# Can the real numbers be embedded into all non-Archimedean real closed fields?

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?

• 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. – nombre May 13 '19 at 12:29

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.)