4
ideal
I
of
R
. Let
X
be a submodule of
M
such that
N
X
= 0. Since
M
is a multiplication module, there exists an ideal
J
of
R
such that
X
=
JM
. Thus, (
I
J
)
M
= 0 by Lemma 0.1 (1). This implies
I
J
Ann(
M
) = 0 and hence
J
= 0 because
I
is
an essential in
R
. Therefore,
X
= 0 and
N
is an essential submodule of
M
.
□
Corollary 2.2
Let
M
be a faithful multiplication
R
-module and
N
a submodule of
M
. Then
N
is essential in
M
if and
only if (
N
:
M
) is essential in
R
.
Lemma 2.3
Let
M
be a faithful multiplication
R
-module and
N
a submodule of
M
. Then Ann(
N
) = Ann(
N
:
M
). In
particular,
N
is a faithful submodule of
M
if and only if (
N
:
M
) is a faithful ideal of
R
.
Proof
The proof is straightforward.
□
Proposition 2.4
Let
M
be a faithful multiplication
R
-module,
N
a submodule of
M
and
I
an ideal of
R
. Then the
following statements are satisfied.
(1) If
IN
is essential in
M
, then
I
is essential in
R
and
N
is essential in
M
.
(2) If
N
is a faithful submodule of
M
, then
N
is essential in
M
.
(3) If
N
is essential in
M
and
I
is faithful, then
IN
is essential in
M
.
(4) If
I
is essential in
R
and
N
is faithful, then
IN
is essential in
M
.
Proof
(1) The proof is straightforward.
(2) Suppose that
N
is a faithful submodule of
M
. Let
X
be a submodule of
M
such that
N
X
= 0. Since
M
is a multiplication module,
N
= (
N
:
M
)
M
and
X
= (
X
:
M
)
M
. Thus, [(
N
:
M
)
(
X
:
M
)]
M
= 0 and this implies that
(
N
:
M
)(
X
:
M
) = 0 because
M
is a faithful. It follows that (
X
:
M
)
Ann(
N
:
M
) = Ann(
N
) = 0, by Lemma 2.3. So that
(
X
:
M
) = 0 and we obtain
X
= 0. According,
N
is essential in
M
.
(3) Suppose
N
is essential in
M
and
I
is faithful. Let
X
be a submodule of
M
such that
IN
X
= 0. Then
I
(
N
X
) = 0 and hence
N
X
Ann(
I
) = 0. By hypothesis,
X
= 0 and
IN
is essential in
M
.
(4) Suppose
I
is essential in
R
and
N
is faithful. Let
X
be a submodule of
M
such that
IN
X
= 0. Then
(
IM
X
)
N
= 0. This implies that
IM
X
= 0 since
N
is faithful. Since
I
is essential in
R
and by Proposition 2.1,
IM
is
essential in
M
and hence,
X
= 0. Consequently,
IN
is essential in
M
.
□
In the next proposition, we present some results which are dually properties of essential submodules of
multiplication modules.
Proposition 2.5
Let
M
be a faithful multiplication
R
-module. Then a submodule
N
of
M
is co-essential in
M
if and
only if there exists co-essential ideal
I
of
R
such that
N
=
IM
.
Proof
Suppose that
N
is a co-essential submodule of
M
. Then there exists an ideal
I
of
R
such that
N
=
IM
. It is
sufficient to prove that
I
is a co-essential ideal of
R
. For any ideal
J
of
R
such that
I + J
=
R
. Then, we have
JM
is
submodule of
M
and
N + JM
=
RM
. Since
N
is a co-essential submodule of
M
,
JM = RM
and hence,
J = R
by Lemma
0.3 (2). Therefore,
I
is a co-essential ideal of
R
. Conversely, suppose that
N
=
IM
for some co-essential ideal
I
of
R
.
Let
X
be a submodule of
M
such that
N + X = M.
Since
M
is a multiplication module, there exists an ideal
J
of
R
such
that
X
=
JM
. Thus, (
I + J
)
M
=
M
and this implies that
I + J = R
by Lemma 0.3 (2), and hence
J = R
because
I
is
co-essential in
R
. Therefore,
X = M
and
N
is a co-essential submodule of
M
.
□
