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What is the normal closure of $\mathbb{Q}_p \cap \bar{\mathbb{Q}}$ over $\mathbb{Q}$? Is it $\bar{\mathbb{Q}}$?

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    $\begingroup$ It's worth pointing out that the normal closure of $\mathbb{R}\cap\bar{\mathbb{Q}}$ over $\mathbb{Q}$ is $\bar{\mathbb{Q}}$ (because it contains the three roots of $t^3-2$, so it contains their quotients, i.e., the primitive cube roots of unity, which gives us what we need to generate $\bar{\mathbb{Q}}$). $\endgroup$ – Gro-Tsen Feb 14 '17 at 13:40
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    $\begingroup$ For any global field $K$, place $v$ of $K$, and embedding of $K_s$ into a separable closure of $K_v$, the normal closure of $K_v \cap K_s$ over $K$ is $K_s$. Indeed, such an intersection corresponds to the decomposition group of ${\rm{Gal}}(K_s/K)$ at a place of $K_s$, and the intersection of decomposition groups for even just two distinct places of $K_s$ (such as two over a place of $K$) is trivial. See 12.1.3 (and 12.1.9 and 12.1.11) in the book Cohomology of Number Fields by Neukirch, Schmidt, and Wingberg. $\endgroup$ – nfdc23 Feb 14 '17 at 15:50
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    $\begingroup$ @nfdc23 You should post this as an answer. $\endgroup$ – Gro-Tsen Feb 14 '17 at 18:17
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(Copied from the @nfdc23's comment, because it's an answer.)

For any global field $K$, place $v$ of $K$, and embedding of $K_s$ into a separable closure of $K_v$, the normal closure of $K_v \cap K_s$ over $K$ is $K_s$. Indeed, such an intersection corresponds to the decomposition group of $\operatorname{Gal}(K_s/K)$ at a place of $K_s$, and the intersection of decomposition groups for even just two distinct places of $K_s$ (such as two over a place of $K$) is trivial. See 12.1.3 (and 12.1.9 and 12.1.11) in the book Cohomology of Number Fields by Neukirch, Schmidt, and Wingberg.

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    $\begingroup$ When I copy a comment by another user to make it an answer, I tick cw. You can do so by editing. $\endgroup$ – YCor Jan 11 '18 at 13:58
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    $\begingroup$ @YCor Thanks, that's not a feature I had noticed before. Makes sense. $\endgroup$ – Doris Jan 11 '18 at 15:48

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