Let $C$ be a smooth, projective curve of genus 2. I want to show that there exists a non-constant rational function $f \in k(C)$ having divisor of the form $$(f) = P_1 + P_2 - P_3 - P_4 $$for points $P_1,P_2,P_3,P_4$. I am given the hint to consider two rational functions $f_1, f_2 \in L(K)$ (the Riemann-Roch space) where $K$ is a canonical divisor on the curve. I know that $\dim L(K) = 2$, so these non-constant $f_1,f_2$ exist. Moreover, I can derive that $\deg(K)=2$. How do I use the latter to solve this? I thought about taking $K = P_3 + P_4$ as that would give me $f_1$ and $f_2$ both with simple poles at $P_3$ and $P_4$. Does this mean that I can take $f = \frac{f_1}{f_2}$ since it doesn't say that the $P_i$ should be different?

  • $\begingroup$ Of course the points $P_i$ should be different: else you could take $div(1)=0=P+P-P-P$ ! $\endgroup$ – Georges Elencwajg Jan 25 '16 at 22:45
  • $\begingroup$ Is it there exist points P or for all P's there is a function ? $\endgroup$ – Rene Schipperus Jan 25 '16 at 22:49
  • $\begingroup$ Rene, it is "there exist points P". Georges, you are absolutely right. Unfortunately that only shows how bad my understanding of these concepts is. Any hint on how I should approach this problem? $\endgroup$ – user307988 Jan 25 '16 at 22:52

Take two linearly independent rational differential forms $\omega_1, \omega_2\in L(K)=\Omega^1(C)$ with divisors $\operatorname {div} (\omega_1)=P_1+P'_1$ and $\operatorname {div} (\omega_2)=P_2+P'_2$ .
The rational function $f=\frac {\omega_1}{\omega_2}\in \operatorname {Rat}(C)$ then has a divisor of the required form $$\operatorname {div}(f)=P_1+P'_1-P_2-P'_2 $$

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    $\begingroup$ How do you know all your P's are different ? $\endgroup$ – Rene Schipperus Jan 25 '16 at 22:55
  • $\begingroup$ What confuses me still is why we can take $div(\omega_1) = P_1 + P_1'$. Is it because a differential on $C$ has the form $g \omega$ for some $g$ and $K = div(\omega)$ so that $\deg(div(\omega_1)) = \deg(div(g \omega)) = \deg(div(g)) + \deg(div(\omega)) = 2$ implies that two such points exist? $\endgroup$ – user307988 Jan 25 '16 at 23:06
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    $\begingroup$ Because $l(K)=2$ one of the points of the canonical series can be chosen arbitrary. $\endgroup$ – Rene Schipperus Jan 25 '16 at 23:09
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    $\begingroup$ Oh I see, essentially we are saying that the canonical series has no base point, and your proof is that if so all points would be equivalent and the curve would be rational. (or a contradiction with R-R). $\endgroup$ – Rene Schipperus Jan 25 '16 at 23:13
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    $\begingroup$ Dear @Rene: with my construction it might indeed happen that $P_i=P'_i$ but I'm sure that $P_1\neq P_2$ because else $P'_1\sim P'_2$, which is impossible for a curve of positive genus. But I can also arrange that all four points $P_1,P_2,P'_1,P'_2$ are different by regarding $f$ as a map $F:C\to \mathbb P^1$ and taking the divisor $D=F^*(p)-F^*(p')$ , where $p\neq p'\in \mathbb P^1$ are two non-branch points for $F$. If $g$ is a rational function on $\mathbb P^1$ with a zero at $p$ and a pole at $p'$ then for the the lift of $g$ to $C$ we have $\operatorname {div}(F^*g)=D=P_1+P_2-P'_1-P'_2$ $\endgroup$ – Georges Elencwajg Jan 25 '16 at 23:21

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