6Further ramification theory

III Local Fields 6.2 Multiple extensions
Suppose we have tower
M/L/K
of finite extensions of local fields. How do the
ramification groups of the different extensions relate? We first do the easy case.
Proposition.
Let
M/L/K
be finite extensions of local fields, and
M/K
Galois.
Then
G
s
(M/K) Gal(M/L) = G
s
(M/L).
Proof. We have
G
s
(M/K) = {σ Gal(M/L) : v
M
(σx x) s + 1} = G
s
(M/K) Gal(M/L).
This is trivial, because the definition uses the valuation
v
M
of the bigger field
all the time. What’s more difficult and interesting is quotients, namely going
from M/K to L/K.
We want to prove the following theorem:
Theorem
(Herbrand’s theorem)
.
Let
M/L/K
be finite extensions of local fields
with M/K and L/K Galois. Then there is some function η
M/L
such that
G
t
(L/K)
=
G
s
(M/K)
G
s
(M/L)
for all s, where t = η
M/L
(s).
To better understand the situation, it helps to provide an alternative charac-
terization of the Galois group.
Definition (i
L/K
). We define
i
L/K
(σ) = min
x∈O
L
v
L
(σ(x) x).
It is then immediate that
G
s
(L/K) = {σ Gal(L/K) : i
L/K
(σ) s + 1}.
This is not very helpful. We now claim that we can compute
i
L/K
using the
following formula:
Proposition.
Let
L/K
be a finite Galois extension of local fields, and pick
α O
L
such that O
L
= O
K
[α]. Then
i
L/K
(σ) = v
L
(σ(α) α).
Proof.
Fix a
σ
. It is clear that
i
L/K
(
σ
)
v
L
(
σ
(
α
)
α
). Conversely, for any
x O
L
, we can find a polynomial g O
K
[t] such that
x = g(α) =
X
b
i
α
i
,
where b
i
O
K
. In particular, b
i
is fixed by σ.
Then we have
v
L
(σ(x) x) = v
L
(σg(α) g(α))
= v
L
n
X
i=1
b
i
(σ(α)
i
α
i
)
!
v
L
(σ(α) α),
using the fact that σ(α) α | σ(α)
i
α
i
for all i. So done.
Now if
M/L/K
are finite Galois extensions of local fields, then
O
M
=
O
K
[
α
]
implies O
M
= O
L
[α]. So for σ Gal(M/L), we have
i
M/L
(σ) = i
M/K
(σ).
Going in the other direction is more complicated.
Proposition.
Let
M/L/K
be a finite extension of local fields, such that
M/K
and L/K are Galois. Then for σ Gal(L/K), we have
i
L/K
(σ) = e
1
M/L
X
τGal(M/K)
τ|
L
=σ
i
M/K
(τ).
Here
e
M/L
is just to account for the difference between
v
L
and
v
M
. So the
real content is that the value of
i
L/K
(
σ
) is the sum of the values of
i
M/K
(
τ
) for
all τ that restrict to σ.
Proof.
If
σ
= 1, then both sides are infinite by convention, and equality holds.
So we assume
σ 6
= 1. Let
O
M
=
O
L
[
α
] and
O
L
=
O
K
[
β
], where
α O
M
and
β O
L
. Then we have
e
M/L
i
L/K
(σ) = e
M/L
v
L
(σβ β) = v
M
(σβ β).
Now if τ Gal(M/K), then
i
M/K
(τ) = v
M
(τα α)
Now fix a τ such that τ|
L
= σ. We set H = Gal(M/L). Then we have
X
τ
0
Gal(M/K)
0
|
L
=σ
i
M/K
(τ
0
) =
X
gH
v
M
(τg(α) α) = v
M
Y
gH
(τg(α) α)
.
We let
b = σ(β) β = τ(β) β
and
a =
Y
gH
(τg(α) α).
We want to prove that v
M
(b) = v
M
(a). We will prove that a | b and b | a.
O
L
. Given
z O
L
,
we can write
z =
n
X
i=1
z
i
β
i
, z
i
O
K
.
Then we know
τ(z) z =
n
X
i=1
z
i
(τ(β)
i
β
i
)
is divisible by τ(β) β = b.
Now let
F
(
x
)
O
L
[
x
] be the minimal polynomial of
α
over
L
. Then explicitly,
we have
F (x) =
Y
gH
(x g(α)).
Then we have
(τF )(x) =
Y
gH
(x τ g(α)),
where
τF
is obtained from
F
by applying
τ
to all coefficients of
F
. Then all
coefficients of
τF F
are of the form
τ
(
z
)
z
for some
z O
L
. So it is divisible
by b. So b divides every value of this polynomial, and in particular
b | (τF F )(α) =
Y
gH
(α g(α)) = ±a,
So b | a.
In other direction, we pick
f O
K
[
x
] such that
f
(
α
) =
β
. Then
f
(
α
)
β
= 0.
