3Discretely valued fields

III Local Fields

3.2 Witt vectors*

We are now going to look at the mixed characteristic analogue of this result. We

want something that allows us to go from characteristic

p

to characteristic 0.

This is known as Witt vectors, which is non-examinable.

We start with the notion of a strict

p

-ring. Roughly this is a ring that satisfies

all the good properties whose name has the word “p” in it.

Definition

(Strict

p

-ring)

.

Let

A

be a ring. A is called a strict

p

-ring if it is

p-torsion free, p-adically complete, and A/pA is a perfect ring.

Note that a strict

p

-ring in particular satisfies the conditions for the Te-

ichm¨uller lift to exist, for x = p.

Example. Z

p

is a strict p-ring.

The next example we are going to construct is more complicated. This is in

some sense a generalization of the usual polynomial rings

Z

[

x

1

, ··· , x

n

], or more

generally,

Z[x

i

| i ∈ I],

for

I

possibly infinite. To construct the “free” strict

p

-ring, after adding all these

variables

x

i

, to make it a strict

p

-ring, we also need to add their

p

th roots, and

the p

2

th roots etc, and then take the p-adic completion, and hope for the best.

Example. Let X = {x

i

: i ∈ I} be a set. Let

B = Z[x

p

−∞

i

| i ∈ I] =

∞

[

n=0

Z[x

p

−n

i

| i ∈ I].

Here the union on the right is taken by treating

Z[x

i

| i ∈ I] ⊆ Z[x

p

−1

i

| i ∈ I] ⊆ ···

in the natural way.

We let

A

be the

p

-adic completion of

B

. We claim that

A

is a strict

p

-ring

and A/pA

∼

=

F

p

[x

p

−∞

i

| i ∈ I].

Indeed, we see that

B

is

p

-torsion free. By Exercise 13 on Sheet 1, we know

A is p-adically complete and torsion free. Moreover,

A/pA

∼

=

B/pB

∼

=

F

p

[x

p

−∞

i

| i ∈ I],

which is perfect since every element has a p-th root.

If A is a strict p-ring, then we know that we have a Teichm¨uller map

[−] : A/pA → A,

Lemma.

Let

A

be a strict

p

-ring. Then any element of

A

can be written

uniquely as

a =

∞

X

n=0

[a

n

]p

n

,

for a unique a

n

∈ A/pA.

Proof. We recursively construct the a

n

by

a

0

= a (mod p)

a

1

≡ p

−1

(a − [a

0

]) (mod p)

.

.

.

Lemma.

Let

A

and

B

be strict

p

-rings and let

f

:

A/pA → B/pB

be a ring

homomorphism. Then there is a unique homomorphism

F

:

A → B

such that

f = F mod p, given by

F

X

[a

n

]p

n

=

X

[f(a

n

)]p

n

.

Proof sketch.

We define

F

by the given formula and check that it works. First of

all, by the formula,

F

is

p

-adically continuous, and the key thing is to check that

it is additive (which is slightly messy). Multiplicativity then follows formally

from the continuity and additivity.

To show uniqueness, suppose that we have some

ψ

lifting

f

. Then

ψ

(

p

) =

p

.

So ψ is p-adically continuous. So it suffices to show that ψ([a]) = [ψ(a)].

We take α

n

∈ A lifting a

p

−n

∈ A/pA. Then ψ(α

n

) lifts f (a)

p

−n

. So

ψ([a]) = lim ψ(α

p

−n

n

) = lim ψ(α

n

)

p

−n

= [f(a)].

So done.

There is a generalization of this result:

Proposition.

Let

A

be a strict

p

-ring and

B

be a ring with an element

x

such that

B

is

x

-adically complete and

B/xB

is perfect of characteristic

p

. If

f

:

A/pA → B/xB

is a ring homomorphism. Then there exists a unique ring

homomorphism

F

:

A → B

with

f

=

F mod x

, i.e. the following diagram

commutes:

A B

A/pA B/xB

F

f

.

Indeed, the conditions on

B

are sufficient for Teichm¨uller lifts to exist, and

we can at least write down the previous formula, then painfully check it works.

We can now state the main theorem about strict p-rings.

Theorem.

Let

R

be a perfect ring. Then there is a unique (up to isomorphism)

strict p-ring W (B) called the Witt vectors of R such that W (R)/pW (R)

∼

=

R.

Moreover, for any other perfect ring

R

, the reduction mod

p

map gives a

bijection

Hom

Ring

(W (R), W (R

0

)) Hom

Ring

(R, R

0

)

∼

.

Proof sketch.

If

W

(

R

) and

W

(

R

0

) are such strict

p

-rings, then the second part

follows from the previous lemma. Indeed, if

C

is a strict

p

-ring with

C/pC

∼

=

R

∼

=

W

(

R

)

/pW

(

R

), then the isomorphism

¯α

:

W

(

R

)

/pW

(

R

)

→ C/pC

and its

inverse

¯α

−1

have unique lifts

γ

:

W

(

R

)

→ C

and

γ

−1

:

C → W

(

R

), and these

are inverses by uniqueness of lifts.

To show existence, let R be a perfect ring. We form

F

p

[x

p

−∞

r

| r ∈ R] → R

x

r

7→ r

Then we know that the

p

-adic completion of

Z

[

x

p

−∞

r

| r ∈ R

], written

A

, is a

strict p-ring with

A/pA

∼

=

F

p

[x

p

−∞

r

| r ∈ R].

We write

I = ker(F

p

[x

p

−∞

r

| r ∈ R] → R).

