149
Solution.
If
K
/
k
is a
fi
eld extension and
α
∈
K
, then
k
(α)
is the
sub
fi
eld of
K
generated by
k
and
α
. In particular,
F
p
(α)
=
F
q
because every nonzero element in
F
q
is a power of
α
.
(ii)
Prove that the irreducible polynomial
p
(
x
)
∈
F
p
[
x
]
of
α
has de
gree
n
.
Solution.
If deg
(
p
)
=
d
, then Proposition 4.30 says that
[
F
p
(α)
:
F
p
] =
d
; it follows that

F
p
(α)
 =
p
d
. But

F
p
(α)
 =

F
p
n
 =
p
n
, and so
d
=
n
.
(iii)
Prove that if
G
=
Gal
(
F
q
/
F
p
)
, then

G
 ≤
n
.
Solution.
Every nonzero element of
F
q
has the form
α
i
for some
i
, and
σ(α
i
)
= [
σ(α)
]
i
. Now
σ
permutes the roots of
p
(
x
)
, the
irreducible polynomial of
α
. Since deg
(
p
)
=
n
, there are at most
n
choices for
σ(α)
, and so

G
 ≤
n
.
(iv)
Prove that Gal
(
F
q
/
F
p
)
is cyclic of order
n
with generator the
Frobenius
F
.
Solution.
If
F
j
=
1 for some
j
<
n
, then
a
p
j
=
a
for all
a
∈
F
q
.
This says that the polynomial
x
p
j
−
x
has
p
n
>
p
j
roots, and this
is a contradiction.
5.23
Given
f
(
x
)
=
ax
2
+
bx
+
c
∈
Q
[
x
]
, prove that the following statements
are equivalent.
(i)
f
(
x
)
is irreducible.
(ii)
√
b
2
−
4
ac
is not rational.
(iii)
Gal
(
Q
(
√
b
2
−
4
ac
)/
Q
)
has order 2.
Solution.
(i)
⇒
(ii). A quadratic is irreducible if and only if it has no rational roots.
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 Fall '11
 KeithCornell
 Group Theory, Galois theory, fQ

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