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I've read in some references (some of them important, like page $328$ from Heights in Diophantine Geometry, by Bombieri and Gubler) that André Weil proved in his PhD thesis that the rank of an abelian variety (over a number field) is finite.

I've read Weil's thesis and what I found was this (I'm paraphrasing):

If $C$ is a plane algebraic curve over a number field $K$, with arbitrary genus, then its jacobian variety $J(C)$ has a structure of a finitelly generated abelian group.

I don't know much about abelian varieties, but I have the feeling that what Weil proved was not in that level of generality mentioned in the book (i.e., for any abelian variety).

Is it true that the case of abelian varieties can be reduced to that of the Jacobian of a plane curve?

If it is, how so?

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It is true that every abelian variety $A$ over a number field $K$ is a quotient $J\to A$ of a Jacobian, but this doesn't show that $J(K)$ finitely generated implies $A(K)$ finitely generated because the map $J(K)\to A(K)$ will not in general be surjective. However, every abelian variety $A$ is isogenous to a direct factor of a Jacobian $J$ (Poincare reducibility), and this can be used to deduce the Mordell-Weil theorem for $A$ from $J$. Normally the Mordell-Weil theorem is proved directly without using Jacobians. The statement in Bombieri and Gubler is inaccurate (but common).

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Every abelian variety over a number field is a quotient of a Jacobian. The basic idea is that for a curve $C$ lying on the abelian variety $A$, the inclusion $C \subset A$ induces (via Albanese functoriality) a surjection $J(C) \twoheadrightarrow A$. More details can be found in Chapter III.10 of Milne's notes on abelian varieties.

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  • $\begingroup$ But the curve $C\subset A$ may not be plane, right? The curves mentioned in Weil's thesis are all plane $\endgroup$ – rmdmc89 Dec 15 '17 at 12:50
  • $\begingroup$ Could we argue that $C$ is always birrationally equivalent to some plane curve? $\endgroup$ – rmdmc89 Dec 15 '17 at 13:57

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