The Stacks project

Proof. First we write $\mathcal{A} = \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i$ as a directed colimit of finitely presented quasi-coherent sheaves as in Lemma 69.9.1. For each $i$ let $\mathcal{B}_ i = \text{Sym}(\mathcal{F}_ i)$ be the symmetric algebra on $\mathcal{F}_ i$ over $\mathcal{O}_ X$. Write $\mathcal{I}_ i = \mathop{\mathrm{Ker}}(\mathcal{B}_ i \to \mathcal{A})$. Write $\mathcal{I}_ i = \mathop{\mathrm{colim}}\nolimits _ j \mathcal{F}_{i, j}$ where $\mathcal{F}_{i, j}$ is a finite type quasi-coherent submodule of $\mathcal{I}_ i$, see Lemma 69.9.2. Set $\mathcal{I}_{i, j} \subset \mathcal{I}_ i$ equal to the $\mathcal{B}_ i$-ideal generated by $\mathcal{F}_{i, j}$. Set $\mathcal{A}_{i, j} = \mathcal{B}_ i/\mathcal{I}_{i, j}$. Then $\mathcal{A}_{i, j}$ is a quasi-coherent finitely presented $\mathcal{O}_ X$-algebra. Define $(i, j) \leq (i', j')$ if $i \leq i'$ and the map $\mathcal{B}_ i \to \mathcal{B}_{i'}$ maps the ideal $\mathcal{I}_{i, j}$ into the ideal $\mathcal{I}_{i', j'}$. Then it is clear that $\mathcal{A} = \mathop{\mathrm{colim}}\nolimits _{i, j} \mathcal{A}_{i, j}$. $\square$