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Jun 3, 2011 at 23:43 comment added Sándor Kovács No. $C'\times C'$ clearly maps to that Kummer. I was confused by something else, but I see the mistake I made. Sorry.
Jun 3, 2011 at 23:41 comment added Jorge Vitório Pereira Very nice! It is perhaps simpler to start with a genus 2 curve with Jacobian isogeneous to the square of an elliptic curve. Since these abelian varieties have infinitely many elliptic curves with self-intersection zero through 0, theirs blow-ups have infinitely many negative elliptic curves.
Jun 3, 2011 at 21:40 history answered Dmitri Panov CC BY-SA 3.0