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Jan 11, 2016 at 23:57 comment added roy smith technically we need also to rule out a hyperelliptic curve, since then there would be g(1,3)'s with base point. But among the curve of g(1,4)'s there is one whose general divisor does not contain a divisor of the g(1,2) so we get again that the genus is ≤ 3.
Jan 9, 2016 at 18:00 comment added roy smith at least for base point free pencils, how's this? If C admits maps of degrees both 3 and 4 to P^1, then the product map to P^1xP^1 is generically injective, so the genus of C is ≤ 6. The same argument (of Castelnuovo) shows a curve with base point free pencils of relatively prime degrees d,e has genus ≤ (d-1)(e-1).
Dec 19, 2015 at 11:26 vote accept Heitor
Dec 18, 2015 at 13:08 answer added Jason Starr timeline score: 5
Dec 18, 2015 at 12:15 history asked Heitor CC BY-SA 3.0