Weierstrass points on algebraic curves

Question

We are considering projective algebraic curves over a closed field $\mathbb{k}$. Let $X$ be such a curve, $\mathbb{k}(X)$ - the field of rational functions of $X$, $D$ - some divisor on $X$. We introduce the space $\mathfrak{L}(D)$ to be $\{f\in\mathbb{k}(X):\mathrm{div}(f)+D\ge 0\}$, $l(D)=\mathrm{dim}\mathfrak{L}(D)$. Given any point $p\in X$ we introduce the spaces $\mathfrak{L}(p),\mathfrak{L}(2p),\mathfrak{L}(3p)\ldots$ and the sequence of their dimensions $l(p),l(2p),l(3p)\ldots$. It can be proven with Riemann-Roch theorem that $l(kp)=k-g+1$, for $k\ge 2g-1$, where $g$ is the genus of $X$. But what we know about first $2g-2$ terms? It is a non decreasing sequence.... If it looks like $1,1,\ldots 1,2,3\ldots g-1$ we say that $p$ is a non-Weierstrass point. Otherwise, of course, $p$ is said to be a Weierstrass point. Is there some algorithm to find such points for given curve? It seems to be rather difficult. I guess (I heard) that there is if $X$ is a hyperelliptic curve...

Answer 1

If $X$ is hyperelliptic of genus at least two, then the Weierstrass points of $X$ are the ramification points of the (unique) hyperelliptic map.

In general you can write down the Wronskian differential of your Riemann surface (which only requires you to give an atlas) and compute its divisor.