Topology

A valuation on a field induces a topology in which a basis for the neighborhoods of are the open balls

for .

Lemma 14.1.1   Equivalent valuations induce the same topology.

Proof. If , then if and only if if and only if so . Thus the basis of open neighborhoods of for and are identical.

A valuation satisfying the triangle inequality gives a metric for the topology on defining the distance from to to be . Assume for the rest of this section that we only consider valuations that satisfy the triangle inequality.

Lemma 14.1.2   A field with the topology induced by a valuation is a topological field, i.e., the operations sum, product, and reciprocal are continuous.

Proof. For example (product) the triangle inequality implies that

is small when and are small (for fixed ).

Lemma 14.1.3   Suppose two valuations and on the same field induce the same topology. Then for any sequence in  we have

Proof. It suffices to prove that if then , since the proof of the other implication is the same. Let . The topologies induced by the two absolute values are the same, so can be covered by open balls . One of these open balls contains 0. There is such that

Since , there exists such that for we have . For such , we have , so , so . Thus .

Proposition 14.1.4   If two valuations and on the same field induce the same topology, then they are equivalent in the sense that there is a positive real such that .

Proof. If and , then if and only if , which is the case if and only if . Thus Lemma 14.1.3 implies that if and only if . On taking reciprocals we see that if and only if , so finally if and only if .

Let now be nonzero elements with and . On applying the foregoing to

we see that

if and only if

Dividing through by , and rearranging, we see that for every rational number ,

Thus

so

Since this equality does not depend on the choice of , we see that there is a constant ( ) such that for all . Thus , so , which implies that is equivalent to .

William Stein 2012-09-24