This is an exercise given during the course in Riemann Surfaces that I attended this year. Let $X$ be a compact Riemann Surface that is a degree $3$ cover of $\mathbb{P}^1(\mathbb{C})$ given by $y^3=f(x)$ with $f\in\mathbb{C}[x]$ a polynomial of degree $5$ or $6$ with distinct zeroes. Determine the genus of $X$ and the fiber over $(0:1) \in\mathbb{P}^1(\mathbb{C})$. Surely I know that I can compute the genus of the curve using Riemann-Hurwitz, but I don't know how to compute the ramification. Surely, over the affine part of $\mathbb{P}^1(\mathbb{C})$, every root of $f$ determines a totally ramified point, but what about the fiber of $(0:1)$. Is it a branch point?

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The genus will be $4$.

There will be points of ramification at the roots of the polynomial (and at infinity in the case of degree $5$). Each point will have a three fold ramification, $r_i=3$ so by the Riemann-Hurewitz formula we have

$$2(1-\gamma)=2(1-g)n-\sum(r_i-1)=2\cdot 3-2\cdot 6=-6$$ so $\gamma =4$ where of course $n=3$ and $g=0$.

Remark that if $f(z)$ is an analytic function and $f(0)\neq 0$ then $\sqrt[n]{f(z)}$ has $n$ distinct well defined branches about zero, corresponding to the $n$ roots of $f(0)$.

Now if $f(z)= (z-a)(z-b)(z-c)(z-d)(z-e)(z-f)$ then we can set $z=a+w^3$ and we will get

$$\sqrt[3]{f(z)}=w\sqrt[3]{ (w^3+(a-b)) (w^3+(a-c)) (w^3+(a-d)) (w^3+(a-e)) (w^3+(a-f)) }$$ and the root is well defined about zero by the above remark. so with this coordinatization $y$ is well defined so the ramification is $3$. So the roots of the polynomial are all ramification points, as remarked above.

To investigate infinity set $z=\frac{1}{w}$ then we have

$$y=\frac{1}{w^2}\sqrt[3]{(1-az)(1-bz)(1-cz)(1-dz)(1-ez)(1-fz)}$$ and this is well defined so $z=\frac{1}{w}$ is a cordinatization at $\infty$ and $\infty$ is unramified. The function has a pole of order $2$ here.

If we try the same trick with a $5$th degree polynomial we get

$$y=\frac{\sqrt[3]{w}}{w^2}\sqrt[3]{(1-aw)(1-bw)(1-cw)(1-dw)(1-ew)}$$ we see this is ramified, so we must set $z=\frac{1}{w^3}$ and get

$$y=\frac{1}{w^5}\sqrt[3]{(1-aw^3)(1-bw^3)(1-cw^3)(1-dw^3)(1-ew^3)}$$ which is well defined with a pole of order $5$. In the case of a fifth degree polynomial the point at infinity is ramified, it takes the role of one of the roots.

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  • $\begingroup$ The notation (0:1) stands for projective coordinates on the projective line. If you use standard identification with the Riemann sphere then thus point is called $\infty$. $\endgroup$ – Moishe Kohan Jul 15 '14 at 17:22
  • $\begingroup$ Yes, $(0:1)$ is the point at infinity of the projective line. Why is the genus 2? I guess it's by Riemann-Hurwitz but I don't understand why you say that infinity ramifies only in case of degree $5$. $\endgroup$ – Simone Jul 15 '14 at 17:54
  • $\begingroup$ I thought the y was squared, ill submit an improved answer in an hour or so. $\endgroup$ – Rene Schipperus Jul 15 '14 at 19:59
  • $\begingroup$ I edited to give a detailed answer. $\endgroup$ – Rene Schipperus Jul 15 '14 at 21:40

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