How to show equivalence of divisors involving points at infinity? Consider the function $y^2=x^3+1$, and its projective closure $y^2z=x^3+z^3$ with the point $[0:1:0]$ added. Then we can show equivalences between certain divisors by considering functions on $\mathbb{P}(x,y,z)$. For example, the divisor $2(1,0)\sim 2(w,0)$, for $w^3=1$ by considering the function $f=(x+z)/(x+zw)$. How can we show equivalences between divisors including $[0:1:0]$? For example, how can one show $2(1,0)\sim 2[0:1:0]$? One cannot consider the divisor $(g)$ given by $g=(x+z)/(y-1)^2$, because $y-1=0$ is solved by more points than $[0:1:0]$, some of which lie on $y^2z=x^3+z^3$, so $(g)$ includes more points than just $2(1,0)-2[0:1:0]$. How does one rectify this?
Note this is a question from Geometry of Algebraic Curves by Harris.
 A: Let me point out a couple errors before you begin: you are missing minus signs from lots of $x$-coordinates in your post. For instance $(1,0)$ and $(w,0)$ are not even on your curve, but $(-1,0)$ and $(-w,0)$ are. We'll assume the equation $y^2=x^3+1$ is correct and adjust based off of that without any further comment for the rest of the answer. You also have an error when talking about $(g)$: $g$ should be homogeneous, but the denominator isn't. It's also not necessarily a problem if your numerator and denominator of $g$ vanish in other places other than your end-goal points: they could cancel each other out if chosen properly.
Now on to the solution. There's no real difference between the example you've written in your post and the problem you're trying to attack - in order to show that $2[-1:0:1]\sim 2[0:1:0]$, you need to find a rational function on your curve which has a zero of order 2 at $[-1:0:0]$ and a pole of order 2 at $[0:1:0]$. This is provided by $(x+z)/z$: $V(x+z)$ intersects $V(y^2z=x^3+z^3)$ with multiplicity 2 at $[-1:0:0]$ (it's a simple tangent) and multiplicity one at $[0:1:0]$ (it's a transverse intersection), and $V(z)$ intersects $V(y^2z=x^3+z^3)$ with multiplicity 3 at $[0:1:0]$. So the divisor associated to $(x+z)/z$ is $2[-1:0:1]+[0:1:0]-3[0:1:0]=2[-1:0:1]-2[0:1:0]$.
Another way to attack this problem is to exploit the fact that $V(y^2z=x^3+z^3)$ is an elliptic curve: then addition in the divisor class group corresponds to addition on the curve, and both $[-1:0:1]$ and $[0:1:0]$ are 2-torsion points (the latter is actually the identity) so $2[-1:0:1]=O=2[0:1:0]$.
