Let $X$ be an irreducible variety, and let $\mathscr L$ be a locally-coherent sheaf of rank 1. How do I prove that there exists a coherent subsheaf $\mathcal I$ of the constant sheaf $k(X)$ such that $\mathcal I \cong \mathscr L$?
Kempf gives a proof that I do not understand:
Fix an open dense subset $V$;
let $\sigma \in \mathscr L(V)$, and define $\mathcal I(U) = \{f \in k(X) | f \cdot \sigma \text{ comes from an element of } \mathscr L(U)\}$.
(a) Why not just take $X$ itself as $V$, if we just want an open dense subset?
(b) What is the definition of $\mathcal I(U)$ exactly, i.e. what's the meaning of "comes from"?
(c) Why is this sheaf isomorphic to the sheaf $\mathscr L$?
I assume the coherence of $\mathcal I$ is automatic because it is isomorphic to $\mathscr L$.
(a) Because $\mathcal{L}$ might not have any global sections, i.e. $\mathcal{L}(X)$ might be empty.
(b) Taking the stalk at the generic point, you can view $\sigma$ as an element of $\mathrm{Frac}(\mathcal{L}(U)) = k(X)$. In particular, it makes sense to ask if $f\cdot \sigma \in \mathcal{L}(U)$.
(c) One way is to write down an isomorphism of sheaves $\mathcal{I}\rightarrow \mathcal{L}$ where on each open $U$, we map $\mathcal{I}(U) \rightarrow \mathcal{L}(U)$ by sending $f\mapsto f\sigma$. By definition of $\mathcal{I}(U)$, this is well-defined is an isomorphism.