3Discretely valued fields

III Local Fields 3.1 Teichm¨uller lifts
Take our favorite discretely valued ring
Z
p
. This is
p
can write each element as
x = a
0
+ a
1
p + a
2
p
2
+ ··· ,
where each
a
i
is in
{
0
,
1
, ··· , p
1
}
. The reason this works is that 0
,
1
, ··· , p
1 are coset representatives of the ring
Z
p
/pZ
p
=
Z/pZ
. While these coset
representatives might feel like a “natural” thing to do in this context, this is
because we have implicitly identified with
Z
p
/pZ
p
=
Z/pZ
as a particular subset
of
Z Z
p
. However, this identification respects effectively no algebraic structure
at all. For example, we cannot multiplying the cosets simply by multiplying the
representatives as elements of
Z
p
, because, say, (
p
1)
2
=
p
2
2
p
+ 1, which is
not 1. So this is actually quite bad, at least theoretically.
It turns out that we can actually construct “natural” lifts in a very general
scenario.
Theorem.
Let
R
be a ring, and let
x R
. Assume that
R
is
x
complete and that
R/xR
is perfect of characteristic
p
. Then there is a unique
map [] : R/xR R such that
[a] a mod x
and
[ab] = [a][b].
for all
a, b R/xR
. Moreover, if
R
has characteristic
p
, then [
] is a ring
homomorphism.
Definition
(Teichm¨uller map)
.
The map [
] :
R/xR R
is called the Te-
ichm uller map. [x] is called the Teichm¨uller lift or representative of x.
The idea of the proof is as follows: suppose we have an
a R/xR
. If we
randomly picked a lift
α
, then chances are it would be a pretty “bad” choice,
since any two such choices can differ by a multiple of x.
p
th root of
a
to
R
, and then take the
p
th power
of it. We claim that this is a better way of picking a lift. Suppose we have picked
two lifts of a
p
1
, say, α
1
and α
0
1
. Then α
0
1
= xc + α
1
for some c. So we have
(α
0
1
)
p
α
p
1
= α
p
1
+ pxc + O(x
2
) α
p
1
= pxc + O(x
2
),
where we abuse notation and write
O
(
x
2
) to mean terms that are multiples of
x
2
.
We now recall that
R/xR
has characteristic
p
, so
p xR
. Thus in fact
pxc = O(x
2
). So we have
(α
0
1
)
p
α
p
1
= O(x
2
).
So while the lift is still arbitrary, any two arbitrary choices can differ by at most
x
2
. Alternatively, our lift is now a well-defined element of R/x
2
R.
We can, of course, do better. We can lift the p
2
th root of a to R, then take
the
p
2
th power of it. Now any two lifts can differ by at most
O
(
x
3
). More
generally, we can try to lift the
p
n
th root of
a
, then take the
p
n
th power of
it. We keep picking a higher and higher
n
, take the limit, and hopefully get
something useful out!
To prove this result, we will need the following messy lemma:
Lemma.
Let
R
be a ring with
x R
such that
R/xR
has characteristic
p
. Let
α, β R be such that
α = β mod x
k
()
Then we have
α
p
= β
p
mod x
k+1
.
Proof.
It is left as an exercise to modify the proof to work for
p
= 2 (it is actually
easier). So suppose p is odd. We take the pth power of () to obtain
α
p
β
p
+
p1
X
i=1
p
i
α
pi
β
i
x
p(k+1)
R.
We can now write
p1
X
i=1
(1)
i
p
i
α
pi
β
i
=
p1
2
X
i=1
(1)
i
p
i
(αβ)
i
α
p2i
β
p2i
= p(α β)(something).
Now since
R/xR
has characteristic
p
, we know
p xR
. By assumption, we know
α β x
k+1
R. So this whole mess is in x
k+2
R, and we are done.
Proof of theorem.
Let
a R/xR
. For each
n
, there is a unique
a
p
n
R/xR
.
