So
σα
j
=
α
i
for some
i
. Since
σ
is one–to–one, it follows that
σ
permutes the set
{
α
1
, α
2
,
· · ·
, α
n
}
.
As these generate
K/k
, we have
σ
(
K
) =
K
.
Definition 3.3
An extension
K/k
which is both separable and normal is said to be
Galois
. The
group
G
of
k
–automorphisms of
K
is called the
Galois group
of
K/k
and is written Gal(
K/k
).
Theorem 2.5 implies that

G

=

K
:
k

.
Associated with any subgroup
H
of
G
is the subfield of
K
given by
Fix(
H
) =
{
x
∈
K
:
σx
=
x
∀
σ
∈
H
}
.
Associated with any intermediate field
L
(i.e.
k
⊆
L
⊆
K
) is a subgroup
Gal(
K/L
) =
{
σ
∈
G
:
σ

L
=
id
L
}
.
Theorem 3.4
(The Fundamental Theorem of Galois Theory) Let
K/k
be a finite Galois extension
with Galois group
G
.
1. The maps
I
:
H
→
Fix(
H
) and
J
:
L
→
Gal(
K/L
) between subgroups of
G
and fields
intermediate between
k
and
K
are inverse bijections.
2. The maps
I
and
J
induce a correspondence between normal subgroups of
G
and normal
extensions of
k
(contained in
K
). If
L/k
is normal, with
k
⊆
L
⊆
K
, then Gal(
L/k
)
G/
(
L
).
3.
I
and
J
are lattice–preserving isomorphisms. This means
(a)
H
⊂
K
iff
I
(
H
)
⊃
I
(
K
).
(b)
I
(
H
∩
K
) =
I
(
H
)
I
(
K
).
(c)
I H, K
=
I
(
H
)
∩
I
(
K
).
Proof
17
1. First observe that clearly
H
⊂
(
J
◦
I
)(
H
) and
L
⊂
(
I
◦
J
)(
L
) for each
H
and
L
. So

(
J
◦
I
)(
H
)
 ≥

H

, and [
K
:
L
]
≥
[
K
: (
I
◦
J
)(
L
)].
Theorem 2.5 implies that

J
(
L
)

= [
K
:
L
].
Theorem
2.4 implies that

H
 ≤
[
K
:
I
(
H
)]. To establish the reverse inequality, we use the Primitive
Element Theorem. Let
α
be a primitive element for
K/k
. Consider the polynomial
f
(
x
) =
σ
∈
H
(
x

σα
)
.
If
τ
∈
H
, the map
σ
→
στ
permutes the elements of
H
. Hence this map preserves
f
. So
f
has
coefficients in
I
(
H
) = Fix(
H
), and thus [
K
:
I
(
H
)]
≤ 
H

, as required. So

H

= [
K
:
I
(
H
)].
Hence we have

(
J
◦
I
)(
H
)

= [
K
:
I
(
H
)] =

H

and [
K
: (
J
◦
I
)(
L
)] =

J
(
L
)

= [
K
:
L
], so
J
◦
I
and
I
◦
J
are both identity maps, as required.
2. Suppose first that
H
G
,
x
∈
Fix(
H
), and
τ
∈
G
.
For each
σ
∈
H
, we have
στ
(
x
) =
τ
(
τ

1
στ
)(
x
) =
τ
(
x
), so
τ
(
x
)
∈
Fix(
H
).
Hence
τ
(Fix(
H
)) = Fix(
H
) for each
τ
(
G
), and so
Fix(
H
)
/k
is a normal extension. Conversely, suppose that
L/k
is normal and
τ
∈
Gal(
K/L
).
If
l
∈
L
and
σ
∈
G
, then
σ
=
m
∈
L
. But
σ

1
τσ
(
l
) =
σ

1
τ
(
m
) =
σ

1
(
m
) =
l.
Hence
σ

1
τσ
∈
Gal(
K/l
), and so Gal(
K/L
)
G
.
The map
σ
→
σ

L
induces the required
isomorphism.
3. Exercise, e.g.
I
(
H
)
∩
I
(
K
)
=
{
x

σx
=
x
for
σ
∈
H
∪
K
}
=
{
x

σx
=
x
for
σ
=
H, K
}
=
I H, K
,
etc.
Example
Let
K
denote the splitting field of
x
4

2 over
Q
.
Determine Gal(
K/
Q
).
Find all the
subfields of
K
and all inclusion relations between them, identifying which ones are normal over
Q
.
Let
α
∈
R
+
satisfy
α
4
= 2, and let
i
∈
C
satisfy
i
2
=

1. Since
x
4

2 = (
x

α
)(
x
+
α
)(
x

iα
)(
x
+
iα
)
,
we have
K
=
Q
(
α, i
).
Eisenstein implies that
x
4

2 is irreducible over
Q
, so [
Q
(
α
) :
Q
] = 4.
Moreover,
i
∈
R
⊃
Q
(
α
), so [
Q
(
α, i
) :
Q
(
α
)] = 2, and hence [
K
:
Q
] = 8. Let
X
=
{
α,

α, iα,

iα
}
.
Then
G
= Gal(
K/
Q
) permutes
X
, and

G

= 8. Let
S
be a square with vertices labelled by the
elements of
X
, and let
D
8
be the group of rigid motions of
S
:
iα
α

α

iα
18
If
σ
∈
G
, then
σ
(

α
) =
σα
, so
σ
(

α
) and
σα
are “opposite” one another. Thus
G
⊂
D
8
, and so
G
=
D
8
, as

G

=

D
8

.
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 Spring '08
 Morrison,D
 Algebra, Galois theory, Root of unity