Let $H$ be a hyperellipitic curve of genus $g$ defined over $\mathbb{F}_q$. The Frobenius endomorphism operates on the divisor class group of $H$ and satisfies a characteristic polynomial $P\in\mathbb{Z}[T]$ of degree $2g$. This polynomial $P$ fulfils the equation $$P(T) = T^{2g} L(1/T),$$ i.e. $P$ is the reciprocal polynomial of $L$, where $L(T)$ denotes the numerator of the zeta-function of $H$ over $\mathbb{F}_q$.

Where do I find that statement (for citation)?

FYI: The statement can be found (without a proof) for example, as preprint of T. Lange, page 23, Theorem 2.35. In the case of elliptic curves ($g=1$), it can be found in the book of Koblitz "A course in number theory and cryptography" or Silverman "The Arithmetic of Elliptic Curves".

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    $\begingroup$ Weil, Courbes algébriques et Variétés abéliennes. $\endgroup$ – Felipe Voloch Nov 8 '11 at 11:56
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    $\begingroup$ And, you don't need to assume that the curve is hyperelliptic. $\endgroup$ – Felipe Voloch Nov 8 '11 at 12:01
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    $\begingroup$ E. Artin's thesis (cf. Collected Works) treated the hyperelliptic case, several years prior to Weil. One convenience there is quadratic reciprocity (at least in char not 2). $\endgroup$ – paul garrett Nov 8 '11 at 13:54
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    $\begingroup$ @paul: Artin certainly proved that $L(1)$ is the class number, but I am not sure about the full statement above which requires the action of Frobenius on the Jacobian, which may need the structure of the Jacobian as an abelian variety. $\endgroup$ – Felipe Voloch Nov 8 '11 at 17:50
  • $\begingroup$ @Felipe Ah, yes, what you say is surely right. Indeed, beyond genus 1, people should be clear on the fact that Weil certainly does deserve credit for legitimizing "Jacobians" in an algebraic setting (as opposed to the "classical" setting, whether the Italian school or Abel-Jacobi). $\endgroup$ – paul garrett Nov 11 '11 at 23:31

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