TITLE: What is $\mathbb{C}^{Aut(\mathbb{C}/\mathbb{Q})}$?
QUESTION [12 upvotes]: Let $Aut(\mathbb{C}/\mathbb{Q})$ be the field automorphisms of $\mathbb{C}$, and $\mathbb{C}^{Aut(\mathbb{C}/\mathbb{Q})}$ the subfield of $\mathbb{C}$ fixed by this group. I supsect that it is equal to $\mathbb{Q}$ but I have difficulty proving it.
Here is what I do : let $x \notin \mathbb{Q}$. We have to find an automorphism not fixing $x$. It is easy to find an embeding $\mathbb{Q}(x) \rightarrow \mathbb{C}$ not sending $x$ on $x$, and use Zorn's lemma to extend it to a maximal subfield (of $\mathbb{C}$) $K \supset \mathbb{Q}(x)$. But it is not true that $K=\mathbb{C}$. So how to handle it ?

REPLY [4 votes]: The trick is to pick a transcendence basis.  For instance, suppose $x\in\mathbb{C}$ is transcendental, and pick a transcendence basis $B$ for $\mathbb{C}$ which contains $x$.  Let $y\in B$ be some element different from $x$.  Then there is an isomorphism $f:\mathbb{Q}(B)\to\mathbb{Q}(B)$ which swaps $x$ and $y$ and fixes every other element of $B$.  The isomorphism $f$ then extends to an isomorphism $\mathbb{C}\to\mathbb{C}$, since $\mathbb{C}$ is an algebraic closure of $\mathbb{Q}(B)$.
The case that $x$ is algebraic but irrational is similar.  In that case we can first take a finite Galois extension $K$ of $\mathbb{Q}$ containing $x$ and an automorphism $f:K\to K$ which does not fix $x$.  Picking a transcendence basis $B$ for $\mathbb{C}$, we can then extend $f$ to an isomorphism $K(B)\to K(B)$ which fixes each element of $B$.  Finally, we can extend to $\mathbb{C}\to\mathbb{C}$, since $\mathbb{C}$ is an algebraic closure of $K(B)$.
(Here we are using the fact that if $k$ and $\ell$ are fields, $f:k\to\ell$ is an isomorphism, and $K$ and $L$ are algebraic closures of $k$ and $\ell$ respectively, then $f$ extends to an isomorphism $K\to L$.  Indeed, we can take a maximal extension $f_0:K_0\to L_0$ of $f$ to an isomorphism between subfields of $K$ and $L$.  If $a\in K\setminus K_0$, then $a$ is algebraic over $K_0$, so since $L$ is algebraically closed we can find $b\in L$ such that $f_0$ extends to $K_0(a)\to L_0(b)$.  So we must have $K=K_0$, and a similar argument shows $L=L_0$.)