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There is the following result of Deuring that goes as follows:

Let $E/L$ be an elliptic curve defined over a number field $L$ with complex multiplication by an order $\mathcal{O}$ in an imaginary quadratic field $K$, $\mathfrak{p}$ a prime in $\overline{\mathbb{Q}}$ over a rational prime $p$, and $\tilde{E}$ the reduction of $E$ modulo $\mathfrak{p}$. Then $p$ is a supersingular prime for $E$ if and only if $p$ is ramified or inert in $K$.

I have two questions:

  • What is an intuitive way to see that this result is true?
  • What is the quickest way to prove this result?

Many thanks in advance. Perhaps this question is a bit silly because it (is not very hard assuming) assumes the main results of complex multiplication...

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  • $\begingroup$ I think the intuition is just that when you have CM, the isomorphism between your endomorphism ring and your CM order allows you to compute the trace of Frobenius in terms of the arithmetic of the splitting of $p$ in $K$. Since $K$ is quadratic, there's only three splitting types, and so the relationship $N=p+1-\text{tr}(\mathfrak{p})$ only leaves you having to check three cases as to whether or not $p\mid \text{tr}(\mathfrak{p})$. Apologies if this is not at all what you're looking for. $\endgroup$
    – Cam McLeman
    Commented Jul 15, 2015 at 17:35
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    $\begingroup$ Note that $V_p(E)$ is free of rank 1 over the rank-2 $\mathbf{Q}_p$-algebra $K_p:=(O_{K,p})[1/p]$ (exercise!), so in ramified & inert cases it is a line over the field $K_p$. But the $D_{\mathfrak{p}}$-action on $V_p(E)$ is $K_p$-linear, so in ordinary cases the unique "ramified" $\mathbf{Q}_p$-line for the $D_{\mathfrak{p}}$-action is $K_p$-stable, impossible in ramified & inert cases. Thus, such cases are ss. Conversely, for ss reduction with 1-dimensional height-2 formal group $\Gamma$, $K_p$ injects into ${\rm{End}}(\Gamma)[1/p]$ which is a division algebra, so $K_p$ isn't split! $\endgroup$
    – grghxy
    Commented Jul 16, 2015 at 4:30

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