Timeline for Ker of corestriction of Galois cohomology
Current License: CC BY-SA 4.0
8 events
when toggle format | what | by | license | comment | |
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Jun 20, 2023 at 1:46 | vote | accept | Duality | ||
Jun 18, 2023 at 10:55 | comment | added | Chris Wuthrich | Maybe I should also comment on the less interesting particular case in the question. Let $W'$ be the set of elements of odd order in $H^1(K,E)$ where $K/\mathbb{Q}$ is quadratic. Then $W' = W^{+} \oplus W^{-}$ where the Galois group acts as $+1$ or $-1$. Of course $W^{+}$ is the odd part of $H^1(\mathbb{Q}, E)$. Then the kernel of corestriction on $W'$ is $W^{-}$ and that is huge as it is the odd part of $H^1(\mathbb{Q}, E^D)$ where $E^D$ is the quadratic twist. Similarly all elements of order $2$ in the image of restriction from the $2$-primary part of $H^1(\mathbb{Q}, E)$ are in this kernel. | |
Jun 18, 2023 at 9:44 | vote | accept | Duality | ||
Jun 18, 2023 at 10:29 | |||||
Jun 17, 2023 at 18:18 | comment | added | Chris Wuthrich | I have added to be more precise in the case $M$ is not finite. | |
Jun 17, 2023 at 18:17 | history | edited | Chris Wuthrich | CC BY-SA 4.0 |
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Jun 17, 2023 at 17:12 | comment | added | Duality | Thank you very much. But is your $M$ finite? In the case, like $M=E(\overline{\Bbb{Q}})$, does the same kind of exact sequence exist? | |
Jun 17, 2023 at 13:10 | history | edited | Chris Wuthrich | CC BY-SA 4.0 |
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Jun 17, 2023 at 13:02 | history | answered | Chris Wuthrich | CC BY-SA 4.0 |