Consider the surface $S$ (therefore a $2$-dimensional, proper k-scheme) and
$B$ irreducible curve (" $1$-dim " ).
Let $f: S \to B$ a proper fibration, so $\mathcal{O}_B= f_*\mathcal{O}_S$
Consider the unique generic point $\eta$ of $B$ and the generic fiber $f^{-1}(\eta)$.
Let $\mathcal{F}$ be coherent sheaf on $S$ such that the restriction $\mathcal{F} \vert_{f^{-1}(\eta)}$ is invertible.
Why and how to prove that there exist an open subset $V \subset S$ with $f^{-1}(\eta) \subset V$ such that the restriction
$\mathcal{F} \vert_{V}$ is also invertible?
Remark by the way: I'm not sure if we need the fibration property for the statement below.
Best Answer
Here is a general way of seeing this. Let a coherent sheaf be given as $\mathcal{O}_X^m\stackrel{\phi}{\to}\mathcal{O}_X^n\to F\to 0$. The set of points where $F$ is generated by at most r elements is open. Clearly, if $r\geq n$, then this is $X$ and hence open. So, assume $r<n$. The complement of points where $F$ needs more than $r$ generators can immediately seen to be the closed subset where all the $n-r\times n-r$ minors of the matrix $\phi$ vanish and hence closed.