TITLE: Why is $F^*(\mathcal L) = \mathcal L^{\otimes p}$ where $F$ is the absolute Frobenius and $\mathcal L$ is an invertible sheaf?
QUESTION [13 upvotes]: Suppose $S$ is such that $\mathcal O_S$ is killed by multiplication by $p$. The absolute Frobenius $F: S \to S$ is defined to be be the identity on the underlying points topological space of $S$, with the sheaf map $F^\#: \mathcal O_S \to \mathcal O_S$ given by $x \mapsto x^p$ for any section.
The pullback sheaf is defined to be
$$F^*(\mathcal L) = F^{-1} L \otimes_{F^{-1}\mathcal O_S} \mathcal O_S$$
Where $F^{-1} \mathcal L$ is the sheafification of the presheaf defined by 
$$U \mapsto \lim_{F(V)\supset U}  \mathcal L (V)$$
But since $F$ is the identity on the underlying space, won't we have $F^{-1} \mathcal L = \mathcal L$ and $F^{-1} \mathcal O_S$ = $\mathcal O_S$, and hence $F^* \mathcal L = \mathcal L$? I can't see how to bring $F^\#$ into the calculation.
I know the intuitive answer is that the transition functions are sent to their p-th powers. But I can't work out the details.

REPLY [5 votes]: Even more simple, and more general:

Let $S$ be a ringed space of $\mathbb{F}_p$-algebras, and let $F : S \to S$ be the absolute Frobenius. Then for every line bundle $\mathcal{L}$ on $S$ we have $F^* \mathcal{L} \cong \mathcal{L}^{\otimes p}$.

Proof. Since $F^*$ is left adjoint to $F_*$ (which could be seen as a definition), for every sheaf of modules $M$ on $S$ we have to prove $\hom(\mathcal{L},F_* M) \cong \hom(\mathcal{L}^{\otimes p},M)$, naturally in $M$. For $\mathcal{L}=\mathcal{O}_S$ this is trivial, both sides identify with $\Gamma(M)$, and in general $\mathcal{L}$ looks locally like this. So we only have to observe that the isomorphisms glue, i.e. that for every isomorphism $\gamma : \mathcal{O}_S \to \mathcal{O}_S$ resp. unit $\gamma \in \Gamma(\mathcal{O}_S)^*$ the diagram
$$\begin{matrix}
\hom(\mathcal{O}_S,F_* M) & \cong & \Gamma(F_* M) & = & \Gamma(M) & \cong & \hom(\mathcal{O}_S^{\otimes p},M) \\
\gamma^* \downarrow  ~ & & ~ \downarrow \cdot \gamma ~ & & ~ \downarrow \cdot \gamma^p & & ~ \downarrow (\gamma^{\otimes p})^* \\\hom(\mathcal{O}_S,F_* M) & \cong & \Gamma(F_* M) & = & \Gamma(M) & \cong & \hom(\mathcal{O}_S^{\otimes p},M)
\end{matrix}$$
commutes, which can be checked directly by the definition of $F$.