Timeline for Result of Deuring, intuitive way to see it's true/quickest way to prove?

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Jul 16, 2015 at 4:30	comment	added	grghxy		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!
Jul 15, 2015 at 17:35	comment	added	Cam McLeman		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.
Jul 15, 2015 at 16:03	history	edited	user61522	CC BY-SA 3.0	added 10 characters in body
Jul 15, 2015 at 15:39	history	edited	user61522	CC BY-SA 3.0	added 120 characters in body
Jul 15, 2015 at 14:58	history	asked	user61522	CC BY-SA 3.0