# When an $\mathscr{S}$-closed submodule is a direct summand

Yongduo Wang Dejun Wu

Department of Applied Mathematics, Lanzhou University of Technology

Lanzhou 730050, P. R. China

E-mail<EMAIL_ADDRESS>
E-mail<EMAIL_ADDRESS>

Abstract

It is well known that a direct sum of CLS-modules is not, in general, a CLS-
module. It is proved that if $M=M_{1}\oplus M_{2}$, where $M_{1}$ and $M_{2}$
are CLS-modules such that $M_{1}$ and $M_{2}$ are relatively ojective (or
$M_{1}$ is $M_{2}$-ejective), then $M$ is a CLS-module and some known results
are generalized. Tercan [8] proved that if a module $M=M_{1}\oplus M_{2}$
where $M_{1}$ and $M_{2}$ are CS-modules such that $M_{1}$ is
$M_{2}$-injective, then $M$ is a CS-module if and only if $Z_{2}(M)$ is a CS-
module. Here we will show that Tercan’s claim is not true.

##  1\. Introduction

CS-modules play important roles in rings and categories of modules and their
generalizations have been studied extensively by many authors recently. In
[3], Goodearl defined an _$\mathscr{S}$ -closed_ submodule of a module $M$ is
a submodule $N$ for which $M/N$ is nonsingular. Note that $\mathscr{S}$-closed
submodules are always closed. In general, closed submodules need not be
$\mathscr{S}$-closed. For example, $0$ is a closed submodule of any module
$M$, but $0$ is $\mathscr{S}$-closed in $M$ only if $M$ is nonsingular. As a
proper generalization of CS-modules, Tercan introduced the concept of CLS-
modules. Following [8], a module $M$ is called a _CLS-module_ if every
$\mathscr{S}$-closed submodule of $M$ is a direct summand of $M$. In this
paper, we continue the study of CLS-modules. Some preliminary results on CLS-
modules are given in Section 1. In Section 2, direct sums of CLS-modules are
studied. It is shown that if $M=M_{1}\oplus M_{2}$, where $M_{1}$ and $M_{2}$
are CLS-modules such that $M_{1}$ and $M_{2}$ are relatively ojective, then
$M$ is a CLS-module and some known results are generalized. Tercan [8] proved
that if a module $M=M_{1}\oplus M_{2}$ where $M_{1}$ and $M_{2}$ are CS-
modules such that $M_{1}$ is $M_{2}$-injective, then $M$ is a CS-module if and
only if $Z_{2}(M)$ is a CS-module. It is shown that Tercan’s claim is not true
in this section.

Throughout this paper, $R$ is an associative ring with identity and all
modules are unital right $R$-modules. We use $N\leq M$ to indicate that $N$ is
a submodule of $M$. Let $M$ be a module and $S\leq M$. $S$ is _essential_ in
$M$ (denoted by $S\leq_{e}M$) if for any $T\leq M,S\cap T=0$ implies $T=0$. A
module $M$ is _CS_ if for any submodule $N$ of $M$, there exists a direct
summand $K$ of $M$ such that $N\leq_{e}K$. A submodule $K$ of $M$ closed in
$M$ if $K$ has no proper essential extension in $M$, i.e., whenever $L$ is a
submodule of $M$ such that $K$ is essential in $L$ then $K=L$. It is well
known that $M$ is CS if and only every closed submodule is a direct summand of
$M$. $Z(M)$($Z_{2}(M)$) denotes the (second) singular submodule of $M$. For
standard definitions we refer to [3].

##  2\. Preliminary results

######  Lemma 2.1.

([8, Lemma 7]) Any direct summand of a CLS-module is a CLS-module.

######  Proposition 2.2.

A module $M$ is a CLS-module if and only if for each $\mathscr{S}$-closed
submodule $K$ of $M$, there exists a complement $L$ of $K$ in $M$ such that
every homomorphism $f:K\oplus L\rightarrow M$ can be extended to a
homomorphism $g:M\rightarrow M$.

###### Proof.

This is a direct consequence of [7, Lemma 2]. ∎

Following [1], a module $M$ is _$\mathscr{G}$ -extending_ if for each
submodule $X$ of $M$ there exists a direct summand $D$ of $M$ such that $X\cap
D\leq_{e}X$ and $X\cap D\leq_{e}D$.

