Let $C$ be an arbitrary smooth plane quartic defined over a number field $K$. Assume $C$ is not hyperelliptic, and denote by $J$ the Jacobian of $C$. How does $\text{End}(J)$ look like for a generic curve $C$?
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2$\begingroup$ A smooth plane quartic is never hyperelliptic. $\endgroup$– SashaCommented Jan 11 at 15:22
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2$\begingroup$ (More generally, the gonality of a smooth plane curve of degree $d$ is $d-1$, given by projection from a point on the curve. This is due to Max Noether.) $\endgroup$– R. van Dobben de BruynCommented Jan 11 at 15:51
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2$\begingroup$ $\operatorname{End}(A)=\mathbb{Z} $ for a general principally polarized abelian variety of dimension $g$, hence in particular when $A$ is the Jacobian of a general plane quartic. $\endgroup$– abxCommented Jan 11 at 15:53
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$\begingroup$ Dear @abx, do you have any reference for this statement? I couldn't find one. $\endgroup$– kindasortaCommented Jan 11 at 16:23
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3$\begingroup$ One reference (giving actually the result for a generic Jacobian, in any characteristic) is: S. Mori, The endomorphism rings of some abelian varieties, Japan. J. Math. 2, no. 1 (1976). There is probably a simpler reference for complex abelian varieties. $\endgroup$– abxCommented Jan 11 at 17:33
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1 Answer
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See https://arxiv.org/abs/math/0405156 where Del Pezzo surfaces of degree 2 are used, in order to construct plane quartics, whose jacobians have endomorphism ring $\mathbb{Z}$.
