97
induction on
n
≥
0. De
fi
ne
b
0
=
a
−
1
0
. If
v
exists, then the equa
tion
u
v
=
1 would imply that
∑
i
+
j
=
n
a
i
b
j
=
0 for all
n
>
0.
Assuming that
b
0
, . . . ,
b
n
−
1
have been de
fi
ned, then we have
0
=
a
0
b
n
+
i
+
j
=
n
j
<
n
a
i
b
j
,
and this can be solved for
b
n
because
a
0
is invertible.
(iii)
Prove that if
σ
∈
k
[[
x
]]
and ord
(σ)
=
n
, then
σ
=
x
n
u
,
where
u
is a unit in
k
[[
x
]]
.
Solution.
Since ord
(σ)
=
n
, we have
σ
=
a
n
x
n
+
a
n
+
1
x
n
+
1
+
a
n
+
2
x
n
+
2
+ · · ·
=
x
n
(
a
n
+
a
n
+
1
x
+
a
n
+
2
x
2
+ · · ·
).
As
a
n
=
0, we have
a
n
+
a
n
+
1
x
+
a
n
+
2
x
2
+· · ·
a unit, by part (ii).
3.41
True or false with reasons.
(i)
If
R
and
S
are commutative rings and
f
:
R
→
S
is a ring ho
momorphism, then
f
is also a homomorphism from the additive
group of
R
to the additive group of
S
.
Solution.
True.
(ii)
If
R
and
S
are commutative rings and if
f
is a homomorphism
from the additive group of
R
to the additive group of
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 Fall '11
 KeithCornell
 Ring, ring homomorphism, Ring theory

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