I'm following through Geometry of Algebraic Curves at the moment. I am asked to compactify $y^2=x^3+1$, which becomes the curve $C$ defined by $Y^2Z=X^3+Z^3$ with the additional point at infinity $[0:1:0]$, in projective coordinates. I am now asked to show the linear equivalence between the divisors
$2(-1,0)\sim 2(-w,0)$, where $w^3=1$ is any 3-rd root of unity.
In other words, I have to find a meromorphic function $f$ on $C$ such that $f$ has multiplicity $2$ at precisely $(-1,0)$ and $(-w,0)$. I don't even know where to begin for this. Any tips or references would be welcome.
Consider $f:=\dfrac{(x+z)^2}{(x+wz)^2}=\dfrac{x^2+2xz+z^2}{x^2+2wxz+w^2z^2}$.
We simply have to observe that $f\neq 0$ as element in $k(C)$.
Enough to consider over the open set $z=1$. That is to see $\dfrac{(x+1)^2}{(x+w)^2}=\dfrac{x^2+2x+1}{x^2+2wx+w^2}$ is non-zero.
If not, then there is a polynomial $g$ such that $g(x^2+2x+1)=0$ in $k[C]=\dfrac{k[x, y]}{(y^2-x^3-1)}$.
That is $g(x^2+2x+1)=h(y^2-x^3-1)$ in $k[x,y]$.
Now $g(x, 0)(x^2+2x+1)=h(x, 0)(-x^3-1)$.
So by UFD ness of $k[x]$, $g(x, 0)| h(x, 0)$ and $h(x, 0)| g(x, 0)$.
But $\deg g(x, 0)+1=\deg h(x, 0)$, a contradiction. Hence $f\neq 0$.
Now more to the point: $$div(f)=2(-1, 0, 1)-2(-w, 0, 1)$$
which gives the desired linear equivalence.