Can an orderable field always be ordered in a way that extends a given subfield's ordering?

Question

Let $F$ be an orderable field, and let $\:\langle E,\hspace{-0.03 in}\leq \rangle \:$ be an ordered subfield of $F$.

Does it follow that $F$ can be made into an ordered field in a way that extends the ordering on $E\hspace{.02 in}$?

Answer 1

No. Let $E = \mathbb{R}(t)$ with the ordering extending the unique one on $\mathbb{R}$ and in which $t > \alpha$ for all $\alpha \in \mathbb{R}$. Let $F = \mathbb{R}(\sqrt{-t})$. Clearly the ordering on $E$ does not extend to $F$. But in fact $F \cong E$ as a field, so certainly $F$ can be ordered.

There is quite a rich theory of when orderings on fields extend from subfields. A few results of this kind can be found on the chapter on ordered fields in these notes, but these don't even properly scratch the surface. I recommend the book Orderings, Valuations and Quadratic Forms by T. Y. Lam as being a good place to start. (Yes, there is more than a whole book's worth of material here.)