5Ramification theory for local fields

III Local Fields


5.2 Unramified extensions
Unramified extensions are easy to classify, since they just correspond to extensions
of the residue field.
Theorem.
Let
K
be a local field. For every finite extension
/k
K
, there is a
unique (up to isomorphism) finite unramified extension
L/K
with
k
L
∼
=
over
k
K
. Moreover, L/K is Galois with
Gal(L/K)
∼
=
Gal(/k
K
).
Proof.
We start with existence. Let
¯α
be a primitive element of
/k
K
with
minimal polynomial
¯
f ∈ k
K
[
x
]. Take a monic lift
f ∈ O
K
[
x
] of
¯
f
such that
deg f
=
deg
¯
f
. Note that since
¯
f
is irreducible, we know
f
is irreducible. So we
can take L = K(α), where α is a root of f (i.e. L = K[x]/f). Then we have
[L : K] = deg f = deg(
¯
f) = [ : k
K
].
Moreover,
k
L
contains a root of
¯
f
, namely the reduction
α
. So there is an
embedding → k
L
, sending ¯α to the reduction of α. So we have
[k
L
: k
K
] ≥ [ : k
L
] = [L : K].
So L/K must be unramified and k
L
∼
=
over k
K
.
Uniqueness and the Galois property follow from the following lemma:
Lemma.
Let
L/K
be a finite unramified extension of local fields and let
M/K
be a finite extension. Then there is a natural bijection
Hom
K - Alg
(L, M) ←→ Hom
k
K
- Alg
(k
L
, k
M
)
given in one direction by restriction followed by reduction.
Proof.
By the uniqueness of extended absolute values, any
K
-algebra homomor-
phism
ϕ
:
L → M
is an isometry for the extended absolute values. In particular,
we have
ϕ
(
O
L
)
⊆ O
M
and
ϕ
(
m
L
)
⊆ m
M
. So we get an induced
k
K
-algebra
homomorphism ¯ϕ : k
L
→ k
M
.
So we obtain a map
Hom
K-Alg
(L, M) → Hom
k
K
-Alg
(k
L
, k
M
)
To see this is bijective, we take a primitive element
¯α ∈ k
L
over
k
K
, and take a
minimal polynomial
¯
f ∈ k
K
[
x
]. We take a monic lift of
¯
f
to
O
k
[
x
], and
α ∈ O
L
the unique root of
f
which lifts
¯α
, which exists by Hensel’s lemma. Then by
counting dimensions, the fact that the extension is unramified tells us that
k
L
= k
K
(¯α), L = K(α).
So we can construct the following diagram:
ϕ Hom
K-Alg
(L, M) Hom
k
K
-Alg
(k
L
, k
M
) ¯ϕ
ϕ(α) {x ∈ M : f(x) = 0} {¯x ∈ k
M
:
¯
f(¯x) = 0} ¯ϕ(¯α)
∼
=
reduction
∼
=
reduction
But the bottom map is a bijection by Hensel’s lemma. So done.
Alternatively, given a map
¯ϕ
:
k
L
→ k
M
, we can lift it to the map
ϕ
:
L → M
given by
ϕ
X
[a
n
]π
n
k
=
X
[ ¯ϕ(a
n
)]π
n
k
,
using the fact that
π
n
k
is a uniformizer in
L
since the extension is unramified.
So we get an explicit inverse.
Proof of theorem (continued).
To finish off the proof of the theorem, we just
note that an isomorphism
¯ϕ
:
k
L
∼
=
k
M
over
k
K
between unramified extensions.
Then
¯ϕ
lifts to a
K
-embedding
ϕ
:
L → M
and [
L
:
K
] = [
M
:
K
] implies that
ϕ is an isomorphism.
To see that the extension is Galois, we just notice that
|Aut
K
(L)| = |Aut
k
K
(k
L
)| = [k
L
: k
K
] = [L : K].
So
L/K
is Galois. Moreover, the map
Aut
K
(
L
)
→ Aut
k
K
(
k
L
) is really a
homomorphism, hence an isomorphism.
Proposition.
Let
K
be a local field, and
L/K
a finite unramified extension,
and
M/K
finite. Say
L, M
are subfields of some fixed algebraic closure
¯
K
of
K
.
Then
LM/M
is unramified. Moreover, any subextension of
L/K
is unramified
over K. If M/K is unramified as well, then LM/K is unramified.
Proof.
Let
¯α
be a primitive element of
k
K
/k
L
, and
¯
f ∈ k
K
[
x
] a minimal polyno-
mial of
¯α
, and
f ∈ O
k
[
x
] a monic lift of
¯
f
, and
α ∈ O
L
a unique lift of
f
lifting
¯α. Then L = K(α). So LM = M(α).
Let
¯g
be the minimal polynomial of
¯α
over
k
M
. Then
¯g |
¯
f
. By Hensel’s
lemma, we can factorize
f
=
gh
in
O
M
[
x
], where
g
is monic and lifts
¯g
. Then
g
(
α
) = 0 and
g
is irreducible in
M
[
x
]. So
g
is the minimal polynomial of
α
over
M. So we know that
[LM : M ] = deg g = deg ¯g ≤ [k
LM
: k
M
] ≤ [LM : M ].
So we have equality throughout and LM/M is unramified.
The second assertion follows from the multiplicativity of
f
L/K
, as does the
third.
Corollary.
Let
K
be a local field, and
L/K
finite. Then there is a unique
maximal subfield
K ⊆ T ⊆ L
such that
T/K
is unramified. Moreover, [
T
:
K
] =
f
L/K
.
Proof.
Let
T/K
be the unique unramified extension with residue field extension
k
L
/k
K
. Then
id
:
k
T
=
k
L
→ k
L
lifts to a
K
-embedding
T → L
. Identifying
T
with its image, we know
[T : K] = f
L/K
.
Now if
T
0
is any other unramified extension, then
T
0
T
is an unramified extension
over K, so
[T : K] ≤ [T T
0
: K] ≤ f
L/K
= [T : K].
So we have equality throughout, and T
0
⊆ T . So this is maximal.