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109
Solution.
There are
q
,
r
∈
R
with
b
i
=
qb
i
+
1
+
r
, where either
r
=
0or
∂(
r
)<∂(
b
i
+
1
)
.I
f
r
=
0, then
b
i
=
qb
i
+
1
; canceling, 1
=
qb
, and this
contradicts the hypothesis that
b
is not a unit. Therefore,
r
±=
0. But
r
=
b
i
−
qb
i
+
1
=
b
i
(
1
−
qb
),
so that
∂(
b
i
)
≤
∂(
b
i
(
1
−
qb
))
=
∂(
r
)
(1
−
qb
is nonzero because
b
is not a unit). Thus,
∂(
b
i
)
≤
∂(
r
)<∂(
b
i
+
1
).
3.72
If
k
is a
f
eld, prove that the ring of formal power series
k
[[
x
]]
is a PID.
Solution.
Let
I
be an ideal in
k
[[
x
]]
; we may assume that
I
±={
0
}
because
{
0
}=
(
0
)
. Choose
τ
∈
I
of smallest order, say,
τ
=
a
m
x
m
+
a
m
+
1
x
m
+
1
+
···=
x
m
v
, where
v
=
a
m
+
a
m
+
1
x
+···
; note that
v
is a unit in
k
[[
x
]]
,
by Exercise 3.40. We claim that
I
=
(
x
m
)
.Now
x
m
=
v
−
1
τ
∈
I
, so that
(
x
m
)
⊆
I
. For the reverse inclusion, if
σ
∈
k
[[
x
]]
is nonzero, then
σ
=
b
n
x
n
+
b
n
+
1
x
n
+
1
+···=
x
n
w
, where
n
≥
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.
 Fall '11
 KeithCornell

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