What does
$\mathrm{Gal}(\overline{\mathbb{Q}}\_{p}/\mathbb{Q})$ mean? ($p$ is a prime number.)


If it is defined as $\mathrm{Aut}(\overline{\mathbb{Q}}\_{p}/\mathbb{Q})$, then does it have any property like usual galois groups?


For example, is the following statement true? :


Let $\alpha \in \overline{\mathbb{Q}}\_{p}$, and assume that for any $\sigma \in \mathrm{Aut} (\overline{ \mathbb{Q} } \_{p} / \mathbb{Q})$, $\sigma (\alpha) = \alpha$. Then, $\alpha \in \mathbb{Q}$.


If this is true, then how can I prove it?


Please give me any advice.


Thanks!
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What does Gal(Q_p/Q) mean?
Yes, this is true. A way to prove this is to use the existence of [transcendence basis](http://en.wikipedia.org/wiki/Transcendence_degree) for field extensions, and the fact that $\overline{\mathbf{Q}\_p}$ is algebraically closed.


First, assume $\alpha$ is transcendental. Then there exists a transcendence basis $S$ of $\overline{\mathbf{Q}\_p}/\mathbf{Q}$ containing $\alpha$. By permuting the elements of $S$, there exists an automorphism $\sigma$ of $\mathbf{Q}(S)$ such that $\sigma(\alpha) \neq \alpha$. Since $\overline{\mathbf{Q}\_p}$ is an algebraic closure of $\mathbf{Q}(S)$, you can extend $\sigma$ to an automorphism of $\overline{\mathbf{Q}\_p}$, which gives a contradiction.


Finally, if $\alpha \in \overline{\mathbf{Q}}$ but $\alpha \not\in \mathbf{Q}$, then you can take the Galois closure $L$ of $\mathbf{Q}(\alpha)$ and find an automorphism $\sigma$ of $L$ such that $\sigma(\alpha) \neq \alpha$. Then, repeating the above argument, you can extend $\sigma$ to $\overline{\mathbf{Q}\_p}$, which gives a contradiction.


Note that the argument works replacing $\overline{\mathbf{Q}\_p}$ by any algebraically closed field of characteristic $0$.


