Is $\mathbb R$ terminal among Archimedean fields?

Question

I was wondering why metrics and norms are always defined to be real, rather than generalized to some other fields (or whatever). The best guess I have so far is:

Because every Archimedean ordered field is (up to unique isomorphism) a subfield of $\mathbb R$ anyway.

But is that actually true? And if it is, can it be strengthened to "every Achimedean ordered ring"? Or even semiring?

I know $\mathbb R$ is the only complete Archimedean field. But a priori, I suppose there could be non-complete examples that cannot be completed (without losing the Archimedean property).

Answer 1

I defer to Proposition 12 (well... the second Proposition 12...) and Theorems 14 and 15 in this answer of mine.

It is not hard to construct an argument that $\mathbb{R}$ is a final object in the category of Archimedean fields from these results. For example see the notes that Pete L. Clark links to on the same page.

Answer 2

I wrote this answer to a related question. I think one can show as a consequence of this stuff that the Dedekind completion of a non-Archimedean ordered field is actually not a field.