Timeline for Galois cohomology of Tate modules
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 5, 2023 at 16:26 | vote | accept | kindasorta | ||
| May 4, 2023 at 22:32 | answer | added | Chris Wuthrich | timeline score: 5 | |
| May 4, 2023 at 16:24 | comment | added | Chris Wuthrich | $0\to T \to T \to E[p^k] \to 0 $ is exact and $T^G=0$ gives $H^1(G,T)[p^k] = E(\mathbb{Q})[p^k]$. | |
| May 4, 2023 at 14:40 | comment | added | kindasorta | Could you please sketch out how the computation of the torsion subgroup is conducted? I mean the torsion subgroup of the cohomology group. | |
| May 4, 2023 at 14:26 | comment | added | Chris Wuthrich | The torsion subgroup is isomorphic to $E(\mathbb{Q})[p^\infty]$ in both cases and that won't change between $E$ and $E'$ if $p>2$. And often, but not always, it is torsion-free. | |
| May 4, 2023 at 12:09 | comment | added | kindasorta | Yeah, I do. Are you saying that if I take $H^1(G_S, T_pE)$, where $S$ is a finite set of primes then it should be torsion free? Do you have a reference for this claim? | |
| May 4, 2023 at 12:04 | comment | added | kindasorta | What about $p\neq 2$? | |
| May 4, 2023 at 11:51 | comment | added | Chris Wuthrich | Not if $p=2$ as then the torsion subgroup of $H^1(\mathbb{Q},T_pE)$, which is isomorphic to $E(\mathbb{Q})[p^{\infty}]$, might be different even if the $p$-torsion $E[p]$ are both trivial. | |
| May 4, 2023 at 11:27 | history | asked | kindasorta | CC BY-SA 4.0 |