I am struggling to prove the following statements
- A nonsingular plane curve of degree $5$ has no $g^1_3$.
- There are nonhyperelliptic curves of genus $6$ which cannot be represented as a nonsingular plane quintic curve.
I assume the second statement is shown by proving that there exists a nonhyperelliptic curve of genus $6$ with a $g^1_3$, but I do not know how to construct such a curve.
Best Answer
First, I answer the second part following your suggestion: by using the first part, one only needs to construct a trigonal curve of genus 6. It is well known that there are trigonal curves for every genus $\ge 3$: for example by considering cyclic covers of $\mathbb{P}^1$, i.e. given by an equation of the form $$y^3= \prod_{i=1}^s (x-\alpha_i)^{r_i},$$ where $r_i=1$ or $2$. If $\sum_{i=1}^s r_i\equiv 0 \pmod{3}$, then the place at infinity does not ramify, and the ramification index at $x=\alpha_i$ is always $e_i=3$. Hurwitz formulae gives then that $g=s-2$. For example, in our case we can take the equation $$y^3=(x^7+1)(x-1)^2.$$
Now for the first part: suppose we have $X$ a smooth planar curve of degree 5. Then we have that $\omega_X \cong \mathcal{O}_X(2)$, so $H^0(\mathbb{P}^2, \mathcal{O}(2)) \to H^0(X, \omega_X)$ is an isomorphism. So the canonical embedding of $X$ is isomorphic to the composition $X \to \mathbb{P}^2 \stackrel{\phi}{\longrightarrow} \mathbb{P}^{5}$ where $\phi$ is the $2$-fold Veronese, given as $$\phi([x_0:x_1:x_2])=[x_0^2: x_0x_1: x_1^2: x_0x_2: x_1x_2: x_2^2].$$
Now, suppose we have $\psi: X\to \mathbb{P}^1$ be a degree 3 morphism, and consider a "general" fiber $(p_1,p_2,p_3)$ (so $\psi(p_i)=\psi(p_j) \forall i,j$). Denote by $D=p_1+p_2+p_3$ the effective divisor; we have then that $\dim |D|=1$, so by Riemann-Roch, $\dim |K-D|=3$ ($K$ denotes a canonical divisor). Since $\phi(X)$ is canonical, this means that the the dimension of the linear system of hyperplanes in $\mathbb{P}^5$ that contains $\phi(p_1)$, $\phi(p_2)$ and $\phi(p_3)$ is 3, so the three points are colinear (compare with Hartshorne, Chapter IV, Example 5.5.3).
But for a "generic" fiber this is not possible: we can choose one such that $p_i=[1:p_{i 1}: p_{i 2}]$ and such that $p_{i 1}\ne 0$ for all $i=1,2,3$ and distinct. But then $$\phi(p_i)=[1:p_{i 1}:p_{i 1}^2:\cdots]$$ and this points are clearly not colinear (by "Vadermonde").