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Let $E/ \Bbb{Q}_p$ be an elliptic curve over $ \Bbb{Q}_p$. $\hat{E}$ denote the corresponding formal group of $E$. I want to prove $\Bbb{Q}_p(E[p])＝\Bbb{Q}_p(\hat{E}[p])$.

$ \hat{E}[p]$ denotes $p$ division point of formal group $\hat{E}$, this is subset of $\Bbb{Z}_p$. To prove this, I think we need to find some relation between $E[p]$ and $\hat{E}$, thank you for your help.

Motivation behind this question : $\Bbb{Q}_p(\hat{E}[p])/ \Bbb{Q}_p$ is Lubin Tate extension of degree $p-1$, so we can conclude $\Bbb{Q}_p(E[p])/ \Bbb{Q}_p$ is degree $p-1$ totally ramified extension.

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    $\begingroup$ 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 $\hat E$. $\endgroup$
    – Chris Wuthrich
    Commented Aug 2, 2022 at 23:51
  • $\begingroup$ Oh, thank you. But is $ \Bbb{Q}_p(E[p])⊇\Bbb{Q}_p( \hat{E}[p])$ true in general? $\endgroup$
    – Duality
    Commented Aug 3, 2022 at 9:03
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    $\begingroup$ Sure, look at chapter VI and Proposition VII.2.2 in Silverman. $\endgroup$
    – Chris Wuthrich
    Commented Aug 3, 2022 at 10:38
  • $\begingroup$ @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])$. $\endgroup$
    – Duality
    Commented Aug 29, 2022 at 14:28
  • $\begingroup$ @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. $\endgroup$
    – Chris Wuthrich
    Commented Aug 30, 2022 at 21:03

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