next, we compare quasi-dual Baer modules and dual Baer modules. Recall that a module M is called a duo module provided every submodule of M is fully invariant. A ring R is called a right duo ring if the right R-module $R_R$ is a duo module; equivalently, every right ideal of R is a two-sided ideal.

Remark 2.8

Let M be an R-module such that $S={\text {End}}_R(M)$ is a right duo ring (e.g., S is a commutative ring). From Proposition 2.4 and [4, Theorem 2.1], it follows that M is a quasi-dual Baer module if and only if M is a dual Baer module. In particular, if M is an indecomposable injective R-module over a commutative noetherian ring R, then M is quasi-dual Baer if and only if M is dual Baer (see [8, Proposition 5.13]).

In [1], there are no examples showing that the class of quasi-dual Baer modules contains properly that of dual Baer modules. In the next example, we provide some quasi-dual Baer modules which are not dual Baer.

Example 2.9

• (i) Let R be a ring. It is shown in [4, Corollary 2.9] that the R-module $R_R$ is dual Baer if and only if R is a semisimple ring. It follows that for any simple ring R which is not semisimple, the R-module $R_R$ is quasi-dual Baer but $R_R$ is not dual Baer (see Example 2.3(iii)). For example, we can take the following rings:

• (a) The differential polynomial ring $R=k[x; \delta ]$, where k is a simple ring of characteristic 0 and $\delta$ is a non-inner derivation on k (see [5, Corollary 3.16]).

• (b) The Weyl algebra $R=A_n(k_0)$, where $k_0$ is a simple ring of characteristic 0 and n is a positive integer (see [5, Corollary 3.17]).

• (c) The ring of skew Laurent polynomials $R=k[x, x^{-1}; \sigma ]$, where k is a field with an automorphism $\sigma$ of infinite order (see [5, Example 1.8 and Corollary 3.19]).

• (d) See also other examples in [5, p. 40].

• (ii) Let S be a simple ring and let $_SN_S$ be an S-S-bimodule. As in Example 2.7(ii), consider the ring $R = \left( \begin{array}{rr} S &{} N \\ N&{} S \\ \end{array} \right)$ and the right R-module $M=N \oplus S$. Assume that S is a domain that is not a division ring (we can take $S=A_n(k)$, the n-th Weyl algebra over a field k of characteristic zero). Let $t \in S$ be a nonzero non-invertible element. Consider the map $f_t:M \rightarrow M$ defined by $f_t((n, s))=(tn, ts)$ for all $(n, s) \in M$. Then $f_t$ is an R-endomorphism of M. As S is a domain, we have $tS \not = S$ (see, e.g., [5, Chapter 1, §1, p. 22, Exercise 1.4]). It follows that $f_t$ is not an epimorphism. Note that the rings ${\text {End}}_R(M)$ and S are isomorphic (see [7, p. 1278]). Then ${\text {End}}_R(M)$ is a domain. This implies that M is an indecomposable R-module. Using [4, Corollary 2.2], we conclude that M is not a dual Baer R-module. However, M is a quasi-dual Baer R-module by Proposition 2.4.

• (iii) Let S be a simple Ore domain which is not a division ring. Consider the ring $R=\left( \begin{array}{cc} S &{} K/S \\ O&{} S \\ \end{array} \right)$, where K is the classical right ring of quotients of S. Take the right R-module $M=\left( \begin{array}{cc} S &{} K/S \\ 0&{} 0 \\ \end{array} \right)$. By similar arguments as in (ii), we prove that M is a quasi-dual Baer module which is not dual Baer. We include this proof for the convenience of the reader: By [6, Example 12], we have ${\text {End}}_R(M) \cong S$. So ${\text {End}}_R(M)$ is a simple domain. Hence M is an indecomposable R-module which is quasi-dual Baer.

On the other hand, S contains a nonzero non-invertible element t as S is not a division ring. Set $r = \left( \begin{array}{rr} t &{} 0 \\ 0&{} 0 \\ \end{array} \right)$ and consider the R-endomorphism $f_t$ of M defined by $f_t(x)=rx$ for all $x \in M$. It is easily seen that

$$f_t(M) = \left\{ \left( \begin{array}{rr} ts &{} \overline{tk} \\ 0&{} 0 \\ \end{array} \right) \mid (s, k) \in S \times K \right\} .$$

Since t is non-invertible in S and S is a domain, we have $ts \not = 1$ for every $s \in S$ (see [5, Chapter 1, §1, p. 22, Exercise 1.4]). Hence $f_t$ is not an epimorphism. As M is indecomposable, M is not a dual Baer R-module by [4, Corollary 2.2].

