I am not an expert in elliptic curves at all, so my question may naive and/or obvious. Let $E$ be an (affine) elliptic curve defined over a finite (or perfect) field of characteristic $p$. Since its module of derivations is free of rank one over its coordinate ring $R$, we can choose a generator (unique up to a constant) $\delta$, what I call the `invariant derivative'. In particular, $\delta^p$, being again a derivative, must be a multiple of $\delta$. My question is, when is $\delta^p$ actually zero. I verified this for $j=0$ and $j=1728$, and it seems in either case, this exactly happens when $E$ is supersingular, respectively when $p\equiv 2\mod 3$ and $p\equiv3\mod 4$, at least for some low values of $p\geq 5$. Is this the pattern in general?
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$\delta^p = A\delta$ where $A$ is the Hasse invariant. In particular $\delta^p = 0$ if and only if $A=0$, i.e. $E$ is supersingular.
answered Sep 9, 2015 at 20:13
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1$\begingroup$ Strictly speaking, $A$ is the Hasse invariant attached to the pair $(E, \omega)$ where $\omega$ is the global 1-form dual to $\delta$. $\endgroup$– grghxyCommented Sep 10, 2015 at 0:52
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$\begingroup$ I guessed that something like that must hold, but I couldn't find a reference. Do you have any? $\endgroup$ Commented Sep 10, 2015 at 2:45
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$\begingroup$ @Hans Schoutens: Isn't that just the definition of the Hasse invariant? $\endgroup$– ericCommented Sep 10, 2015 at 10:24