We define two important properties of valuations, both of which
apply to equivalence classes of valuations (i.e., the property
holds for
if and only if it holds for a valuation
equivalent to
).
Definition 13.2.1 (Discrete)
A valuation
is
discrete
if there is a
such that for any
Thus the absolute values are bounded away from
.
To say that
is discrete is the same as saying
that the set
forms a discrete subgroup of the reals under addition (because
the elements of the group are bounded away from 0).
Proof.
Since
is discrete there is a positive
such that for any positive
we have
.
Suppose
is an arbitrary positive element.
By subtracting off integer multiples of
, we
find that there is a unique
such that
Since
and
, it follows
that
, so
is a multiple of
.
By Proposition 13.2.2, the set
of
for nonzero
is free on one generator, so there
is a such that
, for ,
runs precisely through the set
(Note: we can replace by to see that we
can assume that ).
Definition 13.2.3 (Order)
If
, we call
the
order
of
.
Axiom (2) of valuations
translates into
Definition 13.2.4 (Nonarchimedean)
A valuation
is
nonarchimedean
if we can take
in Axiom (3), i.e., if

(13.3) 
If
is not nonarchimedean then
it is
archimedean.
Note that if we can take for
then we can take for any valuation equivalent to
.
To see that (13.2.1) is equivalent to Axiom (3) with
, suppose
. Then
, so
Axiom (3) asserts that
, which implies
that
, and conversely.
We note at once the following consequence:
Lemma 13.2.5
Suppose
is a nonarchimedean valuation.
If with , then
Proof.
Note that
, which
is true even if
. Also,
where for the last equality we have used that
(if
, then
,
a contradiction).
Definition 13.2.6 (Ring of Integers)
Suppose
is a nonarchimedean absolute
value on a field
. Then
is a ring called the
ring of integers of
with respect to
.
Lemma 13.2.7
Two nonarchimedean valuations
and
are equivalent if and only if they
give the same .
We will prove this modulo the claim (to
be proved later in Section 14.1) that
valuations are equivalent if (and only if) they induce the
same topology.
Proof.
Suppose suppose
is equivalent to
, so
,
for some
. Then
if and only if
, i.e., if
.
Thus
.
Conversely, suppose
.
Then
if and only if
and
, so

(13.4) 
The topology induced by
has as basis
of open neighborhoods the set of open balls
for
, and likewise for
. Since
the absolute values
get arbitrarily close
to
0, the set
of open balls
also
forms a basis of the topology induced
by
(and similarly for
).
By (
13.2.2) we have
so the two topologies both have
as
a basis, hence are equal. That equal topologies
imply equivalence of the corresponding valuations
will be proved in Section
14.1.
The set of with forms an ideal
in . The
ideal
is maximal, since if and
then
, so
, hence , so is a unit.
Lemma 13.2.8
A nonarchimedean valuation
is
discrete if and only if
is a principal ideal.
Proof.
First suppose that
is discrete.
Choose
with
maximal, which
we can do since
so the discrete set
is bounded above.
Suppose
. Then
so
.
Thus
Conversely, suppose
is principal. For any
we have with . Thus
Thus
is bounded away from
,
which is exactly the definition of discrete.
Example 13.2.9
For any prime
, define the
adic valuation
as follows. Write a nonzero
as
, where
. Then
This valuation is both discrete and nonarchimedean.
The ring
is the local ring
which has maximal ideal generated by
. Note that
We will using the following lemma later (e.g., in
the proof of Corollary 14.2.4 and Theorem 13.3.2).
Lemma 13.2.10
A valuation
is nonarchimedean if and only if for all in the ring generated by in .
Note that we cannot identify the ring generated by with
in general, because might have characteristic .
Proof.
If
is nonarchimedean, then
,
so by Axiom (3) with
, we have
. By
induction it follows that
.
Conversely, suppose for all integer multiples of .
This condition is also true if we replace
by
any equivalent valuation, so replace
by
one with , so that the triangle inequality holds.
Suppose with . Then
by the triangle inequality,
Now take
th roots of both sides to get
and take the limit as
to see
that
. This proves that one
can take
in Axiom (3), hence that
is nonarchimedean.
William Stein
20120924