8Lubin–Tate theory
III Local Fields
8.1 Motivating example
We will work out the details of local Artin reciprocity in the case of
Q
p
as a
motivating example for the proof we are going to come up with later. Here we
will need the results of local class field theory to justify our claims, but this is
not circular since this is not really part of the proof.
Lemma. Let L/K be a finite abelian extension. Then we have
e
L/K
= (O
×
K
: N
L/K
(O
×
L
)).
Proof.
Pick
x ∈ L
×
, and
w
the valuation on
L
extending
v
K
, and
n
= [
L
:
K
].
Then by construction of w, we know
v
K
(N
L/K
(x)) = nw(x) = f
L/K
v
L
(x).
So we have a surjection
K
×
N(L/K)
Z
f
L/K
Z
v
K
.
The kernel of this map is equal to
O
×
K
N(L/K)
N(L/K)
∼
=
O
×
K
O
×
K
∩ N(L/K)
=
O
×
K
N
L/K
(O
×
L
)
.
So by local class field theory, we know
n = (K
×
: N(L/K)) = f
L/K
(O
×
K
: N
L/K
(O
×
L
)),
and this implies what we want.
Corollary.
Let
L/K
be a finite abelian extension. Then
L/K
is unramified if
and only if N
L/K
(O
×
L
) = O
×
K
.
Now we fix a uniformizer
π
K
. Then we have a topological group isomorphism
K
×
∼
=
hπ
K
i × O
×
K
.
Since we know that the finite abelian extensions correspond exactly to finite
index subgroups of
K
×
by taking the norm groups, we want to understand
subgroups of K
×
. Now consider the subgroups of K
×
of the form
hπ
m
K
i × U
(n)
K
.
We know these form a basis of the topology of
K
×
, so it follows that finite-index
open subgroups must contain one of these guys. So we can find the maximal
abelian extension as the union of all fields corresponding to these guys.
Since we know that
N
(
LM/K
) =
N
(
L/K
)
∩ N
(
M/K
), it suffices to further
specialize to the cases
hπ
K
i × U
(n)
K
and
hπ
m
K
i × O
K
separately. The second case is easy, because this corresponds to an unramified
extension by the above corollary, and unramified extensions are completely
characterized by the extension of the residue field. Note that the norm group
and the extension are both independent of the choice of uniformizer. The
extensions corresponding to the first case are much more difficult to construct,
and they depend on the choice of
π
K
. We will get them from Lubin–Tate theory.
Lemma.
Let
K
be a local field, and let
L
m
/K
be the extension corresponding
to hπ
m
K
i × O
K
. Let
L =
[
m
L
m
.
Then we have
K
ab
= K
ur
L,
Lemma. We have isomorphisms
W (K
ab
/K)
∼
=
W (K
ur
L/K)
∼
=
W (K
ur
/K) × Gal(L/K)
∼
=
Frob
Z
K
× Gal(L/K)
Proof.
The first isomorphism follows from the previous lemma. The second
follows from the fact that
K
ur
∩ L
=
K
as
L
is totally ramified. The last
isomorphism follows from the fact that
T
K
ur
/K
=
K
ur
trivially, and then by
definition W (K
ur/K
)
∼
=
Frob
Z
K
.
Example. We consider the special case of K = Q
p
and π
K
= p. We let
L
n
= Q
p
(ζ
p
n
),
where
ζ
p
n
is the primitive
p
n
th root of unity. Then by question 6 on example
sheet 2, we know this is a field with norm group
N(Q
p
(ζ
p
n
)/Q
p
) = hpi × (1 + p
n
Z
p
) = hpi × U
(n)
Q
p
,
and thus this is a totally ramified extension of Q
p
.
We put
Q
p
(ζ
p
∞
) =
∞
[
n=1
Q
p
(ζ
p
n
).
Then again this is totally ramified extension, since it is the nested union of
totally ramified extensions.
Then we have
Gal(Q
p
(ζ
p
∞
)/Q
p
)
∼
=
lim
←−
n
Gal(Q
p
(ζ
p
n
)/Q
p
)
= lim
←−
n
(Z/p
n
Z)
×
= Z
×
p
.
Note that we are a bit sloppy in this deduction. While we know that it is true
that
Z
×
p
∼
=
lim
←−
n
(
Z/p
n
Z
)
×
, the inverse limit depends not only on the groups
(
Z/p
n
Z
)
×
themselves, but also on the maps we use to connect the groups together.
Fortunately, from the discussion below, we will see that the maps
Gal(Q
p
(ζ
p
n
)/Q
p
) → Gal(Q
p
(ζ
p
n−1
)/Q
p
)
indeed correspond to the usual restriction maps
(Z/p
n
Z)
×
→ (Z/p
n−1
Z)
×
.
It is a fact that this is the inverse of the Artin map of
Q
p
restricted to
Z
×
p
.
