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!

  • $\begingroup$ 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. $\endgroup$ Dec 22, 2018 at 7:18
  • $\begingroup$ Yes, I got it. Thanks! $\endgroup$ Dec 22, 2018 at 7:20
  • $\begingroup$ @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 $\mathbb(Q)$? $\endgroup$ Dec 22, 2018 at 8:07
  • 1
    $\begingroup$ It's the minimal ordered field. $\endgroup$
    – Slade
    Dec 22, 2018 at 8:14


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