Map from X to $\mathbb P^1_k$ of degree $3$ with $X$ curve of genus $3$. Let $X$ be a curve over $k$ algebraically closed field. Here $X$ is a proper, smooth, connected, $\text{dim}(X)=1$ scheme over $k$.
Suppose that $g(X) = 3$ with $g(X) = \text{dim}_k H^1(X,\mathcal O_X)$ the genus of $X$.
We say that $X$ is hyperelliptic if $g(X)>1$ and there exists a map $X \rightarrow \mathbb P^1_k$ of degree $2$.
Prove the following:
$X$ is not hyperelliptic $\Longleftrightarrow$ there exists a map $X\rightarrow \mathbb P^1_k$ of degree $3$.
($\Rightarrow)$ Take the canonical embedding $X\rightarrow \mathbb P^2_k$ of degree $4$ associated to the canonical divisor. Then $X$ is a quartic in $\mathbb P^2_k$ and we can find a morphism from $X$ to $\mathbb P^1_k$ of degree $3$ using the derivatives of the quartic polynomial that defines $X$.
Any ideas for the other implication?
 A: By contrapositive, it is enough to show that if $X$ is hyperelliptic, then there is no map $X\to\Bbb P^1$ of degree 3. Suppose that there were a map of degree three, and let $D$ be the pullback of the hyperplane class along this map. Then $\deg D=3$ and $\dim |D|=1$, so by Riemann-Roch we would have $l(K-D)=1$, which means that there's a hyperplane in $\Bbb P^{g-1}$ which vanishes on the image of $D$ under the canonical map. But for hyperelliptic curves, the canonical map has image a rational normal curve, which in this case is a conic in $\Bbb P^2$, and by Bezout there is no hyperplane which intersects it in three points.
In general, one may use the base-point free pencil trick (see section 3 of this blog post by Akhil Mathew) to show that a curve of genus $\geq 3$ cannot be simultaneously trigonal and hyperelliptic.
(PS: let me point out that it's perhaps more usual to consider projection from a point as the degree 3 map $X\to\Bbb P^1$ for the first part of the question.)
A: SKETCH: Suppose there exists a map from $X$ to $\Bbb P^1$ of degree $3$ and also a map to $\Bbb P^1$ of degree $2$. This means we have a map from $X$ to $\Bbb P^1\times\Bbb P^1$ and $X$ is an element of the linear system $|3H_1+2H_2|$ on $\Bbb P^1\times\Bbb P^1$ (where $H_i$ is the hyperplane bundle on the respective $\Bbb P^1_i$). Now apply the adjunction formula on $\Bbb P^1\times\Bbb P^1$ to conclude that the genus of $X$ is $2$.
