Let $X$ be a variety over a finite field $k = \mathbb{F}_q$ which is nice enough that its Picard variety exists and $\mathrm{Pic}^0_{X/k}$ is of finite type (e.g. $X$ projective with $X(k) \neq \emptyset$ works, and probably the argument works a bit more generally). Then the divisor class group of $X$ is $\mathrm{Pic}^0_{X/k}(k)$, which is finite because $k$ is finite and $\mathrm{Pic}^0_{X/k}$ is of finite type over $k$.

This argument is short and conceptual, but it's hiding a lot of non-trivial work in constructing the Picard variety. Is there an easier proof of this useful fact? (I'd be happy even if the proof just works for curves).

Let $X$ be a geometrically integral (or geometrically reduced and geometrically connected) proper scheme over a finite field $k = \mathbb{F}_q$, so its Picard scheme exists and $\mathrm{Pic}^0_{X/k}$ is of finite type (e.g. $X$ projective with $X(k) \neq \emptyset$ works, and probably the argument works a bit more generally). By definition the divisor class group of $X$ is $\mathrm{Pic}^0_{X/k}(k)$, which is finite because $k$ is finite and $\mathrm{Pic}^0_{X/k}$ is of finite type over $k$.

This argument is short and conceptual, but it's hiding a lot of non-trivial work in constructing the Picard scheme. Is there an easier proof of this useful fact? (I'd be happy even if the proof just works for smooth curves).