######  Proposition 2.3.

Let $M$ be a $\mathscr{G}$-extending module. Then $M$ is a CLS-module.

###### Proof.

Let $N$ be an $\mathscr{S}$-closed submodule of $M$. There exists a direct
summand $D$ of $M$ such that $N\cap D\leq_{e}N$ and $N\cap D\leq_{e}D$. Note
that $D/(N\cap D)$ is both singular and nonsingular. Then $D=N\cap D$ and so
$N=D$. Therefore, $M$ is a CLS-module. ∎

In general, a CLS-module need not be a $\mathscr{G}$-extending module as the
following example shows.

######  Example 2.4.

Let $K$ be a field and $V=K\times K$. Consider the ring $R$ of $2\times 2$
matrix of the form $(a_{ij})$ with $a_{11},a_{22}\in K,a_{12}\in V,a_{21}=0$
and $a_{11}=a_{22}$. Following [8, Example 14], $R_{R}$ is a CLS module which
is not a module with $(C_{11})$. Therefore, $R_{R}$ is not a
$\mathscr{G}$-extending module by [1, Proposition 1.6].

Applying Proposition 2.3, we will give some examples which are CLS modules,
but not CS-modules as follows.

######  Example 2.5.

Let $M_{1}$ and $M_{2}$ be Abelian groups (i.e., $\mathbb{Z}$-modules) with
$M_{1}$ divisible and $M_{2}=\mathbb{Z}_{p^{{}^{n}}}$, where $p$ is a prime
and $n$ is a positive integer. Since $M=M_{1}\oplus M_{2}$ is
$\mathscr{G}$-extending by [1, Example 3.4], it is a CLS module by Proposition
2.3. But $M$ is not CS, when $M_{1}$ is torsion-free. In particular,
$\mathbb{Q}\oplus\mathbb{Z}_{p^{{}^{n}}}(n\geq 2,p=prime)$ is a CLS module,
but not CS.

######  Example 2.6.

Let $M_{1}$ be a $\mathscr{G}$-extending module with a finite composition
series, $0=X_{0}\leq X_{1}\leq\cdot\cdot\cdot\leq X_{m}=M_{1}$. Let
$M_{2}=X_{m}/X_{m-1}\oplus\cdot\cdot\cdot\oplus X_{1}/X_{0}$. Since
$M=M_{1}\oplus M_{2}$ is $\mathscr{G}$-extending by [1, Example 3.4], it is a
CLS module by Proposition 2.3. But $M$ is not CS in general. In particular,
$M\oplus(U/V)$ is a CLS module, but not CS, where $M$ is a uniserial module
with unique composition series $0\neq V\subset U\subset M$.

######  Proposition 2.7.

Let $M$ be a nonsingular module. Then the following conditions are equivalent.

  1. (i)

$M$ is a CS-module.

  2. (ii)

$M$ is a $\mathscr{G}$-extending module.

  3. (iii)

$M$ is a CLS-module.

###### Proof.

By [1, Proposition 1.8] and [8, Corollary 5]. ∎

######  Proposition 2.8.

Let $M$ be a CLS-module and $X$ be a submodule of $M$. If $Z(M/X)=0$, then
$M/X$ is a CS-module.

###### Proof.

Since $M$ is a CLS-module, $X$ is a direct summand of $M$. Write $M=X\oplus
X^{\prime},X^{\prime}\leq M$. Then $M/X$ is a CS-module by Lemma 2.1 and
Proposition 2.7. ∎

######  Corollary 2.9.

(1, Proposition 1.9) If $M$ is $\mathscr{G}$-extending, $X\unlhd M$, and
$Z(M/X)=0$, then $M/X$ is a CS-module.

######  Corollary 2.10.

(1, Corollary 3.11(i)) Let $M$ be a $\mathscr{G}$-extending module. If $D$ is
a direct summand of $M$ such that $Z(D)=0$, then $D$ is a CS-module.

######  Proposition 2.11.

Let $K\leq_{e}M$ such that $K$ is a CLS-module and for each
$e^{2}=e\in$End$(K)$ there exists $\bar{e}^{2}=\bar{e}\in$End$(M)$ such that
$\bar{e}|_{K}=e$. Then $M$ is a CLS-module.

