I have been asked to show that a smooth plane quartic is never hyperelliptic. I know that
i) The genus of any such curve is 3
ii) The statements of Riemann-Roch and Riemann-Hurwitz
iii) A curve is hyperelliptic if there's a degree 2 map from it to P1.
Can I have a hint about what to do please? Cheers.
Best Answer
A more elementary way of seeing it is that if $f(x,y)=0$ is an affine equation for a smooth projective plane curve $X$ of degree $d\geq3$, then $$\left\{\frac{x^ry^sdx}{\partial f/\partial y}:0\leq r+s\leq d-3\right\}$$ is a basis for the holomorphic differential forms of $X$. Therefore the canonical map $X\to\mathbb{P}^{g-1}$ where $g=\frac{(d-1)(d-2)}{2}$ is the genus of $X$ can be seen as the map $[x:y:1]\mapsto[x^ry^s:0\leq r+s\leq d-3]$. In your case when $d=4$, this map is exactly (after a possible reordering) $[x:y:1]\mapsto[x:y:1]$; that is, the identity.
Now prove the above facts and use this to show that your curve cannot be hyperelliptic.