On strongly coseparable modules

Abstract

A module M is called $\mathfrak {s}$-coseparable if for every nonzero submodule U of M such that M/U is finitely generated, there exists a nonzero direct summand V of M such that $V \subseteq U$ and M/V is finitely generated. It is shown that every non-finitely generated free module is $\mathfrak {s}$-coseparable but a finitely generated free module is not, in general, $\mathfrak {s}$-coseparable. We prove that the class of $\mathfrak {s}$-coseparable modules over a right noetherian ring is closed under finite direct sums. We show that the class of commutative rings R for which every cyclic R-module is $\mathfrak {s}$-coseparable is exactly that of von Neumann regular rings. Some examples of modules M for which every direct summand of M is $\mathfrak {s}$-coseparable are provided.

1 Introduction

Throughout this paper, all rings are associative with an identity element and R denotes such a ring. All modules considered are unital right R-modules, unless otherwise specified. Let M be an R-module. The injective hull of M will be denoted by E(M). The notation $N \subseteq M$ (respectively, $N \le M$) means that N is a subset (respectively, a submodule) of M. By ${\mathbb {N}}$ and ${\mathbb {Z}}$ we denote the set of natural numbers and the ring of integers, respectively.

A module M is called coseparable if every submodule U of M with M/U finitely generated contains a direct summand V of M such that M/V is finitely generated. This concept was initially appeared in 1968 [9] for abelian groups. Then the concept was generalized to modules in 1978 [12] by Hiremath and was further studied by other authors (see e.g., [8, 19]). In this paper, we focus on a subclass of coseparable modules. A module M is said to be strongly coseparable (or ${\mathfrak {s}}$-coseparable) if for every nonzero submodule U of M with the property that M/U is finitely generated, there exists a nonzero direct summand $U^{\prime }$ of M such that $U^{\prime }\subseteq U$ and $M/U^{\prime }$ is finitely generated.

In Sect. 2, we study some basic properties of ${\mathfrak {s}}$-coseparable modules. We begin by giving many examples belonging to this class of modules. In particular, we show that the class of ${\mathfrak {s}}$-coseparable modules is a proper subclass of that of coseparable modules. But we prove that for a non-finitely generated module M, M is coseparable if and only if M is ${\mathfrak {s}}$-coseparable. It is shown that a direct summand of an ${\mathfrak {s}}$-coseparable module is not, in general, ${\mathfrak {s}}$-coseparable. On the other hand, we show that the class of finitely generated ${\mathfrak {s}}$-coseparable modules is closed under direct summands. Moreover, we prove that the class of ${\mathfrak {s}}$-coseparable modules over a right noetherian ring is closed under finite direct sums.

Section 3 is devoted to shed some light on a proper subclass of that of ${\mathfrak {s}}$-coseparable modules. A module M is called completely ${\mathfrak {s}}$-coseparable if every direct summand of M is ${\mathfrak {s}}$-coseparable. Some examples of completely ${\mathfrak {s}}$-coseparable modules are provided.

In Sect. 4, we characterize several classes of rings in terms of ${\mathfrak {s}}$-coseparable modules. The class of rings R for which every finitely generated projective R-module is ${\mathfrak {s}}$-coseparable, is shown to be exactly that of the weakly regular semiprimitive rings; while the class of commutative rings R for which every cyclic R-module is ${\mathfrak {s}}$-coseparable is characterized as that of the von Neumann regular rings.

2 Some examples and properties of strongly coseparable modules

Recall that a module M is called coseparable if for every submodule U of M such that M/U is finitely generated, there exists a direct summand V of M such that $V \subseteq U$ and M/V is finitely generated (see e.g., [9, 12, 19]). It is clear that a finitely generated module is always coseparable. In this paper, we introduce and study a subclass of coseparable modules as follows:

Definition 2.1

A module M is said to be strongly coseparable (or ${\mathfrak {s}}$-coseparable for short) if for every nonzero submodule U of M with the property that M/U is finitely generated, there exists a nonzero direct summand $U^{\prime }$ of M such that $U^{\prime }\subseteq U$ and $M/U^{\prime }$ is finitely generated.

A submodule of a module M is called cofinite if M/N is finitely generated.

Remark 2.2

It is easily seen that a finitely generated module is ${\mathfrak {s}}$-coseparable if and only if every nonzero submodule of M contains a nonzero direct summand. A submodule N of a module M is said to be small in M, written $N<< M$, provided $N + X, \ne M$ for any proper submodule X of M.

It is easily seen that any semisimple module is ${\mathfrak {s}}$-coseparable. In addition, any module M with $Rad(M)=M$ is ${\mathfrak {s}}$-coseparable. In this section, we present more examples of ${\mathfrak {s}}$-coseparable modules. Recall that a module M is called regular if every finitely generated submodule of M is a direct summand.

Example 2.3

Let $M=x_1R + \cdots +x_nR$ be a finitely generated regular module. Let U be a nonzero cofinite submodule of M. Since $Rad(M)=0$, there exists a proper submodule N of M such that $M=U+N$. For each $i \in \{1, \ldots , n\}$, there exist $u_i \in U$ and $y_i \in N$ such that $x_i=u_i+y_i$. Let $U^{\prime }$ be the submodule of U generated by $\{u_1, \ldots , u_n\}$. Since $M=U^{\prime } + N$ and $N \ne M$, we have $U^{\prime } \ne 0$. Moreover, $U^{\prime }$ is a direct summand of M as M is regular. Clearly, $M/U^{\prime }$ is finitely generated. Therefore, M is ${\mathfrak {s}}$-coseparable.

In particular, the right R-module $R_R$ is ${\mathfrak {s}}$-coseparable for every von Neumann regular ring R.

A module M is said to be A-projective if for every submodule X of A, any homomorphism $\varphi :M \rightarrow A/X$ can be lifted to a homomorphism $\psi :M \rightarrow A$.

