# Extensions of Normalized Valuations

Let be a complete field with valuation . We consider the following three cases:
(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 .
(Alternatively, these cases can be subsumed by the hypothesis that the completion of is locally compact.)

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

on multiplies any choice of Haar measure by , and this characterizes the normalized valuations among equivalent ones.

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:

The constant in axiom (3) of a valuation for the ordinary absolute value on is , so the constant for the normalized valuation is :

Note that implies

since .

Lemma 17.2.1   Suppose  is a field that is complete with respect to a normalized valuation  and let  be a finite extension of  of degree . Then the normalized valuation on  which is equivalent to the unique extension of to  is given by the formula

 all (17.3)

Proof. Let be the normalized valuation on that extends . Our goal is to identify , and in particular to show that it is given by (17.2.1).

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

Thus all we have to do is prove that . In case 2 the only nontrivial situation is and , in which case , which is the normalized valuation on defined above.

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

is an isomorphism between the additive group and the direct sum , and this is a homeomorphism if the right hand side is given the product topology. In particular, the Haar measures on and on are the same up to a multiplicative constant in .

Let . Then the left-multiplication-by- map

on is the same as the map

on , so it multiplies the Haar measure by , since on  is assumed normalized (the measure of each factor is multiplied by , so the measure on the product is multiplied by ). Since is assumed normalized, so multiplication by  rescales by , we have

But , so . Since is nontrivial and for we have

so we must have in (17.2.1), as claimed.

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

Theorem 17.2.2   Suppose is a (nontrivial as always) normalized valuation of a field  and let  be a finite extension of . Then for any ,

where the are the normalized valuations equivalent to the extensions of  to .

Proof. Let denote the completion of with respect to . Write

Then Theorem 17.2.2 asserts that

 (17.4)

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