6Further ramification theory

III Local Fields

6.1 Some filtrations

If we have a field

K

, then we have a unit group

U

K

=

O

×

K

. We would like to

come up with a filtration of subgroups of the unit group, namely a sequence

··· ⊆ U

(2)

K

⊆ U

(1)

K

⊆ U

(0)

K

= U

K

of subgroups that tells us how close a unit is to being 1. The further down we

are in the chain, the closer we are to being 1.

Similarly, given a field extension

L/K

, we want a filtration on the Galois

group (the indexing is conventional)

··· ⊆ G

2

(L/K) ⊆ G

1

(L/K) ⊆ G

0

(L/K) ⊆ G

−1

(L/K) = Gal(L/K).

This time, the filtration tells us how close the automorphisms are to being the

identity map.

The key thing about these filtrations is that we can figure out information

about the quotients

U

(s)

K

/U

(s+1)

K

and

G

s

(

L/K

)

/G

s+1

(

L/K

), which is often easier.

Later, we might be able to patch these up to get more useful information about

U

K

and Gal(L/K).

We start with the filtration of the unit group.

Definition (Higher unit groups). We define the higher unit groups to be

U

(s)

K

= U

(s)

= 1 + π

s

K

O

K

.

We also put

U

K

= U

(0)

K

= U

(0)

= O

×

K

.

The quotients of these units groups are surprisingly simple:

Proposition. We have

U

K

/U

(1)

K

∼

=

(k

×

K

, ·),

U

(s)

K

/U

(s+1)

K

∼

=

(k

K

, +).

for s ≥ 1.

Proof.

We have a surjective homomorphism

O

×

K

→ k

×

K

which is just reduction

mod

π

K

, and the kernel is just things that are 1 modulo

π

K

, i.e.

U

(1)

K

. So this

gives the first part.

For the second part, we define a surjection U

(s)

K

→ k

K

given by

1 + π

s

K

x 7→ x mod π

k

.

This is a group homomorphism because

(1 + π

s

K

x)(1 + π

s

K

y) = 1 + π

S

(x + y + π

s

xy),

and this gets mapped to

x + y + π

s

x + y

∼

=

x + y mod π

K

.

Then almost by definition, the kernel is U

(s+1)

K

.

The next thing to consider is a filtration of the Galois group.

Definition (Higher ramification group). Let L/K be a finite Galois extension

of local fields, and v

L

the normalized valuation of L.

Let s ∈ R

≥−1

. We define the sth ramification group by

G

s

(L/K) = {σ ∈ Gal(L/K) : v

L

(σ(x) − x) ≥ s + 1 for all x ∈ O

L

}.

So if you belong to

G

s

for a large

s

, then you move things less. Note that we

could have defined these only for

s ∈ Z

≥−1

, but allowing fractional indices will

be helpful in the future.

Now since σ(x) − x ∈ O

L

for all x ∈ O

L

, we know

G

−1

(L/K) = Gal(L/K).

We next consider the case of G

0

(L/K). This is, by definition

G

0

(L/K) = {σ ∈ Gal(L/K) : v

L

(σ(x) − x) ≥ 1 for all x ∈ O

L

}

= {σ ∈ Gal(L/K) : σ(x) ≡ x mod m for all x ∈ O

L

}.

In other words, these are all the automorphisms that reduce to the identity when

we reduce it to Gal(k

L

/k

K

).

Definition

(Inertia group)

.

Let

L/K

be a finite Galois extension of local fields.

Then the inertia group of L/K is the kernel of the natural homomorphism

Gal(L/K) → Gal(k

L

/k

K

)

given by reduction. We write this as

I(L/K) = G

0

(L/K).

Proposition.

Let

L/K

be a finite Galois extension of local fields. Then the

homomorphism

Gal(L/K) → Gal(k

L

/k

K

)

given by reduction is surjective.

Proof.

Let

T/K

be maximal unramified subextension. Then by Galois theory,

the map

Gal

(

L/K

)

→ Gal

(

T/K

) is a surjection. Moreover, we know that

k

T

= k

L

. So we have a commutative diagram

Gal(L/K) Gal(k

L

/k

K

)

Gal(T/K) Gal(k

T

/k

K

).

∼

So the map Gal(L/K) → Gal(k

L

/k

K

) is surjective.

Then the inertia group is trivial iff

L/K

is unramified. The field

T

is

sometimes called the inertia field.

Lemma.

Let

L/K

be a finite Galois extension of local fields, and let

σ ∈ I

(

L/K

).

Then σ([x]) = [x] for all x.

More generally, let

x ∈ k

L

and

σ ∈ Gal

(

L/K

) with image

¯σ ∈ Gal

(

k

L

/k

K

).

Then we have

[¯σ(x)] = σ([x]).

Proof. Consider the map k

L

→ O

L

given by

f : x 7→ σ

−1

([¯σ(x)]).

