Frobenius action on the Picard group

Question

I have been concerned for the better part of today with the following problem: let $X/\mathbb{F}_q$ be a geometrically connected smooth projective curve. The trinity of Frobenii on $X_{\overline{\mathbb{F}_q}}$ consists of

1. The absolute Frobenius $F$
2. The relative Frobenius $F_r=F_{X/\mathbb{F}_q}\times\text{id}$
3. The geometric Frobenius $F_g=\text{id}\times\text{Frob}^{-1}$

It is known that $F$ acts trivially on all étale cohomology groups of $X_{\overline{\mathbb{F}_q}}$; in particular, it acts trivially on $\text{Pic}(X_{\overline{\mathbb{F}_q}})$. Moreover, $F=F_r\circ F_g=F_g\circ F_r$, so the pullback $F_r^*:\text{Pic}(X_{\overline{\mathbb{F}_q}})\rightarrow\text{Pic}(X_{\overline{\mathbb{F}_q}})$ should be an isomorphism. This bugs me because I thought that $F_r$ was a degree $q$ map between curves! This would entail that no degree $1$ divisor is in the image of $F_r^*$. Surely there is some mistake in my argument, but I haven't managed to spot it for myself :(

Answer 1

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