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Jun 28, 2013 at 3:08 vote accept Akhil Mathew
Jun 27, 2013 at 11:19 answer added Francesco Polizzi timeline score: 4
Jun 26, 2013 at 20:46 comment added Yusuf Mustopa Perhaps more seriously, the singular fibers in the construction I wrote down do not seem to give the union of two elliptic curves with a common point. If the plane cubic in question avoids the node of the singular conic, then the branched double cover I cooked up actually has 2 nodes and not 1, and if it does hit said node, then the preimage of each line cannot be an elliptic curve. It seems I was being quite careless :)
Jun 26, 2013 at 20:40 answer added Jason Starr timeline score: 4
Jun 26, 2013 at 20:13 history edited Akhil Mathew CC BY-SA 3.0
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Jun 26, 2013 at 20:12 answer added Dan Petersen timeline score: 2
Jun 26, 2013 at 20:12 comment added Akhil Mathew @YusufMustopa: I don't think this can work, because the two-fold cover from $C$ to two $\mathbb{P}^1$'s glued together at a node is not flat, which seems to make it difficult to construct such families by such a method. (I'll add this to the original question.)
Jun 26, 2013 at 19:45 comment added Yusuf Mustopa Have you tried taking a linear subspace $\Lambda$ of the $\mathbb{P}^{5}$ of plane conics which avoids double lines (e.g. $\Lambda$ a general element of $\mathbb{G}(2,|\mathcal{O}_{\mathbb{P}^{2}}(2)|)$), looking at the total family $\mathcal{C} \rightarrow \Lambda,$ and taking a double cover of $\mathcal{C}$ branched over a section of the divisor that comes from the intersection of all these plane conics with a fixed plane cubic? I'm pretty sure you could arrange for the latter to be tangent to some of the conics at exactly one point, and to miss the node of most of the singular conics.
Jun 26, 2013 at 19:31 history asked Akhil Mathew CC BY-SA 3.0