###### Proof.

Assume $K$ is a CLS-module. Let $X$ be an $\mathscr{S}$-closed submodule of
$M$. Then $K=(X\cap K)\oplus K^{\prime},K^{\prime}\leq K$. Let $X\cap K=eK$,
where $e^{2}=e\in$End$(K)$. By hypothesis, there exists
$\bar{e}^{2}=\bar{e}\in$End$(M)$ such that $\bar{e}|_{K}=e$. Since
$K\leq_{e}M$, $\bar{e}K\leq_{e}\bar{e}M$. Observe that $\bar{e}M\cap
X\leq_{e}\bar{e}M$. But $\bar{e}M/(\bar{e}M\cap X)$ is nonsingular. Hence
$\bar{e}M\leq X$. Thus $X=\bar{e}M$ as $\bar{e}K\leq_{e}X$. Therefore, $M$ is
a CLS-module. ∎

By analogy with the proof of [2, Corollary 3.14], we can obtain

######  Corollary 2.12.

Let $M$ be a module. If $M$ is CLS, then so is the rational hull of $M$.

##  3\. Direct sums of CLS modules

It is well known that a direct sum of CLS- modules is not, in general, a CLS-
module (see [8]). In this section, direct sums of CLS-modules are studied. It
is shown that if $M=M_{1}\oplus M_{2}$, where $M_{1}$ and $M_{2}$ are CLS-
modules and $M_{1}$ and $M_{2}$ are relatively ojective, then $M$ is a CLS-
module and some known results are generalized. Tercan [8] proved that if a
module $M=M_{1}\oplus M_{2}$ where $M_{1}$ and $M_{2}$ are CS-modules such
that $M_{1}$ is $M_{2}$-injective, then $M$ is a CS-module if and only if
$Z_{2}(M)$ is a CS-module. It is shown that Tercan’s claim is not true in this
section.

Let $A,~{}B$ be right $R$-modules. Recall that $B$ is $A$-ojective [6] if and
only if for any complement $C$ of $B$ in $A\oplus B,A\oplus B$ decomposes as
$A\oplus B=C\oplus A^{\prime}\oplus B^{\prime}$ with $A^{\prime}\leq A$ and
$B^{\prime}\leq B$. $A$ and $B$ are relatively ojective if $A$ is $B$-ojective
and $B$ is $A$-ojective.

######  Lemma 3.1.

Let $M=A\oplus B$, where $B$ is $A$-ojective and $A$ is a CLS-module. If $X$
is an $\mathscr{S}$-closed submodule of $M$ such that $X\cap B=0$, then $M$
decomposes as $M=D\oplus A^{\prime}\oplus B^{\prime}$, where $A^{\prime}\leq
A,B^{\prime}\leq B$.

###### Proof.

Let $X$ be an $\mathscr{S}$-closed submodule of $M$ with $X\cap B=0$. Then
$M/X$ is nonsingular. Note that $X\cap A$ is an $\mathscr{S}$-closed submodule
of $A$. Hence $X\cap A$ is a direct summand of $A$. Write $A=(X\cap A)\oplus
A_{1},A_{1}\leq A$. By Lemma 2.1 and Proposition 2.7, $A_{1}$ is a CS-module.
Let $K=(X\oplus B)\cap A$. Then $X\oplus B=K\oplus B$ and $K=(X\cap
A)\oplus(K\cap A_{1})$. There exists a closed submodule $A_{1}^{\prime}$ of
$A_{1}$ such that $K\cap A_{1}\leq_{e}A_{1}^{\prime}$. Then $A_{1}^{\prime}$
is a direct summand of $A_{1}$. Write $A_{1}=A_{1}^{\prime}\oplus
A_{1}{{}^{\prime\prime}},A_{1}{{}^{\prime\prime}}\leq A_{1}$. Now $X\oplus
B=K\oplus B=(X\cap A)\oplus(K\cap A_{1})\oplus B\leq_{e}(X\cap A)\oplus
A_{1}^{\prime}\oplus B$. Let $N=(X\cap A)\oplus A_{1}^{\prime}\oplus B$. Then
$X$ is a complement of $B$ in $N$. Now $B$ is $(X\cap A)\oplus
A_{1}^{\prime}$-ojective by [6, Proposition 8]. By [6, Theorem 7], $N=X\oplus
A^{\prime}\oplus B^{\prime}$, where $A^{\prime}\leq(X\cap A)\oplus
A_{1}^{\prime}$ and $B^{\prime}\leq B$. Therefore, $M=X\oplus A^{\prime}\oplus
A_{1}{{}^{\prime\prime}}\oplus B^{\prime}$, as required. ∎

