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Let X a smooth projective curve over $\mathbb{C}$. We fix $d$ distinct closed points $x_{1},\dots,x_{d}$.

Can we find a finite surjective morphism $\pi:X\rightarrow\mathbb{P}^{1}$ and local uniformizers on $\mathbb{P}^{1}$, $t_{1},\dots,t_{d}$

such that $\forall i, k[t_{i}^{-1}]\subset k[X-x_{i}]$


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1 Answer 1

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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 $\mathcal O(1) = \mathcal O(n x_i)$.

Thus $\mathcal O(nx_i)=\mathcal O(nx_j)$, so $x_i-x_j$ is an $n$-torsion divisor class, for all $i$, $j$.

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.

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  • $\begingroup$ Thanks, and if I allow two poles? Say I add one point $y$ to the d-uple. Can we have $k[t_{i}^{-1}]\subset k[X-\{y,x_{i}\}]$? $\endgroup$
    – prochet
    Nov 29, 2012 at 0:16
  • $\begingroup$ When is $y$ chosen? $\endgroup$
    – Will Sawin
    Nov 29, 2012 at 0:42
  • $\begingroup$ yes. Or even, can we take $y$ is one of the $x_{i}$ $\endgroup$
    – prochet
    Nov 29, 2012 at 1:07
  • $\begingroup$ You cannot take $y$ to be in $x_i$. Just take two independent points on an elliptic curve. $\endgroup$
    – Will Sawin
    Nov 29, 2012 at 4:09

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