Example 2.4

(i) Let a module $M=M_1 \oplus M_2$ be a direct sum of a finitely generated submodule $M_1$ and a semisimple submodule $M_2$ which is not finitely generated. Assume that $M_2$ is $M_1$-projective (e.g., $M_2$ can be projective). We want to prove that M is ${\mathfrak {s}}$-coseparable. To do this, let U be a cofinite submodule of M. Then $M=U+K$ for some finitely generated submodule K of M. Note that $K \subseteq M_1 \oplus K_2$ for some finitely generated direct summand $K_2$ of $M_2$. Hence, $M = U + (M_1 \oplus K_2)$. Moreover, we have $M=L_2 \oplus (M_1 \oplus K_2)$ for some submodule $L_2$ of $M_2$. Using [6, 4.3], it follows that $L_2$ is $(M_1 \oplus K_2)$-projective. Therefore, $M=U^{\prime } \oplus (M_1 \oplus K_2)$ for some submodule $U^{\prime }$ of U by [6, 4.12]. Clearly, $U^{\prime } \ne 0$ as M is not finitely generated. In addition, $M/U^{\prime }$ is finitely generated. Consequently, M is an ${\mathfrak {s}}$-coseparable module.

(ii) Let T be a commutative ring and let ${\mathfrak {m}}$ be a maximal ideal of T. Consider the ring $R=T \times T/{\mathfrak {m}}$. Taking (i) into account, it follows that for any finitely generated R-module $M_1$, the R-module $M_1 \oplus (0 \oplus T/{\mathfrak {m}})^{({\mathbb {N}})}$ is ${\mathfrak {s}}$-coseparable.

In the following proposition, we determine the structure of the indecomposable ${\mathfrak {s}}$-coseparable modules.

Proposition 2.5

Let M be an indecomposable module. Then the following statements are equivalent:

(i) M is an ${\mathfrak {s}}$-coseparable module;

(ii) $Rad(M)=M$ or M is a simple module.

Proof

(i)$\Rightarrow$(ii) Assume that $Rad(M) \ne M$. Then M contains a maximal submodule N. Suppose that $N \ne 0$. Since M is ${\mathfrak {s}}$-coseparable, there exists a nonzero direct summand $N^{\prime }$ of M such that $N^{\prime } \subseteq N$. Hence, $N^{\prime } = M$ as M is indecomposable. This contradicts the fact that $N \ne M$. Therefore, $N=0$ and hence M is a simple module.

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

It is easily seen that every ${\mathfrak {s}}$-coseparable module is coseparable. But the converse is not true, in general, as illustrated in the next example.

Example 2.6

Let M be an indecomposable finitely generated module which is not simple (e.g., we can take $M=R$ where R is any commutative integral domain which is not a field or $M={\mathbb {Z}}/p^{n}{\mathbb {Z}}$ for some prime number p and an integer $n \ge 2$). Then M is not ${\mathfrak {s}}$-coseparable by Proposition 2.5. On the other hand, M is coseparable since M is finitely generated.

A module M is called weakly regular (or an $I_0$-module) if every submodule, which is not contained in Rad(M), contains a nonzero direct summand of M (see, e.g., [1, 16]). A ring R is called weakly regular (or an $I_0$-ring as in [16, p. 131]) if the right R-module $R_R$ is weakly regular. Equivalently, if the left R-module $_RR$ is weakly regular. Many examples of weakly regular rings are given in [16].

The following characterization provides a rich source of examples of ${\mathfrak {s}}$-coseparable modules (see [8]).

Theorem 2.7

The following conditions are equivalent for an R-module M:

(i):

M is ${\mathfrak {s}}$-coseparable;

(ii):

Either (a) M is coseparable and not finitely generated, or (b) $Rad(M)=0$ and M is a finitely generated module which is weakly regular.

Proof

(i) $\Rightarrow$ (ii) Clearly, M is coseparable. Assume that M is finitely generated. Suppose that $Rad(M) \ne 0$. By hypothesis, Rad(M) contains a nonzero direct summand U of M. But $Rad(M) \ll M$. Therefore, $U \ll M$ and hence $U=0$, a contradiction. It follows that $Rad(M)=0$. It is clear that M is weakly regular.

(a) $\Rightarrow$ (i) Let U be a nonzero cofinite submodule of M. As M is coseparable, U contains a direct summand $U^{\prime }$ of M such that $M/U^{\prime }$ is finitely generated. Thus $U^{\prime } \not = 0$ since M is not finitely generated. It follows that M is ${\mathfrak {s}}$-coseparable.

(b) $\Rightarrow$ (i) This is obvious. $\square$

Following [6, 11.26], a module M is called refinable if, for any submodules N and L of M with $N+L=M$, there exists a direct summand $N^{\prime }$ of M with $N^{\prime } \subseteq N$ and $N^{\prime }+L=M$.

Proposition 2.8

Let M be a refinable module. Assume that $Rad(M)=0$ or M is not finitely generated. Then M is ${\mathfrak {s}}$-coseparable.

Proof

Case 1: Assume that $Rad(M)=0$. Let U be a nonzero cofinite submodule of M. Since $Rad(M)=0$, U is not small in M. Thus $M=U+N$ for some proper submodule N of M. Note that $N/(U\cap N) \cong M/U$ is finitely generated. Then $N=(U\cap N)+ K$ for some finitely generated submodule K of N. Therefore, $M=U + K$. Note that $K\ne M$ since otherwise $N=M$. Since M is a refinable module, there exist a direct summand $U^{\prime }$ of M such that $U^{\prime } \subseteq U$ and $M=U^{\prime }+K$. Notice that $U^{\prime } \ne 0$ and $M/U^{\prime }$ is finitely generated. It follows that M is ${\mathfrak {s}}$-coseparable.

