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

http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Ostrowski.htmlOstrowski 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.

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

*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 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

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

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

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

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

where and with and .

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.

William Stein 2012-09-24