Note that we have
W
(
Q
p
(
ζ
p
∞
)
/Q
p
) =
Gal
(
Q
p
(
ζ
p
∞
)
/Q
p
) because its maximal
unramified subextension is trivial.
We can trace through the above chains of isomorphisms to figure out what
the Artin map does. Let m = Z
×
p
. Then we can write
m = a
0
+ a
1
p + ··· ,
where a
i
∈ {0, ··· , p − 1} and a
0
6= 0. Now for each n, we know
m ≡ a
0
+ a
1
p + ··· + a
n−1
p
n−1
mod p
n
.
By the usual isomorphism Gal(Q
p
(ζ
p
n
)/Q
p
)
∼
=
Z/p
n
Z, we know m acts as
ζ
p
n
7→ ζ
a
0
+a
1
p+...+a
n−1
p
n−1
p
n
“=” ζ
m
p
n
on
Q
p
(
ζ
p
n
), where we abuse notation because taking
ζ
p
n
to powers of
p
greater
than
n
gives 1. It can also be interpreted as (1 +
λ
p
n
)
m
, where
λ
p
n
=
ζ
p
n
−
1 is
a uniformizer, which makes sense using binomial expansion.
So the above isomorphisms tells us that
Art
Q
p
restricted to
Z
×
p
acts on
Q
p
(ζ
p
∞
) as
Art
Q
p
(m)(ζ
p
n
) ≡ σ
m
−1
(ζ
p
n
) = ζ
m
−1
p
n
.
The full Artin map can then be read off from the following diagram:
Q
×
p
W (Q
ab
p
/Q
p
)
hpi × Z
×
p
W (Q
ur
p
/Q
p
) × Gal(Q
p
(ζ
p
∞
)/Q
p
)
∼
=
Art
Q
p
restriction
∼
where the bottom map sends
hp
n
, mi 7→ (Frob
n
Q
p
, σ
m
−1
).
In fact, we have
Theorem (Local Kronecker-Weber theorem).
Q
ab
p
=
[
n∈Z
≥1
Q
p
(ζ
n
),
Q
ur
p
=
[
n∈Z
≥1
(n,p)=1
Q
p
(ζ
n
).
Not a proof.
We will comment on the proof of the generalized version later.
Remark.
There is another normalization of the Artin map which sends a
uniformizer to the geometric Frobenius, defined to be the inverse of the arithmetic
Frobenius. With this convention, Art
Q
p
(m)|
Q
p
(ζ
p
∞
)
is σ
m
.
We can define higher ramification groups for general Galois extensions.
Definition
(Higher ramification groups)
.
Let
K
be a local field and
L/K
Galois.
We define, for s ∈ R
≥−1
G
s
(M/K) = {σ ∈ Gal(M/K) : σ|
L
∈ G
s
(L/K) for all finite
Galois subextension M/K}.
This definition makes sense, because the upper number behaves well when
we take quotients. This is one of the advantages of upper numbering. Note that
we can write the ramification group as the inverse limit
G
s
(M/K)
∼
=
lim
←−
L/K
G
s
(L/K),
as in the case of the Galois group.
Example.
Going back to the case of
K
=
Q
p
. We write
Q
p
n
for the unramified
extension of degree n of Q
p
. By question 11 of example sheet 3, we know that
G
s
(Q
p
n
(ζ
p
m
)/Q
p
) =
Gal(Q
p
n
(ζ
p
m
)/Q
p
) s = −1
Gal(Q
p
n
(ζ
p
m
)/Q
p
n
) −1 < s ≤ 0
Gal(Q
p
n
(ζ
p
m
)/ζ
p
k
) k − 1 < s ≤ k ≤ m − 1
1 s > m − 1
,
which corresponds to
hpi × U
(0)
hp
n
i × U
(m)
s = −1
hp
n
i × U
(0)
hp
n
i × U
(m)
−1 < s ≤ 0
hp
n
i × U
(k)
hp
n
i × U
(m)
k − 1 < s ≤ k ≤ m − 1
1 s > m − 1
under the Artin map.
By taking the limit as n, m → ∞, we get
Theorem. We have
G
s
(Q
ab
p
/Q
p
) = Art
Q
p
(1 + p
k
Z
p
) = Art
Q
p
(U
(k)
),
where k is chosen such that k − 1 < s ≤ k, k ∈ Z
≥0
.
Corollary. If L/Q
p
is a finite abelian extension, then
G
s
(L/Q
p
) = Art
Q
p
N(L/Q
p
)(1 + p
n
Z
p
)
N(L/Q
p
)
,
where n − 1 < s ≤ n.
Here Art
Q
p
induces an isomorphism
Q
×
p
N(L/Q
p
)
→ Gal(L/Q
p
).
So it follows that
L ⊆ Q
p
(
ζ
p
m
) for some
n
if and only if
G
s
(
L/Q
p
) = 1 for all
s > m − 1.