This is multiplicative, because every term is multiplicative, and

σ

−1

([¯σ(x)]) ≡ x mod π

L

.

So this map f has to be the Teichm¨uller lift by uniqueness.

That’s all we’re going to say about the inertia group. We now consider the

general properties of this filtration.

Proposition.

Let

L/K

be a finite Galois extension of local fields, and

v

L

the

normalized valuation of

L

. Let

π

L

be the uniformizer of

L

. Then

G

s+1

(

L/K

) is

a normal subgroup of G

s

(L/K) for s ∈ Z

≥0

, and the map

G

s

(L/K)

G

s+1

(L/K)

→

U

(s)

L

U

(s+1)

L

given by

σ 7→

σ(π

L

)

π

L

is a well-defined injective group homomorphism, independent of the choice of

π

L

.

Proof. We define the map

φ : G

s

(L/K) →

U

(s)

L

U

(s+1)

L

σ 7→ σ(π

L

)/π

L

.

We want to show that this has kernel G

s+1

(L/K).

First we show it is well-defined. If σ ∈ G

s

(L/K), we know

σ(π

L

) = π

L

+ π

s+1

L

x

for some x ∈ O

L

. So we know

σ(π

L

)

π

L

= 1 + π

s

L

x ∈ U

(s)

L

.

So it has the right image. To see this is independent of the choice of

π

L

, we let

u ∈ O

×

L

. Then σ(u) = u + π

s+1

L

y for some y ∈ O

L

.

Since any other uniformizer must be of the form π

L

u, we can compute

σ(π

L

u)

π

L

u

=

(π

L

+ π

s+1

L

)(u + π

s+1

L

y)

π

L

u

= (1 + π

s

L

x)(1 + π

s+1

L

u

−1

y)

≡ 1π

s

L

x (mod U

s+1

L

).

So they represent the same element in in U

(s)

L

/U

(s+1)

L

.

To see this is a group homomorphism, we know

φ(στ) =

σ(τ(π

L

))

π

L

=

σ(τ(π

L

))

τ(π

L

)

τ(π

L

)

π

L

= φ(σ)φ(t),

using the fact that τ (π

L

) is also a uniformizer.

Finally, we have to show that ker φ = G

s+1

(L/K). We write down

ker φ = {σ ∈ G

s

(L/K) : v

L

(σ(π

L

) − π

L

) ≥ s + 2}.

On the other hand, we have

G

s+1

(L/K) = {σ ∈ G

s

(L/K) : v

L

(σ(z) − z) ≥ s + 2 for all z ∈ O

L

}.

So we trivially have

G

s+1

(

L/K

)

⊆ ker φ

. To show the converse, let

x ∈ O

L

and

write

x =

∞

X

n=0

[x

n

]π

n

L

.

Take σ ∈ ker φ ⊆ G

s

(L/K) ⊆ I(L/K). Then we have

σ(π

L

) = π

L

+ π

s+2

L

y, y ∈ O

L

.

Then by the previous lemma, we know

σ(x) − x =

∞

X

n=1

[x

n

] ((σ(π

L

))

n

− π

n

L

)

=

∞

X

n=1

[x

n

]

(π

L

+ π

s+2

L

y)

n

− π

n

L

= π

s+2

L

(things).

So we know v

L

(σ(x) − x) ≥ s + 2.

Corollary. Gal(L/K) is solvable.

Proof. Note that

\

s

G

s

(L/K) = {id}.

So (

G

s

(

L/K

))

s∈Z

≥−1

is a subnormal series of

Gal

(

L/K

), and all quotients are

abelian, because they embed into

U

(s)

L

U

(s+1)

L

∼

=

(

k

K

,

+) (and

s

=

−

1 can be checked

separately).

Thus if

L/K

is a finite extension of local fields, then we have, for

s ≥

1,

injections

G

s

(L/K)

G

s+1

(L/K)

→ k

L

.

Since k

L

is a p-group, it follows that

|G

s

(L/K)|

|G

s+1

(L/K)|

is a pth power. So it follows that for any t, the quotient

|G

1

(L/K)|

|G

t

(L/K)|

is also a

p

th power. However, we know that the intersection of all

G

s

(

L/K

)

is

{id}

, and also

Gal

(

L/K

) is finite. So for sufficiently large

t

, we know that

|G

t

(L/K)| = 1. So we conclude that

Proposition. G

1

(L/K) is always a p-group.

We now use the injection

G

0

(L/K)

G

1

(L/K)

→ k

×

L

,

and the fact that

k

×

L

has order prime to

p

. So

G

1

(

L/K

) must be the Sylow

p-subgroup of G

0

(L/K). Since it is normal, it must be the unique p-subgroup.

Definition

(Wild inertia group and tame quotient)

. G

1

(

L/K

) is called the wild

inertia group, and the quotient G

0

(L/K)/G

1

(L/K) is the tame quotient.