We lift this arbitrarily to some α
n
R such that
α
n
a
p
n
mod x.
We define
β
n
= α
p
n
n
.
The claim is that
[a] = lim
n→∞
β
n
exists and is independent of the choices.
Note that if the limit exists no matter how we choose the
α
n
, then it
must be independent of the choices. Indeed, if we had choices
β
n
and
β
0
n
,
then
β
1
, β
0
2
, β
3
, β
0
4
, β
5
, β
0
6
, ···
is also a respectable choice of lifts, and thus must
converge. So β
n
and β
0
n
must have the same limit.
Since the ring is
x
-adically complete and is discretely valued, to show the
limit exists, it suffices to show that β
n+1
β
n
β
n+1
β
n
= (α
p
n+1
)
p
n
α
p
n
n
.
We now notice that
α
p
n+1
(a
p
n1
)
p
= a
p
n
α
n
mod x.
So by applying the previous the lemma many times, we obtain
(α
p
n+1
)
p
n
α
p
n
n
mod x
n+1
.
So β
n+1
β
n
x
n+1
R. So lim β
n
exists.
To see [a] = a mod x, we just have to note that
lim
n→∞
α
p
n
n
lim
n→∞
(a
p
n
)
p
n
= lim a = a mod x.
(here we are using the fact that the map
R R/xR
is continuous when
R
is
given the x-adic topology and R/xR is given the discrete topology)
The remaining properties then follow trivially from the uniqueness of the
above limit.
For multiplicativity, if we have another element
b R/xR
, with
γ
n
R
lifting b
p
n
for all n, then α
n
γ
n
lifts (ab)
p
n
. So
[ab] = lim α
p
n
n
γ
p
n
n
= lim α
p
n
n
lim γ
p
n
n
= [a][b].
If R has characteristic p, then α
n
+ γ
n
lifts a
p
n
+ b
p
n
= (a + b)
p
n
. So
[a + b] = lim(α
n
+ γ
n
)
p
n
= lim α
p
n
n
+ lim γ
p
n
n
= [a] + [b].
Since 1 is a lift of 1 and 0 is a lift of 0, it follows that this is a ring homomorphism.
Finally, to show uniqueness, suppose
φ
:
R/xR R
is a map with these
properties. Then we note that
φ
(
a
p
n
)
a
p
n
mod x
, and is thus a valid choice
of α
n
. So we have
[a] = lim
n→∞
φ(a
p
n
)
p
n
= lim φ(a) = φ(a).
Example. Let R = Z
p
and x = p. Then [] : F
p
Z
p
satisfies
[x]
p1
= [x
p1
] =  = 1.
So the image of [
x
] must be the unique
p
1th root of unity lifting
x
(recall we
proved their existence via Hensel’s lemma).
When proving theorems about these rings, the Teichm¨uller lifts would be
very handy and natural things to use. However, when we want to do actual
computations, there is absolutely no reason why these would be easier!
As an application, we can prove the following characterization of equal
characteristic complete DVF’s.
Theorem.
Let
K
be a complete discretely valued field of equal characteristic
p
,
and assume that k
K
is perfect. Then K
=
k
K
((T )).
Proof.
Let
K
be a complete DVF. Since every DVF the field of fractions of
its valuation ring, it suffices to prove that
O
K
=
k
K
[[
T
]]. We know
O
K
has
characteristic
p
. So [
] :
k
K
O
K
is an injective ring homomorphism. We
choose a uniformizer π O
K
, and define
k
K
[[T ]] O
K
by
X
n=0
a
n
T
n
7→
X
n=0
[a
n
]π
n
.
Then this is a ring homomorphism since [
] is. The bijectivity follows from
property (v) in our list of properties of complete DVF’s.
Corollary.
Let
K
be a local field of equal characteristic
p
. Then
k
K
=
F
q
for
some q a power of p, and K
=
F
q
((T )).