Homework #11 Solutions
p 348, #12
Since
π
3
∈
F
,
π
is a root of the polynomial
x
3

π
3
∈
F
[
x
]. This polynomial is
irreducible over
F
since the only possible root in
F
would be
π
itself, and it is easy to show
that
π
∈
F
(if it were,
π
would be algebraic over
Q
). Therefore a basis for
F
(
π
) over
F
is
{
1
, π, π
2
}
.
p 348, #14
Let
F
=
Q
(
3
√
5) =
{
a
+
b
3
√
5 +
c
(
3
√
5)
2

a, b, c
∈
Q
}
and
φ
:
F
→
F
be an
automorphism. Arguing as we have several times before, we can show that
φ
(
r
) =
r
for all
r
∈
Q
.
It follows that 5 =
φ
(5) =
φ
((
3
√
5)
3
) =
φ
(
3
√
5)
3
.
Since
φ
(
3
√
5)
∈
F
⊂
R
, we can
therefore conclude that
φ
(
3
√
5) =
3
√
5. Finally, this means that
φ
(
a
+
b
3
√
5 +
c
(
3
√
5)
2
)
=
φ
(
a
) +
φ
(
b
)
φ
(
3
√
5) +
φ
(
c
)
φ
(
3
√
5)
2
=
a
+
b
3
√
5 +
c
(
3
√
5)
2
for any rational
a, b, c
. But every element of
F
can be written in the form above, and so it
must be that
φ
(
α
) =
α
for every
α
∈
F
. That is, the only automorphism of
F
is the identity.
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 Spring '07
 RyanDaileda
 Algebra, Complex number, zp, positive degree factors

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