Show that a smooth plane quartic is never hyperelliptic

Question

I have been asked to show that a smooth plane quartic is never hyperelliptic. I know that

1. The genus of any such curve is 3
2. The statements of Riemann-Roch and Riemann-Hurwitz
3. A curve is hyperelliptic if there's a degree 2 map from it to $\Bbb P^1$.

Can I have a hint about what to do please? Cheers.

Answer 1

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.

Answer 2

This is exercise IV.3.2 in Hartshorne. There are two main steps:

1. Show that the effective canonical divisors are hyperplane sections. (Hint: $X$ is a complete intersection.)
2. Show that if $D$ is an effective, degree $2$ divisor, then $\dim |D| = 0$. (Hint: consider the line through the two points in the support of $D$, or the tangent line if $D = 2P$. Use the criterion for very ampleness at the beginning of the chapter combined with #1.)

Answer 3

Thanks all, I have been able to answer this now, based upon the hints of bothe Robert and Andrew. My strategy was:

1. Quote the following theorem: A smooth projective curve $V$ of genus $g>1$ is non-hyperelliptic if the canonical map $f: V \mapsto P^{g-1}$ is an embedding.

2. Observe that $K_v$ in this case is $(d-3)H$ for some hyperplane $H$ so the linear system $L(K_V)$ has basis ${1, X1/X0, X2/X0}$, the genus of the curve being 3. This induces a morphism into $P^2$ which is seen to essentially be the identity, so we are done by the theorem in part 1.

Answer 4

It suffices to show that the plane quartic curve $C$ is a canonical curve. Using adjunction formula, $K_C=i^*_C(\mathcal{O}_{\mathbb{P}^2}(-3)\otimes\mathcal{O}_{\mathbb{P}^2}(4))=i_C^*\mathcal{O}_{\mathbb{P}^2}(1)$. So $i_C: C\hookrightarrow\mathbb{P}^2$ is a canonical embedding.