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Let $E/L$ be an elliptic curve defined over a number field $L$. Assume moreover that $E$ has complex multiplication by an imaginary quadratic field $K$. Let $\mathfrak{p}$ be a prime ideal of $K$. Assume that $E[\mathfrak{p}]\subseteq E(L)$ and that $E$ has good reduction at $\wp$ where $\wp$ is a prime ideal of $L$ above $\mathfrak{p}\cap \mathbb{Z}$. Consider the reduction map $$ \pi: E(L)\rightarrow \tilde{E}(\mathcal{O}_L/\wp), $$ where $\tilde{E}$ denotes the reduction of $E$ modulo $\wp$.

Consider the following two statements:

S1: $\ker(\pi)\supseteq E[\mathfrak{p}]$

S2: $\tilde{E}[\mathfrak{p}](\mathcal{O}_L/\wp)=\{0\}$

Note that S2 implies S1.

Q: Is S2 true, if so, how does one prove it ?

This does not seem to follow directly from the theory of formal groups.

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  • $\begingroup$ 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. $\endgroup$
    – Jeff Yelton
    Commented Jun 9, 2016 at 10:17
  • $\begingroup$ 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 ? $\endgroup$
    – Hugo Chapdelaine
    Commented Jun 9, 2016 at 10:27
  • $\begingroup$ 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 $p$. $\endgroup$
    – Hugo Chapdelaine
    Commented Jun 9, 2016 at 22:02
  • $\begingroup$ ....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 $K$. $\endgroup$
    – Hugo Chapdelaine
    Commented Jun 9, 2016 at 22:08
  • $\begingroup$ 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. $\endgroup$
    – Hugo Chapdelaine
    Commented Jun 10, 2016 at 7:49

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