- for all nonzero (positivity).
- for all (triangle inequality).
- for all and (homogeneity).

and

where .

To finish the proof, we show that there is a such that for all ,

By what we have already shown, the function is continuous in the -topology, so by local compactness it attains its lower bound on the unit circle . (Why is the unit circle compact? With respect to , the topology on is the same as that of a product of copies of . If the valuation is archimedean then or with the standard topology and the unit circle is compact. If the valuation is non-archimedean, then we saw (see Remark 15.1.7) that if is locally compact, then the valuation is discrete, in which case we showed that the unit disc is compact, hence the unit circle is also compact since it is closed.) Note that by part 1 of Definition 16.1.1. Also, by definition of , for any there exists such that (just take the max coefficient in our basis). Thus we can write any as where and with . We then have

William Stein 2012-09-24