I have a question about an argument in the proof Prop. 2.3 in Hartshorne's "Algebraic Geometry" (chapter V, page 370/371):
That's not clear to me why the self intersection of the (closed) fiber $f^2$ should be vanish. What is meant here by two distinct fibers? Honestly, it's not clear how this argument works.
Remark: $f= \pi^{-1}(b)$ for a closed $b$.
According to page 358 the self intersection can be calculated via $f^2 = deg(\mathcal{L}(f) \vert _f)$ where $\mathcal{L}(f)$ is of course the invertible sheaf corresponding to $f$ interpreted as Cartier divisor.
$X$ is a $\Bbb P^1$-bundle over $C$, and for $b\ne c\in C$, $\pi^{-1}(b)\cap \pi^{-1}(c) = \emptyset$, just by set theory. (Note that any two fibers are homologous — or numerically equivalent.)