Timeline for Relation between division point of elliptic curve and formal group of elliptic curve, $\Bbb{Q}_p(E[p])＝\Bbb{Q}_p(\hat{E}[p])$
Current License: CC BY-SA 4.0
10 events
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| Aug 30, 2022 at 21:03 | comment | added | Chris Wuthrich | @dandelion This is trivial: The composition of $\hat E[p] \cong E_1[p] \subset E[p]$ is an injective $\mathbb{Q}_p$-equivariant map. I actually view the isomorphism between $E_1(\mathbb{Q}_p)$ and $\hat E(p\mathbb{Z}_p)$ as an identification. | |
| Aug 29, 2022 at 14:28 | comment | added | Duality | @Chris Wuthrich : $E_1[p] \cong \hat{E}[p]$ as a group is true, but from here, could you tell me how you prove set theoretic inclusion $E[p]⊇\hat{E}[p]$? Without proving this set theoretic inclusion, we cannot say $ \Bbb{Q}_p(E[p])⊇\Bbb{Q}_p( \hat{E}[p])$. | |
| Aug 3, 2022 at 10:38 | comment | added | Chris Wuthrich | Sure, look at chapter VI and Proposition VII.2.2 in Silverman. | |
| Aug 3, 2022 at 9:03 | comment | added | Duality | Oh, thank you. But is $ \Bbb{Q}_p(E[p])⊇\Bbb{Q}_p( \hat{E}[p])$ true in general? | |
| Aug 2, 2022 at 23:51 | comment | added | Chris Wuthrich | If $E$ has good supersingular reduction, yes, $E[p]=\hat E[p]$ as Galois modules. If the reduction is good ordinary, then $0\to \hat E[p]\to E[p]\to \tilde E[p]\to 0$ is exact as Galois modules and so $\mathbb{Q}_p(E[p])$ is a bigger extension than the one for $\hat E$. | |
| Aug 2, 2022 at 20:05 | history | edited | Duality | CC BY-SA 4.0 |
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| Aug 2, 2022 at 20:01 | history | edited | YCor | CC BY-SA 4.0 |
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| Aug 2, 2022 at 19:38 | history | edited | Duality |
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| S Aug 2, 2022 at 15:56 | review | First questions | |||
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| S Aug 2, 2022 at 15:56 | history | asked | Duality | CC BY-SA 4.0 |