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Mar 28, 2018 at 8:17 comment added Martin Bright (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.
S Mar 28, 2018 at 1:27 history suggested nfdc23 CC BY-SA 3.0
inserted missing hypotheses for existence of Picard scheme; be careful!
Mar 28, 2018 at 1:27 comment added dorebell 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).
Mar 28, 2018 at 1:08 comment added nfdc23 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.
Mar 28, 2018 at 1:05 review Suggested edits
S Mar 28, 2018 at 1:27
Mar 28, 2018 at 1:04 comment added nfdc23 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.
Mar 28, 2018 at 0:17 history asked dorebell CC BY-SA 3.0