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.
Thanks in advance!