3Discretely valued fields

III Local Fields 3 Discretely valued fields
We are now going to further specialize. While a valued field already has some
nice properties, we can’t really say much more about them without knowing
Recall our previous two examples of valued fields:
Q
p
and
F
p
((
T
)). The
valuations had the special property that they take values in
Z
. Such fields are
known as discretely valued fields.
Definition (Discretely valued field). Let K be a valued field with valuation v.
We say
K
is a discretely valued field (DVF) if
v
(
k
×
)
R
is a discrete subgroup
of R, i.e. v(k
×
) is infinite cyclic.
Note that we do not require the image to be exactly
Z R
. So we allow
scaled versions of the valuation. This is useful because the property of mapping
into
Z
is not preserved under field extensions in general, as we will later see. We
will call those that do land in Z normalized valuations.
Definition
(Normalized valuation)
.
Let
K
be a DVF. The normalized valuation
V
K
on
K
is the unique valuation on
K
in the given equivalence class of valuations
whose image is Z.
Note that the normalized valuation does not give us a preferred choice of
absolute value, since to obtain an absolute value, we still have to arbitrarily pick
the base c > 1 to define |x| = c
v(x)
.
Definition
(Uniformizer)
.
Let
K
be a discrete valued field. We say
π K
is
uniformizer if
v
(
π
)
>
0 and
v
(
π
) generates
v
(
k
×
) (iff
v
(
π
) has minimal positive
valuation).
So with a normalized valuation, we have v
K
(π) = 1.
Example.
The usual valuation on
Q
p
is normalized, and so is the usual valuation
on k((T )). p is a uniformizer for Q
p
and T is a uniformizer for k((T )).
The kinds of fields we will be interested are local fields. The definition we
have here might seem rather ad hoc. This is just one of the many equivalent
characterizations of a local field, and the one we pick here is the easiest to state.
Definition
(Local field)
.
A local field is a complete discretely valued field with
a finite residue field.
Example. Q
and
Q
p
with
v
p
are both discretely valued fields, and
Q
p
is a local
field. p is a uniformizer.
Example. The Laurent series field k((T )) with valuation
v
X
a
n
T
n
= inf{n : a
n
6= 0}
is a discrete valued field, and is a local field if and only if
k
is finite field, as the
residue field is exactly k. We have
O
k((T ))
= k[[T ]] =
(
X
n=0
a
n
T
n
: a
n
k
)
.
Here T is a uniformizer.
These discretely valued field are pretty much like the p-adic numbers.
Proposition.
Let
K
be a discretely valued field with uniformizer
π
. Let
S O
K
be a set of coset representatives of O
k
/m
k
= k
K
containing 0. Then
(i) The non-zero ideals of O
K
are π
n
O
K
for n 0.
(ii)
The ring
O
K
is a PID with unique prime
π
(up to units), and
m
K
=
πO
K
.
(iii)
The topology on
O
K
induced by the absolute value is the
π
(iv) If K is complete, then O
K
(v) If K is complete, then any x K can be written uniquely as
x =
X
n−∞
a
n
π
n
,
where a
n
S, and
|x| = |π|
inf{n:a
n
6=0}
.
(vi)
The completion
ˆ
K
is also discretely valued and
π
is a uniformizer, and
moreover the natural map
O
k
π
n
O
k
O
ˆ
K
π
n
O
ˆ
K
is an isomorphism.
Proof. The same as for Q
p
and Z
p
, with π instead of p.
Proposition.
Let
K
be a discretely valued field. Then
K
is a local field iff
O
K
is compact.
Proof.
If
O
K
is compact, then
π
n
O
K
is compact for all
n
0 (where
π
is the
uniformizer), and in particular complete. So
K =
[
n0
π
n
O
K
is complete, as this is an increasing union, and Cauchy sequences are bounded.
Also, we know the quotient map
O
K
k
K
is continuous when
k
K
is given the
discrete topology, by definition of the
π
k
K
is compact and
discrete, hence finite.
In the other direction, if
K
is local, then we know
O
K
n
O
K
is finite for
all
n
0 (by induction and finiteness of
k
K
). We let (
x
i
) be a sequence in
O
K
.
Then by finiteness of
O
K
O
K
, there is a subsequence (
x
1,i
) which is constant
modulo
π
. We keep going, choosing a subsequence (
x
n+1,i
) of (
x
n
i
) such that
(
x
n+1,i
) is constant modulo
π
n+1
. Then (
x
i,i
)
i=1
converges, since it is Cauchy as
|x
ii
x
jj
| |π|
j
for j i. So O
K
is sequentially compact, hence compact.
Now the valuation ring
O
K
inherits a valuation from
K
, and this gives it a
structure of a discrete valuation ring. We will define a discrete valuation ring in
a funny way, but there are many equivalent definitions that we will not list.
Definition
(Discrete valuation ring)
.
A ring
R
is called a discrete valuation
ring (DVR) if it is a PID with a unique prime element up to units.
Proposition. R is a DVR iff R
=
O
K
for some DVF K.
Proof.
We have already seen that valuation rings of discrete valuation fields are
DVRs. In the other direction, let
R
be a DVR, and
π
a prime. Let
x R \{
0
}
.
Then we can find a unique unit
u R
×
and
n Z
0
such that
x
=
π
n
u
(say,
by unique factorization of PIDs). We define
v(x) =
(
n x 6= 0
x = 0
This is then a discrete valuation of
R
. This extends uniquely to the field of
fractions K. It remains to show that R = O
K
. First note that
K = R
1
π
.
This is since any non-zero element in
R
1
π
looks like
π
n
u, u R
×
, n Z
, and
is already invertible. So it must be the field of fractions. Then we have
v(π
n
u) = n Z
0
π
n
u R.
So we have R = O
K
.
Now recall our two “standard” examples of valued fields
F
p
((
T
)) and
Q
p
. Both of their residue fields are
F
p
, and in particular has characteristic
p
.
However,
F
p
((
T
)) itself is also of characteristic
p
, while
Q
p
has characteristic 0.
It would thus be helpful to split these into two different cases:
Definition
(Equal and mixed characteristic)
.
Let
K
be a valued field with
residue field k
K
. Then K has equal characteristic if
char K = char k
K
.
Otherwise, we have K has mixed characteristic.
If
K
has mixed characteristic, then necessarily
char K
= 0, and
char k
K
>
0.
Example. Q
p
has mixed characteristic, since
char Q
p
= 0 but
char k
Q
p
=
Z/pZ = p.
We will also need the following definition:
Definition
(Perfect ring)
.
Let
R
be a ring of characteristic
p
. We say
R
is
perfect if the Frobenius map
x 7→ x
p
is an automorphism of
R
, i.e. every element
of R has a pth root.
Fact.
Let
F
be a field of characteristic
p
. Then
F
is perfect if and only if every
finite extension of F is separable.
Example. F
q
is perfect for every q = p
n
.