It is clear that for a duo module M, M is quasi-dual Baer if and only if M is a dual Baer module. In particular, for a right duo ring R (e.g., R is a commutative ring), the R-module $R_R$ is quasi-dual Baer if and only if $R_R$ is a dual Baer R-module if and only if R is a semisimple ring by [4, Corollary 2.9]. Next, we characterize the class of rings R for which the right R-module $R_R$ is quasi-dual Baer. A ring R is called a right quasi-dual Baer ring if the right R-module $R_R$ is a quasi-dual Baer R-module. Left quasi-dual Baer rings are defined similarly.

Proposition 2.10

The following are equivalent for a ring R:

• (i) R is a right quasi-dual Baer ring;

• (ii) For every left ideal I of R, the right ideal IR is a direct summand of the right R-module $R_R$;

• (iii) For every two-sided ideal I of R, the right ideal I is a direct summand of the right R-module $R_R$;

• (iv) R is a finite product of simple rings.

Proof

Given $a \in R$, let $\varphi _a$ denote the R-endomorphism of the R-module $R_R$ defined by $\varphi _a(x)=ax$ for all $x \in R$.

(i) $\Rightarrow$ (ii) Let I be a left ideal of R. It is not hard to see that the set $\mathfrak {I} = \{\varphi _a \mid a \in I\}$ is a left ideal of the ring $S={\text {End}}_R(R_R)$. By Proposition 2.4, the right ideal $\mathfrak {I}(R_R) = \sum _{a \in I} aR = IR$ is a direct summand of $R_R$.

(ii) $\Rightarrow$ (iii) is immediate.

(iii) $\Rightarrow$ (iv) We claim that R is a semiprime ring. Suppose, contrary to our claim, that R contains a nonzero two-sided ideal I with $I^2=0$. By assumption, there exists a right ideal K of R such that $I \oplus K = R$. This implies that $I^2 \oplus KI = I$. Hence $KI=I \subseteq K$. Therefore, $I=0$, a contradiction. Applying [3, Lemma 3.1], we deduce that every two-sided ideal of R is generated by a central idempotent. From [10, 3.5], it follows that R is a finite product of simple rings.

(iv) $\Rightarrow$ (i) Let I be a fully invariant submodule of the R-module $R_R$. Then, I is a two-sided ideal of R. By [10, 3.5], I is a direct summand of $R_R$. The result follows from Corollary 2.2. $\square$

Applying Proposition 2.10, we obtain the following result which shows that the quasi-dual Baer property is left–right symmetric for any ring.

Corollary 2.11

A ring R is right quasi-dual Baer if and only if it is left quasi-dual Baer.

In the next proposition, we provide a characterization of indecomposable quasi-dual Baer modules over any ring R.

Proposition 2.12

Let M be an indecomposable R-module. Then, the following are equivalent:

• (i) M is quasi-dual Baer;

• (ii) For every fully invariant proper submodule N of M, ${\text {Hom}}_R(M, N)=0$.

Proof

(i) $\Rightarrow$ (ii) This follows from Proposition 2.1.

(ii) $\Rightarrow$ (i) This follows from the fact that $\mathfrak {D}(M)={\text {End}}_R(M)$ and $\mathfrak {D}(N)=0$ for every fully invariant proper submodule N of M. $\square$

Corollary 2.13

Let R be a local ring with maximal right ideal m. Assume that $M=E(R/m)$ is a quasi-dual Baer module. Then R/m is injective (i.e., R is a division ring) or ${\text {Rad}}(M)=M$.

Proof

Put $S=R/m$. Assume that S is an injective R-module. Then R is a right V-ring. This yields $m=0$ and hence R is a division ring. Now assume that $E(S) \not = S$. Suppose that M has a maximal submodule L. As R is a local ring, there exists an R-isomorphism $\lambda :M/L \rightarrow S$. Let $\pi :M \rightarrow M/L$ denote the natural epimorphism. Then, $\lambda \pi \in {\text {Hom}}_R(M, S)$ and $\lambda \pi \not = 0$. Therefore, ${\text {Hom}}_R(M, S) \not = 0$. From Proposition 2.12, it follows that S is not fully invariant in M, a contradiction. Consequently, ${\text {Rad}}(M)=M$. $\square$

A ring R is called right Bass if ${\text {Rad}}(M) \not = M$ for every nonzero right R-module M. It is well known that every right perfect ring is right Bass (see [2, Remark 28.5(3)]).

