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Jun 10, 2016 at 7:50 comment added Hugo Chapdelaine There should still exist some kind of uniform proof involving the theory of formal groups...
Jun 10, 2016 at 7:49 comment added Hugo Chapdelaine In order to treat the case where $p O_K=\mathfrak{p}\bar{\mathfrak{p}}$ one may use the following ad-hoc observation: The map $E----> E/E[\mathfrak{p}] \pmod{\wp}$ is the Frobenius post composed with an isomorphism with kernel $E[\mathfrak{p}]$. S2 follows from that.
Jun 10, 2016 at 7:13 history edited Hugo Chapdelaine CC BY-SA 3.0
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Jun 9, 2016 at 22:08 comment added Hugo Chapdelaine ....Therefore, $\tilde{E}(O_L/\wp)$ has size $p^n+1$ which is coprime to $p$. So essentially it remains to treat the special case where $p$ splits in $K$.
Jun 9, 2016 at 22:02 comment added Hugo Chapdelaine Now that I'm thinking about it, if $\mathfrak{p}\cap \mathbf{Z}=p\mathbf{Z}$ is inert in $K$ then, for $p\geq 5$, the eigenvalues of $Fr_{\mathfrak{p}}$ are associated algebraic numbers which must differ by a sign, and therefore the coefficien $Tr(Fr_{\mathfrak{p}})=a_{\mathfrak{p}}=0$. Therefore, $\tilde{E(O_L/\wp)$ has size p^2+1 which is coprime to $p$.
Jun 9, 2016 at 16:45 history edited Hugo Chapdelaine
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Jun 9, 2016 at 10:27 comment added Hugo Chapdelaine If E is defined over $Q$ and has CM by $K$, then E will be supersingular at p if and only if p is inert in K. Does it help ?
Jun 9, 2016 at 10:22 history edited Hugo Chapdelaine CC BY-SA 3.0
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Jun 9, 2016 at 10:17 comment added Jeff Yelton Maybe I'm missing something, but it seems to me that $\mathrm{ker}(\pi) \supseteq E[\mathfrak{p}]$ implies that $E$ has supersingular reduction at the prime lying above $\mathfrak{p}$, which is not necessarily the case.
Jun 9, 2016 at 9:51 history asked Hugo Chapdelaine CC BY-SA 3.0