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Mar 22, 2010 at 18:20 history edited Robin Chapman CC BY-SA 2.5
minor correction and slight rewording
Mar 22, 2010 at 18:18 comment added Bjorn Poonen I thought I would give the precise reference being discussed above: the error pointed out by Robin is in the corollary following Theorem 6 on page 176 of the second edition of Lang, Elliptic functions, and the error in the proof is where he says "it follows from Appendix 1 that the only automorphisms of A are...".
Mar 22, 2010 at 15:05 comment added Tommaso Centeleghe I agree with you, Lang contains a mistake. Thanks for having thought about the question.
Mar 22, 2010 at 15:01 vote accept Tommaso Centeleghe
Mar 22, 2010 at 15:01
Mar 22, 2010 at 12:19 comment added Robin Chapman I now think Lang is wrong, since $E$ could have an automorphism of order $3$, $4$ or $6$. But then $E$ has $j$-invariant $0$ or $1728$ and so has a model over $\mathbb{F}_p$. Over $\mathbb{F}_{p^2}$ this model will have $(p+1)^2$ points.
Mar 22, 2010 at 12:12 comment added Robin Chapman The proof is in Lang's Elliptic Functions. Briefly speaking, given a supersingular $E$ defined over $\mathbb{F}_{p^2}$ the Frobenius endomorphism and $p$ are both totally inseparable and so differ by a factor of an automorphism. If $p\ge 5$ the only automorphism is $\pm1$ (according to Lang who doesn't quite convince we :-( ). So the Frobenius is $+p$ or $-p$ and that of the quadratic twist is the other.
Mar 22, 2010 at 11:37 comment added Tommaso Centeleghe Thank you for your answer. Even though I do not see a complete proof of the existence of this k-structure I was asking. Given the fact that there exists a model E' of E over k, how do you show that you can pick E' so that the relative Frobenius is -p?
Mar 22, 2010 at 11:33 history edited Robin Chapman CC BY-SA 2.5
spelling correction
Mar 22, 2010 at 11:28 history answered Robin Chapman CC BY-SA 2.5