Are $\mathbb{R}$ and $\mathbb{Q}$ the only subfields of $\mathbb{C}$ with natural structure as ordered fields?

Question

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 $\mathbb{Q}$?

Answer 1

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 $\mathbb{R}$.

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I hope this helps ^_^

Answer 2

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.

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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 $\mathbb{C}$.