Does hyperelliptic curve has a base point free $g_3^1$? Let $X$ be a projective hyperelliptic smooth curve over an algebraically closed field $k$ (which means there is a finite morphism $f:X\to \mathbb P^1$ of degree $2$). My question is that: does there exist a finite morphism $g:X\to \mathbb P^1$ of degree $3$? (The title of the question is just a translation of this question to linear system.) I think the answer is no. But how to prove it?
 A: Take a curve of bidegree $(2,3)$ on $\mathbb{P}^1 \times \mathbb{P}^1$ (hence of genus 2). The two projections to the factors are coverings over $\mathbb{P}^1$ of degree 2 and 3, respectively.
A: There are two excellent answers for the question. I just wanted to give a slightly different argument for the fact that if $g>2$, then the curve has no $g^1_3$. If we had $f, g$ as in the question, we get a morphism $\pi=(f,g):X\to \mathbb{P}^1\times\mathbb{P}^1$. Let $Y=\pi(X)$. Then $Y$ is a $(2,3)$ curve and so by genus formula, its arithmetic genus is $2$. Since we have a factorization $X\to Y\stackrel{f}{\to} \mathbb{P}^1$ of $\pi$, we see that $\deg \pi$ divides 2 and similarly it also divides 3, using $g$. So, $\deg \pi=1$ and so $\pi$ is birational and thus it is just the normalization of $Y$. Since geometric genus is always less than or equal to the arithmetic genus, we get $g\leq 2$.
A: While Sasha has pointed out that there are curves of genus 2 which are both hyperelliptic and trigonal, these are kind of the final examples.
Theorem. If $X$ is a curve of genus $>2$, then $X$ cannot be both hyperelliptic and trigonal.
To prove this theorem, we'll need some standard results you can find in any textbook plus one preliminary result we'll prove ourselves.
Theorem (Riemann-Roch). Let $X$ be a curve of genus $g$, let $D$ be a divisor on $X$, let $K$ be the canonical divisor on $X$, and let $l(D)=h^0(\mathcal{O}_X(D))$. Then $l(D)-l(K-D)=\deg D + 1 -g$.
Theorem (Clifford's theorem). Let $X$ be a curve and let $D$ be an effective special divisor on $X$ (special $\Leftrightarrow$ $l(K-D)>0$). Then  $2(l(D)-1)\leq \deg D$ with equality iff $D=0$, $D=K$, or $X$ hyperelliptic and $D$ a multiple of a $g_2^1$.
Lemma (Base-point free pencil trick). Let $\mathcal{L}$ and $\mathcal{M}$ be line bundles on a curve $X$. Let $s_1,s_2$ be sections of $H^0(\mathcal{L})$ which have no common zero. The map $H^0(\mathcal{M})\oplus H^0(\mathcal{M}) \to H^0(\mathcal{L}\otimes\mathcal{M})$ by $(t_1,t_2)\mapsto (s_1t_1+s_2t_2)$ has kernel $H^0(\mathcal{L}^{-1}\otimes\mathcal{M})$.
Proof. In fact, we have an exact sequence of sheaves $$0\to \mathcal{L}^{-1}\otimes\mathcal{M} \to \mathcal{M}\oplus\mathcal{M} \to \mathcal{L}\otimes\mathcal{M}\to 0$$ where the maps are $t\mapsto (s_2t,-s_1t)$ and $(t_1,t_2)\mapsto (s_1t_1+s_2t_2)$. One can verify exactness locally from the fact that $s_1$ and $s_2$ have no common zeroes. $\blacksquare$.
Now to prove the theorem: let $\mathcal{L}$ be the line bundle corresponding to a trigonal map and $\mathcal{M}$ be the line bundle corresponding to a hyperelliptic map (that is, $\deg \mathcal{L} = 3$, $\deg \mathcal{M}=2$, and $l(\mathcal{L})=l(\mathcal{M})=2$). Then $\mathcal{L}^{-1}\otimes\mathcal{M}$ has degree $-1$ and therefore no global sections, $\mathcal{L}\otimes\mathcal{M}$ is of degree 5, and by the base-point free pencil trick we see that $h^0(\mathcal{L}\otimes\mathcal{M})\geq 4$. If $\mathcal{L}\otimes\mathcal{M}$ is special, this will contradict Clifford's theorem, so we're reduced to figuring out when $\mathcal{L}\otimes\mathcal{M}$ is special.
Let $D$ be an effective divisor corresponding to $\mathcal{L}\otimes\mathcal{M}$. Then $l(D)\geq 4$, so by Riemann-Roch $l(K-D) = l(D)+g-1-\deg D \geq 4+g-1-5 = g-2$, so as soon as $g>2$ we have $D$ special, showing that no curve of genus $>2$ is both hyperelliptic and trigonal.

Let me just quickly point out that one can always find a $g_3^1$ on a curve of genus 1 or 2: for a curve of genus 1, embed it as a plane cubic and project from a point not on the curve; for a curve of genus 2, embed it as a plane quartic with a node and project from a smooth point on the curve.
