Automorphisms of $\mathbb{C}$ Is it true that $G_{\mathbb{Q}}$, the absolute Galois group of $\mathbb{Q}$, is a subgroup of $Aut(\mathbb{C})$ ?
Or a simpler question: can any automorphism of $\overline{\mathbb{Q}}$ be extended to an automorphism of $\mathbb{C}$?
 A: There is the more general fact that any automorphism of any subfield of $\mathbb{C}$ can be extended to an automorphism of $\mathbb{C}$. For a proof, see the paper Automorphisms of the Complex Numbers by Paul Yale of Pomona College. Here is a JSTOR link. In general, if $k$ is an arbitrary (EDIT: algebraically closed) field, my guess would be Yale' argument could be easily extended to show that any automorphism of a subfield $h\subset k$ can be extended to an automorphism of $k$.
A: Your question depends on the axiom of choice.  As noted in other comments, if you assume AC, then the complex numbers have crazy automorphisms, and any automorphism of any subfield of $\mathbb{C}$ can be extended to an automorphism of all of $\mathbb{C}$.
However!  If you do not assume the axiom of choice, then it is consistent to say that the only automorphisms of the complex numbers are the identity and conjugation (this is consistent with ZF and implied by additional axioms such as the axiom of determinacy [which implies among other things that all sets are measurable]).  Note that this is inconsistent with AC, but it is not the negation of it.
In this case (without AC), the only automorphisms that can be extended to automorphisms of $\mathbb{C}$ are the identity and complex conjugation!
So with the axiom of choice, the automorphisms are really crazy and any automorphism of a subfield can be extended to an automorphism of all of $\mathbb{C}$.  But without the axiom of choice (and WITH some other consistent axioms), the automorphisms are all really boring!
Interesting stuff.
-Pat Devlin
