I need to find divisor of functon $f=y$ i.e. $\frac{y}{z}$ for projective curve $y^2z=x^3-xz^2$ and I have some questions:
For example, I have pole: $(0:1:0)$ and zeros $(0:0:1),(1:0:-i),(1:0:1)$. It is easy to see that in all zero points, uniformazier is $y$, so order of function is $1$ and I conclude, since this is principal divisor, that order of pole must be $-3$, so $$divf=(0:0:1)+(1:0:-1)+(1:0:1)-3(0:1:0).$$
Is this correct?
I tried to find that order of pole is $-3$, but uniformizer at pole is $y-1$, and I'm not sure how to find order of function in this case? Any hints?
You have a typo when you list your zeroes for the first time: Your second zero should be $(1:0:-1)$ instead of $(1:0:-i)$.
Your argument is correct. But if you want to compute the order of $f$ at the point $(0:1:0)$ directly, it is easiest to work in the affine chart given by $y=1$ where $(x,z)=(0,0)$. In this chart $f=\frac{1}{z}$ and our curve is given by $z=x^3-xz^2$. Let $x$ be a uniformizer. Then manipulating the equation of our curve, we get $\frac{1}{z}=\frac{1+xz}{x^3} $. Taking orders at the point $P=(x,z)=(0,0)$, and using that order changes products into sums,
$$\text{ord}_P(f)=\text{ord}_P(\frac{1}{z})=\text{ord}_P(1+xz)-3\text{ord}_P(x). $$
Since $1+xz$ does not vanish at $P$, it's order is zero. But $x$ being the uniformizer has order $1$. We conclude that $\text{ord}_P(f)=-3$ directly.