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
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
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
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
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 =
e1
M
i=0
π
i
W (k).
Looking at valuations, one sees that 1
, π, π, ··· , π
e1
are linearly independent
over W (k). So we can form
M =
e1
M
i=0
π
i
W (k) R.
We consider R/pR. Looking at Teichm¨uller expansions
X
n=0
[x
n
]π
n
e1
X
n=0
[x
n
]π
n
mod pR,
we see that 1
, π, ··· , π
e1
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
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
.