What is the normal closure of $\mathbb{Q}_p \cap \bar{\mathbb{Q}}$ over $\mathbb{Q}$? Is it $\bar{\mathbb{Q}}$?
3
-
4$\begingroup$ 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-TsenCommented Feb 14, 2017 at 13:40
-
8$\begingroup$ 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. $\endgroup$– nfdc23Commented Feb 14, 2017 at 15:50
-
2$\begingroup$ @nfdc23 You should post this as an answer. $\endgroup$– Gro-TsenCommented Feb 14, 2017 at 18:17
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
(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.
-
1$\begingroup$ When I copy a comment by another user to make it an answer, I tick cw. You can do so by editing. $\endgroup$– YCorCommented Jan 11, 2018 at 13:58
-
1$\begingroup$ @YCor Thanks, that's not a feature I had noticed before. Makes sense. $\endgroup$– DorisCommented Jan 11, 2018 at 15:48