5Ramification theory for local fields
III Local Fields
5.1 Ramification index and inertia degree
Suppose we have an extension
L/K
of local fields. Then since
m
K
⊆ m
L
, and
O
L
⊆ O
L
, we obtain an injection
k
K
=
O
K
m
k
→
O
L
m
L
= k
L
.
So we also get an extension of residue fields
k
L
/k
K
. The question we want to ask
is how much of the extension is “due to” the extension of residue fields
k
L
/k
K
,
and how much is “due to” other things happening.
It turns out these are characterized by the following two numbers:
Definition
(Inertia degree)
.
Let
L/K
be a finite extension of local fields. The
inertia degree of L/K is
f
L/K
= [k
L
: k
K
].
Definition
(Ramification index)
.
Let
L/K
be a finite extension of local fields,
and let
v
L
be the normalized valuation of
L
and
π
K
a uniformizer of
K
. The
integer
e
L/K
= v
L
(π
K
)
is the ramification index of L/K.
The goal of the section is to show the following result:
Theorem. Let L/K be a finite extension. Then
[L : K] = e
L/K
f
L/K
.
We then have two extreme cases of ramification:
Definition
(Unramified extension)
.
Let
L/K
be a finite extension of local fields.
We say L/K is unramified if e
L/K
= 1, i.e. f
L/K
= [L : K].
Definition
(Totally ramified extension)
.
Let
L/K
be a finite extension of local
fields. We say L/K is totally ramified if f
L/K
= 1, i.e. e
L/K
= [L : K].
In the next section we will, amongst many things, show that every extension
of local fields can be written as an unramified extension followed by a totally
ramified extension.
Recall the following: let
R
be a PID and
M
a finitely-generated
R
-module.
Assume that
M
is torsion-free. Then there is a unique integer
n ≥
0 such that
M
∼
=
R
n
. We say
n
has rank
n
. Moreover, if
N ⊆ M
is a submodule, then
N
is
finitely-generated, so N
∼
=
R
m
for some m ≤ n.
Proposition.
Let
K
be a local field, and
L/K
a finite extension of degree
n
.
Then
O
L
is a finitely-generated and free
O
K
module of rank
n
, and
k
L
/k
K
is
an extension of degree ≤ n.
Moreover, L is also a local field.
Proof.
Choose a
K
-basis
α
1
, ··· , α
n
of
L
. Let
k · k
denote the maximum norm
on L.
n
X
i=1
x
i
α
i
= max
i=1,...,n
|x
i
|
as before. Again, we know that
k · k
is equivalent to the extended norm
| · |
on
L as K-norms. So we can find r > s > 0 such that
M = {x ∈ L : kxk ≤ s} ⊆ O
L
⊆ N = {x ∈ L : kxk ≤ r}.
Increasing r and decreasing s if necessary, we wlog r = |a| and s = |b| for some
a, b ∈ K.
Then we can write
M =
n
M
i=1
O
k
bα
i
⊆ O
L
⊆ N =
n
M
i=1
O
K
aα
i
.
We know that
N
is finitely generated and free of rank
n
over
O
K
, and so is
M
.
So O
L
must be finitely generated and free of rank n over O
K
.
Since m
k
= m
k
∩ O
K
, we have a natural injection
O
K
m
k
→
O
L
m
L
= k
L
.
Since
O
L
is generated over
O
K
by
n
elements, we know that
k
K
is generated by
n elements over k
K
, so it has rank at most n.
To see that
L
is a local field, we know that
k
L
/k
K
is finite and
k
K
is finite,
so
k
L
is finite. It is complete under the norm because it is a finite-dimensional
vector space over a complete field.
Finally, to see that the valuation is discrete, suppose we have a normalized
valuation on K, and w the unique extension of v
K
to L. Then we have
w(α) =
1
n
v
K
(N
L/K
(α)).
So we have
w(L
×
) ⊆
1
n
v(K
×
) =
1
n
Z.
So it is discrete.
Note that we cannot just pick an arbitrary basis of
L/K
and scale it to give
a basis of
O
L
/O
K
. For example,
Q
2
(
√
2
)
/Q
2
has basis 1
,
√
2
, but
|
√
2|
=
1
√
2
and cannot be scaled to 1 by an element in Q
2
.
Even if such a scaled basis exists, it doesn’t necessarily give a basis of the
integral rings. For example,
Q
3
(
√
−1
)
/Q
3
has a
Q
3
-basis 1
,
1 + 3
√
−1
and
|1 + 3
√
−1| = 1, but
√
−1 6∈ Z
3
+ Z
3
(1 + 3
√
−1).
So this is not a basis of O
Q
3
(
√
−1)
over Z
3
.
Theorem. Let L/K be a finite extension. Then
[L : K] = e
L/K
f
L/K
,
and there is some α ∈ O
L
such that O
L
= O
K
[α].