######  Theorem 3.2.

Let $M=M_{1}\oplus M_{2}$, where $M_{1}$ and $M_{2}$ are CLS-modules. If
$M_{1}$ and $M_{2}$ are relatively ojective, then $M$ is a CLS-module.

###### Proof.

Let $X$ be an $\mathscr{S}$-closed submodule of $M$. If $X\cap M_{1}=0$, then
$X$ is a direct summand of $M$ by Lemma 3.1. Let $X\cap M_{1}\neq 0$. Then
$X\cap M_{1}$ is a direct summand of $M_{1}$. Write $M_{1}=(X\cap M_{1})\oplus
M_{1}^{\prime},M_{1}^{\prime}\leq M_{1}$. If $X\cap M_{2}=0$, then the result
follows by Lemma 3.1. Let $X\cap M_{2}\neq 0$. Then $X\cap M_{2}$ is a direct
summand of $M_{2}$. Write $M_{2}=(X\cap M_{2})\oplus
M_{2}^{\prime},M_{2}^{\prime}\leq M_{2}$. Then $X=(X\cap M_{1})\oplus(X\cap
M_{2})\oplus(X\cap(M_{1}^{\prime}\oplus M_{2}^{\prime}))$. Note that
$M_{1}^{\prime}$ and $M_{2}^{\prime}$ are CS-module and $M_{1}^{\prime}$ and
$M_{2}^{\prime}$ are relatively ojective, so $M_{1}^{\prime}\oplus
M_{2}^{\prime}$ are a CS-module by [6, Theorem 7]. Hence
$X\cap(M_{1}^{\prime}\oplus M_{2}^{\prime})$ is a direct summand of
$M_{1}^{\prime}\oplus M_{2}^{\prime}$. Therefore, $M$ is a CLS-module, as
desired. ∎

######  Corollary 3.3.

([8, Theorem 10]) Let $R$ be a ring and $M$ a right $R$-module such that
$M=M_{1}\oplus M_{2}\oplus\cdots\oplus M_{n}$ is a finite direct sum of
relatively injective modules $M_{i},1\leq i\leq n$. Then $M$ is a CLS-module
if and only if $M_{i}$ is a CLS-module for each $1\leq i\leq n$.

Let $M_{1}$ and $M_{2}$ be modules such that $M=M_{1}\oplus M_{2}$. Recall
that $M_{1}$ is $M_{2}$-ejective [1] if and only if for every submodule $K$ of
$M$ with $K\cap M_{1}=0$ there exists a submodule $M_{3}$ of $M$ such that
$M=M_{1}\oplus M_{3}$ and $K\cap M_{3}\leq_{e}K$.

######  Lemma 3.4.

Let $A_{1}$ be a direct summand of $A$ and $B_{1}$ a direct summand of $B$. If
$A$ is $B$-ejective, then $A_{1}$ is $B_{1}$-ejective.

###### Proof.

Write $M=A\oplus B,\ A=A_{1}\oplus A_{2}$ and $B=B_{1}\oplus B_{2}$. First we
prove that $A_{1}$ is $B$-ejective. Write $N=A_{1}\oplus B$. Let $X$ be a
submodule of $N$ with $X\cap A_{1}=0$. Then $X\cap A=0$. Since $A$ is
$B$-ejective, there is a submodule $C$ of $M$ such that $M=A\oplus C$ and
$X\cap C\leq_{e}X$. Hence $N=A_{1}\oplus(N\cap(A_{2}\oplus C))$. Clearly,
$X\cap(N\cap(A_{2}\oplus C))\leq_{e}X$. Therefore, $A_{1}$ is $B$-ejective.
Next we prove that $A$ is $B_{1}$-ejective. Write $L=A\oplus B_{1}$. Let $Y$
be a submodule of $L$ with $Y\cap A=0$. Since $A$ is $B$-ejective, there
exists a submodule $D$ of $M$ such that $M=A\oplus D$ and $D\cap Y\leq_{e}Y$.
Then $L=A\oplus(L\cap D)$. Clearly, $Y\cap(L\cap D)\leq_{e}Y$. Therefore, $A$
is $B_{1}$-ejective. Thus $A_{1}$ is $B_{1}$-ejective. ∎

######  Theorem 3.5.

Let $M=M_{1}\oplus M_{2}$, where $M_{1}$ and $M_{2}$ are CLS-modules. If
$M_{1}$ is $M_{2}$-ejective, then $M$ is a CLS-module.

