I searched for it for a long time, but it seems that everybody is taking this for granted and does not bother to point out a proof. Would it be possible that someone points me to a proof or makes me see the obvious.
Thank you very much.
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$\begingroup$ What is your definition of "generic"? I expect that there are countably many subvarieties in the moduli space parameterizing curves that admit a positive degree morphism to some genus 1 curve. $\endgroup$– Jason StarrCommented Feb 23, 2015 at 16:26
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6$\begingroup$ Of course, Jason is correct. However, a very general point of $M_g$ (by which I mean, exclude a countable union of proper sub varieties) corresponds to a curve with a simple Jacobian. This curve can't map onto anything of smaller genus. (This is over $\mathbb{C}$.) $\endgroup$– Donu ArapuraCommented Feb 23, 2015 at 16:47
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$\begingroup$ @DonuArapura Could you give me a reference published somewhere? Thanks a lot. $\endgroup$– SyedCommented Feb 23, 2015 at 17:19
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$\begingroup$ @DonuArapura because the set of morphism is a map between this union of sub varieties to $\cup M_{g_i<g}$ ? Is this the classical proof? $\endgroup$– SyedCommented Feb 23, 2015 at 19:13
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1$\begingroup$ For your first question, it follows from from Cor. 17.5.2 of Birkenhake and Lange's book on Complex Abelian Varieties (2nd ed), but I'm sure there are more suitable references. $\endgroup$– Donu ArapuraCommented Feb 23, 2015 at 19:42
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OK, here's an answer. As I said in the comments, if the $C$ is a (smooth projective) curve with a simple Jacobian, then it can't map onto a curve of smaller genus. Since I was a bit curious myself, I found a reference for the next step: Koizumi "The ring of correspondences on a generic curve of genus g" Nagoya (1976). The result is better than I expected! Koizumi proves that if $C$ is the geometric generic curve of $M_g$ over the algebraic closure of the prime field, then $End(J( C) =\mathbb{Z}$; so in particular, it is simple.
answered Feb 23, 2015 at 21:26