Here is something I've been wondering about recently.
Suppose you have an arbitrary ordered field $F$, and let $F(\sqrt{a})$ be a field extension with $a>0$ in $F$. Is there then some way to order $F(\sqrt{a})$ such that $F(\sqrt{a})$ is also an ordered field? Here $F(\sqrt{a})$ is the set of all $p+q\sqrt{a}$ for $p,q,a\in F$. I'm hoping to satisfy existence of a positive cone in $F(\sqrt{a})$.
I was interested because for field extensions like $\mathbb{Q}(\sqrt{2})$, we know that $\mathbb{Q}(\sqrt{2})$ is a subfield of $\mathbb{R}$, which itself is ordered, so we can take the positive cone of $\mathbb{Q}(\sqrt{2})$ to be those elements which are positive in $\mathbb{R}$. But if we don't know a larger ordered field containing the extension, what do we do instead? Thanks.
The answer is yes if and only if $a$ is positive in $F$. For the a non-example consider $\mathbb{C}/\mathbb{R}$. Here $\mathbb{C} = \mathbb{R}(\sqrt{-1})$ and $\mathbb{C}$ is not an ordered field