Math 5126
Wednesday, January 23
First Homework Solutions
1. Since
R
/
I
is a Noetherian ring, it is Noetherian as an
R
/
I
module. The submodules
of
R
/
I
considered as an
R
module are precisely the same as the submodules of
R
/
I
considered as an
R
/
I
module; thus
R
/
I
is a Noetherian
R
module. Obviously
I
is a
Noetherian
R
module, because

I

is ﬁnite. Therefore
R
is a Noetherian
R
module (
N
and
M
/
N
are Noetherian if and only if
M
is Noetherian) and the result is proven.
2. If
M
is free (and nonzero), then
M
has a submodule isomorphic to
R
and so certainly
M
is not a torsion module. Conversely suppose
M
is not free. Since
M
is a nonzero
cyclic
R
module, it is isomorphic to
R
/
I
for some ideal
I
of
R
(reason: if
M
=
Rm
,
then we can deﬁne an
R
module epimorphism
R
±
M
by
r
7→
rm
, and then we let
I
be the kernel). Note that
I
6
=
0 because
M
is not free, so there exists 0
6
=
x
∈
I
. Then
x
(
R
/
I
) =
0 which shows that
M
is a torsion module.
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This note was uploaded on 01/29/2009 for the course MATH 5125 taught by Professor Palinnell during the Fall '07 term at Virginia Tech.
 Fall '07
 PALinnell
 Math, Algebra

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