Math 5126
Monday, January 28
Second Homework Solutions
1. Section 12.1, Exercise 21 on page 472. Prove that a ﬁnitely generated module over a
PID is projective if and only if it is free.
Free modules are projective, even if the ring is not a PID. Conversely suppose
P
is a ﬁnitely generated projective module over a PID. Since
P
is a submodule of a
free module, it is torsion free and so by the structure theorem for ﬁnitely generated
modules over a PID, we see that
P
is free.
2. (a) If
M
∩
N
=
0, then there exist 0
6
=
m
∈
M
and 0
6
=
n
∈
N
. We now have
nm
∈
M
because
M
is an
R
submodule, and
mn
∈
N
because
N
is an
R
submodule. Since
M
∩
N
=
0, we deduce that
mn
=
0, which contradicts the fact that
R
is an integral
domain.
(b) Suppose
I
is cyclic, say
I
=
R f
. Then
x
=
r f
and
y
=
s f
for some
r
,
s
∈
R
. Note
that
f
is not a unit, because
R f
6
=
R
. Furthermore
x
and
y
are irreducible in the
UFD
R
, hence
r
and
s
are units. We deduce that
x
and
y
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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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