TITLE: Are $\mathbb{R}$ and $\mathbb{Q}$ the only subfields of $\mathbb{C}$ with natural structure as ordered fields?
QUESTION [8 upvotes]: 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 $\mathbb{Q}$?

REPLY [7 votes]: An ordered field satisfying "For every $x$, either $x$ or $-x$ has a square root" is uniquely orderable and this ordering is definable from the algebraic structure alone (set $a\le b$ iff $b-a$ has a square root). Since this property is expressible by a first-order sentence, we have a model-theoretic way to generate uniquely orderable subfields of $\mathbb{R}$, namely elementary extensions: if $A\preccurlyeq\mathbb{R}$ (indeed if $A$ is a $\forall\exists$-elementary substructure of $\mathbb{R}$) then $A$ is uniquely orderable. Combining this observation with the downward Lowenheim-Skolem theorem, we get results like the following:

There is a countable uniquely orderable subfield of $\mathbb{R}$ containing $\pi$.

Suppose the continuum hypothesis fails. Then there is a uniquely orderable subfield $A$ of $\mathbb{R}$ of cardinality strictly between $\aleph_0$ and $2^{\aleph_0}$.


Of course results like this can be gotten without model theory, but it's neat to observe that model theory provides a simple machine for generating them.

Perhaps a bit more interestingly, there are non-Archimedean uniquely orderable fields of every infinite cardinality (we can construct one by hand, or if we're feeling lazy we can use compactness). Since any field of characteristic $0$ and cardinality $\le 2^{\aleph_0}$ can be embedded into $\mathbb{C}$ (note that the algebraic closure will again have cardinality $\le 2^{\aleph_0}$, and the isomorphism type of an algebraically closed field is determined by its characteristic and transcendence dimension), this means that there are uniquely orderable non-Archimedean subfields of $\mathbb{C}$.