2Field extensions
II Galois Theory
2.8 The fundamental theorem of Galois theory
Finally, we can get to the fundamental theorem of Galois theory. Roughly, given
a Galois extension
K ⊆ L
, the fundamental theorem tell us there is a one-to-one
correspondence between intermediate field extensions
K ⊆ F ⊆ L
and subgroups
of the automorphism group Gal(L/K).
Given an intermediate field
F
, we can obtain a subgroup of
Gal
(
L/K
) by
looking at the automorphisms that fix
F
. To go the other way round, given a
subgroup
H ≤ Gal
(
L/K
), we can obtain a corresponding field by looking at the
field of elements that are fixed by everything in
H
. This is known as the fixed
field, and can in general be defined even for non-Galois extensions.
Definition (Fixed field). Let
L/K
be a field extension,
H ≤ Aut
K
(
L
) a
subgroup. We define the fixed field of H as
L
H
= {α ∈ L : ϕ(α) = α for all ϕ ∈ H}.
It is easy to see that L
H
is an intermediate field K ⊆ L
H
⊆ L.
Before we get to the fundamental theorem, we first prove Artin’s lemma.
This in fact proves part of the results in the fundamental theorem, but is also
useful on its own right.
Lemma (Artin’s lemma). Let
L/K
be a field extension and
H ≤ Aut
K
(
L
) a
finite subgroup. Then L/L
H
is a Galois extension with Aut
L
H
(L) = H.
Note that we are not assuming that L/K is Galois, or even finite!
Proof. Pick any α ∈ L. We set
{α
1
, ··· , α
n
} = {ϕ(α) : ϕ ∈ H},
where
α
i
are distinct. Here we are allowing for the possibility that
ϕ
(
α
) =
ψ
(
α
)
for some distinct ϕ, ψ ∈ H.
By definition, we clearly have n < |H|. Let
f =
n
Y
1
(t − α
i
) ∈ L[t].
We know that any
ϕ ∈ H
gives an homomorphism
L
[
t
]
→ L
[
t
], and any such
map fixes
f
because
ϕ
just permutes the
α
i
. Thus, the coefficients of
f
are in
L
H
, and thus f ∈ L
H
[t].
Since
id ∈ H
, we know that
f
(
α
) = 0. So
α
is algebraic over
L
H
. Moreover,
if q
α
is the minimal polynomial of α over L
H
, then q
α
| f in L
H
[t]. Hence
[L
H
(α) : L
H
] = deg q
α
≤ deg f ≤ |H|.
Further, we know that
f
has distinct roots. So
q
α
is separable, and so
α
is
separable. So it follows that L/L
H
is a separable extension.
We next show that
L/L
H
is simple. This doesn’t immediately follow from
the primitive element theorem, because we don’t know it is a finite extension
yet, but we can still apply the theorem cleverly.
Pick
α ∈ L
such that [
L
H
(
α
) :
L
H
] is maximal. This is possible since
[L
H
(α) : L
H
] is bounded by |H|. The claim is that L = L
H
(α).
We pick an arbitrary
β ∈ L
, and will show that this is in
L
H
(
α
). By the
above arguments,
L
H
⊆ L
H
(
α, β
) is a finite separable extension. So by the
primitive element theorem, there is some
λ ∈ L
such that
L
H
(
α, β
) =
L
H
(
λ
).
Note that we must have
[L
H
(λ) : L
H
] ≥ [L
H
(α) : L
H
].
By maximality of [
L
H
(
α
) :
L
H
], we must have equality. So
L
H
(
λ
) =
L
H
(
α
). So
β ∈ L
H
(α). So L = L
H
(α).
Finally, we show it is a Galois extension. Let L = L
H
(α). Then
[L : L
H
] = [L
H
(α) : L
H
] ≤ |H| ≤ |Aut
L
H
(L)|
Recall that we have previously shown that for any extension
L/L
H
, we have
|Aut
L
H
(L)| ≤ [L : L
H
]. Hence we must have equality above. So
[L : L
H
] = |Aut
L
H
(L)|.
So the extension is Galois. Also, since we know that
H ⊆ Aut
L
H
(
L
), we must
have H = Aut
L
H
(L).
