I have often seen the assertion that for a smooth plane curve $C$ of degree $d$ the gonality of $C$ is $d-1$ and each gonality pencil is obtained by projection from a point of $C$ onto a line.


(let me recall that the gonality of $C$ is by definition the minimal degree $d$ of divisors $D$ on $C$ with $r(D)=1$, those divisors which attain the minimum being called gonality pencils)


Is there a relatively simple proof for this fact?



The gonality of smooth plane curves
Let $X \subset \mathbb{P}^2$ be a smooth planar curve of degree $d$. It is fairly well known that $\Omega^1(X) \cong \mathcal{O}(d-3)|\_{X}$, and $H^0(\mathbb{P}^2, \mathcal{O}(d-3)) \to H^0(X, \Omega^1)$ is an isomorphism. So the canonical embedding of $X$ is isomorphic to the composition $X \to \mathbb{P}^2 \stackrel{\phi}{\longrightarrow} \mathbb{P}^{\binom{d-1}{2}-1}$ where $\phi$ is the $(d-3)$-fold Veronese.


Suppose that $X$ is $k$-gonal for $k \leq d-1$. Let $(x\_1, x\_2, \ldots, x\_k)$ be a fiber of the map to $\mathbb{P}^1$, and let $D$ be the divisor $\sum x\_i$. (If our map is separable, then a generic fiber has distinct points. In general, our map is the composition of a power of Frobenius and a separable map, so we can reduce to the separable case.) Then $\dim H^0(X, D) \geq 2$ so (by Riemann-Roch) $\dim H^0(X, K-D) > \binom{d-1}{2}-k$. This says that the points $\phi(x\_1)$, $\phi(x\_2)$, ..., $\phi(x\_k)$ are linearly dependent in $\mathbb{P}^{\binom{d-1}{2}-1}$.


Choose a generic coordinate system $(u:v:w)$ on $\mathbb{P}^2$, and write $x\_i = (u\_i: v\_i : w\_i)$. By genericity, we can assume that the $(u\_i:v\_i)$ are distinct and all $u\_i$ and $v\_i$ are nonzero. Then $d-2$ of the coordinates of $\phi(x\_i)$ are $(u\_i^{d-3}: u\_i^{d-4} v\_i : \cdots : v\_i^{d-3})$. If $k<d-1$, this is a $k \times (d-2)$ Vandermonde matrix and hence of rank $k$, a contradiction.


If $k=d-1$, our Vandermonde is $(d-1) \times (d-2)$, so it is fine for it to have rank $d-2$. But we will show that, in this case, the points $x\_i$ are forced to be on a line.


In this case, we are requiring that the whole $(d-1) \times \binom{d-1}{2}$ matrix has rank $(d-2)$. The $(d-2)$ columns mentioned above already have rank $d-2$, since they are Vandermonde, so every other column is in their linear span. In particular, for every $j$, there is a degree $d-3$ polynomial $F\_j(u,v)$ such that $F\_j(u\_i,v\_i) = u\_i^j v\_i^{d-4-j} w\_i$ for all $i$. 


Then $u\_i F\_j(u\_i, v\_i) = v\_i F\_{j+1} (u\_i, v\_i)$. But $u F\_j - v F\_{j+1}$ is a degree $d-2$ homogenous polynomial so, if it vanishes at $d-1$ points, it is zero. We deduce that $u F\_j = v F\_{j+1}$ for $1 \leq j \leq d-5$, and thus that there is a degree $1$ polynomial $G$ such that $F\_j(u,v) = u^j v^{d-4-j} G(u,v)$. Thus $w\_i = G(u\_i, v\_i)$ for all $i$, and we have shown that the points lie on a line.


