Every Archimedean real closed field is isomorphic to a subfield of $\mathbb{R}$. But I’m wondering if something in the opposite direction is true.

Suppose that $F$ is a non-Archimedean real closed field. Then my question is, does $F$ necessarily have a subfield isomorphic to $\mathbb{R}$? If not, does anyone know of a counterexample?

  • $\begingroup$ On the other hand, the field of real algebraic numbers, which is the smallest real-closed field, does embed uniquely into every real-closed field. $\endgroup$ – nombre May 13 '19 at 12:29

The answer is no.

For an easy counterexample, we just use Lowenheim-Skolem: take your favorite non-Archimedean real closed field $F$, let $c\in F$ be an infinite element, and let $K$ be a countable elementary substructure of $F$ with $c\in K$. Then $K$ is a countable non-Archimedean real closed field.

Another way to get these is by the omitting types theorem - we can e.g. whip up a real closed field of arbitrarily large cardinality which omits the type of $\pi$. In fact, we can do better: there are non-Archimedean real closed fields of arbitrarily large cardinality which don't contain any finite element whose "standard part" is non-algebraic! This really kills any hope for a positive answer to a question along the lines of the OP.

(One key point here is that there is at most "one way up to standardness" to embed $\mathbb{R}$ in a field of characteristic $0$, since the rationals are dense in $\mathbb{R}$ and individually definable; this means that we don't have to work too hard to rule out embeddings.)


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