Is $\mathbb Q(\sqrt2)$ the fixed field of some automorphism of $\overline{\mathbb Q}$? Is is possible that $\mathbb{Q}(\sqrt{2})$ is the fixed field of an automorphism $\sigma$ of $\overline{\mathbb{Q}}$?
Thanks in advance for your help.
 A: No.  Let $H = \langle \sigma \rangle$.  The field extension $\overline{\mathbb{Q}}/\overline{\mathbb{Q}}^{H}$ has automorphism group the closure $\overline{H}$ of the cyclic subgroup $H$ in the Krull topology, hence an abelian profinite group.  So if the fixed field were $\mathbb{Q}(\sqrt{2})$, then the absolute Galois group of $\mathbb{Q}$ would be an extension of a group of order $2$ by an abelian group, hence every finite quotient would be solvable.  But since e.g. $S_5$ is a Galois group over $\mathbb{Q}$, this is not the case.
This argument does use something about the complicatedness of the absolute Galois group of $\mathbb{Q}$: compare e.g. that the fixed field of the Frobenius automorphism in $\overline{\mathbb{F}}_p$ is $\mathbb{F}_p$: here the absolute Galois group is abelian.
An even more boundary case is the formal Laurent series field $\mathbb{R}((t))$, in which case the algebraic closure is the field of Puiseux series $\bigcup_{n \geq 1} \mathbb{C}((t^{\frac{1}{n}}))$.  Here the natural automorphism $t^{\frac{1}{n}} \mapsto \zeta_n t^{\frac{1}{n}}$ has fixed field $\mathbb{C}((t)) = \mathbb{R}((t))(\sqrt{-1})$, so this gives a Galois theoretic context in which the extension of a group of order $2$ by an infinite cyclic group really does occur.
Added: The absolute Galois group of a field $k$ is the automorphism group $\operatorname{Aut}(\overline{k}/k)$ of its separable (= algebraic, in characteristic $0$) closure over $k$.  The Krull topology on Galois groups is what I defined in my number theory class earlier today -- which is not helpful, I know, but an interesting coincidence.  It is described for instance in this note...but that is a fancier presentation than is probably helpful for a first reading (this is a survey/research paper looking at Galois theory of arbitrary field extensions).
Perhaps we should take the topology out of the picture: first do the case in which $H$ is finite cyclic.  This would imply that the algebraic closure of $\mathbb{Q}$ has finite degree over $\mathbb{Q}$, which you should agree is absurd.  Then if it is infinite cyclic this means that there will be for each $n$ one subextension $\overline{\mathbb{Q}}^H \subset K_n \subset \overline{\mathbb{Q}}$ with $\operatorname{Aut}(K_n/\overline{\mathbb{Q}}^H)$ cyclic of order $n$ and $\overline{\mathbb{Q}} = \bigcup_n K_n$.  Again this would mean -- maybe a little more concretely this time -- that every finite Galois group over $\mathbb{Q}$ would have an index $2$ cyclic subgroup, and that certainly is not the case!
A: First note that $\sqrt{3}, \sqrt{5}, \sqrt{15} \not\in \mathbf{Q}(\sqrt{2})$.
Assume, if possible, that the fixed field of $\sigma$ is $\mathbf{Q}(\sqrt{2})$. Then $\sigma(\sqrt{3}) \ne \sqrt{3}$, so we must have $\sigma(\sqrt{3}) = -\sqrt{3}$. Similarly, $\sigma(\sqrt{5}) = -\sqrt{5}$. Therefore $\sigma(\sqrt{15}) = \sqrt{15}$, contradicting the assumption.
A: Here's a partial answer:
If the group $H=\langle \sigma \rangle$ is a finite subgroup of $\operatorname{Aut}(\overline{\mathbb{Q}})$, the extension $\overline{\mathbb{Q}}/K$, where $K$ is the fixed field of $H$, is necessarily finite. Since $\overline{\mathbb Q}/\mathbb{Q}(\sqrt{2})$ is not finite, $\mathbb{Q}(\sqrt{2})$ can't be the fixed field of $H$. 
(In fact, given any field $F$ and any finite subgroup $H$ of the automorphism group of $F$, the extension $F/K$, where $K$ is the fixed field of $H$, is not only finite but also Galois.)
So, all that remains is to look at the automorphisms of infinite order. 
