1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Starr Feb 23 '15 at 16:26
  • 5
    $\begingroup$ 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 Arapura Feb 23 '15 at 16:47
  • $\begingroup$ @DonuArapura Could you give me a reference published somewhere? Thanks a lot. $\endgroup$ – Syed Feb 23 '15 at 17:19
  • $\begingroup$ @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? $\endgroup$ – Syed Feb 23 '15 at 19:13
  • 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 Arapura Feb 23 '15 at 19:42
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.