8
(3)
(4). Suppose
that
S
is a CS-ring. Since
M
is multiplication
R
-module and by Lemma 0.4,
M
is a
faithful multiplication
S
-module. Let
N
be a closed
S
-submodule of
M
. Then (
N
:
M
) is closed in
S
by Corollary 4.3.
Since
S
is a CS-ring, there is an ideal
J
of
S
such that
S =
(
N
:
M
)
J
. By Lemma 4.1,
M =
(
N
:
M
)
M
JM
. Thus
M
= N
JM
. Therefore,
M
is a CS-module as an
S
-module.
(4)
(1). Suppose that
M
is a CS-module as an
S
-module. Similar with the proof of (1)
(2), we have
S
is a
CS-ring. By Lemma 0.5,
M
is finitely generated faithful multiplication
R
-module and hence
S
R
by Lemma 0.6. Since
S
is a CS-ring, thus
R
is a CS-ring. Similar with the proof of (3)
(4), we have
M
is a CS-module as an
R
-module.
□
We close this section with the addition property of CS-module which is multiplication module.
Proposition 5.2
Let
M
be a multiplication
R
-module. If
M
is a CS-module, then so is any submodule
N
of
M
.
Proof
Suppose that
M
is a CS-module and
N
is a submodule of
M
. Let
A
be a submodule of
N
. By hypothesis,
M = K
T
for some submodules
K
,
T
of
M
where
A
is essential in
K
. Since
N
is a submodule of
M
,
N =
(
N
:
M
)
M
(
N
K
) + (
N
T
)
N.
So that
N =
(
N
K
) + (
N
T
), and hence (
N
K
)
(
N
T
) = 0 since
M = K
T
.
Thus,
N =
(
N
K
)
(
N
T
). It is easy to see that
A
is essential in
N
K
. Therefore,
N
is a CS-module.
□
Acknowledgement
First of all I would like to thank my advisor Dr.Sarapee Chairat for guidance, encouragement and support
throughout the process of this work. I would like to express my gratitude to Department of Mathematics, Thaksin
University, for the necessary facility, for warm hospitality and kind friendship.
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