# Valuations

Definition 15.1.1 (Valuation)   A on a field is a function defined on with values in satisfying the following axioms:
(1)
if and only if ,
(2)
, and
(3)
there is a constant such that whenever .

The is the valuation for which for all . We will often tacitly exclude the trivial valuation from consideration.

From (2) we have

so by (1). If and , then by (2). In particular, the only valuation of a finite field is the trivial one. The same argument shows that , so

all

Definition 15.1.2 (Equivalent)   Two valuations and on the same field are if there exists such that

for all .

Note that if is a valuation, then is also a valuation. Also, equivalence of valuations is an equivalence relation.

If is a valuation and is the constant from Axiom (3), then there is a such that (i.e., ). Then we can take as constant for the equivalent valuation . Thus every valuation is equivalent to a valuation with . Note that if , e.g., if is the trivial valuation, then we could simply take in Axiom (3).

Proposition 15.1.3   Suppose is a valuation with . Then for all we have

 (triangle inequality) (15.1)

Proof. Suppose with . Then satisfies . By Axiom (3) we have , so multiplying by we see that

Also we have

and inductively we have for any that

If is any positive integer, let be such that . Thenn

since . In particular,

 (for ) (15.2)

Applying (15.1.2) to and using the binomial expansion, we have for any that

Now take th roots of both sides to obtain

We have by elementary calculus that

so . (The elementary calculus'': We instead prove that , since the argument is the same and the notation is simpler. First, for any we have , since upon taking th powers this is equivalent to , which is true by hypothesis. Second, suppose there is an such that for all . Then taking logs of boths sides we see that . But , so there is no such . Thus as .)

Note that Axioms (1), (2) and Equation (15.1.1) imply Axiom (3) with . We take Axiom (3) instead of Equation (15.1.1) for the technical reason that we will want to call the square of the absolute value of the complex numbers a valuation.

Lemma 15.1.4   Suppose , and is a valuation on with . Then

(Here the big absolute value on the outside of the left-hand side of the inequality is the usual absolute value on real numbers, but the other absolute values are a valuation on an arbitrary field .)

Proof. We have

so . The same argument with and swapped implies that , which proves the lemma.

William Stein 2004-05-06