# Countable ordered subfield of any Ordered Field

When we talk about any ordered field($$\mathbb{k}$$), we can always generate the rational elements of the field($$\mathbb{Q_{k}}$$) by first generating the natural elements($$\mathbb{N_{k}}$$) and the integer elements($$\mathbb{Z_{k}}$$) entirely from the multiplicative identity($$\mathbb{1_{k}}$$). We all know that $$\mathbb{Q_{k}}$$ is a countable subfield $$\mathbb{k}$$.

Now I want to ask this question that is it possible to find another countable subfield of $$\mathbb{k}$$ other than that of $$\mathbb{Q_{k}}$$? If that is possible to find, will it be order isomorphic to $$\mathbb{Q_{k}}$$?

Will the field $$\mathbb{k}$$ being Archimedean or non-Archimedean alter the results to the previous question?

P.S. I was trying to find this countable subfield in an ordered field both in particular examples such as that of $$\mathbb{R}$$(Archimedean) and Rational functions defined on an integral domain(non-Archimedean). In both the cases I was unable to find so. Although I strongly believe that it is not possible to find another countable field in an Archimedean field, but I am not sure for the non-Archimedean case.

• All number fields are countable, and many of them are contained in $\Bbb R$, hence have an archimedean order. The first non-trivial examples are the quadratic number fields $\Bbb Q (\sqrt d)$ for square-free positive integers $d$. -- As for non-archimedean orders: for any countable field $F$, the rational function field $F (x)$ is also countable. Dec 22, 2018 at 7:18
• @TorstenSchoeneberg, I want to ask another question. Like we can uniquely characterise $\mathbb(R)$ as a complete ordered field, is there a possible unique charactisation of $\mathbb(Q)$? Dec 22, 2018 at 8:07