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.