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Mar 31, 2021 at 10:09 history edited YCor
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Mar 30, 2021 at 7:11 vote accept R.P.
Mar 29, 2021 at 18:39 history edited R.P. CC BY-SA 4.0
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Mar 28, 2021 at 23:09 history became hot network question
Mar 28, 2021 at 20:03 vote accept R.P.
Mar 29, 2021 at 9:11
Mar 28, 2021 at 19:58 answer added Maarten Derickx timeline score: 7
Mar 28, 2021 at 19:05 comment added Maarten Derickx Yes, I was sort of guessing that, hence the comment. If you want to do Brauer-Manin to study $X(\mathbb{Q})$ the right way to go really is to see $X$ as subvariety of $E$ and only use classes coming from $\operatorname{BR}(E)$ exactly for the reason you mention. It would be really magical if by somehow removing a few known rational points from $E$ we could better study the rational points of $E$ by using a larger Brauer group. Sadly this magic doesn't happen. I will post a full answer what you can get using $\operatorname{BR}(E)$.
Mar 28, 2021 at 18:17 comment added R.P. @M.D. The problem is that I am really interested in X(Q). So unless you are telling me that it is hopeless to use Brauer-Manin obstructions for studying rational (as opposed to integral) points on X, this does not precisely answer my question (although I do appreciate your comment).
Mar 28, 2021 at 17:57 comment added Maarten Derickx The reason that you get into problems with the Brauer-Manin obstruction for X non-proper is that you are working with $X/\mathbb{Q}$ and using $X(\mathbb{Q}_p)$. You get something meaningful for $X$ if you instead extend $X$ to $\mathbb{Z}[1/N]$ for some integer $N$ and use Brauer-Manin to study $X(\mathbb{Z}[1/N])$ instead of $X(\mathbb{Q})$. Then you can produce an integral version of the Brauer-Manin obstruction like is done in equation 2.2 of imo.universite-paris-saclay.fr/~harari/articles/dhvol.pdf for open subsets of $\mathbb{P}^1$. More general references undoubtedly exist.
Mar 28, 2021 at 17:01 history edited R.P. CC BY-SA 4.0
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Mar 28, 2021 at 16:55 history edited R.P. CC BY-SA 4.0
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Mar 28, 2021 at 15:07 history asked R.P. CC BY-SA 4.0