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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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closed as too localized by Martin Brandenburg, Alain Valette, Dan Petersen, Charles Siegel, Simon Thomas Nov 13 '11 at 21:04

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Please read mathoverflow.net/faq#whatnot . $\overline{\mathbb{Q}_p} / \mathbb{Q}_p$ is an infinite Galois extension. –  Martin Brandenburg Nov 13 '11 at 12:21
    
(but over $\mathbb{Q}$ it's not Galois) –  Martin Brandenburg Nov 13 '11 at 12:22
1  
@Hiro: One can use the axiom of choice to construct an isomorphism of abstract fields (i.e. ignoring the topology) $\overline{\mathbb{Q}_p} \cong \overline{\mathbb{Q}}$. –  Daniel Loughran Nov 13 '11 at 12:32
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@Daniel: No, one cannot. $\overline{\mathbb{Q}}$ is countable, $\overline{\mathbb{Q}_p}$ is not. –  Robert Kucharczyk Nov 13 '11 at 12:48
    
Thanks for your comments! –  Hiro Nov 13 '11 at 13:07
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1 Answer 1

up vote 6 down vote accepted

Yes, this is true. A way to prove this is to use the existence of transcendence basis 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 $0$.

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Oh, I did not think that it is such a general fact! And your proof is clear and interesting! Thank you so much! –  Hiro Nov 13 '11 at 13:06
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