2Field extensions

II Galois Theory



2.3 K-homomorphisms and the Galois Group
Usually in mathematics, we not only want to study objects, but maps between
objects. Suppose we have two field extensions
K L
and
K L
. What should
a map between these two objects look like? Obviously, we would like this map
to be a field homomorphisms between
L
and
L
. Moreover, since this is a map
between the two field extensions, and not just the fields themselves, we would
like this map to preserve things in
K
, and is just a map between the “extended
parts” of L and L
.
Definition (
K
-homomorphism). Let
L/K
and
L
/K
be field extensions. A
K
-homomorphism
ϕ
:
L L
is a ring homomorphism such that
ϕ|
K
=
id
, i.e. it
fixes everything in
K
. We write
Hom
K
(
L, L
) for the set of all
K
-homomorphisms
L L
.
A
K
-isomorphism is a
K
-homomorphism which is an isomorphism of rings.
A
K
-automorphism is a
K
-isomorphism
L L
. We write
Aut
K
(
L
) for the set
of all K-automorphism L L.
There are a couple of things to take note of
(i) Given any ϕ Hom
K
(L, L
), we know that
(a)
Since
ϕ|
K
=
id
, we know that
ker ϕ
=
L
. Since we know that
ker ϕ
is
an ideal, and a field only has two ideals, we must have
ker ϕ
= 0. So
ϕ
is injective. It is, in fact, true that any homomorphisms of fields is
injective.
(b) ϕ
gives an isomorphism
L ϕ
(
L
). So
ϕ
(
L
) is a field and we get the
field extensions K ϕ(L) L
.
(ii)
If [
L
:
K
] = [
L
:
K
]
<
, then any homomorphism in
Hom
K
(
L, L
) is in
fact an isomorphism. So
{K-homomorphisms : L L
} = {K-isomorphisms : L L
},
This is since any
K
-homomorphism
ϕ
:
L L
is an injection. So
[
L
:
K
] = [
ϕ
(
L
) :
K
]. Hence we know that [
L
:
K
] = [
ϕ
(
L
) :
K
]. But we
know that
ϕ
(
L
) is a subfield of
L
. This is possible only if
L
=
ϕ
(
L
). So
ϕ is a surjection, and hence an isomorphism.
In particular, Aut
K
(L) = Hom
K
(L, L).
Example. We want to determine Aut
R
(C). If we pick any ψ Aut
R
(C), then
(ψ(
1))
2
+ 1 = ψ(
1
2
+ 1) = ψ(0) = 0.
So under any automorphism
ψ
, the image of
1
is a root of
t
2
+ 1. Therefore
ψ
(
1
) =
1
or
1
. In the first case,
ψ
is the identity. In the second
case, the automorphism is
ϕ
:
a
+
b
1 7→ a b
1
, i.e. the complex conjugate.
So Aut
R
(C) = {id, ϕ}.
Similarly, we can show that
Aut
Q
(
Q
(
2
)) =
{id, ϕ}
, where
ϕ
swaps
2
with
2.
Example. Let
µ
3
= 1 but
µ
= 1 (i.e.
µ
is a third root of unity). We want to
determine A = Hom
Q
(Q(
3
2), C).
First define ϕ, ψ by
ϕ(
3
2) =
3
2µ
ψ(
3
2) =
3
2µ
2
,
We have ϕ, ψ A. Are there more?
Let λ A. Then we must have
(λ(
3
2))
3
2 = 0.
So
λ
(
3
2
) is a root of
t
3
2. So it is either
3
2,
3
2µ
or
3
2µ
2
. So
λ
is either
id
,
ϕ or ψ. So A = {id, ϕ, ψ}.
Note that in general, if
α
is algebraic over
Q
, then
Q
(
α
)
=
Q
[
t
]
/P
α
. Hence
to specify a
Q
-homomorphism from
Q
(
α
), it suffices to specify the image of
t
, or
just the image of α.
We will later see that the number of automorphisms
|Aut
K
(
L
)
|
is bounded
above by the degree of the extension [
L
:
K
]. However, we need not always have
[
L
:
K
] many automorphisms. When we do have enough automorphisms, we call
it a Galois extension.
Definition (Galois extension). Let
L/K
be a finite field extension. This is a
Galois extension if |Aut
K
(L)| = [L : K].
Definition (Galois group). The Galois group of a Galois extension
L/K
is
defined as
Gal
(
L/K
) =
Aut
K
(
L
). The group operation is defined by function
composition. It is easy to see that this is indeed a group.
Example. The extension
Q
(
7
)
/Q
is Galois. The degree [
Q
(
7
) :
Q
] = 2, and
the automorphism group is
Aut
Q
(
Q
(
7
)) =
{id, ϕ}
, where
ϕ
swaps
7
with
7.
Example. The extension
Q
(
3
2
)
/Q
is not Galois. The degree is [
Q
(
3
2
) :
Q
] = 3,
but the automorphism group is Aut
Q
(Q(
3
2)) = {id}.
To show that there is no other automorphism, note that the automorphism
group can be viewed as a subset of
Hom
Q
(
Q
(
3
2
)
, C
). We have just seen that
Hom
Q
(
Q
(
3
2
)
, C
) has three elements, but only the identity maps
Q
(
3
2
) to itself,
while the others map
3
2 to
3
2µ
i
∈ Q(
3
2). So this is the only automorphism.
The way we should think about this is that there is something missing in
Q
(
3
2
), namely
µ
. Without the
µ
, we cannot get the other automorphisms we
need. In fact, in the next example, we will show that Q Q(
3
2, µ) is Galois.
Example.
Q
(
3
2, µ
)
/Q
is a Galois extension. Firstly, we know that [
Q
(
3
2, µ
) :
Q
(
3
2
)] = 2 because
µ
3
1 = 0 implies
µ
2
+
µ
+ 1 = 0. So the minimal
polynomial has degree 2. This also means that
µ ∈ Q
(
3
2
). We also know that
[Q(
3
2) : Q] = 3. So we have
[Q(
3
2, µ) : Q] = 6
by the Tower law.
Now denote
α
=
3
2
,
β
=
3
2µ
and
γ
=
3
2µ
2
. Then
Q
(
3
2, µ
) =
Q
(
α, β, γ
).
Now let
ϕ Aut
Q
(
Q
(
3
2, µ
)), then
ϕ
(
α
),
ϕ
(
β
) and
ϕ
(
γ
) are roots of
t
3
2. These
roots are exactly α, β, γ. So
{ϕ(α), ϕ(β), ϕ(γ)} = {α, β, γ}.
Hence
ϕ
is completely determined by a permutation of the roots of
t
3
2. So
Aut
Q
(
3
2, µ)
=
S
3
and |Aut
Q
(
3
2, µ)| = 6.
Most of the time, we will only be interested in Galois extensions. The main
reason is that Galois extensions satisfy the fundamental theorem of Galois theory,
which roughly says: if
L/K
is a finite Galois extension, then there is a one-to-one
correspondence of the set of subgroups
H Gal
(
L/K
) and the intermediate
fields
K F L
. In particular, the normal subgroups corresponds to the
“normal extensions”, which is something we will define later.
However, just as we have seen, it is not straightforward to check if an extension
is Galois, even in specific cases like the examples above. Fortunately, by the
time we reach the proper statement of the fundamental theorem, we would have
developed enough machinery to decide easily whether certain extensions are
Galois.