###### Proof.

Let $N$ be an $\mathscr{S}$-closed submodule of $M$. If $N\cap M_{1}=0$, then
$M_{1}$ is nonsingular. Since $M_{1}$ is $M_{2}$-ejective, there is a
submodule $M_{3}$ of $M$ such that $M=M_{1}\oplus M_{3}$ and $N\cap
M_{3}\leq_{e}N$. Note that $N/(N\cap M_{3})$ is both singular and nonsingular.
Hence $N=N\cap M_{3}$. Since $M_{3}\cong M_{2}$, $M_{3}$ is a CLS-module.
Clearly, $M_{3}/N$ is nonsingular. Then $N$ is a direct summand of $M$. Let
$N\cap M_{1}\neq 0$. Then $N\cap M_{1}$ is a direct summand of $M_{1}$. Write
$M_{1}=(N\cap M_{1})\oplus M_{1}^{\prime},M_{1}^{\prime}\leq M_{1}$.
Similarly, $M_{2}=(N\cap M_{2})\oplus M_{2}^{\prime},M_{2}^{\prime}\leq
M_{2}$. Then $N=(N\cap M_{1})\oplus(N\cap
M_{2})\oplus(N\cap(M_{1}^{\prime}\oplus M_{2}^{\prime}))$. Since $M_{1}$ is
$M_{2}$-ejective, $M_{1}^{\prime}$ is $M_{2}^{\prime}$-ejective by Lemma 3.4.
Note that $M_{1}^{\prime}$ and $M_{2}^{\prime}$ are $\mathscr{G}$-extending
module. By [1, Theorem 3.1], $M_{1}^{\prime}\oplus M_{2}^{\prime}$ is
$\mathscr{G}$-extending. Hence $N\cap(M_{1}^{\prime}\oplus M_{2}^{\prime})$ is
a direct summand of $M_{1}^{\prime}\oplus M_{2}^{\prime}$. Therefore, $M$ is a
CLS-module, as desired. ∎

######  Corollary 3.6.

([8, Theorem 9]) Let $M=M_{1}\oplus M_{2}$, where $M_{1}$ and $M_{2}$ are CLS-
modules. If $M_{1}$ is $M_{2}$-injective, then $M$ is a CLS-module.

######  Corollary 3.7.

Let $M=M_{1}\oplus M_{2}\oplus\cdots\oplus M_{n}$ be a finite direct sum. If
$M_{i}$ is $M_{j}$-ejective for all $j>i$ and each $M_{i}$ is a CLS-module,
then $M$ is a CLS-module.

###### Proof.

By analogy with the proof of [1, Corollary 3.2]. ∎

######  Corollary 3.8.

Let $M=M_{1}\oplus M_{2}$. Then

  1. (i)

If $M_{1}$ is injective, then $M$ is a CLS-module if and only if $M_{2}$ is a
CLS-module.

  2. (ii)

If $M_{1}$ is a CLS-module and $M_{2}$ is semisimple, then $M$ is a CLS-
module.

######  Corollary 3.9.

A module $M$ is a CLS-module if and only if $M=Z_{2}(M)\oplus
M^{\prime},M^{\prime}\leq M$, where $Z_{2}(M)$ and $M^{\prime}$ are CLS-
modules.

###### Proof.

Let $M$ be a CLS-module. Then $M=Z_{2}(M)\oplus M^{\prime},M^{\prime}\leq M$.
By Lemma 2.1, $Z_{2}(M)$ and $M^{\prime}$ are CLS-modules. Conversely, if
$M=Z_{2}(M)\oplus M^{\prime},M^{\prime}\leq M$, then $M^{\prime}$ is
$Z_{2}(M)$-injective. Now the result follows by Theorem 3.5. ∎

######  Corollary 3.10.

Let $M=M_{1}\oplus M_{2}$, where $M_{1}$ and $M_{2}$ are CS-modules. If $M$ is
nonsingular and $M_{1}$ is $M_{2}$-ejective, then $M$ is a CS-module.

