# Examples of Valuations

The archetypal example of an archimedean valuation is the absolute value on the complex numbers. It is essentially the only one:

Theorem 13.3.1 (Gelfand-Tornheim)   Any field  with an archimedean valuation is isomorphic to a subfield of  , the valuation being equivalent to that induced by the usual absolute value on  .

We do not prove this here as we do not need it. For a proof, see [Art59, pg. 45, 67].

There are many non-archimedean valuations. On the rationals there is one for every prime , the -adic valuation, as in Example 13.2.9.

Theorem 13.3.2 (Ostrowski)   The nontrivial valuations on are those equivalent to , for some prime , and the usual absolute value .

Remark 13.3.3   Before giving the proof, we pause with a brief remark about Ostrowski. According to
   http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Ostrowski.html

Ostrowski was a Ukrainian mathematician who lived 1893-1986. Gautschi writes about Ostrowski as follows: ... you are able, on the one hand, to emphasise the abstract and axiomatic side of mathematics, as for example in your theory of general norms, or, on the other hand, to concentrate on the concrete and constructive aspects of mathematics, as in your study of numerical methods, and to do both with equal ease. You delight in finding short and succinct proofs, of which you have given many examples ...'' [italics mine]

We will now give an example of one of these short and succinct proofs.

Proof. Suppose is a nontrivial valuation on .

Nonarchimedean case: Suppose for all , so by Lemma 13.2.10, is nonarchimedean. Since is nontrivial, the set

is nonzero. Also is an ideal and if , then , so or , so is a prime ideal of  . Thus , for some prime number . Since every element of has valuation at most , if with , then , so . Let , so . Then for any and any with , we have

Thus on , hence on by multiplicativity, so is equivalent to , as claimed.

Archimedean case: By replacing by a power of , we may assume without loss that satisfies the triangle inequality. We first make some general remarks about any valuation that satisfies the triangle inequality. Suppose is greater than . Consider, for any the base- expansion of :

where

and . Since , taking logs we see that , so

Let . Then by the triangle inequality for , we have

where in the last step we use that . Setting , for , in the above inequality and taking th roots, we have

The first factor converges to  as , since (because ). The second factor is

which also converges to , for the same reason that (because as ). The third factor is

Putting this all together, we see that

Our assumption that is nonarchimedean implies that there is with and . Then for all with we have

 (13.5)

so , so as well (i.e., any with automatically satisfies ). Also, taking the power on both sides of (13.3.1) we see that

 (13.6)

Because, as mentioned above, , we can interchange the roll of and to obtain the reverse inequality of (13.3.2). We thus have

Letting and setting , we have

Thus for all integers with we have , which implies that is equivalent to .

Let be any field and let , where is transcendental. Fix a real number . If is an irreducible polynomial in the ring , we define a valuation by

 (13.7)

where and with and .

Remark 13.3.4   This definition differs from the one page 46 of [Cassels-Frohlich, Ch. 2] in two ways. First, we assume that instead of , since otherwise does not satisfy Axiom 3 of a valuation. Also, we write instead of , so that the product formula will hold. (For more about the product formula, see Section 18.1.)

In addition there is a a non-archimedean valuation defined by

 (13.8)

This definition differs from the one in [Cas67, pg. 46] in two ways. First, we assume that instead of , since otherwise does not satisfy Axiom 3 of a valuation. Here's why: Recall that Axiom 3 for a non-archimedean valuation on asserts that whenever and , then . Set , where is an irreducible polynomial. Then , since . However, , since . If we take instead of , as I propose, then , as required.

Note the (albeit imperfect) analogy between and . If , so , the valuation is of the type (13.3.3) belonging to the irreducible polynomial .

The reader is urged to prove the following lemma as a homework problem.

Lemma 13.3.5   The only nontrivial valuations on which are trivial on are equivalent to the valuation (13.3.3) or (13.3.4).

For example, if is a finite field, there are no nontrivial valuations on , so the only nontrivial valuations on are equivalent to (13.3.3) or (13.3.4).

William Stein 2012-09-24