A divisor on a nonsingular irreducible projective curve $X$ is defined as a formal sum $D=\sum_{P\in X}n_P P$, with only finite nonzero $n_P$. If the curve is defined by $f$, and denotes its function field $K$. For any $z\in K$, we can define a divisor div$(z)=\sum_{P\in X} Ord_P(z) P$. And it’s said that, since $z$ has only finite zeros and poles, this is a well defined divisor.
I am not sure why it has finite zeros? If we write $z=g/h$, $g,h \in \Gamma(X)$, then zeros can be written as V$(f,g)$, which is an algebraic set, thus is finite, right? Hope someone could help. Thanks!
First, one should require that the $z\in K$ be nonzero: taking $0$ certainly doesn't give you a finite number of zeros. Secondly, you're missing justification about why $V(f,g)$ should be finite. Clearly it's not true that if you just pick two homogeneous polynomials in three indeterminants, they determine a finite set of points in $\Bbb P^2$: consider $X$ and $XY$, for instance. (You're also missing an explanation of why you can talk about $V(f,g)$ - $f$ and $g$ belong to different rings here, so you need to explain how they end up in the same place. It can be done, but it's important to try to be clear about these things if you're not sure about them.)
The features of this problem do let us make this argument, though. From the condition that $C$ is irreducible, it's defining equation has to be irreducible as a polynomial in $k[x_0,x_1,x_2]$. Then, as $g$ is a nonzero element of the domain $k[x_0,x_1,x_2]/(f)$, it's not a zero-divisor, and any lift of it to a function $g'\in k[x_0,x_1,x_2]$ won't share any common factors with $f$. As $f$ and $g'$ share no common factor, the curves they determine in $\Bbb P^2$ don't share any common components, and by Bezout's theorem, they have a finite number of intersection points. As these intersection points exactly correspond to points on $C$ with $g=0$, there must be finitely many of them and we are finished.