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Homework #7 Solutions
#1
Let
I
=
h
f
(
x
)
, g
(
x
)
i
. Since
F
[
x
] is a PID there is an
h
(
x
)
∈
F
[
x
] so that
I
=
h
h
(
x
)
i
.
But this implies that
h
(
x
) divides both
f
(
x
) and
g
(
x
). As
f
(
x
) and
g
(
x
) are relatively prime
this can only happen if
h
(
x
) is a unit in
F
[
x
]. Hence
h
f
(
x
)
, g
(
x
)
i
=
I
=
h
h
(
x
)
i
=
F
[
x
]. Since
1
∈
F
[
x
] we conclude that there exist
r
(
x
)
, s
(
x
)
∈
F
[
x
] so that
r
(
x
)
f
(
x
) +
s
(
x
)
g
(
x
) = 1.
#2
According to Exercise 1, there are polynomials
r
1
(
x
)
, s
1
(
x
)
∈
F
[
x
] so that
r
(
x
)
f
(
x
) +
s
(
x
)
g
(
x
) = 1.
Multiplying both sides of this equation by
c
(
x
) we obtain
r
2
(
x
)
f
(
x
) +
s
2
(
x
)
g
(
x
) =
c
(
x
), where
r
2
(
x
) =
c
(
x
)
r
1
(
x
) and
s
2
(
x
) =
c
(
x
)
s
1
(
x
). Now apply the divi
sion algorithm to obtain
r
2
(
x
) =
a
(
x
)
g
(
x
) +
r
(
x
)
s
2
(
x
) =
b
(
x
)
f
(
x
) +
s
(
x
)
where
a
(
x
)
, b
(
x
)
, r
(
x
)
, s
(
x
)
∈
F
[
x
],
r
(
x
) = 0 or deg
r
(
x
)
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 Spring '07
 RyanDaileda

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