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Let $S$ be an affine scheme of characteristic $p > 0$, let $E \rightarrow S$ be an elliptic curve over $S$, and let $F$ denote the absolute Frobenius. Since $E$ is its own $\mathrm{Pic}^0$ there is an identification $\mathrm{Lie}(E/S) = H^1(E, \mathscr{O}_E)$; suppose that both of these are free $\mathscr{O}_S$-modules of rank $1$. Let $D \in \mathrm{Lie}(E/S)$ be a translation invariant derivation and let $x \in H^1(E, \mathscr{O}_E)$ be the corresponding cohomology class. I would like to interpret $F^*(x) \in H^1(E, \mathscr{O}_E)$ in terms of $D$: is $$ F^*(x) = D^p, $$ where $D^p$ denotes the $p$-fold composition of $D$ with itself? Note that $D^p$ is again an invariant derivation because we are in characteristic $p$. It seems natural to guess that this equality should hold, but how does one prove it?

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  • $\begingroup$ Is $S$ affine? / Shouldn't $H^1(E, \mathcal{O}_E)$ be $R^1 f_* \mathcal{O}_E$? If $E=S$, we should have ${\rm Lie}(E/S)=0$... $\endgroup$ Apr 11, 2015 at 2:33
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    $\begingroup$ @PiotrAchinger: Sorry, I forgot the hypothesis you mention (which is non-essential if one is thinking about everything Zariski locally). Choosing $E = S$ as you suggest doesn't make sense though. $\endgroup$
    – Lisa S.
    Apr 11, 2015 at 3:01

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This is explained in the proof of Theorem 3 of section 15 of Mumford's "Abelian varieties" (pages 138-140 in the new edition) in the case when the base is an algebraically closed field of characteristic $p$. Suitably interpreted, the argument there continues to work over an arbitrary $\mathbb{F}_p$-scheme $S$.

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