Case 2: Assume that M is not finitely generated. Then M is ${\mathfrak {s}}$-coseparable by [8, Example 2.3] and Theorem 2.7. $\square$

A module M is said to have the finite exchange property if for any finite index set I, whenever $M \oplus N = \oplus _{i \in I}A_i$ for modules N and $A_i$, then $M \oplus N=M \oplus (\oplus _{i \in I}B_i)$ for submodules $B_i \le A_i$. A ring R is called an exchange ring if the R-module $R_R$ has the finite exchange property. This is equivalent to saying that for any $r \in R$, there is an idempotent e in R with $e \in rR$ and $1-e \in (1-r)R$ (see [6, 11.16]).

As an application of the previous proposition, we obtain the following example.

Example 2.9

Let M be a quasi-projective module having the finite exchange property with $Rad(M)=0$. For example, we can take the T-module $M=R/Rad(R)$ where R is an exchange ring and $T=R/Rad(R)$. Then M is ${\mathfrak {s}}$-coseparable by Proposition 2.8 and [6, 11.31]. Using [16, Proposition 28.6], it follows that every finitely generated projective module M with $Rad(M)=0$ over an exchange ring is ${\mathfrak {s}}$-coseparable.

Recall that a module M is said to be $\pi$-projective if for every two submodules U, V of M with $U+V=M$ there exists $f \in End_R(M)$ such that $\text{ Im }(f) \subseteq U$ and $\text{ Im }(1-f) \subseteq V$. It is well known that every quasi-projective module is $\pi$-projective (see [18, p. 359]).

Proposition 2.10

Let M be a $\pi$-projective module such that every proper finitely generated submodule is contained in a proper finitely generated direct summand of M. Assume that $Rad(M)=0$ or M is not finitely generated. Then M is an ${\mathfrak {s}}$-coseparable module.

Proof

Case 1: Assume that $Rad(M)=0$. Let U be a nonzero cofinite submodule of M. By similar arguments as in the proof of Proposition 2.8, we have $M=U+K$ such that K is a finitely generated proper submodule of M. By hypothesis, there exists a finitely generated proper direct summand L of M such that $K \subseteq L$. Hence, $M=U+L$. Since M is $\pi$-projective, there exists a submodule $U^{\prime }$ contained in U such that $M=U^{\prime }\oplus L$ by [18, 41.14(3)]. Since $L\ne M$, we have $U^{\prime }\ne 0$. Hence, M is ${\mathfrak {s}}$-coseparable.

Case 2: Assume that M is not finitely generated. Then M is ${\mathfrak {s}}$-coseparable by [8, Proposition 2.5] and Theorem 2.7. $\square$

Example 2.11

(i) Let R be a ring and consider the R-module $M=R^{(I)}$, where I is an infinite index set. From Proposition 2.10, it follows that M is ${\mathfrak {s}}$-coseparable.

(ii) Let M be a $\pi$-projective regular module such that $Rad(M)=0$ or M is not finitely generated. Then M is ${\mathfrak {s}}$-coseparable by Proposition 2.10.

(iii) Let R be a von Neumann regular ring and let P be a projective R-module. By [16, Proposition 6.7(4)], P is a regular module. Moreover, we have $Rad(P)=0$ by [3, Proposition 17.10]. Therefore, P is ${\mathfrak {s}}$-coseparable by (ii).

Next, we exhibit some examples to illustrate that a direct summand of an ${\mathfrak {s}}$-coseparable module is not, in general, ${\mathfrak {s}}$-coseparable.

Example 2.12

Let R be a commutative domain which is not a field.

(i) Assume that $Rad(R)=0$ (e.g., we can take $R={\mathbb {Z}}$). By the preceding Example, $M=R^{({\mathbb {N}})}$ is an ${\mathfrak {s}}$-coseparable R-module. However, the right R-module $R_R$ is not ${\mathfrak {s}}$-coseparable by Proposition 2.5.

(ii) Let ${\mathfrak {m}}$ be a maximal ideal of R. Consider the ring $T=R \times R/{\mathfrak {m}}$. By Example 2.4(ii), the T-module $(R \oplus 0) \oplus (0 \oplus R/{\mathfrak {m}})^{({\mathbb {N}})}$ is ${\mathfrak {s}}$-coseparable. On the other hand, the T-module $R \oplus 0$ is not ${\mathfrak {s}}$-coseparable by Proposition 2.5.

The following two results should be contrasted with Example 2.12.

Proposition 2.13

Let M be an ${\mathfrak {s}}$-coseparable R-module over a right noetherian ring R. Let N be a fully invariant direct summand of M which is not finitely generated. Then N is ${\mathfrak {s}}$-coseparable.

Proof

Note that M is coseparable. Then N is coseparable by [8, Proposition 2.12]. Since N is not finitely generated, it follows that N is ${\mathfrak {s}}$-coseparable by Theorem 2.7. $\square$

Combining [8, Proposition 2.16] and Theorem 2.7, we obtain the following result.

Proposition 2.14

Let M be an R-module and let N be a fully invariant submodule of M. Assume that M is ${\mathfrak {s}}$-coseparable and M/N is not finitely generated. Then M/N is ${\mathfrak {s}}$-coseparable.

Next, motivated by Theorem 2.7, we will be concerned with finitely generated ${\mathfrak {s}}$-coseparable modules.

Definition 2.15

A module M is called ${\mathfrak {f}}{\mathfrak {g}}{\mathfrak {s}}$-coseparable if M is finitely generated and ${\mathfrak {s}}$-coseparable.

Proposition 2.16

(i) Any finitely generated submodule of an ${\mathfrak {f}}{\mathfrak {g}}{\mathfrak {s}}$-coseparable module M is also an ${\mathfrak {f}}{\mathfrak {g}}{\mathfrak {s}}$-coseparable module.

