Non-hyperelliptic Riemann surface

Question

Let $S$ be the Riemann surface of the plane algebraic curve

$XYZ^3+X^5+Y^5 = 0$.

How can I prove that $S$ isn't a hyperelliptic Riemann surface?

Answer 1

For any closed subvariety $Y \subset \mathbb{P}^N$ which is a smooth complete intersection of hypersurfaces $F_1,\ldots,F_r$ of degrees $d_1,\ldots,d_r$, then there is a formula for the canonical bundle $\omega_Y$:

$\omega_Y = \mathcal{O}_Y(\sum_{i=1}^r d_i-N-1)$.

In particular, whenever the canonical bundle is positive, $\omega_Y$ is very ample. If we take the case $N = 2$, $r = 1$ then we see that this happens for a plane curve of degree $d$ if and only if $d \geq 4$. You have a smooth plane curve of degree $5$, so the canonical bundle is very ample.

On the other hand, a hyperelliptic curve (of genus $g \geq 2$) has a canonical bundle which is ample but not very ample: e.g. this is Proposition IV.5.2 in Hartshorne's Algebraic Geometry.

Thus no smooth plane curve of degree $d \geq 4$ is hyperelliptic.

I realize that your question is couched in the language of Riemann surface theory rather than algebraic geometry, so maybe you don't want an answer in such sheaf-theoretic language. If so, no problem -- please say so, and someone else will surely come along to leave another answer.