1Basic theory
III Local Fields
1.1 Fields
Definition
(Absolute value)
.
Let
K
be a field. An absolute value on
K
is a
function | · | : K → R
≥0
such that
(i) |x| = 0 iff x = 0;
(ii) |xy| = |x||y| for all x, y ∈ K;
(iii) |x + y| ≤ |x| + |y|.
Definition (Valued field). A valued field is a field with an absolute value.
Example.
The rationals, reals and complex numbers with the usual absolute
values are absolute values.
Example
(Trivial absolute value)
.
The trivial absolute value on a field
K
is the
absolute value given by
|x| =
(
1 x 6= 0
0 x = 0
.
The only reason we mention the trivial absolute value here is that from
now on, we will assume that the absolute values are not trivial, because trivial
absolute values are boring and break things.
There are some familiar basic properties of the absolute value such as
Proposition. ||x| − |y|| ≤ |x − y|
. Here the outer absolute value on the left
hand side is the usual absolute value of
R
, while the others are the absolute
values of the relevant field.
An absolute value defines a metric d(x, y) = |x − y| on K.
Definition
(Equivalence of absolute values)
.
Let
K
be a field, and let
| · |, | · |
0
be absolute values. We say they are equivalent if they induce the same topology.
Proposition.
Let
K
be a field, and
| · |, | · |
0
be absolute values on
K
. Then
the following are equivalent.
(i) | · | and | · |
0
are equivalent
(ii) |x| < 1 implies |x|
0
< 1 for all x ∈ K
(iii) There is some s ∈ R
>0
such that |x|
s
= |x|
0
for all x ∈ K.
Proof.
(i)
⇒
(ii) and (iii)
⇒
(i) are easy exercises. Assume (ii), and we shall
prove (iii). First observe that since
|x
−1
|
=
|x|
−1
, we know
|x| >
1 implies
|x|
0
>
1, and hence
|x|
= 1 implies
|x|
0
= 1. To show (iii), we have to show that
the ratio
log |x|
log |x
0
|
is independent of x.
Suppose not. We may assume
log |x|
log |x|
0
<
log |y|
log |y|
0
,
and moreover the logarithms are positive. Then there are
m, n ∈ Z
>0
such that
log |x|
log |y|
<
m
n
<
log |x|
0
log |y|
0
.
Then rearranging implies
x
n
y
m
< 1 <
x
n
y
m
0
,
a contradiction.
Exercise.
Let
K
be a valued field. Then equivalent absolute values induce the
same the completion
ˆ
K
of
K
, and
ˆ
K
is a valued field with an absolute value
extending | · |.
In this course, we are not going to be interested in the usual absolute values.
Instead, we are going to consider some really weird ones, namely non-archimedean
ones.
Definition
(Non-archimedean absolute value)
.
An absolute value
| · |
on a field
K
is called non-archimedean if
|x
+
y| ≤ max
(
|x|, |y|
). This condition is called
the strong triangle inequality.
An absolute value which isn’t non-archimedean is called archimedean.
Metrics satisfying
d
(
x, z
)
≤ max
(
d
(
x, y
)
, d
(
y, z
)) are often known as ultra-
metrics.
Example. Q, R and C under the usual absolute values are archimedean.
In this course, we will only consider non-archimedean absolute values. Thus,
from now on, unless otherwise mentioned, an absolute value is assumed to be
non-archimedean. The metric is weird!
We start by proving some absurd properties of non-archimedean absolute
values.
Recall that the closed balls are defined by
B(x, r) = {y : |x − y| ≤ r}.
Proposition.
Let (
K, | · |
) be a non-archimedean valued field, and let
x ∈ K
and r ∈ R
>0
. Let z ∈ B(x, r). Then
B(x, r) = B(z, r).
So closed balls do not have unique “centers”. Every point can be viewed as
the center.
Proof. Let y ∈ B(z, r). Then
|x − y| = |(x − z) + (z − y)| ≤ max(|x − z|, |z − y|) ≤ r.
So y ∈ B(x, r). By symmetry, y ∈ B(x, r) implies y ∈ B(z, r).
Corollary. Closed balls are open.
Proof. To show that B(x, r) is open, we let z ∈ B(x, r). Then we have
{y : |y − z| < r} ⊆ B(z, r) = B(x, r).
So we know the open ball of radius
r
around
z
is contained in
B
(
x, r
). So
B
(
x, r
)
is open.
Norms in non-archimedean valued fields are easy to compute:
Proposition.
Let
K
be a non-archimedean valued field, and
x, y ∈ K
. If
|x| > |y|, then |x + y| = |x|.
More generally, if
x
=
P
∞
c=0
x
i
and the non-zero
|x
i
|
are distinct, then
|x| = max |x
i
|.
Proof.
On the one hand, we have
|x
+
y| ≤ max{|x|, |y|}
. On the other hand,
we have
|x| = |(x + y) − y| ≤ max(|x + y|, |y|) = |x + y|,
since we know that we cannot have
|x| ≤ |y|
. So we must have
|x|
=
|x
+
y|
.
Convergence is also easy for valued fields.
Proposition. Let K be a valued field.
(i) Let (x
n
) be a sequence in K. If x
n
− x
n+1
→ 0, then x
n
is Cauchy.
If we assume further that K is complete, then
(ii)
Let (
x
n
) be a sequence in
K
. If
x
n
− x
n+1
→
0, then a sequence (
x
n
) in
K converges.
(iii) Let
P
∞
n=0
y
n
be a series in K. If y
n
→ 0, then
P
∞
n=0
y
n
converges.
The converses to all these are of course also true, with the usual proofs.
Proof.
(i)
Pick
ε >
0 and
N
such that
|x
n
− x
n+1
| < ε
for all
n ≥ N
. Then given
m ≥ n ≥ N, we have
|x
m
− x
n
| = |x
m
− x
m−1
+ x
m−1
− x
m−2
+ ··· − x
n
|
≤ max(|x
m
− x
m−1
|, ··· , |x
n+1
− x
n
|)
< ε.
So the sequence is Cauchy.
(ii) Follows from (1) and the definition of completeness.
(iii) Follows from the definition of convergence of a series and (2).
The reason why we care about these weird non-archimedean fields is that
they have very rich algebraic structure. In particular, there is this notion of the
valuation ring.
Definition
(Valuation ring)
.
Let
K
be a valued field. Then the valuation ring
of K is the open subring
O
K
= {x : |x| ≤ 1}.
We prove that it is actually a ring
Proposition. Let K be a valued field. Then
O
K
= {x : |x| ≤ 1}
is an open subring of
K
. Moreover, for each
r ∈
(0
,
1], the subsets
{x
:
|x| < r}
and {x : |x| ≤ r} are open ideals of O
K
. Moreover, O
×
K
= {x : |x| = 1}.
Note that this is very false for usual absolute values. For example, if we take
R with the usual absolute value, we have 1 ∈ O
R
, but 1 + 1 6∈ O
R
.
Proof. We know that these sets are open since all balls are open.
To see
O
K
is a subring, we have
|
1
|
=
|−
1
|
= 1. So 1
, −
1
∈ O
K
. If
x, y ∈ O
K
,
then
|x
+
y| ≤ max
(
|x|, |y|
)
≤
1. So
x
+
y ∈ O
K
. Also,
|xy|
=
|x||y| ≤
1
·
1 = 1.
So xy ∈ O
K
.
That the other sets are ideals of O
K
is checked in the same way.
To check the units, we have
x ∈ O
×
K
⇔ |x|, |x
−1
| ≤
1
⇔ |x|
=
|x|
−1
= 1.