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?



Interpreting Frobenius pullback as an invariant differential in the case of an elliptic curve
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 $S$.


