Timeline for Calculating the Galois cohomology group $H^1(K_v, \, E[p^{\infty}])$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 11, 2022 at 14:48 | history | edited | Dietrich Burde | CC BY-SA 4.0 |
edited body
|
| Dec 12, 2021 at 7:22 | comment | added | ali | tate duality is for multiplicative groups, this duality is for abelian varieties | |
| Dec 11, 2021 at 18:09 | vote | accept | Adithya Chakravarthy | ||
| Dec 11, 2021 at 18:09 | comment | added | Adithya Chakravarthy | Ah, I see now. Apologies, I got the exact sequence you wrote mixed up with Tate duality, which relates $H^i$ with $H^{2-i}$, whereas your exact sequence relates $H^i$ with $H^{1-i}$. This sequence is very helpful, thanks! | |
| Dec 11, 2021 at 16:39 | comment | added | ali | The prefect pairing says that $H^1(K,E)$ is dual to $H^0(K,E)$ which is by definition the galios invariants of $E(K^{sep})$, which is $E(K)$ | |
| Dec 11, 2021 at 16:36 | history | edited | ali | CC BY-SA 4.0 |
added 1 character in body
|
| Dec 11, 2021 at 14:41 | comment | added | Adithya Chakravarthy | Hmm, I'm not sure I follow. How would knowing $E(K_v)$ tell us what $H^1(K, E)$ is? I'm thinking one can try finding the image of $E(K_v)$ under the Kummer map, but that would only give us part of the full group $H^1(K_v, E)$, not the whole thing, right? | |
| Dec 11, 2021 at 13:37 | comment | added | ali | I think your aproach could also work using some class field theory, and if you look at the proof in the milne book it is basically using simmilar ideas | |
| Dec 11, 2021 at 13:34 | history | answered | ali | CC BY-SA 4.0 |