For any field $K$ of characteristic zero, its prime subfield $Q(K)$ has a natural ordering inherited from $\mathbb{Q}$. I am interested in finding out to what extent (if at all) this natural ordering can be extended to larger subfields of $K$, in a natural way. (Of course, many constraints on $K$ may be required for this to be natural).
My motivating example is $\mathbb{C}$, and I want to generalise the role that $\mathbb{R}$ plays as the 'natural' choice of ordered subfield of $\mathbb{C}$. One can check that $\mathbb{R}$ has a unique structure as an ordered field and that $\mathbb{C}$ cannot be realised as an ordered field. Since there are no fields strictly between $\mathbb{R}$ and $\mathbb{C}$, we have that $\mathbb{R}$ is maximal as an ordered subfield of $\mathbb{C}$. Note that the ordering of proper subfields of $\mathbb{R}$ is not necessarily unique.
There are certainly maximal ordered subfields for any $K$ (Zorn's Lemma), but no guarantee of uniqueness. Any ideas?
Best Answer
There are automorphisms of $\mathbb{C}$ which do not send reals to reals; consequently, there are subfields of $\mathbb{C}$ which are isomorphic to $\mathbb{R}$ (hence are maximal orderable subfields of $\mathbb{C}$) but are not equal to $\mathbb{R}$.
In fact, I believe that there are in fact maximal uniquely orderable subfields of $\mathbb{C}$ which are non-Archimedean according to their unique order, hence a fortiori not isomorphic to $\mathbb{R}$ (let alone equal to $\mathbb{R}$); however, I don't have a complete argument for this at the moment. Below is my incomplete argument - the gap is at the end:
First, we isolate a strengthening of unique orderability: say that a field $F$ is nice iff $F$ has characteristic $0$ and for each nonzero $x\in F$ exactly one of $x$ and $-x$ has a square root in $F$. Every nice field is uniquely orderable, with unique ordering given by $$a\le b\leftrightarrow \exists x(x^2+a=b).$$
Now it's not hard to show that there is a countable nice field which is non-Archimedean according to its unique ordering. For example, we can prove this via compactness + downward Lowenheim-Skolem after observing that niceness is given by a conjunction of first-order sentences. Fix such a field $F$.
The algebraic closure $\overline{F}$ of $F$ is again countable. This means that $\overline{F}$, and hence $F$ itself, embeds into $\mathbb{C}$: for algebraically closed fields $A,B$ of the same characteristic, there is an embedding of $A$ into $B$ iff the transcendence dimension of $A$ is $\le$ that of $B$. (To prove this, just pick transcendence bases of $A$ and $B$.) So WLOG we can assume that $F$ is literally a subfield of $\mathbb{C}$.
A natural idea at this point is to apply Zorn's lemma to the poset of uniquely orderable subfields of $\mathbb{C}$ which contain $F$. However, it's not obvious to me that this poset satisfies the hypothesis of Zorn's lemma: why should the union of a chain of uniquely orderable fields be uniquely orderable?
Instead, we turn to niceness again. Let $\mathbb{P}$ be the poset of nice subfields of $\mathbb{C}$ which contain $F$ as a subfield. By Zorn, $\mathbb{P}$ has a maximal element $G$. However, we're not quite done: while $G$ is a maximal nice subfield of $\mathbb{C}$, it's not immediately obvious that $G$ is a maximal uniquely orderable subfield of $\mathbb{C}$. Keep in mind that there are uniquely orderable subfields of $\mathbb{C}$ which are not nice (e.g. $\mathbb{Q}$)!
It's at this point that I get stuck: basically, I need to be able to add square roots to an orderable subfield of $\mathbb{C}$ in a way which results in a still-orderable subfield of $\mathbb{C}$, and I don't immediately see how to do this. However, I also don't see an obstacle to this, so I tentatively suspect that this gap can be closed.