Corollary 2.14

Let R be a local ring with maximal right ideal m and $S=R/m$. Assume that ${\text {Rad}}(E(S)) \not = E(S)$ (e.g., R is a local right perfect ring). Then, the following are equivalent:

• (i) E(S) is quasi-dual Baer;

• (ii) R is a division ring.

Proof

This follows from Corollary 2.13. $\square$

Next, we deal with some special cases of direct sums of quasi-dual Baer modules.

Proposition 2.15

Let an R-module $M=M_1 \oplus M_2$ be a direct sum of submodules $M_1$ and $M_2$ such that ${\text {Rad}}(M_1)=M_1$ and $M_2$ is semisimple. Let $S={\text {End}}_R(M)$. Then, the following are equivalent:

• (i) M is quasi-dual Baer;

• (ii) $M_1$ is quasi-dual Baer and ${\mathfrak {I}}(M_2) \cap M_1 \subseteq {\mathfrak {I}}(M_1)$ for every left ideal $\mathfrak {I}$ of S;

• (iii) $M_1$ is quasi-dual Baer and ${\mathfrak {I}}(M_2) \cap M_1 \subseteq {\mathfrak {I}}(M_1)$ for every two-sided ideal $\mathfrak {I}$ of S.

Proof

• (i) $\Rightarrow$ (ii) Clearly, $M_1$ is a quasi-dual Baer module by Corollary 2.5. Now let $\mathfrak {I}$ be a left ideal of S. Therefore, $\mathfrak {I}(M)$ is a fully invariant submodule of M. Then $\mathfrak {I}(M) = (\mathfrak {I}(M) \cap M_1) \oplus (\mathfrak {I}(M)\cap M_2)$. Note that $\mathfrak {I}(M) = K_1 + K_2$, where $K_1={\mathfrak {I}}(M_1)$ and $K_2={\mathfrak {I}}(M_2)$. Moreover, $M_1={\text {Rad}}(M)$ is a fully invariant submodule of M. Therefore, $K_1 \subseteq M_1$. By modularity, we have $\mathfrak {I}(M) \cap M_1 = K_1 + (K_2\cap M_1)$. This yields $\mathfrak {I}(M) = (K_1 + (K_2\cap M_1)) \oplus (\mathfrak {I}(M)\cap M_2)$. Since $K_2\cap M_1$ is semisimple, there exists a semisimple submodule N of $K_2\cap M_1$ such that $K_1 + (K_2\cap M_1) = K_1 \oplus N$. Therefore, $\mathfrak {I}(M)=(K_1 \oplus N) \oplus (\mathfrak {I}(M)\cap M_2)$. As M is quasi-dual Baer, $\mathfrak {I}(M)$ is a direct summand of M by Proposition 2.4. This implies that $K_1 \oplus N$ is a direct summand of $M_1$. Thus ${\text {Rad}}(N)=N$ as ${\text {Rad}}(M_1)=M_1$. But N is semisimple. Then $N=0$. It follows that $\mathfrak {I}(M)=K_1 \oplus (\mathfrak {I}(M)\cap M_2)$. Hence, $K_1+(K_2\cap M_1) = K_1$. We thus get $K_2 \cap M_1 \subseteq K_1$.

• (ii) $\Rightarrow$ (iii) This is clear.

• (iii) $\Rightarrow$ (i) Let $\mathfrak {I}$ be a nonzero two-sided ideal of $S={\text {End}}_R(M)$. Let $\pi _1:M \rightarrow M_1$ be the projection map and $\mu _1: M_1 \rightarrow M$ be the inclusion map. Consider the subset $\mathfrak {I}_1=\{\pi _1f\mu _1 \mid f\in \mathfrak {I}\}$ of $S_1={\text {End}}_R(M_1)$. It is easily seen that $\mathfrak {I}_1$ is a left ideal of $S_1$. Since $M_1$ is quasi-dual Baer, $N_1={\mathfrak {I}}(M_1)=\mathfrak {I}_1(M_1)$ is a direct summand of $M_1$ by Proposition 2.4. Since $\mathfrak {I}(M)$ is a fully invariant submodule of M, we have $\mathfrak {I}(M) = (\mathfrak {I}(M) \cap M_1) \oplus (\mathfrak {I}(M) \cap M_2)$. As $M_2$ is semisimple, there exists a semisimple submodule $L \le {\mathfrak {I}}(M_2)$ such that $\mathfrak {I}(M) = N_1 + {\mathfrak {I}}(M_2) = N_1 \oplus L$. By modularity, we have $\mathfrak {I}(M) \cap M_1 = N_1 \oplus (L \cap M_1)$. By hypothesis, $L \cap M_1 \subseteq N_1$. Thus $L \cap M_1 = 0$. Therefore, $\mathfrak {I}(M) \cap M_1 = N_1$ is a direct summand of $M_1$. Since $\mathfrak {I}(M) \cap M_2$ is a direct summand of $M_2$, $\mathfrak {I}(M)$ is a direct summand of M. Using again Proposition 2.4, it follows that M is quasi-dual Baer. $\square$

