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
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).
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
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
)
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
))
sZ
≥−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.