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.
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
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
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
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
.