The next result provides a necessary condition for a direct sum of a module $M_1$ with ${\text {Rad}}(M_1)=M_1$ and a semisimple module $M_2$ to be quasi-dual Baer.

Corollary 2.16

Let an R-module $M=M_1 \oplus M_2$ be a direct sum of submodules $M_1$ and $M_2$ such that ${\text {Rad}}(M_1)=M_1$ and $M_2$ is semisimple. Assume that M is quasi-dual Baer. Then, $Im \varphi$ is not fully invariant in $M_1$ for all $0 \not = \varphi \in Hom_R(M_2, M_1)$.

Proof

Suppose to the contrary that there exists a nonzero $\varphi \in {\text {Hom}}_R(M_2, M_1)$ such that $Im \varphi$ is fully invariant in $M_1$. Consider the endomorphism $\psi$ of M defined by $\psi (x_1)=0$ for every $x_1 \in M_1$ and $\psi (x_2)=\varphi (x_2)$ for every $x_2 \in M_2$. Let $S={\text {End}}_R(M)$ and consider the left ideal $\mathfrak {I}=S\psi$ of S. By Proposition 2.15, we have ${\mathfrak {I}}(M_2) \cap M_1 \subseteq {\mathfrak {I}}(M_1)$. Since $\varphi (M_2)$ is fully invariant in $M_1$ and $M_1={\text {Rad}}(M)$ is fully invariant in M, it follows that $\varphi (M_2)=\psi (M)$ is fully invariant in M. Note that ${\mathfrak {I}}(M_2) = \varphi (M_2)$ and ${\mathfrak {I}}(M_1) = 0$. So $\varphi (M_2) \cap M_1 = 0$; that is $\varphi (M_2) = 0$. This contradicts the choice of $\varphi$. $\square$

Example 2.17

Let L be a simple R-module such that ${\text {Rad}}(E(L))=E(L)$ (e.g., we can take a simple module L over a Dedekind domain R). Consider the R-module $M=E(L) \oplus L$ and let $\mu :L \rightarrow E(L)$ be the inclusion map. Note that $Im \mu$ is fully invariant in E(L). By Corollary 2.16, M is not quasi-dual Baer.

The next corollary is a direct consequence of Proposition 2.15.

Corollary 2.18

Let an R-module $M=M_1 \oplus M_2$ be a direct sum of submodules $M_1$ and $M_2$ such that ${\text {Rad}}(M_1)=M_1$, $M_2$ is semisimple and ${\text {Hom}}_R(M_2, M_1)=0$. Then, the following are equivalent:

• (i) M is quasi-dual Baer;

• (ii) $M_1$ is quasi-dual Baer.

We conclude this section by providing conditions under which a direct sum of quasi-dual Baer modules is quasi-dual Baer.

Proposition 2.19

Let $M=\oplus _{\lambda \in \Lambda } M_{\lambda }$ such that each $M_{\lambda }$ $($ $\lambda \in \Lambda$ $)$ is a fully invariant submodule of M which is a quasi-dual Baer R-module. Then M is quasi-dual Baer.