(ii) Any direct summand of an ${\mathfrak {f}}{\mathfrak {g}}{\mathfrak {s}}$-coseparable module M is also an ${\mathfrak {f}}{\mathfrak {g}}{\mathfrak {s}}$-coseparable module.

Proof

(i) Let N be a finitely generated submodule of an ${\mathfrak {f}}{\mathfrak {g}}{\mathfrak {s}}$-coseparable module M. Let U be a nonzero submodule of N. Since M is ${\mathfrak {s}}$-coseparable, there exists a nonzero direct summand $U^{\prime }$ of M such that $U^{\prime } \subseteq U$. It is clear that $U^{\prime }$ is a direct summand of N. Thus N is ${\mathfrak {f}}{\mathfrak {g}}{\mathfrak {s}}$-coseparable.

(ii) This follows easily from (i). $\square$

Corollary 2.17

Let R be a right noetherian ring and let M be an ${\mathfrak {s}}$-coseparable R-module. Then every cofinite submodule of M is also ${\mathfrak {s}}$-coseparable.

Proof

If M is finitely generated, then the result follows from Proposition 2.16(i). Now assume that M is not finitely generated. Let N be a cofinite submodule of M. Then clearly N is not finitely generated. Note that N is coseparable by [8, Proposition 2.15]. Therefore, N is ${\mathfrak {s}}$-coseparable by Theorem 2.7. $\square$

Proposition 2.18

Let M be an ${\mathfrak {s}}$-coseparable module. Then the following hold:

(i) Every indecomposable cofinite submodule of M is a direct summand of M.

(ii) If M is finitely generated, then every indecomposable submodule of M is a simple direct summand of M.

Proof

(i) Let U be a nonzero indecomposable cofinite submodule of M. Since M is ${\mathfrak {s}}$-coseparable, there exists a nonzero direct summand $U^{\prime }$ of M such that $U^{\prime } \subseteq U$. By modularity, $U^{\prime }$ is a direct summand of U. But U is indecomposable. Then $U=U^{\prime }$ and hence U is a direct summand of M.

(ii) Assume that M is ${\mathfrak {f}}{\mathfrak {g}}{\mathfrak {s}}$-coseparable. Let U be an indecomposable submodule of M. Then U is a direct summand of M by (i). Therefore, U is an ${\mathfrak {f}}{\mathfrak {g}}{\mathfrak {s}}$-coseparable module by Proposition 2.16(ii). Therefore, M is a simple module by Proposition 2.5. $\square$

Next, we provide some sufficient conditions for an ${\mathfrak {f}}{\mathfrak {g}}{\mathfrak {s}}$-coseparable module to be semisimple. The following corollary is a direct consequence of Proposition 2.18.

Corollary 2.19

Let M be a finitely generated module which is a sum of indecomposable submodules. Then M is ${\mathfrak {f}}{\mathfrak {g}}{\mathfrak {s}}$-coseparable if and only if M is semisimple.

A module M is said to be nitely cogenerated (or nitely embedded) if there exist nitely many simple modules $S_1, S_2, \ldots , S_k\, \mathrm{such}\ \mathrm{that}\ E(M)\cong E(S_1) \oplus E(S_2) \oplus ... \oplus E(S_k).$

Corollary 2.20

Let M be a finitely generated R-module. Assume that one of the following conditions is fulfilled:

(i) M has either the ascending chain or the descending chain condition on direct summands, or

(ii) M is noetherian or artinian, or

(iii) M is finitely cogenerated, or

(iv) R is a noetherian ring, or

(v) R is a semiperfect ring.

Then M is an ${\mathfrak {f}}{\mathfrak {g}}{\mathfrak {s}}$-coseparable module if and only if M is semisimple.

Proof

Applying [3, Proposition 10.14] and [18, 21.3(2), 41.6 and 42.6], we conclude that M is a sum of indecomposable submodules. Now the result follows from Corollary 2.19. $\square$

Proposition 2.21

Let M be a finitely generated R-module with essential socle. Then the following statements are equivalent:

(i) M is an ${\mathfrak {s}}$-coseparable module;

(ii) $Rad(M)=0$;

(iii) Every simple submodule of M is a direct summand of M.

Proof

(i) $\Rightarrow$ (ii) By Theorem 2.7.

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

(iii) $\Rightarrow$ (i) Let U be a nonzero submodule of M. Since Soc(M) is essential in M, U contains a simple submodule T. By (iii), T is a nonzero direct summand of M. This completes the proof. $\square$

A module M is called semiartinian if every nonzero factor module of M has a nonzero socle. A ring R is called a right semiartinian ring if the right R-module $R_R$ is semiartinian. The following corollary follows directly from Proposition 2.21.

Corollary 2.22

Let M be a finitely generated R-module which is a semiartinian module with $Rad(M)=0$. Then M is ${\mathfrak {f}}{\mathfrak {g}}{\mathfrak {s}}$-coseparable.

Recall that a ring R is called a right V-ring if every simple right R-module is an injective module.

Example 2.23

Let R be a right semiartinian right V-ring. Using Proposition 2.21, it follows that every finitely generated R-module is ${\mathfrak {f}}{\mathfrak {g}}{\mathfrak {s}}$-coseparable.

Next, we focus on the question of when is the direct sum of two ${\mathfrak {s}}$-coseparable modules, ${\mathfrak {s}}$-coseparable?

Combining Theorem 2.7 and [1, Theorem 2.9], we obtain the following proposition.

Proposition 2.24

Any finite direct sum of ${\mathfrak {f}}{\mathfrak {g}}{\mathfrak {s}}$-coseparable modules is ${\mathfrak {f}}{\mathfrak {g}}{\mathfrak {s}}$-coseparable.

