# What is the degree of a real closure of an ordered field?

Given a subfield $F$ of $\Bbb R$, let the real-closure of $F$ be the smallest subfield of $\Bbb R$ which is real-closed and extends $F$. Since the theory of real-closed fields is complete, this is in fact equivalent to the elementary submodel of $\Bbb R$ generated by $F$.

And in fact if $K$ is the real-closure of $F$, then $K$ is an algebraic extension of $F$, since we only need to add roots for polynomials. Recall that one of the characterizations of real-closed fields is that $[\bar K:K]=2$, and therefore $[K:F]$ is either $1$ or infinite, otherwise $\bar K=\bar F$ is a finite degree extension of $F$ which is not $2$.

So, as a set theorist I am not satisfied by "infinite", of course. So I want to know what sort of infinities can be obtained.

1. The real closure of a countable field is a countable field, so it has to have degree $\aleph_0$. And in general, if $K$ is the real-closure of $F$, then $|K|=|F|$.

2. If $B$ is a transcendence basis for $\Bbb R$ over $\Bbb Q$, and $B'\subseteq B$, then the real-closure of $\Bbb Q(B')$ has degree $|B'|\cdot\aleph_0$, since for each $b\in B'$ we need to add countably many roots, but the cardinality of the real-closure equals to that of the field. In particular, it follows that the degree can be any infinite cardinal between $\aleph_0$ and $2^{\aleph_0}$ (including these two cases, of course).

Question I. Can $\Bbb R$ be the real-closure of a [proper] subfield $F$, such that $[\Bbb R:F]=\aleph_0$, or generally less than $2^{\aleph_0}$?

More specifically, consider $B$ a transcendence basis for $\Bbb R$ over $\Bbb Q$, without loss of generality $\pi\in B$. Consider $F_0=\Bbb Q(B\setminus\{\pi\})$ and $K$ the real-closure of $F_0$. Let $F=K(\pi)$, then $\Bbb R$ is the real-closure of $F$, since $B\subseteq F$. What is $[\Bbb R:F]$?

Question II. We can define real-closure for arbitrary ordered fields, not just subfields of $\Bbb R$, and it is unique up to isomorphism (since adding $\sqrt{-1}$ generates the algebraic closure).

Given an ordered field $F$, and its real-closure $K$, how can we determine $[K:F]$ from the properties of $F$?