###### Proof.

By Proposition 2.7 and Theorem 3.5. ∎

######  Corollary 3.11.

Let $M=M_{1}\oplus M_{2}$ be a direct sum of CS-modules $M_{1}$ and $M_{2}$,
where $M_{2}$ is nonsingular. If $M_{1}$ is $M_{2}$-ejective and
$Z_{2}(M_{1})$ is $M_{2}$-injective, then $M$ is a CS-module.

###### Proof.

By analogy with the proof of [8, Corollary 11]. ∎

######  Corollary 3.12.

([4, Theorem 4]) Let $M=M_{1}\oplus M_{2}$ be a direct sum of CS-modules
$M_{1}$ and $M_{2}$, where $M_{2}$ is nonsingular. If $M_{1}$ is
$M_{2}$-injective, then $M$ is a CS-module.

######  Corollary 3.13.

Let $M=M_{1}\oplus M_{2}$ be a direct sum of CS-modules $M_{1}$ and $M_{2}$.
If $M_{1}$ is $M_{2}$-ejective, $Z_{2}(M_{1})$ is $M_{2}$-injective and
$Z_{2}(M_{2})$ is $M_{1}$-injective, then $M$ is a CS-module if and only if
$Z_{2}(M)$ is a CS-module.

###### Proof.

Let $Z_{2}(M)$ be a CS-module. Then $M=Z_{2}(M_{1})\oplus Z_{2}(M_{1})\oplus
M_{1}^{\prime}\oplus M_{2}^{\prime}$, where $M_{1}^{\prime}\leq M_{1}$ and
$M_{2}^{\prime}\leq M_{2}$. By [6, Theorem 1], $Z_{2}(M_{1})$ is
$M_{1}^{\prime}$-injective and $Z_{2}(M_{2})$ is $M_{2}^{\prime}$-injective.
Then $Z_{2}(M)$ is $M_{1}^{\prime}\oplus M_{2}^{\prime}$-injective. Since
$M_{1}$ is $M_{2}$-ejective, $M_{1}^{\prime}\oplus M_{2}^{\prime}$ is a CS-
module by Corollary 3.10. Hence $M$ is a CS-module by [6, Theorem 1]. ∎

######  Corollary 3.14.

Let $M=M_{1}\oplus M_{2}$ be a direct sum of CS-modules $M_{1}$ and $M_{2}$
such that $M_{1}$ is $M_{2}$-injective and $Z_{2}(M_{2})$ is
$M_{1}$-injective. Then $M$ is a CS-module if and only if $Z_{2}(M)$ is a CS-
module.

We close this paper with the following.

A. Tercan [8, Corollary 13] showed that if a module $M=M_{1}\oplus M_{2}$
where $M_{1}$ and $M_{2}$ are CS-modules such that $M_{1}$ is
$M_{2}$-injective, then $M$ is a CS-module if and only if $Z_{2}(M)$ is a CS-
module. The following example shows that this claim is not true.

######  Example 3.15.

Let $R=\mathbb{Z}$ and
$M_{\mathbb{Z}}=\mathbb{Q}\oplus\mathbb{Z}_{p^{n}}(n\geq 2,p=prime)$. We know
that $\mathbb{Q}$ is $\mathbb{Z}_{p^{n}}$-injective and $\mathbb{Q}$,
$\mathbb{Z}_{p^{n}}$ are uniform modules. Following by [1, Example 3.4], $M$
is not CS. Next we show that $Z_{2}(M)$ is CS. Since $\mathbb{Q}_{\mathbb{Z}}$
is nonsingular, it is easy to see that $Z_{2}(M)=Z_{2}(\mathbb{Z}_{p^{n}})$.
Since $\mathbb{Z}_{p^{n}}$ is CS, $Z_{2}(\mathbb{Z}_{p^{n}})$, as a direct
summand of $\mathbb{Z}_{p^{n}}$, is CS.

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