My question refers to following argument in the proof of a statement in Hartshornes AG (chapter V, p 370):
We consider the fiber $f= \pi^{-1}(b)$ of a closed point $b$.
How he argue here with "distinct" fibers? Is here used implicitely something like "moving lemma", such that to calculate the self intersectivion one shifts $f$ to another disjunct fiber?
Why it can be done or why does it work here? Wikis description of moving lemma doen't help me to decide why it is here possible.
Are all fibers linear equivalent such that they can be shifted without changing intersection number?
Or is here used the property that $f$ is a Cartier divisor with corresponding ample invertible sheaf such that every global section of it can "replace" $f$ for calculation?
You do not need the moving lemma here. One approach is to find other points which will be linearly equivalent to $b$, or at least some multiple of $b$.
In particular, by Riemann-Roch (or exercise 4.1.2 in Hartshorne) there exists a rational function $f\in k(C)$ such that $\mathrm{div}(f) = \sum n_ip_i - nb$, where $n_i \ge 0$ and for some $n>0$. In particular we have that $nb \sim n_ip_i \in \mathrm{Pic}(C)$ .
It then follows that $n \pi^{-1}(b) \sim \sum n_i \pi^{-1}(p_i)$ (pullback $\mathrm{div}(f)$ and check that its divisor is $n \pi^{-1}(b) -\sum n_i \pi^{-1}(p_i)$). In particular, you see that $nf$ is linearly equivalent to $\sum n_i f_i$, where $f_i$ doesn't intersect $f$. So now you have $n^2f^2 = nf (\sum n_if_i) = 0$.