Proof

Let $S={\text {End}}_R(M)$ and for each $\lambda \in \Lambda$, let $S_\lambda ={\text {End}}_R(M_\lambda )$. Let $\mathfrak {I}$ be a left ideal of the ring S. Since $M_{\lambda }$ is fully invariant in M for all $\lambda \in \Lambda$, we obtain $\mathfrak {I}(M)=\oplus _{\lambda \in \Lambda } \mathfrak {I}(M_{\lambda })$. For each $\lambda \in \Lambda$ and $f \in S$, let $f_\lambda$ denote the endomorphism of $M_{\lambda }$ defined by $f_\lambda (x)=f(x)$ for all $x \in M_{\lambda }$. It is not hard to see that $\mathfrak {I}_\lambda = \{f_\lambda \mid f \in \mathfrak {I}\}$ is a left ideal of $S_\lambda$. Since each $M_{\lambda }$ $($ $\lambda \in \Lambda$ $)$ is quasi-dual Baer, it follows that $\mathfrak {I}_\lambda (M_{\lambda })$ is a direct summand of $M_{\lambda }$ for every $\lambda \in \Lambda$ (see Proposition 2.4). Therefore $\mathfrak {I}(M)$ is a direct summand of M. Hence, by Proposition 2.4, we conclude that M is a quasi-dual Baer module. $\square$

3 Modules over commutative rings

This section is devoted to the study of some properties of quasi-dual Baer modules over commutative rings. Throughout this section, R will denote a commutative ring, unless otherwise stated.

Proposition 3.1

Let R be a commutative ring and let M be a cyclic R-module. Then M is quasi-dual Baer if and only if M is semisimple.

Proof

Without loss of generality, we can assume that $M=R/I$ for some ideal I of R. It is easily seen that the module $M_R$ is quasi-dual Baer if and only if the module $M_{R/I}$ is quasi-dual Baer. The result follows from Proposition 2.10. $\square$

A module M is said to have the strong summand sum property (SSSP for short) if the sum of any family of direct summands of M is a direct summand of M. The class of dual Baer modules was characterized in [4, Theorem 2.1] as that of modules M having the SSSP such that for every $\varphi : M \rightarrow M$, $Im \varphi$ is a direct summand of M. To obtain a partial analogue of this characterization for quasi-dual Baer modules, Amouzegar and Talebi [1] introduced the FI-strong summand sum property (FI-SSSP for short). A module M is said to have the FI-SSSP if the sum of any family of fully invariant direct summands of M is a direct summand of M. The next result is taken from [1, Lemmas 2.1 and 2.2].

Lemma 3.2

Let R be a ring (not necessarily commutative) and let M be a quasi-dual Baer module. Then:

• (1) M has the FI-SSSP, and

• (2) for every endomorphism $\varphi$ of M such that $Im \varphi$ is fully invariant in M, $Im \varphi$ is a direct summand of M.

Proposition 3.3

Let M be a module over a commutative ring R. If M is quasi-dual Baer, then $M\mathfrak {a}$ is a direct summand of M for any ideal $\mathfrak {a}$ of R.

Proof

Let $\mathfrak {a}$ be an ideal of R and let $r \in \mathfrak {a}$. It is clear that $Mr=Im f_r$ where $f_r$ is the R-endomorphism of M defined by $f_r(x)=xr$ for every $x \in M$. Moreover, it is easily seen that Mr is a fully invariant submodule of M. Now using Lemma 3.2, it follows that $Im f_r$ is a direct summand of M for all $r \in \mathfrak {a}$. Note that M has the FI-SSSP (see Lemma 3.2). Therefore, $M\mathfrak {a} = \sum _{r \in \mathfrak {a}}Mr = \sum _{r \in \mathfrak {a}} Im f_r$ is a direct summand of M. $\square$

The converse of Proposition 3.3 is not true, in general, as shown in the following example. This example shows also that the class of quasi-dual Baer modules is not closed under direct products.

Example 3.4

Let $\mathbf {P}$ denote the set of all prime numbers. Consider the $\mathbb {Z}$-module $M=\prod _{p \in \mathbf {P}} \mathbb {Z}/p\mathbb {Z}$. Then the torsion submodule of M is $T(M) = \oplus _{p \in \mathbf {P}} \mathbb {Z}/p\mathbb {Z}$. For each $p \in \mathbf {P}$, let $f_p$ denote the endomorphism of M defined by $(x_p)_{p \in \mathbf {P}} \mapsto (y_p)_{p \in \mathbf {P}}$ such that $y_q=0$ for every $q \not = p$ and $y_p=x_p$. Let $\mathfrak {I}$ denote the left ideal of $S={\text {End}}_{\mathbb {Z}}(M)$ generated by all the $f_p$ ($p \in \mathbf {P}$). Then $\mathfrak {I}(M)=T(M)$ is not a direct summand of M. Therefore M is not quasi-dual Baer by Proposition 2.4. On the other hand, it is easily seen that $M(n\mathbb {Z})=Mn$ is a direct summand of M for every integer n.

Corollary 3.5

Let M be a nonzero projective module over a commutative ring R. If M is a quasi-dual Baer R-module, then ${\text {Rad}}(M)=0$.

