3Discretely valued fields

III Local Fields

3.1 Teichm¨uller lifts

Take our favorite discretely valued ring

Z

p

. This is

p

-adically complete, so we

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

-adically

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.

Suppose we instead lifted a

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

+

p−1

X

i=1

p

i

α

p−i

β

i

∈ x

p(k+1)

R.

We can now write

p−1

X

i=1

(−1)

i

p

i

α

p−i

β

i

=

p−1

2

X

i=1

(−1)

i

p

i

(αβ)

i

α

p−2i

− β

p−2i

= 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

→ 0 x-adically. Indeed, we have

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

−n−1

)

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]

p−1

= [x

p−1

] = [1] = 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 )).