Skip to main content
8 events
when toggle format what by license comment
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
added 1203 characters in body
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
edited body
Jun 17, 2023 at 13:02 history answered Chris Wuthrich CC BY-SA 4.0