Let consider the ruled surface $X$ with morphism $\pi: X \to C$ to a curve.
Why then all fibers (of closed points) are as divisors algebraically equivalent?
Hartshorne explains it using an argument which isn't clear to me:
Why it suffice to be parametrised by curve $C$? Algebraically equivalent divisors have isomorphic invertible ideal sheaves on X, so why does the argument come from $C$?