Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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 $\mathbb{C}$.

share|improve this answer
great answer. thanks. –  KotelKanim Jan 13 '13 at 19:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.