This is lemma V.2.1 of Hartshorne's Algebraic Geometry. Surfaces are all nonsingular and projective over an algebraically closed field.
A ruled surface is a surface $X$, a nonsingular curve $C$, and a surjective morphism $f : X \rightarrow C$ such that for all closed points $y \in C$, $X \times_C Spec(k(y)) \sim \mathbb{P}^1_{k(y)}$, and that $f$ has a right inverse $g$.
Now suppose $X$ is a ruled surface. $D$ is a divisor on $X$, and suppose that for all fibers $F$ of $f$, $D . f = n \geq 0$, where $D . f$ indicates intersection number.
If $y$ is a closed point of $C$, then $O_X(D)$ pulls back to a sheaf of degree $n$ (and thus independent of $y$) on $X \times_C Spec(k(y))$. Hartshorne then uses Grauert's theorem to conclude that $f_* O_X(D)$ is locally free of rank $n+1$.
The part that I'm stuck on is that this proof doesn't cover the generic point of $C$. Is it sufficient to check only closed points for Grauert's theorem to apply? The generic point isn't a divisor on $C$ so we can't pull it back to $X$ like we did with closed points.
Best Answer
Yes, it is sufficient to check at closed points. If $\mathcal{O}_X(D)$ pushes forward to a coherent sheaf on $C$, then the rank of $f_*\mathcal{O}_X(D)$ is an upper semi-continuous function (see exercise II.5.8(a), for instance). This means that the set where $f_*\mathcal{O}_X(D)$ is of rank $>n$ is closed, and if it is non-empty, must have a closed point since $C$ is quasi-compact (i.e. it can't just be the generic point of $C$). But we've verified that the rank of $f_*\mathcal{O}_X(D)$ at each closed point is $n$, so we're good.