During the lecture we defined the principal divisor of a rational function on a smooth curve as it follows:
Consider the smooth curve $C\subseteq\mathbb{P}^2$. Take $g\in{K(C)^*}$. Then the principal divisor is $(g)=\sum{\nu_P(g)P}\in Div^0_{C}$. Where $\nu_P(g)\in\mathbb{Z}$ is the order of the zero, if $P$ is a zero, or minus the order of the pole, if $P$ is a pole.
I have some troubles computing it, in particular computing $\nu_P(g)$. A particular case can be: I have the smooth curve $C:y^2=x^3+1$ over $\mathbb{F}_{13}$ and I have to compute the principal divisor of the rational function $g=\frac{x^2}{y}$.
What I tried:
I homogenized both the curve and the function. So I get: $g=\frac{x^2}{yz}$ and $C:y^2z=x^3+z^3$
I want to find the zeros, so I have to find the points that satisfy: $x^2=0$ and $y^2z=x^3+z^3$.
So I found $(0:1:0)$ and $(0:1:1)$ and $(0:-1:1)$. How do I find the order of these zeros?
I want to find the poles, so I have to find the points that satisfy: $yz=0$ and $y^2z=x^3+z^3$. I have found $(0:1:0)$, $(-1:0:1)$, $(4:0:1)$, $(-3:0:1)$. How can I find the orders?
What I know is that the sum of the orders of the points has to be (if i'm not wrong) $6$.
The divisor of the rational function $g=\frac{x^2}{yz}$ on the curve $C$ given by $ y^2z=x^3+z^3$ is $$2\cdot(0:1:1) +2\cdot(0:-1:1)-(-1:0:1)-(4:0:1)-(-3:0:1)-(0:1:0) $$
Justification: an example of calculation
On the affine part $z=1$ of $C$, the curve has equation $f=y^2-x^3+1=0$ and the rational function is $g=\frac{x^2}{y}$.
The point $A=(0:1:1)$ has affine cordinates $(0,1)$ and at that point $x$ is a uniformizing parameter (because $\frac {\partial f}{\partial y}(A)\neq0$ ).
Thus the order at $A$ of $g=\frac{x^2}{y}$ is $2$ (notice that $y$ neither vanishes nor has a pole at $A$ so that its order is $0$ at $A$)