Lemma 69.9.6. Let $S$ be a scheme. Let $X$ be a quasi-compact and quasi-separated algebraic space over $S$. Let $\mathcal{A}$ be a finite quasi-coherent $\mathcal{O}_ X$-algebra. Then $\mathcal{A} = \mathop{\mathrm{colim}}\nolimits \mathcal{A}_ i$ is a directed colimit of finite and finitely presented quasi-coherent $\mathcal{O}_ X$-algebras with surjective transition maps.

Proof. By Lemma 69.9.3 there exists a finitely presented $\mathcal{O}_ X$-module $\mathcal{F}$ and a surjection $\mathcal{F} \to \mathcal{A}$. Using the algebra structure we obtain a surjection

$\text{Sym}^*_{\mathcal{O}_ X}(\mathcal{F}) \longrightarrow \mathcal{A}$

Denote $\mathcal{J}$ the kernel. Write $\mathcal{J} = \mathop{\mathrm{colim}}\nolimits \mathcal{E}_ i$ as a filtered colimit of finite type $\mathcal{O}_ X$-submodules $\mathcal{E}_ i$ (Lemma 69.9.2). Set

$\mathcal{A}_ i = \text{Sym}^*_{\mathcal{O}_ X}(\mathcal{F})/(\mathcal{E}_ i)$

where $(\mathcal{E}_ i)$ indicates the ideal sheaf generated by the image of $\mathcal{E}_ i \to \text{Sym}^*_{\mathcal{O}_ X}(\mathcal{F})$. Then each $\mathcal{A}_ i$ is a finitely presented $\mathcal{O}_ X$-algebra, the transition maps are surjective, and $\mathcal{A} = \mathop{\mathrm{colim}}\nolimits \mathcal{A}_ i$. To finish the proof we still have to show that $\mathcal{A}_ i$ is a finite $\mathcal{O}_ X$-algebra for $i$ sufficiently large. To do this we choose an étale surjective map $U \to X$ where $U$ is an affine scheme. Take generators $f_1, \ldots , f_ m \in \Gamma (U, \mathcal{F})$. As $\mathcal{A}(U)$ is a finite $\mathcal{O}_ X(U)$-algebra we see that for each $j$ there exists a monic polynomial $P_ j \in \mathcal{O}(U)[T]$ such that $P_ j(f_ j)$ is zero in $\mathcal{A}(U)$. Since $\mathcal{A} = \mathop{\mathrm{colim}}\nolimits \mathcal{A}_ i$ by construction, we have $P_ j(f_ j) = 0$ in $\mathcal{A}_ i(U)$ for all sufficiently large $i$. For such $i$ the algebras $\mathcal{A}_ i$ are finite. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).