# Completeness

We recall the definition of metric on a set .

Definition 16.2.1 (Metric)   A on a set  is a map

such that for all ,
1. and if and only if ,
2. , and
3. .

A is a sequence in such that for all there exists  such that for all we have . The of  is the set of Cauchy sequences in  modulo the equivalence relation in which two Cauchy sequences and are equivalent if . A metric space is if every Cauchy sequence converges, and one can show that the completion of  with respect to a metric is complete.

For example, (usual archimedean absolute value) defines a metric on  . The completion of with respect to this metric is the field of real numbers. More generally, whenever is a valuation on a field that satisfies the triangle inequality, then defines a metric on . Consider for the rest of this section only valuations that satisfy the triangle inequality.

Definition 16.2.2 (Complete)   A field is with respect to a valuation if given any Cauchy sequence , ( ), i.e., one for which

there is an such that

w.r.t.

(i.e., ).

Theorem 16.2.3   Every field with valuation can be embedded in a complete field with a valuation extending the original one in such a way that is the closure of with respect to . Further is unique up to a unique isomorphism fixing .

Proof. Define to be the completion of with respect to the metric defined by . Thus is the set of equivalence classes of Cauchy sequences, and there is a natural injective map from to sending an element to the constant Cauchy sequence . Because the field operations on are continuous, they induce well-defined field operations on equivalence classes of Cauchy sequences componentwise. Also, define a valuation on by

and note that this is well defined and extends the valuation on .

To see that is unique up to a unique isomorphism fixing , we observe that there are no nontrivial continuous automorphisms that fix . This is because, by denseness, a continuous automorphism is determined by what it does to , and by assumption  is the identity map on . More precisely, suppose and  is a positive integer. Then by continuity there is (with ) such that if and then . Since is dense in , we can choose the above to be an element of . Then by hypothesis , so . Thus .

Corollary 16.2.4   The valuation is non-archimedean on if and only if it is so on . If is non-archimedean, then the set of values taken by on and are the same.

Proof. The first part follows from Lemma 15.2.10 which asserts that a valuation is non-archimedean if and only if for all integers . Since the valuation on extends the valuation on , and all are in , the first statement follows.

For the second, suppose that is non-archimedean (but not necessarily discrete). Suppose with . First I claim that there is such that . To see this, let , where  is some element of with , note that , and choose such that , so

Since is non-archimedean, we have

where in the last equality we use that . Also,

so , which is in the set of values of on .

Subsections
William Stein 2004-05-06