109
Solution.
There are
q
,
r
∈
R
with
b
i
=
qb
i
+
1
+
r
, where either
r
=
0 or
∂(
r
) < ∂(
b
i
+
1
)
. If
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
fi
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
≥
m
and
w
=
b
n
+
b
n
+
1
x
+ · · ·
.
Hence,
σ
=
x
m
x
n
−
m
w
∈
(
x
m
)
, so that
I
⊆
(
x
n
)
. Therefore, every ideal in
k
[[
x
]]
is principal.
3.73
Let
k
be a
fi
eld, and let polynomials
a
1
(
x
)
,
a
2
(
x
)
,
. . .
,
a
n
(
x
)
in
k
[
x
]
be
given.
(i)
Show that the greatest common divisor
d
(
x
)
of these polynomials
has the form
∑
t
i
(
x
)
a
i
(
x
)
, where
t
i
(
x
)
∈
k
[
x
]
for 1
≤
i
≤
n
.
Solution.
The proof is the same as for Exercise 1.65(i), mutatis
mutandis.
(ii)
Prove that if
c
(
x
)
is a monic common divisor of these polynomials,
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 Fall '11
 KeithCornell
 Ring, Greatest common divisor, Integral domain, Formal power series, Principal ideal domain

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