Let $S$ be a 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 $\mathrm{Lie}(E/S) = H^1(E, \mathscr{O}_E)$. 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?

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?