Proposition 2.25

Let $M=M_1 \oplus M_2$ be a direct sum of ${\mathfrak {s}}$-coseparable submodules $M_1$ and $M_2$ such that $M_2$ is finitely generated and semisimple. Then M is ${\mathfrak {s}}$-coseparable.

Proof

Let U be a nonzero cofinite submodule of M. Then

$$\begin{aligned} M/U = ((M_1+U)/U) + ((M_2+U)/U). \end{aligned}$$

Since $(M_2+U)/U$ is semisimple, it follows that $(M_1+U)/U$ is a direct summand of M/U. Hence, $M_1/(U \cap M_1)$ is finitely generated.

Case 1: Assume that $U \cap M_1 \ne 0$. Since $M_1$ is ${\mathfrak {s}}$-coseparable, there exists a nonzero direct summand $U_1$ of $M_1$ such that $U_1 \subseteq U \cap M_1 \subseteq U$ and $M_1/U_1$ is finitely generated. This clearly implies that $U_1$ is a direct summand of M and $M/U_1$ is finitely generated.

Case 2: Assume that $U\cap M_1=0$. Then $M_1$ is finitely generated. Using Proposition 2.24, it follows that U contains a nonzero cofinite direct summand of M.

Consequently, M is ${\mathfrak {s}}$-coseparable. $\square$

Corollary 2.26

Let R be a right noetherian ring. Then any finite direct sum of ${\mathfrak {s}}$-coseparable R-modules is ${\mathfrak {s}}$-coseparable.

Proof

Let $M=M_1\oplus M_2$ be a direct sum of two ${\mathfrak {s}}$-coseparable submodules.

Case 1: Assume that $M_1$ is finitely generated. Then $M_1$ is semisimple by Corollary 2.20. Applying Proposition 2.25, we conclude that M is ${\mathfrak {s}}$-coseparable.

Case 2: Assume that both $M_1$ and $M_2$ are not finitely generated. Then M is ${\mathfrak {s}}$-coseparable by [8, Corollary 3.6] and Theorem 2.7. $\square$

A module M is called a duo module if for every submodule N of M and any endomorphism f of M, we have $f(N) \subseteq N$.

Proposition 2.27

Let $M=\oplus _{i\in I} M_i$ be a direct sum of nonzero submodules $M_i$ $(i\in I)$ such that for every submodule N of M, we have $N=\oplus _{i\in I}(N \cap M_i)$ (e.g., M is a duo module). Assume that $M_i$ is ${\mathfrak {s}}$-coseparable for all $i \in I$. Then M is ${\mathfrak {s}}$-coseparable.

Proof

Case 1: Assume that M is finitely generated. Then I is a finite set and each $M_i$ ($i \in I$) is finitely generated. Using Proposition 2.24, we see that M is ${\mathfrak {s}}$-coseparable.

Case 2: Assume that M is not finitely generated. Note that M is coseparable by [8, Proposition 3.16]. Hence, M is ${\mathfrak {s}}$-coseparable by Theorem 2.7. $\square$

3 Completely ${\mathfrak {s}}$-coseparable modules

Definition 3.1

A module M is called completely ${\mathfrak {s}}$ -coseparable if every direct summand of M is ${\mathfrak {s}}$-coseparable.

It is clear that semisimple modules are completely ${\mathfrak {s}}$-coseparable. In addition, every module M with $Rad(M)=M$ is completely ${\mathfrak {s}}$-coseparable.

Example 3.2

(i) From Proposition 2.16(ii), it follows that every ${\mathfrak {f}}{\mathfrak {g}}{\mathfrak {s}}$-coseparable module is completely ${\mathfrak {s}}$-coseparable.

(ii) By Example 2.12, the ${\mathbb {Z}}$-module $M={\mathbb {Z}}^{({\mathbb {N}})}$ is ${\mathfrak {s}}$-coseparable. On the other hand, M is not completely ${\mathfrak {s}}$-coseparable since ${\mathbb {Z}}$ is not ${\mathfrak {s}}$-coseparable by Proposition 2.5.

In the next proposition, we provide more examples of completely ${\mathfrak {s}}$-coseparable modules.

Proposition 3.3

Let $M=M_1\oplus M_2$ be a direct sum of a semisimple submodule $M_1$ and a submodule $M_2$ with $Rad(M_2)=M_2$. Then, M is completely ${\mathfrak {s}}$-coseparable.

Proof

If $M_2=0$, it is clear that M is completely ${\mathfrak {s}}$-coseparable. Now assume that $M_2 \ne 0$. Then, M is not finitely generated. Using Theorem 2.7, the proof is completed by showing that M is coseparable. Let U be a nonzero submodule of M with $M/U = ((M_1+U)/U) + ((M_2+U)/U)$ finitely generated. Since $(M_1+U)/U$ is semisimple, it follows that $(M_2+U)/U$ is a direct summand of M/U. Hence, $(M_2+U)/U$ is finitely generated. But $Rad((M_2+U)/U) = (M_2+U)/U$. Therefore, $(M_2+U)/U = 0$. This yields $M_2 \subseteq U$. By modularity, we have $U=(U \cap M_1) \oplus M_2$. Since $U \cap M_1$ is a direct summand of $M_1$, it follows that U is a direct summand of M. This completes the proof. $\square$

Recall that a module M is said to have finite uniform dimension if M does not contain an infinite independent set of submodules. Dually, a module M is said to have finite hollow dimension if M does not contain an infinite coindependent family of submodules; that is, for some $n \in {\mathbb {N}}$, there exists an epimorphism from M to a direct sum of n nonzero modules but no epimorphism from M to a direct sum of more than n nonzero modules (see, for example, [6, p. 47]). Next, we present some special cases where the converse of Proposition 3.3 holds.