Proof.
We will be lazy and write
e
=
e
L/K
and
f
=
f
L/K
. We first note that
k
L
/k
K
is separable, so there is some
¯α ∈ k
L
such that
k
L
=
k
K
(
¯α
) by the
primitive element theorem. Let
¯
f(x) ∈ k
K
[x]
be the minimal polynomial of
¯α
over
k
K
and let
f ∈ O
L
[
x
] be a monic lift of
¯
f
with deg f = deg
¯
f.
We first claim that there is some
α ∈ O
L
lifting
¯α
such that
v
L
(
f
(
α
)) = 1
(note that it is always
≥
1). To see this, we just take any lift
β
. If
v
L
(
f
(
β
)) = 1,
then we are happy and set
α
=
β
. If it doesn’t work, we set
α
=
β
+
π
L
, where
π
L
is the uniformizer of L.
Then we have
f(α) = f(β + π
L
) = f(β) + f
0
(β)π
L
+ bπ
2
L
for some
b ∈ O
L
, by Taylor expansion around
β
. Since
v
L
(
f
(
β
))
≥
2 and
v
L
(
f
0
(
β
)) = 0 (since
¯
f
is separable, we know
f
0
(
β
) does not vanish when we
reduce mod m), we know v
L
(f(α)) = 1. So f(α) is a uniformizer of L.
We now claim that the elements
α
i
π
j
for
i
= 0
, ··· , f −
1 and
j
= 0
, ··· , e −
1
are an O
K
-basis of O
L
. Suppose we have
X
i,j
a
ij
α
i
π
j
= 0
for some a
ij
∈ K not all 0. We put
s
j
=
f−1
X
i=0
a
ij
α
i
.
We know that 1
, α, ··· , α
f−1
are linearly independent over
K
since their re-
ductions are linearly independent over
k
K
. So there must be some
j
such that
s
j
6= 0.
The next claim is that if
s
j
6
= 0, then
e | v
L
(
s
j
). We let
k
be an index for
which |a
kj
| is maximal. Then we have
a
−1
kj
s
j
=
f−1
X
i=0
a
−1
kj
a
ij
α
i
.
Now note that by assumption, the coefficients on the right have absolute value
≤ 1, and is 1 when i = k. So we know that
a
−1
kj
s
j
6≡ 0 mod π
L
,
because 1, ¯α, ··· , ¯α
f−1
are linearly independent. So we have
v
L
(a
−1
kj
s
j
) = 0.
So we must have
v
L
(s
j
) = v
L
(a
kj
) + v
L
(a
−1
kj
s
j
) ∈ v
L
(K
×
) = ev
L
(L
×
) = eZ.
Now we write
X
a
ij
α
i
π
j
=
e−1
X
j=0
s
j
π
j
= 0.
If
s
j
6
= 0, then we have
v
L
(
s
j
π
j
) =
v
L
(
s
j
) +
j ∈ j
+
eZ
. So no two non-zero
terms in
P
e−1
j=0
s
j
π
j
have the same valuation. This implies that
P
e−1
j=0
s
j
π
j
6
= 0,
which is a contradiction.
We now want to prove that
O
L
=
M
i,j
O
K
α
i
π
j
.
We let
M =
M
i,j
O
K
α
i
π
j
,
and put
N =
f−1
M
i=0
O
L
α
i
.
Then we have
M = N + πN + π
2
N + ··· + π
e−1
N.
We are now going to use the fact that 1
, ¯α, ··· , ¯α
f−1
span
k
L
over
k
K
. So we
must have that O
L
= N + πO
L
. We iterate this to obtain
O
L
= N + π(N + O
L
)
= N + πN + π
2
O
L
= ···
= N + πN + π
2
N + ··· + π
e−1
N + π
n
O
L
= M + π
K
O
L
,
using the fact that
π
K
and
π
e
have the same valuation, and thus they differ by
a unit in O
L
. Iterating this again, we have
O
L
= M + π
n
k
O
L
for all
n ≥
1. So
M
is dense in
O
L
. But
M
is the closed unit ball in the subspace
M
i,j
Kα
i
π
j
⊆ l
with respect to the maximum norm with respect to the given basis. So it must
be complete, and thus M = O
L
.
Finally, since
α
i
π
j
=
α
i
f
(
α
)
j
is a polynomial in
α
, we know that
O
L
=
O
K
[α].
Corollary. If M/L/K is a tower of finite extensions of local fields, then
f
M/K
= f
L/K
f
M/L
e
M/K
= e
L/K
e
M/L
Proof.
The multiplicativity of
f
M/K
follows from the tower law for the residue
fields, and the multiplicativity of
e
M/K
follows from the tower law for the local
fields and that f
M/K
e
M/K
= [M : K].