Cohomology map induced by Frobenius morphism

Question

Suppose $X$ is an elliptic curve over a field $k$ of characteristic $p>0$. Let $F\colon X\to X$ be the Frobenius morphism. Then $F$ induces a map $F^*\colon H^1(X,\mathcal{O}_X)\to H^1(X,\mathcal{O}_X)$ on cohomology. In Hartshorne chapter 4 page 332, he claimed that this map is not linear, but it is $p$-linear, namely $F^*(\lambda a)=\lambda^pF^*(a)$ for all $\lambda\in k,a\in H^1(X,\mathcal{O}_X)$. Why the map of cohomology is not linear and is $p$-linear?

Answer 1

In general, I think it is a good way to practice with the definition of cohomology. Since your $X$ is a variety over $k$ (no need to assume $X$ to be elliptic here), cohomology of quasi-coherent sheaves can be computed by Cech cohomology. Writing things down and you see that's why $F$ is $p$-linear; for instance, $X = \mathbb{P}^1_k$.

Edit: Thanks to the comment by Hank Scorpio, I just realized that we are using the absolute Frobenius morphism, but the story goes in the same way because affine locally (by making an usage of Cech cohomology), you Frobenius morphism is $A \longrightarrow A, a \longmapsto a^p$