Skip to main content
10 events
when toggle format what by license comment
Jun 12, 2012 at 6:47 comment added Harry The following MO thread might be relevant: mathoverflow.net/questions/77644/…
Jun 12, 2012 at 1:19 comment added Will Sawin One can find a map to $\mathbb P^n$ by choosing a divisor of sufficient degree and applying Riemann-Roch. But of course you really need a $K$-rational map, which requires a $K$-rational divisor of prime degree, which need not exist. So gonality isn't really the issue with this argument.
Jun 12, 2012 at 0:40 vote accept David Zureick-Brown
Jun 12, 2012 at 0:39 comment added David Zureick-Brown @Will: That would do it. What is the map to P^n?
Jun 11, 2012 at 23:36 comment added Will Sawin Are you sure your proof only works for prime gonality, and not when there exists nonconstant map of prime degree to $\mathbb P^1$. All curves have a nonconstant map of prime degree to $\mathbb P^1$, because they have a nonconstant map of prime degree to $\mathbb P^n$ and you can project down to $\mathbb P^1$.
Jun 11, 2012 at 23:35 answer added Olivier Benoist timeline score: 8
Jun 11, 2012 at 22:52 answer added user631 timeline score: 13
Jun 11, 2012 at 20:09 history edited David Zureick-Brown CC BY-SA 3.0
deleted 56 characters in body
Jun 11, 2012 at 19:57 history edited David Zureick-Brown CC BY-SA 3.0
typo, added a short remark
Jun 11, 2012 at 19:51 history asked David Zureick-Brown CC BY-SA 3.0