Proof

Assume that M is quasi-dual Baer. Let $J={\text {Rad}}(R)$. Since M is projective, we have ${\text {Rad}}(M)=MJ$ by [2, Proposition 17.10]. From Proposition 3.3, it follows that ${\text {Rad}}(M)$ is a direct summand of M. So there exists a submodule N of M such that $M={\text {Rad}}(M) \oplus N$. Clearly, we have ${\text {Rad}}(N)=0$. Therefore, ${\text {Rad}}({\text {Rad}}(M))={\text {Rad}}(M)$. Thus, ${\text {Rad}}(M)=0$ by [2, Proposition 17.14]. $\square$

Remark 3.6

Let R be a commutative ring and let M be an indecomposable quasi-dual Baer R-module. Using Proposition 3.3 and applying similar arguments as in the proof of [9, Proposition 4.4], we conclude that either ${\text {Rad}}(M)=M$ or M is a simple R-module. But this is not true, in general, if the ring R is not commutative. Indeed, we can consider the example given in [9, Example 4.10 and Remark 4.11].

Corollary 3.7

Let M be an indecomposable $\mathbb {Z}$-module. Then, the following are equivalent:

• (i) M is quasi-dual Baer;

• (ii) M is dual Baer;

• (iii) $M \cong \mathbb {Q}$ or $M \cong \mathbb {Z}(p^{\infty })$ or $M \cong \mathbb {Z}/p\mathbb {Z}$, where p is a prime number.

Proof

(i) $\Rightarrow$ (iii) Note that for every $\mathbb {Z}$-module N with $Rad(N)=N$, N is an injective module. Now, this implication follows easily from Remark 3.6.

(iii) $\Rightarrow$ (ii) This follows from [4, Corollary 3.5].

(ii) $\Rightarrow$ (i) This is immediate. $\square$

Recall that a ring R is called semilocal if $R/{\text {Rad}}(R)$ is a semisimple ring. The next result characterizes quasi-dual Baer modules over commutative semilocal rings.

Theorem 3.8

Let M be a nonzero module over a commutative semilocal ring R and let $S={\text {End}}_R(M)$. Then, the following conditions are equivalent:

• (i) M is a quasi-dual Baer R-module;

• (ii) (a) $M=M_1 \oplus M_2$ is a direct sum of submodules $M_1$ and $M_2$ such that ${\text {Rad}}(M_1)=M_1$ is quasi-dual Baer and $M_2$ is semisimple, and

• (iii) (b) ${\mathfrak {I}}(M_2) \cap M_1 \subseteq {\mathfrak {I}}(M_1)$ for every two-sided ideal $\mathfrak {I}$ of S.

Proof

(i)$\Rightarrow$ (ii) Let $J={\text {Rad}}(R)$. By [2, Corollary 15.18], we have ${\text {Rad}}(M)=MJ$. Applying Proposition 3.3, we deduce that ${\text {Rad}}(M)$ is a direct summand of M. Set $M_1={\text {Rad}}(M)$ and let $M_2$ be a submodule of M such that $M=M_1 \oplus M_2$. Clearly, ${\text {Rad}}(M_2) = M_2J=0$ (see [2, Corollary 15.18]). Therefore $M_2$ is an R/J-module. Since R/J is a semisimple ring, it follows that $M_2$ is a semisimple R-module. Note that $M_1$ is quasi-dual Baer by Corollary 2.5. This proves (a). Note that (b) follows directly from Proposition 2.15.

(ii) $\Rightarrow$ (i) By Proposition 2.15. $\square$

In the following result, we describe the structure of quasi-dual Baer modules over commutative perfect rings.

Corollary 3.9

Let M be a nonzero module over a commutative perfect ring R. Then the following are equivalent:

• (i) M is a quasi-dual Baer module;

• (ii) M is a dual Baer module;

• (iii) M is a semisimple module.

Proof

(i) $\Rightarrow$ (iii) Since R is a semilocal ring (see [2, Theorem 28.4]), there exists a semisimple submodule N of M such that $M={\text {Rad}}(M) \oplus N$ by Theorem 3.8. But ${\text {Rad}}(M) \ll M$ by [2, Remark 28.5(3)]. Then, ${\text {Rad}}(M)=0$. It follows that $M=N$ is a semisimple R-module.

(iii) $\Rightarrow$ (ii) $\Rightarrow$ (i) These implications are obvious. $\square$