This question comes from Hartshorne's Algebraic Geometry. I'm trying to show that given an elliptic curve $E \to \operatorname{Spec} k$, $\operatorname{char} k = p>0$, that the Hasse invariant of $E$ is one if and only if the dual Frobenius morphism (I think sometimes called the relative Frobenius) $\hat F' :X \to X_{(p)}$ is separable. The hint is to use that the tangent space to an elliptic curve is the tangent space to the Jacobian (isomorphic to $E$), which is just $H^1(X,\mathcal{O}_X)$.
I can use Lemma 50.12.1 from the Stacks Project to conclude that $\hat F' :X \to X_{(p)}$ is separable if and only if $\hat F'^* \Omega_{X_{(p)}} \to \Omega_X$ is nonzero, which now by Hartshorne Proposition IV.2.1 this happens if and only if the sequence $$0 \to \hat F'^* \Omega_{X_{(p)}} \to \Omega_X \to \Omega_{X/X_{(p)}} \to 0$$ is exact.
If I can use this to get a nonzero map on cohomology $H^0(X,\Omega_{X}) \to H^0(X,\Omega_X)$, I can get the desired map using Serre Duality, which is the definition of nonzero Hasse invariant. However I'm not sure how to get the map on cohomology above.