Then define

J =

(

∞

X

n=0

[a

k

]p

n

∈ A : a

n

∈ I for all n

)

.

This turns out to be an ideal.

J A R

0 I A/pA R 0

We put

W

(

R

) =

A/J

. We can then painfully check that this has all the required

properties. For example, if

x =

∞

X

n=0

[a

n

]p

n

∈ A,

and

px =

∞

X

n=0

[a

n

]p

n+1

∈ J,

then by definition of

J

, we know [

a

n

]

∈ I

. So

x ∈ J

. So

W

(

R

)

/J

is

p

-torsion

free. By a similar calculation, one checks that

∞

\

n=0

p

n

W (R) = {0}.

This implies that

W

(

R

) injects to its

p

-adic completion. Using that

A

is

p

-adically

complete, one checks the surjectivity by hand.

Also, we have

W (R)

pW (R)

∼

=

A

J + pA

.

But we know

J + pA =

(

X

n

[a

n

]p

n

| a

0

∈ I

)

.

So we have

W (R)

pW (R)

∼

=

F

p

[x

p

−∞

r

| r ∈ R]

I

∼

=

R.

So we know that W (R) is a strict p-ring.

Example. W

(

F

p

) =

Z

p

, since

Z

p

satisfies all the properties

W

(

F

p

) is supposed

to satisfy.

Proposition.

A complete DVR

A

of mixed characteristic with perfect residue

field and such that p is a uniformizer is the same as a strict p-ring A such that

A/pA is a field.

Proof.

Let

A

be a complete DVR such that

p

is a uniformizer and

A/pA

is

perfect. Then

A

is

p

-torsion free, as

A

is an integral domain of characteristic 0.

Since it is also p-adically complete, it is a strict p-ring.

Conversely, if

A

is a strict

p

-ring, and

A/pA

is a field, then we have

A

×

⊆

A \ pA, and we claim that A

×

= A \ pA. Let

x =

∞

X

n=0

[x

n

]p

n

with

x

0

6

= 0, i.e.

x 6∈ pA

. We want to show that

x

is a unit. Since

A/pA

is a

field, we can multiply by [

x

−1

0

], so we may wlog

x

0

= 1. Then

x

= 1

− py

for

some y ∈ A. So we can invert this with a geometric series

x

−1

=

∞

X

n=0

p

n

y

n

.

So

x

is a unit. Now, looking at Teichm¨uller expansions and factoring out multiple

of

p

, any non-zero element

z

can be written as

p

n

u

for a unique

n ≥ Z

≥0

and

u ∈ A

×

. Then we have

v(z) =

(

n z 6= 0

∞ z = 0

is a discrete valuation on A.

Definition

(Absolute ramification index)

.

Let

R

be a DVR with mixed charac-

teristic

p

with normalized valuation

v

R

. The integer

v

R

(

p

) is called the absolute

ramification index of R.

Corollary.

Let

R

be a complete DVR of mixed characteristic with absolute

ramification index 1 and perfect residue field k. Then R

∼

=

W (k).

Proof.

Having absolute ramification index 1 is the same as saying

p

is a uni-

formizer. So

R

is a strict

p

-ring with

R/pR

∼

=

k

. By uniqueness of the Witt

vector, we know R

∼

=

W (k).

Theorem.

Let

R

be a complete DVR of mixed characteristic

p

with a perfect

residue field k and uniformizer π. Then R is finite over W (k).

Proof.

We need to first exhibit

W

(

k

) as a subring of

R

. We know that

id

:

k → k

lifts to a homomorphism

W

(

k

)

→ R

. The kernel is a prime ideal because

R

is

an integral domain. So it is either 0 or

pW

(

k

). But

R

has characteristic 0. So it

can’t be pW (k). So this must be an injection.

Let e be the absolute ramification index of R. We want to prove that

R =

e−1

M

i=0

π

i

W (k).

Looking at valuations, one sees that 1

, π, π, ··· , π

e−1

are linearly independent

over W (k). So we can form

M =

e−1

M

i=0

π

i

W (k) ⊆ R.

We consider R/pR. Looking at Teichm¨uller expansions

∞

X

n=0

[x

n

]π

n

≡

e−1

X

n=0

[x

n

]π

n

mod pR,

we see that 1

, π, ··· , π

e−1

generate

R/pR

as

W

(

k

)-modules (all the Teichm¨uller

lifts live in W (k)). Therefore R = M + pR. We iterate to get

R = M + p(M + pR) = M + p

2

r = ··· = M + p

m

R

for all

m ≥

1. So

M

is dense in

R

. But

M

is also

p

-adically complete, hence

closed in R. So M = R.

The important statement to take away is

Corollary.

Let

K

be a mixed characteristic local field. Then

K

is a finite

extension of Q

p

.

Proof.

Let

F

q

be the residue field of

K

. Then

O

K

is finite over

W

(

F

q

) by the

previous theorem. So it suffices to show that

W

(

F

q

) is finite over

W

(

F

p

) =

Z

p

.

Again the inclusion

F

p

⊆ F

q

gives an injection

W

(

F

p

)

→ W

(

F

q

). Write

q

=

p

d

,

and let x

1

, ··· , x

d

∈ W (F

q

) be lifts of an F

p

-bases of F

q

.. Then we have

W (F

q

) =

d

M

i=1

x

d

Z

p

+ pW (F

q

),

and then argue as in the end of the previous theorem to get

W (F

q

) =

d

M

i=1

x

d

Z

p

.