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 $\mathbb{C}$?

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    $\begingroup$ Yes, any automorphism of $\overline{\mathbb Q}$ can be extended to an automorphism of $\mathbb C$, hence $\mathrm{Aut}(\overline{\mathbb Q})$ is a quotient of $\mathrm{Aut}(\mathbb C)$. This holds for any extension pair of algebraically closed fields. $\endgroup$ Apr 4 '12 at 16:22
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    $\begingroup$ See e.g. en.wikipedia.org/wiki/Transcendence_degree#Applications . $\endgroup$ Apr 4 '12 at 16:58
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    $\begingroup$ That's in essential Theorem V.2.8 in Lang's algebra book. "in essential", because you first have to choose a transcendence base $T$ of $\mathbb{C}|\bar{\mathbb{Q}}$, extend your automorphism $f$ to $\bar{\mathbb{Q}}(T)$ by $f|T = id_T$ and then you can apply the cited theorem. $\endgroup$
    – Ralph
    Apr 4 '12 at 17:08
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    $\begingroup$ The natural map is from $Aut(\mathbb C)$ to $G_{\mathbb Q}$ (given by restriction), not the other way around. As has now been noted in several comments and answers, this map is surjective. On the other hand, this does not answer your first question, as to whether $G_{\mathbb Q}$ is a subgroup of $Aut(\mathbb C)$. In fact, you probably don't actually care about this question; the surjectivity is what you seemed interested in. Nevertheless, I imagine that the answer is no, in the sense that the surjection $Aut(\mathbb C) \to G_{\mathbb Q}$ presumably doesn't split. Regards, $\endgroup$
    – Emerton
    Apr 5 '12 at 2:57
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    $\begingroup$ @Emerton: you are right that the simpler question was my original one but, nonetheless, the other question is also interesting... ;) $\endgroup$
    – expmat
    Apr 6 '12 at 19:05

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 $k$.

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    $\begingroup$ It certainly does not work for arbitrary fields. For example, let $h=\mathbb Q(\sqrt 2)$, $k=\mathbb Q(\sqrt[4]2)$, and $f$ the automorphism of $h$ such that $f(\sqrt2)=-\sqrt2$. It works when $k$ is algebraically closed, as explained in the comments above. $\endgroup$ Apr 4 '12 at 22:16
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    $\begingroup$ For arbitrary fields k this isn't true... for instance, take h to be any nontrivial real algebraic extension of the rationals, and k to be the reals. (There are also counterexamples if k is taken to be the p-adic numbers, for any prime p.) I think it does extend to any algebraically closed k, though. $\endgroup$
    – zeb
    Apr 4 '12 at 22:20

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

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    $\begingroup$ Axiom of determinacy, whose consistency requires some pretty large cardinals, is an overkill. $|\mathrm{Aut}(\mathbb C)|=2$ is in fact relatively consistent with ZF (with no large cardinal assumptions), since it follows from ZF + DC + “every set of reals has the property of Baire”. $\endgroup$ Apr 6 '12 at 11:09

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