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8.4. FINITELY GENERATED MODULES OVER A PID, PART I
371
Proof.
We prove this by induction on
n
. A submodule (ideal) of
R
is
generated by a single element, since
R
is a PID. This veriﬁes the base case
n
D
1
.
Suppose that
F
has rank
n > 1
and that the assertion holds for free
modules of smaller rank.
Let
f
f
1
;:::;f
n
g
be a basis of
F
, put
F
0
D
span
.
f
f
1
;:::;f
n
±
1
g
/
. Let
N
be a submodule of
F
and let
N
0
D
N
\
F
0
. By the induction hypothesis,
N
0
has a generating set with no more than
n
±
1
elements.
Every element
x
2
F
has a unique expansion
x
D
P
n
i
D
1
˛
i
.x/f
i
.
The map
x
7!
˛
n
.x/
is an
R
–module homomorphism from
F
to
R
. If
˛
n
.N/
D f
0
g
, then
N
D
N
0
, and
N
is generated by no more than
n
±
1
elements. Otherwise, the image of
N
under this map is a nonzero ideal of
R
, so of the form
dR
for some nonzero
d
2
R
. Choose
h
2
N
such that
˛
n
.h/
D
d
.
If
x
2
N
, then
˛
n
.x/
D
rd
for some
r
2
R
. Then
y
D
x
±
rh
satisﬁes
˛
n
.y/
D
0
, so
y
2
N
\
F
0
D
N
0
. Thus
x
D
y
C
rh
2
N
0
C
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This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.
 Fall '08
 EVERAGE
 Algebra

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