Corollary 3.4

Let R be a right noetherian ring and let M be an R-module with $S = End_R(M)$. Assume that one of the following conditions is fulfilled:

(i) The ring S is semilocal;

(ii) The module $S_S$ has finite uniform dimension;

(iii) The module M has finite hollow dimension;

(iv) The module M has finite uniform dimension;

(v) The module M is artinian.

Then, M is completely ${\mathfrak {s}}$-coseparable if and only if $M=M_1 \oplus M_2$ is a direct sum of submodules $M_1$ and $M_2$ such that $Rad(M_1)=M_1$ and $M_2$ is finitely generated and semisimple.

Proof

Note that M satisfies ACC on direct summands by [8, Poof of Proposition 3.13].

$(\Rightarrow )$ By [8, Proposition 3.13], $M=M_1 \oplus M_2$ such that $Rad(M_1)=M_1$ and $M_2$ is finitely generated. Since M is completely ${\mathfrak {s}}$-coseparable, $M_2$ is ${\mathfrak {s}}$-coseparable. Clearly, $M_2$ has ACC on direct summands. Hence, $M_2$ is semisimple by Corollary 2.20.

$(\Leftarrow )$ By Proposition 3.3. $\square$

Proposition 3.5

Let R be a semiperfect ring and let M be a projective R-module. Then, M is completely ${\mathfrak {s}}$-coseparable if and only if M is semisimple.

Proof

The sufficiency follows by Proposition 3.3. For the necessity, suppose that M is completely ${\mathfrak {s}}$-coseparable. By [3, Theorem 27.12], M has a direct decomposition $M=\oplus _{i\in I}M_i$ such that each $M_i$ $(i \in I)$ is indecomposable. Therefore, $M_i$ is ${\mathfrak {s}}$-coseparable and projective for each $i\in I$. Note that $Rad(M_i) \ne M_i$ for each $i\in I$ (see [3, Proposition 17.14]). Using Proposition 2.5, it follows that each $M_i$ $(i \in I)$ is a simple submodule of M. Therefore, M is semisimple. $\square$

Proposition 3.6

Let R be a right noetherian ring and let M be an injective R-module. Then, M is completely ${\mathfrak {s}}$-coseparable if and only if $M=M_1 \oplus M_2$ is a direct sum of submodules $M_1$ and $M_2$ such that $Rad(M_1)=M_1$ and $M_2$ is semisimple.

Proof

The sufficiency is obvious. Conversely, suppose that M is completely ${\mathfrak {s}}$-coseparable. By [3, Theorem 25.6], $M=\oplus _{i\in I}M_i$ is a direct sum of indecomposable submodules $M_i$ $(i \in I)$. It follows that each $M_i$ $(i\in I)$ is ${\mathfrak {s}}$-coseparable. By Proposition 2.5, for each $i \in I$, we have $Rad(M_i)=M_i$ or $M_i$ is simple. The result follows. $\square$

A ring R is said to be an SSI-ring if every semisimple right R-module is injective.

Proposition 3.7

Let R be a ring. Then, the following conditions are equivalent:

(i) Every completely ${\mathfrak {s}}$-coseparable R-module is injective;

(ii) R is a right noetherian right V-ring.

Moreover, in this case, every completely ${\mathfrak {s}}$-coseparable R-module is semisimple.

Proof

(i) $\Rightarrow$ (ii) Let M be a semisimple module. Then, M is completely ${\mathfrak {s}}$-coseparable. By hypothesis, M is injective. Hence, R is an SSI-ring. Thus R is a right noetherian right V-ring by [5, Proposition 1].

(ii) $\Rightarrow$ (i) Let M be a completely ${\mathfrak {s}}$-coseparable module. By [5, Proposition 1], Soc(M) is injective. Then, $M=Soc(M) \oplus N$ for some submodule N of M with $Soc(N)=0$. Suppose $N\ne 0$. As R is a right V-ring, N contains a maximal submodule L. Since N is ${\mathfrak {s}}$-coseparable, $N=U \oplus K$ is a direct sum of submodules U and K such that $U \subseteq L$ and K is a nonzero finitely generated submodule of M. Note that K is ${\mathfrak {s}}$-coseparable. As R is a right noetherian ring, K is semisimple by Corollary 2.20. This implies that $Soc(N) \ne 0$, a contradiction. It follows that $N=0$ and $M=Soc(M)$ is injective. $\square$

Corollary 3.8

The following assertions are equivalent for a commutative ring R:

(i) Every completely ${\mathfrak {s}}$-coseparable R-module is injective;

(ii) R is a semisimple ring.

Proof

This is a consequence of [5, p. 236 Proposition 1 and Corollary] and Proposition 3.7. $\square$

Next, we present a special subclass of completely ${\mathfrak {s}}$-coseparable modules. Following [4], a module M is called an md-module if every maximal submodule of M is a direct summand of M. This is equivalent to say that every cofinite submodule of M is a direct summand of M by [2, Lemma 2.7]. Therefore, every md-module is ${\mathfrak {s}}$-coseparable. Moreover, the class of md-modules is closed under direct summands by [4, Proposition 2.1]. It follows that every md-module is completely ${\mathfrak {s}}$-coseparable.

Example 3.9

(i) Let R be a commutative noetherian ring and let $\lbrace U_\lambda \rbrace _{\lambda \in \Lambda }$ be a collection of simple R-modules. Then, $M=\prod _ {\lambda \in \Lambda } U_\lambda$ is an md-module by [4, Theorem 6.2]. This implies that M is completely ${\mathfrak {s}}$-coseparable.

(ii) Let a module $M=M_1 \oplus M_2$ be a direct sum of a submodule $M_1$ with $Rad(M_1)=M_1$ and a semisimple submodule $M_2$. From the proof of Proposition 3.3, it follows that M is an md-module.

