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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).

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    $\begingroup$ For smooth curves, by removing a closed point it is (easily seen to be) equivalent to show finiteness of the class group of the Dedekind domain $O_{K,S}$: the ring of $S$-integers of a global field $K$ with $S$ any non-empty finite set of places containing the archimedean places. But such a class group admits an idelic description, and so as such the desired finiteness (and finite generation of $O_{K,S}^{\times}$ with rank $\#S - 1$) is immediate from the compactness of the norm-1 idele class group that is proved uniformly over any global field in many books. $\endgroup$
    – nfdc23
    Commented Mar 28, 2018 at 1:04
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    $\begingroup$ Why would you expect an elementary proof beyond curves when there's no other definition than (rational points of the) "identity component of Picard scheme"? Only for curves is there an alternative concrete definition in terms of degree-0 Weil divisors up to linear equivalence. $\endgroup$
    – nfdc23
    Commented Mar 28, 2018 at 1:08
  • $\begingroup$ thanks! I'd be happy to accept either of these comments as an answer. (I was thinking in terms of coordinates in projective space and missed the adelic interpretation). $\endgroup$
    – dorebell
    Commented Mar 28, 2018 at 1:27
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    $\begingroup$ (I came across this argument on an exercise sheet here in Leiden.) For a smooth projective curve: you can use Riemann-Roch to show that every class is represented by a combination of prime divisors of bounded degree; such prime divisors correspond to points over an extension of bounded degree, so there are only finitely many of them. $\endgroup$
    – Martin Bright
    Commented Mar 28, 2018 at 8:17

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