Theorem. Let
L/K
be a finite field extension. Then
L/K
is Galois if and only
if L
H
= K, where H = Aut
K
(L).
Proof.
(
⇒
) Suppose
L/K
is a Galois extension. We want to show
L
H
=
K
.
Using Artin’s lemma (and the definition of H), we have
[L : K] = |Aut
K
(L)| = |H| = |Aut
L
H
(L)| = [L : L
H
]
So [L : K] = [L : L
H
]. So we must have L
H
= K.
(⇐) By the lemma, K = L
H
⊆ L is Galois.
This is an important theorem. Given a Galois extension
L/K
, this gives us
a very useful test of when elements of
α ∈ L
are in fact in
K
. We will use this a
lot.
Finally, we get to the fundamental theorem.
Theorem (Fundamental theorem of Galois theory). Assume
L/K
is a (finite)
Galois extension. Then
(i) There is a one-to-one correspondence
H ≤ Aut
K
(L) ←→ intermediate fields K ⊆ F ⊆ L.
This is given by the maps
H 7→ L
H
and
F 7→ Aut
F
(
L
) respectively.
Moreover, |Aut
K
(L) : H| = [L
H
: K].
(ii) H ≤ Aut
K
(
L
) is normal (as a subgroup) if and only if
L
H
/K
is a normal
extension if and only if L
H
/K is a Galois extension.
(iii)
If
H ◁ Aut
K
(
L
), then the map
Aut
K
(
L
)
→ Aut
K
(
L
H
) by the restriction
map is well-defined and surjective with kernel isomorphic to H, i.e.
Aut
K
(L)
H
= Aut
K
(L
H
).
Proof. Note that since L/K is a Galois extension, we know
|Aut
K
(L)| = |Hom
K
(L, L)| = [L : K],
By a previous theorem, for any intermediate field
K ⊆ F ⊆ L
, we know
|Hom
K
(
F, L
)
|
= [
F
:
K
] and the restriction map
Hom
K
(
L, L
)
→ Hom
K
(
F, L
)
is surjective.
(i)
The maps are already well-defined, so we just have to show that the maps
are inverses to each other. By Artin’s lemma, we know that
H
=
Aut
L
H
(
L
),
and since
L/F
is a Galois extension, the previous theorem tells that
L
Aut
F
(L)
=
F
. So they are indeed inverses. The formula relating the index
and the degree follows from Artin’s lemma.
(ii)
Note that for every
ϕ ∈ Aut
K
(
L
), we have that
L
ϕHϕ
−1
=
ϕL
H
, since
α ∈ L
ϕHϕ
−1
iff
ϕ
(
ψ
(
ϕ
−1
(
α
))) =
α
for all
ψ ∈ H
iff
ψ
(
ϕ
−1
(
α
)) =
ϕ
−1
(
α
)
for all ψ ∈ H iff α ∈ ϕL
H
. Hence H is a normal subgroup if and only if
ϕ(L
H
) = L
H
for all ϕ ∈ Aut
K
(L). (∗)
Assume (
∗
). We want to first show that
Hom
K
(
L
H
, L
H
) =
Hom
K
(
L
H
, L
).
Let
ψ ∈ Hom
K
(
L
H
, L
). Then by the surjectivity of the restriction map
Hom
K
(
L, L
)
→ Hom
K
(
L
H
, L
),
ψ
must be the restriction of some
˜
ψ ∈
Hom
K
(
L, L
). So
˜
ψ
fixes
L
H
by (
∗
). So
ψ
sends
L
H
to
L
H
. So
ψ ∈
Hom
K
(L
H
, L
H
). So we have
|Aut
K
(L
H
)| = |Hom
K
(L
H
, L
H
)| = |Hom
K
(L
H
, L)| = [L
H
: K].
So L
H
/K is Galois, and hence normal.
Now suppose
L
H
/K
is a normal extension. We want to show this implies
(
∗
). Pick any
α ∈ L
H
and
ϕ ∈ Aut
K
(
L
). Let
P
α
be the minimal polynomial
of
α
over
K
. So
ϕ
(
α
) is a root of
P
α
(since
ϕ
fixes
P
α
∈ K
, and hence
maps roots to roots). Since
L
H
/K
is normal,
P
α
splits over
L
H
. This
implies that ϕ(α) ∈ L
H
. So ϕ(L
H
) = L
H
.
