- (1)
- is discrete non-archimedean and the residue class field is finite.
- (2i)
- The completion of with respect to is .
- (2ii)
- The completion of with respect to is .

In case (1) we defined the normalized valuation to
be the one such that if Haar measure of the ring of integers is ,
then
(see Definition 15.1.11).
In case (2i) we say that
is normalized if it is the ordinary
absolute value, and in (2ii) if it is the *square* of the ordinary
absolute value:

(normalized)

In every case, for every , the map

We have already verified the above characterization for non-archimedean valuations, and it is clear for the ordinary absolute value on , so it remains to verify it for . The additive group is topologically isomorphic to , so a choice of Haar measure of is the usual area measure on the Euclidean plane. Multiplication by is the same as rotation followed by scaling by a factor of , so if we rescale a region by a factor of , the area of the region changes by a factor of the square of . This explains why the normalized valuation on is the square of the usual absolute value. Note that the normalized valuation on does not satisfy the triangle inequality:

By the preceding section there is a positive real number such that for all we have

One can argue in a unified way in all cases as follows. Let be a basis for . Then the map

Let . Then the left-multiplication-by- map

In the case when need not be complete with respect to the valuation on , we have the following theorem.

By Theorem 17.1.8, the are exactly the normalizations of the extensions of to the (i.e., the are in bijection with the extensions of valuations, so there are no other valuations missed). By Lemma 17.1.1, the normalized valuation on is . The theorem now follows by taking absolute values of both sides of (17.2.2).

What next?! We'll building up to giving a new proof of finiteness of the class group that uses that the class group naturally has the discrete topology and is the continuous image of a compact group.

William Stein 2012-09-24