Example 3.10

Let M be a regular R-module over a right noetherian ring R. Let U be a nonzero submodule of M with M/U finitely generated. Then, $M=U+K$ for some finitely generated submodule K of M. Since R is right noetherian, $U\cap K$ is a finitely generated submodule of M. But M is regular. Then, $U\cap K$ is a direct summand of M. Therefore, $U\cap K$ is a direct summand of K. Hence, $K=(U\cap K) \oplus L$ for some submodule L of K. Clearly, $M=U\oplus L$. Hence, M is an md-module. Therefore, M is completely ${\mathfrak {s}}$-coseparable.

4 Rings whose modules are ${\mathfrak {s}}$-coseparable

In this section, we characterize some classes of rings via ${\mathfrak {s}}$-coseparable modules. A ring R is called semiprimitive if $Rad(R)=0$.

We begin with the following result which provides a description of all rings R for which the right R-module $R_R$ is ${\mathfrak {s}}$-coseparable.

Theorem 4.1

The following statements are equivalent for a ring R:

(i) The R-module $R_R$ is ${\mathfrak {s}}$-coseparable;

(ii) R is a weakly regular semiprimitive ring;

(iii) For every $0 \not = a \in R$, there is $0 \not = x \in R$ with $x = xax$;

(iv) Every (finitely generated) free R-module F is ${\mathfrak {s}}$-coseparable;

(v) Every finitely generated projective R-module is ${\mathfrak {s}}$-coseparable;

(vi) For every finitely generated projective R-module P, $End_R(P)$ is a weakly regular semiprimitive ring.

Proof

(i) $\Leftrightarrow$ (ii) This follows from Theorem 2.7.

(ii) $\Leftrightarrow$ (iii) By [16, Remark 15.3(3)].

(i) $\Rightarrow$ (iv) Let F be a free R-module. Then, $Rad(F)=0$ by Theorem 2.7. If F is not finitely generated, then F is ${\mathfrak {s}}$-coseparable by Example 2.11(i). Now if F is finitely generated, then F is ${\mathfrak {s}}$-coseparable by Proposition 2.24.

(iv) $\Rightarrow$ (v) Let P be a finitely generated projective R-module. It is well known that P is isomorphic to a direct summand of a finitely generated free R-module F. Therefore, P is ${\mathfrak {s}}$-coseparable by Proposition 2.16.

(v) $\Rightarrow$ (i) This is immediate.

(ii) $\Rightarrow$ (vi) Let P be a finitely generated projective R-module. From [10, Theorem 3.8], it follows that $S=End_R(P)$ is a weakly regular ring. Moreover, by [3, Corollary 17.12], we have $Rad(S)=Hom_R(P, PRad(R))=0$. Thus S is a semiprimitive ring.

(vi) $\Rightarrow$ (ii) This follows from the fact that the rings $End_R(R_R)$ and R are isomorphic. $\square$

The next corollary follows directly from Theorem 4.1.

Corollary 4.2

For any ring R, the right R-module $R_R$ is ${\mathfrak {s}}$-coseparable if and only if it is completely ${\mathfrak {s}}$-coseparable.

Next, we shed some light on the class of rings R for which every cyclic R-module is ${\mathfrak {s}}$-coseparable.

Proposition 4.3

Let R be a ring such that every cyclic R-module is ${\mathfrak {s}}$-coseparable. Then, R is a right V-ring.

Proof

Let I be a right ideal of R. By hypothesis, R/I is an ${\mathfrak {s}}$-coseparable R-module. Hence, $Rad(R/I)=0$ by Theorem 2.7. Therefore, R is a right V-ring by [13, Theorem 3.75]. $\square$

Corollary 4.4

The following statements are equivalent for a commutative ring R:

(i) Every cyclic R-module is ${\mathfrak {s}}$-coseparable;

(ii) R is a von Neumann regular ring.

Proof

(i) $\Rightarrow$ (ii) This follows from Proposition 4.3 and the fact that a commutative ring is von Neumann regular if and only if it is a V-ring (see [13, Corollary 3.73]).

(ii) $\Rightarrow$ (i) Let I be an ideal of R. Note that R/I is a von Neumann regular ring. Then, every cyclic R-submodule of R/I is a direct summand of the R-module R/I. Therefore, R/I is an ${\mathfrak {s}}$-coseparable R-module. $\square$

It is shown in Example 2.12 that the class of ${\mathfrak {s}}$-coseparable modules is not closed under factor modules. In the same vein, the next example shows that a factor module of a cyclic ${\mathfrak {s}}$-coseparable module need not be ${\mathfrak {s}}$-coseparable, in general.

Example 4.5

Let R be a commutative weakly regular semiprimitive ring which is not von Neumann regular. An explicit example of R can be found in [16, Example 15.7(9)]. By Theorem 4.1, the R-module R is ${\mathfrak {s}}$-coseparable. On the other hand, using Corollary 4.4, it follows that the ring R has a cyclic R-module M which is not ${\mathfrak {s}}$-coseparable. It is clear that $M \cong R/I$ for some nonzero ideal I of R.

Recall that a ring R is called right semiartinian if every nonzero cyclic right R-module has a nonzero socle.

Corollary 4.6

Let R be a right semiartinian ring. Then, the following conditions are equivalent:

(i) R is a right V-ring;

(ii) Every finitely generated R-module is ${\mathfrak {s}}$-coseparable;

(iii) Every cyclic R-module is ${\mathfrak {s}}$-coseparable.

Proof

(i) $\Rightarrow$ (ii) Let M be a finitely generated R-module. Then, $Rad(M)=0$ since R is a right V-ring. Moreover, M is a semiartinian R-module by [15, p. 183 Proposition VIII.2.5]. Therefore, M is ${\mathfrak {s}}$-coseparable by Corollary 2.22.

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

(iii) $\Rightarrow$ (i) This follows from Proposition 4.3. $\square$

The next result provides some information about the class of rings R for which every R-module is ${\mathfrak {s}}$-coseparable.

