Subfields of $\mathbb{C}$ which extend to $\mathbb{C}$ via finite extensions

Question

The field Q is a subfield of C but it is in a sense "much smaller" than C. The field R however has a finite extension of order just two to the field C. My question is:

Are there other subfields of C which are "finitely" less than C, or is R the only example?

I know that for such a subfield we would have automorphisms of C that fix the subfield and have finite order, but I have read that there do exist wild automorphisms of C, so we cannot discard that option in this way. I have also considered the intersection of such a subfield with R: it would itself be a subfield of R. If we could conclude the extension from this subfield of R to R is finite we'd be done, because the only possible automorphism of R can be proven to be the trivial one. However, I'm not sure we can state that if F has a finite extension G then for some other field H we have that F $\cap$ H has finite extension G $\cap$ H.

Answer 1

A theorem of Artin and Schreier states that if $K$ is algebraically closed and of finite degree over a field $k$, then (1) the characteristic is zero; (2) the degree is two; and (3) $K=k(i)$, where $i^2=-1$. You can find it in Lang’s Algebra, and many other sources.

Answer 2

Here is a relatively simple example of such a subfield which is not isomorphic to $\mathbb{R}$. Consider the ordered field $\mathbb{R}(x)$, where $\mathbb{R}$ is given the usual order and $x$ is larger than every element of $\mathbb{R}$. Let $K$ be an ordered field which is maximal among all ordered algebraic extensions of $\mathbb{R}(x)$ (such a $K$ clearly exists by Zorn's lemma). It is not hard to show that every odd-degree polynomial over $K$ has a root in $K$, and every positive element of $K$ has a square root in $K$ (to prove this, show that if this failed, you could order an extension of $K$ in which you adjoin such a root). One of the standard proofs of the fundamental theorem of algebra shows that the field $K[\sqrt{-1}]$ is algebraically closed (see this discussion on Wikipedia, for instance). That is, $K[\sqrt{-1}]$ is an algebraic closure of $\mathbb{R}(x)$. It follows that $K[\sqrt{-1}]$ is isomorphic to $\mathbb{C}$ (they are both algebraically closed fields of characteristic zero and transcendence degree $2^{\aleph_0}$).

Finally, I claim $K$ is not isomorphic to $\mathbb{R}$. To see this, note that they are clearly not isomorphic as ordered fields, since $\mathbb{R}$ is archimedean but $K$ is not. But the orderings on $\mathbb{R}$ and $K$ are completely determined by their field structure, since in both fields, an element is positive iff it has a square root. Thus $\mathbb{R}$ and $K$ cannot be isomorphic as fields.

As a final remark, the result alluded to in Lubin's answer says more precisely that every example is of this kind: if $K\subset \mathbb{C}$ is a proper subfield such that $\mathbb{C}$ is finite over $K$, then $K$ can be given a (unique) ordering such that every positive element has a square root and every odd-degree polynomial has a root, and $\mathbb{C}=K[i]$.

Answer 3

Under the axiom of choice there is an isomorphism $\sigma: \bigcup_{n\ge 1} \Bbb{C}((x^{1/n}))\to \Bbb{C}$ so that $K=\sigma(\bigcup_{n\ge 1}\Bbb{R}((x^{1/n})))$ works,

$[\Bbb{C}:K]=2$ and there is no isomorphism $\rho:K\to \Bbb{R}$ since $\rho|_\Bbb{R}$ would be the identity.