Proof.
Let $\mathcal{E}$ be a finite locally free $\mathcal{O}_ S$-module. Since finite locally free modules are of finite presentation we can find an $i$ and an $\mathcal{O}_{S_ i}$-module $\mathcal{E}_ i$ of finite presentation such that $f_ i^*\mathcal{E}_ i \cong \mathcal{E}$, see Lemma 32.10.2. After increasing $i$ we may assume $\mathcal{E}_ i$ is a flat $\mathcal{O}_{S_ i}$-module, see Algebra, Lemma 10.168.1. (Using this lemma is not necessary, but it is convenient.) Then $\mathcal{E}_ i$ is finite locally free by Algebra, Lemma 10.78.2.
If $\mathcal{L}$ is an invertible $\mathcal{O}_ S$-module, then by the above we can find an $i$ and finite locally free $\mathcal{O}_{S_ i}$-modules $\mathcal{L}_ i$ and $\mathcal{N}_ i$ pulling back to $\mathcal{L}$ and $\mathcal{L}^{\otimes -1}$. After possible increasing $i$ we see that the map $\mathcal{L} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes -1} \to \mathcal{O}_ X$ descends to a map $\mathcal{L}_ i \otimes _{\mathcal{O}_{S_ i}} \mathcal{N}_ i \to \mathcal{O}_{S_ i}$. And after increasing $i$ further, we may assume it is an isomorphism. It follows that $\mathcal{L}_ i$ is an invertible module (Modules, Lemma 17.25.2) and the proof of (2) is complete.
Given $\mathcal{I}$ as in (3) we see that $\mathcal{O}_ S \to \mathcal{O}_ S/\mathcal{I}$ is a map of finitely presented $\mathcal{O}_ S$-modules. Hence by Lemma 32.10.2 this is the pullback of some map $\mathcal{O}_{S_ i} \to \mathcal{F}_ i$ of finitely presented $\mathcal{O}_{S_ i}$-modules. After increasing $i$ we may assume this map is surjective (details omitted; hint: use Algebra, Lemma 10.127.5 on affine open cover). Then the kernel of $\mathcal{O}_{S_ i} \to \mathcal{F}_ i$ is a finite type quasi-coherent ideal in $\mathcal{O}_{S_ i}$ whose pullback gives $\mathcal{I}$.
$\square$
Comments (0)