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It seems that a few similar questions where "answered" (or closed as duplicates) saying $\mathbb R$ is the only such subfield, pointing to the Artin-Schreier theorem which states that for every field $F$, such that $0<[\bar{F}:F]<\infty$, we have that $F$ is real-closed, $[\bar{F}:F]=2$, and $\bar{F}=F[i]$ (where $i^2=-1)$. But, as far as I can see this still leaves open (for me) the following questions:

1) Are there any order two subfields of $\mathbb{C}$ other than $\mathbb{R}$?

2) Are all such fields isomorphic?

I think that in terms of the group $G=Aut(\mathbb C)$ this is equivalente to:

1) are there order 2 elements in $G$ other than conjugation?

2) are all such elements conjugate in $G$? (I think this seemingly stronger condition is equivalent to having abstract isomorphism between the subfields)

If I am not mistaken, the answer to (1) should be positive, since applying some "wild" automorphism of $\mathbb C$ to $\mathbb R$ should produce such subfield (it can't stabilize $\mathbb R$ without fixing it, since $\mathbb R$ has no non-trivial automorphism). Question (2) is essentially whether all of them come up in this way.

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up vote 1 down vote accepted

With the axiom of choice, you can indeed show that 'wild' automorphisms exist. (I've read that ZF is consistent with $Aut(\mathbb{C}) \cong \mathbb{Z} / 2$)

For your second question, the answer is obviously yes. But then I stop to think of it, it seems clear there are counter-examples.

My favorite recipe for constructing an 'interesting' real closed field begins with taking some formally real field, such as $\mathbb{Q}$, then ordering $\mathbb{Q}(x)$ such that $x$ is infinite. This clearly embeds (as a field) into $\mathbb{C}$, and thus so does its real closure. By transfinite induction, the following algorithm terminates (after transfinitely many steps):

  • Set $F = \mathbb{Q}^r$
  • Choose an embedding $\phi : \overline{F} \to \mathbb{C}$
  • While $\overline{F} \not\cong \mathbb{C}$:
    • Choose $\zeta \in \mathbb{C} \setminus \phi(\overline{F})$
    • Set $G = F(x)$ (ordered with $x$ larger than every element of $F$)
    • Extend $\phi$ by $\phi(x) = \zeta$
    • Extend $\phi : G \to \mathbb{C}$ to $\overline{G} \to \mathbb{C}$
    • Set $F = G^r$

(basically, just start with the real algebraic numbers, and keep adjoining infinite elements and taking the real closure until the algebraic closure is isomorphic to $\mathbb{C}$. At limit ordinals, you collect everything done so far with a nested union)

The end result is a real closed field $F$ and an isomorphism $\phi : \overline{F} \to \mathbb{C}$. However, $F \not\cong \mathbb{R}$, because $F$ has an element larger than every rational number! (the ordering on a real closed field is unique, so there can't be a 'wild' isomorphism)

I haven't thought it through, but I think the field of Puiseaux series in $x$ over $\mathbb{Q}^r$ (with $x$ infinite) -- i.e. $\mathbb{Q}^r((x))^r$ -- is an explicit field whose algebraic closure is $\mathbb{C}$.

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great answer. thanks. – KotelKanim Jan 13 '13 at 19:43

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