2
CHAPTER 6. GALOIS THEORY
Proof.
(i) Let
F
0
be the fxed feld oF
G
.I
F
σ
is an
F
automorphism oF
E
, then by
defnition oF
F
0
,
σ
fxes everything in
F
0
. Thus the
F
automorphisms oF
G
coincide with
the
F
0
automorphisms oF
G
. Now by (3.4.7) and (3.5.8),
E/F
0
is Galois. By (3.5.9), the
size oF the Galois group oF a fnite Galois extension is the degree oF the extension. Thus
[
E
:
F
]=[
E
:
F
0
], so by (3.1.9),
F
=
F
0
.
(ii) Suppose that
F
=
F
(
H
). By the theorem oF the primitive element (3.5.12), we
have
E
=
F
(
α
) For some
α
∈
E
. Defne a polynomial
f
(
X
)
∈
E
[
X
]by
f
(
X
)=
Y
σ
∈
H
(
X
−
σ
(
α
))
.
IF
τ
is any automorphism in
H
, then we may apply
τ
to
f
(that is, to the coeﬃcients oF
f
;
we discussed this idea in the prooF oF (3.5.2)). The result is
(
τf
)(
X
Y
σ
∈
H
(
X
−
(
τσ
)(
α
))
.
But as
σ
ranges over all oF
H
, so does
, and consequently
=
f
. Thus each coeﬃcient
oF
f
is fxed by
H
, so
f
∈
F
[
X
]. Now
α
isarooto
F
f
, since
X
−
σ
(
α
) is 0 when
X
=
α
and
σ
is the identity. We can say two things about the degree oF
f
:
(1) By defnition oF
f
, deg
f
=

H

<

G

=[
E
:
F
], and, since
f
is a multiple oF the
minimal polynomial oF
α
over
F
,
(2) deg
f
≥
[
F
(
α
):
F
E
:
F
], and we have a contradiction.
♣
There is a converse to the frst part oF (6.1.2).
6.1.3 Proposition
Let
be a fnite extension with Galois group
G
. IF the fxed feld oF
G
is
F
, then
is Galois.
Proof.
Let
G
=
{
σ
1
,...,σ
n
}
, where
σ
1
is the identity. To show that
is normal,
we consider an irreducible polynomial
f
∈
F
[
X
] with a root
α
∈
E
. Apply each au
tomorphism in
G
to
α
, and suppose that there are
r
distinct images
α
=
α
1
=
σ
1
(
α
),
α
2
=
σ
2
(
α
)
,...,α
r
=
σ
r
(
α
). IF
σ
is any member oF
G
, then
σ
will map each
α
i
to some
α
j
, and since
σ
is an injective map oF the fnite set
{
α
1
r
}
to itselF, it is surjective as
well. To put it simply,
σ
permutes the
α
i
. Now we examine what
σ
does to the
elementary
symmetric functions
oF the
α
i
, which are given by
e
1
=
r