finite surjective morphism to the projective line - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T07:55:50Z http://mathoverflow.net/feeds/question/114819 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114819/finite-surjective-morphism-to-the-projective-line finite surjective morphism to the projective line prochet 2012-11-28T22:43:57Z 2012-11-28T23:01:49Z <p>Let X a smooth projective curve over $\mathbb{C}$. We fix $d$ distinct closed points $x_{1},\dots,x_{d}$.</p> <p>Can we find a finite surjective morphism $\pi:X\rightarrow\mathbb{P}^{1}$ and local uniformizers on $\mathbb{P}^{1}$, $t_{1},\dots,t_{d}$</p> <p>such that $\forall i, k[t_{i}^{-1}]\subset k[X-x_{i}]$</p> <hr> http://mathoverflow.net/questions/114819/finite-surjective-morphism-to-the-projective-line/114823#114823 Answer by Will Sawin for finite surjective morphism to the projective line Will Sawin 2012-11-28T23:01:49Z 2012-11-28T23:01:49Z <p>No. This condition implies that $t_i$, pulled back to $X$, vanishes only at $x_i$. Thus, $x_i$ is the entire fiber of the point where $t_i$ vanishes, so if $n$ is the degree of the map, then $\mathcal O(1) = \mathcal O(n x_i)$.</p> <p>Thus $\mathcal O(nx_i)=\mathcal O(nx_j)$, so $x_i-x_j$ is an $n$-torsion divisor class, for all $i$, $j$.</p> <p>To show this is impossible, it suffices to find two points $x_1$ and $x_2$ on a curve, such that $x_1-x_2$ is not a torsion divisor. To do this, let $E$ be an elliptic curve, let $x_1$ be the identity, and let $x_2$ be a non-torsion point. Then $x_1-x_2$ is not a torsion divisor class.</p>