We know that $\mathbb{R}$ and $\mathbb{Q}$ have a unique structure as ordered fields with the usual order, and that $\mathbb{C}$ cannot be realised as an ordered field. Various non-trivial subfields of $\mathbb{R}$, such as $\mathbb{Q}(\sqrt 2)$, have multiple orderings, but I can't yet completely rule out the possibility of there being other subfields of $\mathbb{C}$ with unique order structure.
Clearly any such subfield cannot contain $bi$ for any $b\in\mathbb{R}^\times$. How can I determine whether there are any other possibilities than $\mathbb{R}$ and $\mathbb{Q}$?
Best Answer
What about the field $\mathbb{Q} \left ( \sqrt{2}, \sqrt{\sqrt{2}}, \sqrt{\sqrt{\sqrt{2}}}, \ldots \right )$?
Now every element we've added is a square of something (it's the square of the next number we add), so each element we add must be positive. This eliminates the choice we have in ordering, and so the ordering should be unique.
This is an adaptation of the usual proof that $\mathbb{R}$ has a unique ordering: There we use the fact that every positive number has a square root, while here we just enforce this property for a generating set.
As an aside, we need to do something aggressive like this (adding infinitely many elements), since any $\mathbb{Q}(\alpha)$ admits an ordering for each real root of the minimal polynomial of $\alpha$.
Edit:
I thought harder about my aside: That means $\mathbb{Q} \left ( \sqrt[3]{2} \right )$ will work too, since there's only one way to embed it into $\mathbb{R}$.
I hope this helps ^_^