2Field extensions
II Galois Theory
2.4 Splitting fields
As mentioned in the introduction, one major motivation for Galois theory is to
study the roots of polynomials. So far, we have just been talking about field
extensions. The idea here is given a field
K
and a polynomial
f ∈ K
[
t
], we
would like to study the field extension obtained by adding all roots of
f
. This is
known as the splitting field of f (over K).
Notation. Let
L/K
be a field extension,
f ∈ K
[
t
]. We write
Root
f
(
L
) for the
roots of f in L.
First, we establish a correspondence between the roots of a polynomial and
K-homomorphisms.
Lemma. Let
L/K
be a field extension,
f ∈ K
[
t
] irreducible,
deg f >
0. Then
there is a 1-to-1 correspondence
Root
f
(L) ←→ Hom
K
(K[t]/⟨f⟩, L).
Proof.
Since
f
is irreducible,
⟨f⟩
is a maximal ideal. So
K
[
t
]
/⟨f⟩
is a field. Also,
there is a natural inclusion
K → K
[
t
]
/⟨f⟩
. So it makes sense to talk about
Hom
K
(K[t]/⟨f⟩, L).
To any
β ∈ Root
f
(
L
), we assign
φ
:
K
[
t
]
/⟨f⟩ → L
where we map
¯
t 7→ β
(
¯
t
is
the equivalence class of
t
). This is well defined since if
¯
t
=
¯g
, then
g
=
t
+
hf
for
some h ∈ K[t]. So φ(¯g) = φ(t + hf ) = β + h(β)f(β) = β.
Conversely, given any
K
-homomorphism
φ
:
K
[
t
]
/⟨f⟩ → L
, we assign
β
=
φ(
¯
t). This is a root since f(β) = f(φ(
¯
t)) = φ(f(
¯
t)) = φ(0) = 0.
This assignments are inverses to each other. So we get a one-to-one corre-
spondence.
Recall that if
K ⊆ F
is a field extension, then for any
α ∈ F
with minimal
polynomial
P
α
, we have
K
[
t
]
/⟨P
α
⟩
∼
=
K
(
α
). Since an irreducible
f
is the minimal
polynomial of its roots, we can view the above lemma as telling us something
about Hom
K
(K(α), L).
Corollary. Let
L/K
be a field extension,
f ∈ K
[
t
] irreducible,
deg f >
0. Then
|Hom
K
(K[t]/⟨f⟩, L)| ≤ deg f.
In particular, if E = K[t]/⟨f⟩, then
|Aut
K
(E)| = |Root
f
(E)| ≤ deg f = [E : K].
So E/K is a Galois extension iff |Root
f
(E)| = deg f .
Proof. This follows directly from the following three facts:
– |Root
f
(L)| ≤ deg f
– Aut
K
(E) = Hom
K
(E, E)
– deg f = [K(α) : K] = [E : K].
Definition (Splitting field). Let
L/K
be a field extensions,
f ∈ K
[
t
]. We say
f
splits over L if we can factor f as
f = a(t −α
1
) ···(t − α
n
)
for some
a ∈ K
and
α
j
∈ L
. Alternatively, this says that
L
contains all roots of
f.
We say
L
is a splitting field of
f
if
L
=
K
(
α
1
, ··· , α
n
). This is the smallest
field where f has all its roots.
Example.
– C is the splitting field of t
2
+ 1 ∈ R[t].
– Q
(
3
√
2, µ
) is a splitting field of
t
3
−
2
∈ Q
[
t
], where
µ
is a third root of
unity.
–
By the fundamental theorem of algebra, for any
K ⊆ C
and
f ∈ K
[
t
],
there is a splitting field L ⊆ C of f.
Note that the degree of the splitting field need not be (bounded by) the
degree of the polynomial. In the second example, we have [
Q
(
3
√
2, µ
) :
Q
] = 6,
but t
3
− 2 only has degree 3.
More generally, we can show that every polynomial has a splitting field, and
this is unique up to isomorphism. This is important, since we would like to talk
about the splitting field of a polynomial all the time.
Theorem. Let K be a field, f ∈ K[t]. Then
(i) There is a splitting field of f.
(ii) The splitting field is unique (up to K-isomorphism).
Proof.
(i)
If
deg f
= 0, then
K
is a splitting field of
f
. Otherwise, we add the roots
of f one by one.
Pick
g | f
in
K
[
t
], where
g
is irreducible and
deg g >
0. We have the field
extension
K ⊆ K
[
t
]
/⟨g⟩
. Let
α
1
=
¯
t
. Then
g
(
α
1
) = 0 which implies that
f
(
α
1
) = 0. Hence we can write
f
= (
t − α
1
)
h
in
K
(
α
1
)[
t
]. Note that
deg h < deg f
. So we can repeat the process on
h
iteratively to get a field
extensions
K ⊆ K
(
α
1
, ··· , α
n
). This
K
(
α
1
, ··· , α
n
) is a splitting field of
f.
(ii)
Assume
L
and
L
′
are both splitting fields of
f
over
K
. We want to find a
K-isomorphism from L to L
′
.
Pick largest
F, F
′
such that
K ⊆ F ⊆ L
and
K ⊆ F
′
⊆ L
′
are field
extensions and there is a
K
-isomorphism from
ψ
:
F → F
′
. By “largest”,
we mean we want to maximize [F : K].
We want to show that we must have
F
=
L
. Then we are done because
this means that F
′
is a splitting field, and hence F
′
= L
′
.
So suppose
F
=
L
. We will try to produce a larger
˜
F
with
K
-isomorphism
˜
F →
˜
F
′
⊆ L
′
.
Since
F
=
L
, we know that there is some
α ∈ Root
f
(
L
) such that
α ∈ F
.
Then there is some irreducible
g ∈ K
[
t
] with
deg g >
0 such that
g
(
α
) = 0
and g | f. Say f = gh.
Now we know there is an isomorphism
F
[
t
]
/⟨g⟩ → F
(
α
) by
¯
t 7→ α
. The
isomorphism
ψ
:
F → F
′
extends to a isomorphism
µ
:
F
[
t
]
→ F
′
[
t
].
Then since the coefficients of
f
are in
K
, we have
f
=
µ
(
f
) =
µ
(
g
)
µ
(
h
).
So
µ
(
g
)
| f
in
F
′
[
t
]. Since
g
is irreducible in
F
[
t
],
µ
(
g
) is irreducible in
F
′
[
t
]. So there is some
α
′
∈ Root
µ(g)
(
L
′
)
⊆ Root
f
(
L
′
) and isomorphism
F
′
[t]/⟨µ(g)⟩ → F
′
(α
′
).
Now
µ
induces a
K
-isomorphism
F
[
t
]
/⟨g⟩ → F
′
[
t
]
/⟨µ
(
g
)
⟩
, which in turn
induces a
K
-isomorphism
F
(
α
)
→ F
′
(
α
′
). This contradicts the maximality
of F . So we must have had F = L.
Note that the splitting is unique just up to isomorphism. We could be
quotienting by different polynomials and still get the same splitting field.
Example.
Q
(
√
7
) is a splitting field of
t
2
−
7
∈ Q
[
t
]. At the same time,
Q
(
√
7
)
is also a splitting field of t
2
+ 3t +
1
2
∈ Q[t].