Timeline for Galois cohomology of Tate modules
Current License: CC BY-SA 4.0
14 events
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| May 6, 2023 at 9:20 | comment | added | Chris Wuthrich | @kindasorta Also, I will stop answering questions like this now. Unfortunately, I don't have time to tutor you on these topics. I recommend texts by Greenberg on the Iwasawa theory of elliptic curves where you will learn more about the Galois cohomology of elliptic curves. He has also a text with many references regarding congruences of elliptic curves. | |
| May 6, 2023 at 9:13 | comment | added | Chris Wuthrich | If you look at the example, then it should be clear that is not the case whatever meaning "Tate dual" has in this case. Instead, $H^2(G_S,T)$ is the cokernel of the last map in the Cassels sequence above and hence it is dual to the fine (or strict) Selmer group of $E$. Its torsion subgroup is a subgroup of the Tate-Shafarevich group as expected in the above example. | |
| May 6, 2023 at 7:09 | comment | added | kindasorta | Does Pouitou-Tate duality tell us something about the $p$-torsion in $H^2(G_S, T)$ in terms of $H^0(G_S, T')$, where $T'$ is the Tate dual of $T$? | |
| May 5, 2023 at 18:54 | comment | added | Chris Wuthrich | @kindasorta So it is just $E[p]$? If you read my answer you should sport that the rank of $H^1(G_S, T)$ is not the dimension of $H^1(G_S,E[p])$ because the term $H^2(G_S,T)[p]$ may be non-zero. | |
| May 5, 2023 at 18:25 | comment | added | kindasorta | @ChrisWuthrich $\overline{T_pE}$ is the associated residual representation, and my $E[p^{\infty}]$ should indeed be $E(\mathbb{Q})[p^{\infty}]$. | |
| May 5, 2023 at 17:41 | comment | added | Chris Wuthrich | @kindasorta What is $\overline{T_pE}$? ($E[p^\infty]$ is never zero, but $E(\mathbb{Q})[p^\infty]$ is often zero.) | |
| May 5, 2023 at 17:40 | history | edited | Chris Wuthrich | CC BY-SA 4.0 |
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| May 5, 2023 at 17:38 | comment | added | Chris Wuthrich | @JeremyRouse Indeed, my bad, the $p^2$ torsion can also appear for $p=3$. I will change that. It doesn't affect the example as there the $3$-adic representation is surjective anyway. | |
| May 5, 2023 at 17:20 | comment | added | Jeremy Rouse | I don't see why the $p$-power torsion subgroups are the same if $\overline{\rho}$ and $\overline{\rho'}$ are isomorphic. Can't you have elliptic curves $E$ and $E'$ that are $3$-congruent where $E$ has a rational point of order $9$ and $E'$ doesn't? | |
| May 5, 2023 at 16:38 | comment | added | kindasorta | Thank you for the incredibly detailed answer! Then, it seems that using Mazur's "Eisenstein ideal" theorem about torsion of Elliptic curves over $\mathbb{Q}$, it is true that for $p > 5$ we have $E[p^{\infty}] = 0$, so that $H^1(\mathbb{Q}, T_pE)$ is torsion free, and hence a free $\mathbb{Z}_p$-module. Its rank then should be the same as the $\mathbb{F}_p$ rank of $H^1(\mathbb{Q}, \overline{T_pE})$. Is this accurate? | |
| May 5, 2023 at 16:26 | vote | accept | kindasorta | ||
| May 4, 2023 at 22:48 | history | edited | Chris Wuthrich | CC BY-SA 4.0 |
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| May 4, 2023 at 22:36 | comment | added | Chris Wuthrich | Actually, the fact that the $\lambda$-invariant for $E$ is $2$ proves that the $3$-primary part of Sha has at most $9$ elements and the argument with the congruence shows provably that this is an equality. No need to rely on BSD. | |
| May 4, 2023 at 22:32 | history | answered | Chris Wuthrich | CC BY-SA 4.0 |