most recent 30 from http://mathoverflow.net 2013-06-18T22:40:07Z http://mathoverflow.net/feeds/question/80811 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80811/what-does-galq-p-q-mean Hiro 2011-11-13T11:39:55Z 2011-11-13T13:08:11Z

<p>What does $\mathrm{Gal}(\overline{\mathbb{Q}}_{p}/\mathbb{Q})$ mean? ($p$ is a prime number.)</p> <p>If it is defined as $\mathrm{Aut}(\overline{\mathbb{Q}}_{p}/\mathbb{Q})$, then does it have any property like usual galois groups?</p> <p>For example, is the following statement true? :</p> <p>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, $\alpha \in \mathbb{Q}$.</p> <p>If this is true, then how can I prove it?</p> <p>Please give me any advice.</p> <h1>Thanks!</h1>

http://mathoverflow.net/questions/80811/what-does-galq-p-q-mean/80815#80815 François Brunault 2011-11-13T12:30:02Z 2011-11-13T12:30:02Z

<p>Yes, this is true. A way to prove this is to use the existence of <a href="http://en.wikipedia.org/wiki/Transcendence_degree" rel="nofollow">transcendence basis</a> for field extensions, and the fact that $\overline{\mathbf{Q}_p}$ is algebraically closed.</p> <p>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.</p> <p>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.</p> <p>Note that the argument works replacing $\overline{\mathbf{Q}_p}$ by any algebraically closed field of characteristic $0$.</p>