Corollary 4.7

The following statements are equivalent for a ring R:

(i) Every R-module is ${\mathfrak {s}}$-coseparable;

(ii) R is a weakly regular semiprimitive ring such that every factor module of an ${\mathfrak {s}}$-coseparable module is also ${\mathfrak {s}}$-coseparable.

Proof

(i) $\Rightarrow$ (ii) By Theorem 4.1.

(ii) $\Rightarrow$ (i) Let M be an R-module. Then, $M \cong F/N$ for some free R-module F and a submodule N of F. By Theorem 4.1, F is ${\mathfrak {s}}$-coseparable. By hypothesis, M is also an ${\mathfrak {s}}$-coseparable module. $\square$

Recall that a ring R is said to be right semihereditary if every finitely generated right ideal of R is projective as a right R-module.

From Proposition 2.5 and [3, Proposition 17.14], it follows that every nonzero indecomposable projective module which is not simple could not be ${\mathfrak {s}}$-coseparable. For example, the ${\mathbb {Z}}$-module ${\mathbb {Z}}$ is not ${\mathfrak {s}}$-coseparable (see also [8, Example 2.11(i)]). Next, we present some examples of classes of rings R for which every projective R-module is ${\mathfrak {s}}$-coseparable.

Proposition 4.8

Let R be a ring. Suppose that one of the following conditions is satisfied:

(i) R is an exchange ring;

(ii) R is a right semihereditary ring.

Then, every projective R-module is ${\mathfrak {s}}$-coseparable if and only if R is a weakly regular semiprimitive ring.

Proof

($\Rightarrow$) This follows from Theorem 4.1.

($\Leftarrow$) By Theorem 4.1, every finitely generated projective module is ${\mathfrak {s}}$-coseparable. Now let P be a projective module which is not finitely generated. Applying [14, Corollary 5.5] and [17, Theorem 1], it follows that P is a direct sum of finitely generated submodules. Moreover, we have $Rad(P)=0$ since $Rad(R)=0$ (see [3, Proposition 17.10]). Therefore, P is ${\mathfrak {s}}$-coseparable by Proposition 2.10. $\square$

As an application of Proposition 4.8, we have the following example.

Example 4.9

(i) Let T be an exchange ring and consider the ring $R=T/Rad(T)$. By [16, Theorem 29.2], R is a semiprimitive exchange ring. Moreover, R is weakly regular by [16, Remark 29.7(1)]. Using Proposition 4.8, it follows that every projective R-module is ${\mathfrak {s}}$-coseparable.

(ii) Note that every von Neumann regular ring is a right semihereditary ring which is weakly regular and semiprimitive.

A ring R is called a right max ring if every nonzero right R-module contains a maximal submodule. It is well known that right V-rings and right perfect rings are right max rings (see [3, Theorem 28.4]).

Proposition 4.10

Let R be a ring such that $Rad(E(S)) \ne E(S)$ for each simple R-module S (for instance, R could be a right max ring). Then, the following statements are equivalent:

(i) Every finitely cogenerated R-module is ${\mathfrak {s}}$-coseparable;

(ii) E(S) is ${\mathfrak {s}}$-coseparable for every simple R-module S;

(iii) R is a right V-ring.

Proof

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

(ii) $\Rightarrow$ (iii) Let S be a simple R-module. Since E(S) is an indecomposable ${\mathfrak {s}}$-coseparable module, it follows that E(S) is simple by Proposition 2.5. Thus $S=E(S)$ is injective. Therefore, R is a right V-ring.

(iii) $\Rightarrow$ (i) Since R is a right V-ring, every finitely cogenerated R-module is semisimple. The result follows. $\square$

A ring R is said to be right small if the right R-module $R_R$ is a small submodule in its injective hull $E(R_R)$. Note that every commutative domain is a (right) small ring (see [11, Proposition 2.4 and Corollary 2.5]). We conclude this paper by two results devoted to the class of rings R for which every injective R-module is ${\mathfrak {s}}$-coseparable. Let us first give an example of a ring belonging to this class.

Example 4.11

Let R be a right small ring. From [11, Proposition 2.4 and Corollary 2.5], it follows that $Rad(E)=E$ for every injective R-module E. Hence, every injective R-module is ${\mathfrak {s}}$-coseparable.

Proposition 4.12

Let R be a ring and let M be an R-module. Consider the following statements:

(i) R is a right max ring and every R-injective module is ${\mathfrak {s}}$-coseparable;

(ii) Every nonzero R-module contains a nonzero finitely generated injective submodule;

(iii) R is a right semiartinian right V-ring.

Then, (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) hold.

Proof

(i) $\Rightarrow$ (ii) Let M be a nonzero R-module and let E(M) be the injective hull of M. Since R is a right max ring, E(M) has a maximal submodule L. By hypothesis, E(M) is ${\mathfrak {s}}$-coseparable. Then, there exist submodules N and K of E(M) such that $E(M)=N\oplus K$, $N \subseteq L$ and K is a nonzero finitely generated submodule of M. But M is essential in E(M). Therefore, $M\cap K \ne 0$. As K is injective, K is ${\mathfrak {s}}$-coseparable. Hence, $M\cap K$ contains a nonzero direct summand U of K. It is clear that U is a finitely generated injective submodule of M.

(ii) $\Rightarrow$ (iii) By [7, Theorem 13]. $\square$

Corollary 4.13

Let R be a right perfect ring. Then, the following statements are equivalent:

(i) Every injective R-module is ${\mathfrak {s}}$-coseparable;

(ii) R is a semisimple ring.

Proof

(i) $\Rightarrow$ (ii) By [3, Theorem 28.4], R is a right max ring and R/Rad(R) is semisimple. By Proposition 4.12, R is a right V-ring. Hence, $Rad(R)=0$. It follows that R is a semisimple ring.

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