Hence,
H
is a normal subgroup if and only if
ϕ
(
L
H
) =
L
H
if and only if
L
H
/K is a Galois extension.
(iii)
Suppose
H
is normal. We know that
Aut
K
(
L
) =
Hom
K
(
L, L
) restricts
to
Hom
K
(
L
H
, L
) surjectively. To show that we in fact have restriction
to
Aut
K
(
L
H
), by the proof above, we know that
ϕ
(
L
H
) =
L
H
for all
ϕ ∈ Aut
K
(
L
H
). So this does restrict to an automorphism of
L
H
. In other
words, the map
Aut
K
(
L
)
→ Aut
K
(
L
H
) is well-defined. It is easy to see
this is a group homomorphism.
Finally, we have to calculate the kernel of this homomorphism. Let
E
be the kernel. Then by definition,
E ⊇ H
. So it suffices to show that
|E|
=
|H|
. By surjectivity of the map and the first isomorphism theorem
of groups, we have
|Aut
K
(L)|
|E|
= |Aut
K
(L
H
)| = [L
H
: K] =
[L : K]
[L : L
H
]
=
|Aut
K
(L)|
|H|
,
noting that
L
H
/K
and
L/K
are both Galois extensions, and
|H|
= [
L
H
:
K] by Artin’s lemma. So |E| = |H|. So we must have E = H.
Example. Let
p
be an odd prime, and
ζ
p
be a primitive
p
th root of unity. Given
a (square-free) integer
n
, when is
√
n
in
Q
(
ζ
p
)? We know that
√
n ∈ Q
(
ζ
p
) if and
only if
Q
(
√
n
)
⊆ Q
(
ζ
p
). Moreover, [
Q
(
√
n
) :
Q
] = 2, i.e.
Q
(
√
n
) is a quadratic
extension.
We will later show that
Gal
(
Q
(
ζ
p
)
/Q
)
∼
=
(
Z/pZ
)
∗
∼
=
C
p−1
. Then by the
fundamental theorem of Galois theory, quadratic extensions contained in
Q
(
ζ
p
)
correspond to index 2-subgroups of
Gal
(
Q
(
ζ
p
)
/Q
). By general group theory,
there is exactly one such subgroup. So there is exactly one square-free
n
such
that
Q
(
√
n
)
⊆ Q
(
ζ
p
) (since all quadratic extensions are of the form
Q
(
√
n
)),
given by the fixed field of the index 2 subgroup of (Z/pZ)
∗
.
Now we shall try to find some square root lying in
Q
(
ζ
p
). We will not fully
justify the derivation, since we can just square the resulting number to see that
it is correct. We know the general element of Q(ζ
p
) looks like
p−1
X
k=0
c
k
ζ
k
p
.
We know
Gal
(
Q
(
ζ
p
)
/Q
)
∼
=
(
Z/pZ
)
∗
acts by sending
ζ
p
7→ ζ
n
p
for each
n ∈
(
Z/pZ
)
∗
,
and the index 2 subgroup consists of the quadratic residues. Thus, if an element
is fixed under the action of the quadratic residues, the quadratic residue powers
all have the same coefficient, and similarly for the non-residue powers.
If we wanted this to be a square root, then the action of the remaining
elements of
Gal
(
Q
(
ζ
p
)
/Q
) should negate this object. Since these elements swap
the residues and the non-residues, we would want to have something like
c
k
= 1 if
k
is a quadratic residue, and
−
1 if it is a non-residue, which is just the Legendre
symbol! So we are led to try to square
τ =
p−1
X
k=1
k
p
ζ
k
p
.
It is an exercise in the Number Theory example sheet to show that the square
of this is in fact
τ
2
=
−1
p
p.
So we have
√
p ∈ Q(ζ
p
) if p ≡ 1 (mod 4), and
√
−p ∈ Q(ζ
p
) if p ≡ 3 (mod 4).