This implies that the polynomial
f
(
x
)
β
divides the minimal polynomial of
α
in O
L
[x]. So we have
f(x) β = F (x)h(x)
for some h O
L
[x].
Then noting that f has coefficients in O
K
, we have
(f τβ)(x) = (τ f τb)(x) = (τ F )(x)(τh)(x).
Finally, set x = α. Then
b = β τβ = ±a(τ h)(α).
So a | b.
Now that we understand how the
i
L/K
behave when we take field extensions,
we should be able to understand how the ramification groups behave!
We now write down the right choice of η
L/K
: [1, ) [1, ):
η
L/K
(s) =
e
1
L/K
X
σG
min(i
L/K
(σ), s + 1)
!
1.
Theorem
(Herbrand’s theorem)
.
Let
M/L/K
be a finite extension of local
fields with M/K and L/K Galois. We set
H = Gal(M/L), t = η
M/L
(s).
Then we have
G
s
(M/K)H
H
= G
t
(L/K).
By some isomorphism theorem, and the fact that
H G
s
(
M/K
) =
G
s
(
M/L
),
this is equivalent to saying
G
t
(L/K)
=
G
s
(M/K)
G
s
(M/L)
.
Proof.
Let
G
=
Gal
(
M/K
). Fix a
σ Gal
(
L/K
). We let
τ Gal
(
M/K
) be an
extension of σ to M that maximizes i
M/K
, i.e.
i
M/K
(τ) i
M/K
(τg)
for all g H. This is possible since H is finite.
We claim that
i
L/K
(σ) 1 = η
M/L
(i
M/K
(τ) 1).
If this were true, then we would have
σ
G
s
(M/K)H
H
τ G
s
(M/K)
i
M/K
(τ) 1 s
Since η
M/L
is strictly increasing, we have
η
M/L
(i
M/K
(τ) 1) η
M/L
(s) = t
i
L/K
(σ) 1 t
σ G
t
(L/K),
and we are done.
To prove the claim, we now use our known expressions for
i
L/K
(
σ
) and
η
M/L
(i
M/K
(τ) 1) to rewrite it as
e
1
M/L
X
gH
i
M/K
(τg) = e
1
M/L
X
gH
min(i
M/L
(g), i
M/K
(τ)).
We then make the stronger claim
i
M/K
(τg) = min(i
M/L
(g), i
M/K
(τ)).
We first note that
i
M/K
(τg) = v
M
(τg(α) α)
= v
M
(τg(α) g(α) + g(α) α)
min(v
M
(τg(α) g(α)), v
M
(g(α) α))
= min(i
M/K
(τ), i
M/K
(g))
We cannot conclude our (stronger) claim yet, since we have a
in the middle.
We now have to split into two cases.
(i)
If
i
M/K
(
g
)
i
M/K
(
τ
), then the above shows that
i
M/K
(
τg
)
i
M/K
(
τ
).
But we also know that it is bounded above by
m
. So
i
M/K
(
τg
) =
i
M/K
(
τ
).
So our claim holds.
(ii)
If
i
M/K
(
g
)
< i
M/K
(
τ
), then the above inequality is in fact an equality as
the two terms have different valuations. So our claim also holds.
So done.
We now prove an alternative characterization of the function
η
L/K
, using a
funny integral.
Proposition. Write G = Gal(L/K). Then
η
L/K
(s) =
Z
s
0
dx
(G
0
(L/K) : G
x
(L/K))
.
When 1 x < 0, our convention is that
1
(G
0
(L/K) : G
x
(L/K))
= (G
x
(L/K) : G
0
(L/K)),
which is just equal to 1 when 1 < x < 0. So
η
L/K
(s) = s if 1 s 0.
Proof.
We denote the RHS by
θ
(
s
). It is clear that both
η
L/K
(
s
) and
θ
(
s
) are
piecewise linear and the break points are integers (since
i
L/K
(
σ
) is always an
integer). So to see they are the same, we see that they agree at a point, and
that they have equal derivatives. We have
η
L/K
(0) =
|{σ G : i
L/K
(σ) 1}|
e
L/K
1 = 0 = θ(0),
since the numerator is the size of the inertia group.
If s [1, ) \ Z, then
η
0
L/K
(s) = e
1
L/K
(|{σ G : i
L/K
(σ) s + 1}|)
=
|G
s
(L/K)|
|G
0
(L/K)|
=
1
(G
0
(L/K) : G
s
(L/K))
= θ
0
(s).
So done.
We now tidy up the proof by inventing a different numbering of the ramifica-
tion groups. Recall that
η
L/K
: [1, ) [1, )
is continuous, strictly increasing, and
η
L/K
(1) = 1, η
L/K
(s) as s .
So this is invertible. We set
Notation.
ψ
L/K
= η
1
L/K
.
Definition
(Upper numbering)
.
Let
L/K
be a Galois extension of local fields.
Then the upper numbering of the ramification groups of L/K is defined by
G
t
(L/K) = G
ψ
L/K
(t)
(L/K)
for t [1, ). The original number is called the lower numbering.
