# Types of Valuations

We define two important properties of valuations, both of which apply to equivalence classes of valuations (i.e., the property holds for if and only if it holds for a valuation equivalent to ).

Definition 13.2.1 (Discrete)   A valuation is discrete if there is a such that for any

Thus the absolute values are bounded away from .

To say that is discrete is the same as saying that the set

forms a discrete subgroup of the reals under addition (because the elements of the group are bounded away from 0).

Proposition 13.2.2   A nonzero discrete subgroup of is free on one generator.

Proof. Since is discrete there is a positive such that for any positive we have . Suppose is an arbitrary positive element. By subtracting off integer multiples of , we find that there is a unique such that

Since and , it follows that , so is a multiple of .

By Proposition 13.2.2, the set of for nonzero is free on one generator, so there is a such that , for , runs precisely through the set

(Note: we can replace by to see that we can assume that ).

Definition 13.2.3 (Order)   If , we call the order of .

Axiom (2) of valuations translates into

Definition 13.2.4 (Non-archimedean)   A valuation is non-archimedean if we can take in Axiom (3), i.e., if

 (13.3)

If is not non-archimedean then it is archimedean.

Note that if we can take for then we can take for any valuation equivalent to . To see that (13.2.1) is equivalent to Axiom (3) with , suppose . Then , so Axiom (3) asserts that , which implies that , and conversely.

We note at once the following consequence:

Lemma 13.2.5   Suppose is a non-archimedean valuation. If with , then

Proof. Note that , which is true even if . Also,

where for the last equality we have used that (if , then , a contradiction).

Definition 13.2.6 (Ring of Integers)   Suppose is a non-archimedean absolute value on a field . Then

is a ring called the ring of integers of with respect to .

Lemma 13.2.7   Two non-archimedean valuations and are equivalent if and only if they give the same .

We will prove this modulo the claim (to be proved later in Section 14.1) that valuations are equivalent if (and only if) they induce the same topology.

Proof. Suppose suppose is equivalent to , so , for some . Then if and only if , i.e., if . Thus .

Conversely, suppose . Then if and only if and , so

 (13.4)

The topology induced by    has as basis of open neighborhoods the set of open balls

for , and likewise for   . Since the absolute values get arbitrarily close to 0, the set of open balls also forms a basis of the topology induced by    (and similarly for   ). By (13.2.2) we have

so the two topologies both have as a basis, hence are equal. That equal topologies imply equivalence of the corresponding valuations will be proved in Section 14.1.

The set of with forms an ideal in . The ideal is maximal, since if and then , so , hence , so is a unit.

Lemma 13.2.8   A non-archimedean valuation is discrete if and only if is a principal ideal.

Proof. First suppose that is discrete. Choose with maximal, which we can do since

so the discrete set  is bounded above. Suppose . Then

so . Thus

Conversely, suppose is principal. For any we have with . Thus

Thus is bounded away from , which is exactly the definition of discrete.

Example 13.2.9   For any prime , define the -adic valuation as follows. Write a nonzero as , where . Then

This valuation is both discrete and non-archimedean. The ring is the local ring

which has maximal ideal generated by . Note that

We will using the following lemma later (e.g., in the proof of Corollary 14.2.4 and Theorem 13.3.2).

Lemma 13.2.10   A valuation is non-archimedean if and only if for all in the ring generated by in .

Note that we cannot identify the ring generated by with  in general, because  might have characteristic .

Proof. If is non-archimedean, then , so by Axiom (3) with , we have . By induction it follows that .

Conversely, suppose for all integer multiples  of . This condition is also true if we replace by any equivalent valuation, so replace by one with , so that the triangle inequality holds. Suppose with . Then by the triangle inequality,

Now take th roots of both sides to get

and take the limit as to see that . This proves that one can take in Axiom (3), hence that is non-archimedean.

William Stein 2012-09-24