2Field extensions
II Galois Theory
2.7 Normal extensions
We are almost there. We will now move on to study normal extensions. Normal
extensions are very closely related to Galois extensions. In fact, we will show
that if an extension is normal and separable, then it is Galois. The advantage
of introducing the idea of normality is that normality is a much more concrete
definition to work with. It is much easier to check if an extension is normal than
to check if
|Aut
K
(
L
)
|
= [
K
:
L
]. In particular, we will shortly prove that the
splitting field of any polynomial is normal.
This is an important result, since we are going to use the splitting field to
study the roots of a polynomial, and since we mostly care about polynomials
over
Q
, this means all these splitting fields are automatically Galois extensions
of Q.
It is not immediately obvious why these extensions are called “normal” (just
like most other names in Galois theory). We will later see that normal extensions
are extensions that correspond to normal subgroups, in some precise sense given
by the fundamental theorem of Galois theory.
Definition (Normal extension). Let
K ⊆ L
be an algebraic extension. We say
L/K
is normal if for all
α ∈ L
, the minimal polynomial of
α
over
K
splits over
L.
In other words, given any minimal polynomial, L should have all its roots.
Example. The extension
Q
(
3
√
2
)
/Q
is not normal since the minimal polynomial
t
3
− 2 does not split over Q(
3
√
2).
In some sense, extensions that are not “normal” are missing something. This
is somewhat similar to how Galois extensions work. Before we go deeper into
this, we need a lemma.
Lemma. Let
L/F/K
be finite extensions, and
¯
K
is the algebraic closure of
K
.
Then any ψ ∈ Hom
K
(F,
¯
K) extends to some φ ∈ Hom
K
(L,
¯
K).
Proof.
Let
ψ ∈ Hom
K
(
F,
¯
K
). If
F
=
L
, then the statement is trivial. So assume
L = F .
Pick
α ∈ L \ F
. Let
q
α
∈ F
[
t
] be the minimal polynomial of
α
over
F
.
Consider
ψ
(
q
α
)
∈
¯
K
[
t
]. Let
β
be any root of
q
α
, which exists since
¯
K
is
algebraically closed. Then as before, we can extend
ψ
to
F
(
α
) by sending
α
to
β. More explicitly, we send
N
X
i=0
a
i
α
i
7→
X
ψ(a
i
)β
i
,
which is well-defined since any polynomial relation satisfied by
α
in
F
is also
satisfied by β.
Repeat this process finitely many times to get some element in
Hom
K
(
L,
¯
K
).
We will use this lemma to characterize normal extensions.
Theorem. Let
L/K
be a finite extension. Then
L/K
is a normal extension if
and only if L is the splitting field of some f ∈ K[t].
Proof.
Suppose
L/K
is normal. Since
L
is finite, let
L
=
K
(
α
1
, ··· , α
n
) for some
α
i
∈ L
. Let
P
α
i
be the minimal polynomial of
α
i
over
K
. Take
f
=
P
α
1
···P
α
n
.
Since
L/K
is normal, each
P
α
i
splits over
L
. So
f
splits over
L
, and
L
is a
splitting field of f.
For the other direction, suppose that
L
is the splitting field of some
f ∈ K
[
t
].
First we wlog assume
L ⊆
¯
K
. This is possible since the natural injection
K →
¯
K
extends to some
φ
:
L →
¯
K
by our previous lemma, and we can replace
L
with
φ(L).
Now suppose
β ∈ L
, and let
P
β
be its minimal polynomial. Let
β
′
be another
root. We want to show it lives in L.
Now consider
K
(
β
). By the proof of the lemma, we can produce an embedding
ι
:
K
(
β
)
→
¯
K
that sends
β
to
β
′
. By the lemma again, this extends to an
embedding of
L
into
¯
K
. But any such embedding must send a root of
f
to a
root of
f
. So it must send
L
to
L
. In particular,
ι
(
β
) =
β
′
∈ L
. So
P
β
splits
over L.
This allows us to identify normal extensions easily. The following theorem
then allows us to identify Galois extensions using this convenient tool.
Theorem. Let L/K be a finite extension. Then the following are equivalent:
(i) L/K is a Galois extension.
(ii) L/K is separable and normal.
(iii) L
=
K
(
α
1
, ··· , α
n
) and
P
α
i
, the minimal polynomial of
α
i
over
K
, is
separable and splits over L for all i.
Proof.
–
(i)
⇒
(ii): Suppose
L/K
is a Galois extension. Then by definition, this
means
|Hom
K
(L, L)| = |Aut
K
(L)| = [L : K].
To show that
L/K
is separable, recall that we proved that an extension is
separable if and only if there is some
E
such that
|Hom
K
(
L, E
)
|
= [
L
:
K
].
In this case, just pick
E
=
L
. Then we know that the extension is separable.
To check normality, let
α ∈ L
, and let
P
α
be its minimal polynomial over
K. We know that
|Root
P
α
(L)| = |Hom
K
(K[t]/⟨P
α
⟩, L)| = |Hom
K
(K(α), L)|.
But since
|Hom
K
(
L, L
)
|
= [
L
:
K
] and
K
(
α
) is a subfield of
L
, this implies
|Hom
K
(K(α), L)| = [K(α) : K] = deg P
α
.
Hence we know that
|Root
P
α
(L)| = deg P
α
.
So P
α
splits over L.
–
(ii)
⇒
(iii): Just pick
α
1
, ··· , α
n
such that
L
=
K
(
α
1
, ··· , α
n
). Then these
polynomials are separable since the extension is separable, and they split
since
L/K
is normal. In fact, by the primitive element theorem, we can
pick these such that n = 1.
–
(iii)
⇒
(i): Since
L
=
K
(
α
1
, ··· , α
n
) and the minimal polynomials
P
α
i
over
K
are separable, by a previous theorem, there are some extension
E
of K such that
|Hom
K
(L, E)| = [L : K].
To simplify notation, we first replace
L
with its image inside
E
under some
K
-homomorphism
L → E
, which exists since
|Hom
K
(
L, E
)
|
= [
L
:
K
]
>
0.
So we can assume L ⊆ E.
We now claim that the inclusion
Hom
K
(L, L) → Hom
K
(L, E)
is a surjection, hence a bijection. Indeed, if
φ
:
L → E
, then
φ
takes
α
i
to
φ
(
α
i
), which is a root of
P
α
i
. Since
P
α
i
splits over
L
, we know
φ
(
α
i
)
∈ L
for all i. Since L is generated by these α
i
, it follows that φ(L) ⊆ L.
Thus, we have
[L : K] = |Hom
K
(L, E)| = |Hom
K
(L, L)|,
and the extension is Galois.
From this, it follows that if
L/K
is Galois, and we have an intermediate field
K ⊆ F ⊆ L, then L/F is also Galois.
Corollary. Let
K
be a field and
f ∈ K
[
t
] be a separable polynomial. Then the
splitting field of f is Galois.
This is one of the most crucial examples.