To rephrase our previous theorem using the upper numbering, we need a
little lemma:
Lemma.
Let
M/L/K
be a finite extension of local fields, and
M/K
and
L/K
be Galois. Then
η
M/K
= η
L/K
η
M/L
.
Hence
ψ
M/K
= ψ
M/L
ψ
L/K
.
Proof.
Let
s
[
1
,
), and let
t
=
η
M/L
(
s
), and
H
=
Gal
(
M/L
). By Her-
brand’s theorem, we know
G
t
(L/K)
=
G
s
(M/K)H
H
=
G
s
(M/K)
H G
s
(M/K)
=
G
s
(M/K)
G
s
(M/L)
.
Thus by multiplicativity of the inertia degree, we have
|G
s
(M/K)|
e
M/K
=
|G
t
(L/K)|
e
L/K
|G
s
(M/L)|
e
M/L
.
By the fundamental theorem of calculus, we know that whenever the derivatives
make sense, we have
η
0
M/K
(s) =
|G
s
(M/K)|
e
M/K
.
So putting this in, we know
η
0
M/K
(s) = η
0
L/K
(t)η
0
M/L
(s) = (η
L/K
η
M/L
)
0
(s).
Since
η
M/K
and
η
L/K
η
M/L
agree at 0 (they both take value 0), we know that
the functions must agree everywhere. So done.
Corollary.
Let
M/L/K
be finite Galois extensions of local fields, and
H
=
Gal(M/L). Let t [1, ). Then
G
t
(M/K)H
H
= G
t
(L/K).
Proof. Put s = η
L/K
(t). Then by Herbrand’s theorem, we have
G
t
(M/K)H
H
=
G
ψ
M/K
(t)
(M/K)H
H
=
G
η
M/L
(ψ
M/K
(t))
(L/K)
= G
s
(L/K)
= G
t
(L/K).
This upper numbering might seem like an unwieldy beast that was invented
just so that our theorem looks nice. However, it turns out that often the upper
numberings are rather natural, as we could see in the example below:
Example.
Consider
ζ
p
n
a primitive
p
n
th root of unity, and
K
=
Q
p
(
ζ
p
n
). The
minimal polynomial of ζ
p
n
is the p
n
th cyclotomic polynomial
Φ
p
n
(x) = x
p
n1
(p1)
+ x
p
n1
(p2)
+ ··· + 1.
It is an exercise on the example sheet to show that this is indeed irreducible.
So
K/Q
p
is a Galois extension of degree
p
n1
(
p
1). Moreover, it is totally
ramified by question 6 on example sheet 2, with uniformizer
π = ζ
p
n
1
is a uniformizer. So we know
O
K
= Z
p
[ζ
p
n
1] = Z
p
[ζ
p
n
].
We then have an isomorphism
Z
p
n
Z
×
Gal(L/Q
p
)
obtained by sending m σ
m
, where
σ
m
(ζ
p
n
) = ζ
m
p
n
.
We have
i
K/Q
p
(σ
m
) = v
K
(σ
m
(ζ
p
n
) ζ
p
n
)
= v
K
(ζ
m
p
n
ζ
p
n
)
= v
K
(ζ
m1
p
n
1)
since
ζ
p
n
is a unit. If
m
= 1, then this thing is infinity. If it is not 1, then
ζ
m1
p
n
is a primitive
p
nk
th root of unity for the maximal
k
such that
p
k
| m
1. So
by Q6 on example sheet 2, we have
v
K
(ζ
m1
p
n
1) =
p
n1
(p 1)
p
nk1
(p 1)
= p
k
.
Thus we have
v
K
(ζ
m1
p
n
1) p
k
m 1 mod p
k
.
It then follows that for
p
k
s + 1 p
k1
+ 1,
we have
G
s
(K/Q
p
)
=
{m (Z/p
n
)
×
: m 1 mod p
k
}.
Now m 1 mod p
k
iff σ
m
(ζ
p
k
) = ζ
p
k
. So in fact
G
s
(K/Q
p
)
=
Gal(K/Q
p
(ζ
p
k
)).
Finally, when s p
n
1, we have
G
s
(K/Q
p
) = 1.
We claim that
η
K/Q
p
(p
k
1) = k.
So we have
G
k
(K/Q
p
) = Gal(K/Q
p
(ζ
p
k
)).
This actually looks much nicer!
To actually compute
η
K/Q
p
, we have notice that the function we integrate to
get η looks something like this (not to scale):
1
p1
1
p(p1)
1
p
2
(p1)
p 1
p
2
1 p
3
1
The jumps in the lower numbering are at
p
k
1 for
k
= 1
, ··· , n
1. So we have
η
K/Q
p
(p
k
1) = (p 1)
1
p 1
+ ((p
2
1) (p 1))
1
p(p 1)
+ ··· + ((p
k
1) (p
k1
1))
1
p
k1
(p 1)
= k.