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Jul 13, 2011 at 13:47 vote accept Álvaro Lozano-Robledo
Jul 10, 2011 at 15:18 comment added Chris Wuthrich Absolutely correct. I am embarrassed about my mistake. The only thing it shows is that the reduction is ordinary.
Jul 9, 2011 at 14:20 comment added Álvaro Lozano-Robledo Chris, if I am not mistaken, the slopes of the Newton polygon will determine the valuations of the x-coordinates of the 37-torsion points. But even if the valuations of the roots of the polynomial of degree $666$ are $0$, that doesn't mean that the extension generated by the roots is unramified, does it? For instance, the slopes of $1+x+x^2$ are $[0,0]$ for $p=3$ but, of course, the prime $3$ ramifies in $\mathbb{Q}(\zeta_3)/\mathbb{Q}$.
Jul 9, 2011 at 8:05 comment added Chris Wuthrich You won't need to factor your diabolic polynomial; the slopes of the Newton polygon in $\mathbb{Q}_p[x]$ are enough to determine the ramification. In your case there are 666 unit roots and 18 of valuation $−1/18$.
Jul 8, 2011 at 20:22 history edited Álvaro Lozano-Robledo CC BY-SA 3.0
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Jul 8, 2011 at 20:20 answer added Felipe Voloch timeline score: 13
Jul 8, 2011 at 20:02 history asked Álvaro Lozano-Robledo CC BY-SA 3.0