Any curve of genus three is either hyperelliptic or trigonal?

Question

A curve $C$ is said to be trigonal if it admits a rational function $f: C \to \mathbb{CP}^1$ of degree $3$. Is any nonsingular plane curve of degree four trigonal? Can the map be chosen so that it has exactly $10$ ramification points?

Answer 1

For the first question, the answer is yes. We can choose a point $P$ on the nonsingular plane curve $C$ of degree 4. Project $C$ from $P$ onto a hyperplane (i.e. a copy of $\mathbb{P}^1$) in $\mathbb{P}^2$. Noting that any line passing through $P$ cuts $C$ at other three points, this projection is a rational map of degree 3.

Answer 2

The answer to both of the questions is yes.

Construct an $($almost everywhere$)$ $3$-sheeted covering map by choosing a point on our curve $C$ and a hyperplane not passing through it, and projecting from the point to the hyperplane. If we worked out the coordinates, this would clearly be given by a rational function, and it has degree $3$ by Bézout's Theorem. By Riemann-Hurwitz, which applies in full generality by using the $$e_\varphi(P) = v_P(\varphi^*t_{\varphi(P)})$$"vanishing of defining polynomial $P$ of $C$ along the projection line" definition, we have that$$\sum e_\varphi(P) - 1 = 4 + 3 \cdot 2 = 10,$$so we have exactly $10$ ramification points when there is no higher order ramification. Ramification points of order higher than $2$ correspond to degree $3$ or higher vanishing of $P$ along a projection line, i.e. tangency of the projection line at an inflection point. There are finitely many of these, so a generic choice of point and hyperplane will not any higher ramification, and the desired properties.