Lemma 68.15.2. Let $S$ be a scheme. Let $f : X \to Y$ be an affine morphism of algebraic spaces over $S$ with $Y$ Noetherian. Then every quasi-coherent $\mathcal{O}_ X$-module is a filtered colimit of finitely presented $\mathcal{O}_ X$-modules.

**Proof.**
Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Write $f_*\mathcal{F} = \mathop{\mathrm{colim}}\nolimits \mathcal{H}_ i$ with $\mathcal{H}_ i$ a coherent $\mathcal{O}_ Y$-module, see Lemma 68.15.1. By Lemma 68.12.2 the modules $\mathcal{H}_ i$ are $\mathcal{O}_ Y$-modules of finite presentation. Hence $f^*\mathcal{H}_ i$ is an $\mathcal{O}_ X$-module of finite presentation, see Properties of Spaces, Section 65.30. We claim the map

is surjective as $f$ is assumed affine, Namely, choose a scheme $V$ and a surjective étale morphism $V \to Y$. Set $U = X \times _ Y V$. Then $U$ is a scheme, $f' : U \to V$ is affine, and $U \to X$ is surjective étale. By Properties of Spaces, Lemma 65.26.2 we see that $f'_*(\mathcal{F}|_ U) = f_*\mathcal{F}|_ V$ and similarly for pullbacks. Thus the restriction of $f^*f_*\mathcal{F} \to \mathcal{F}$ to $U$ is the map

which is surjective as $f'$ is an affine morphism of schemes. Hence the claim holds.

We conclude that every quasi-coherent module on $X$ is a quotient of a filtered colimit of finitely presented modules. In particular, we see that $\mathcal{F}$ is a cokernel of a map

with $\mathcal{G}_ j$ and $\mathcal{H}_ i$ finitely presented. Note that for every $j \in I$ there exist $i \in I$ and a morphism $\alpha : \mathcal{G}_ j \to \mathcal{H}_ i$ such that

commutes, see Lemma 68.5.3. In this situation $\mathop{\mathrm{Coker}}(\alpha )$ is a finitely presented $\mathcal{O}_ X$-module which comes endowed with a map $\mathop{\mathrm{Coker}}(\alpha ) \to \mathcal{F}$. Consider the set $K$ of triples $(i, j, \alpha )$ as above. We say that $(i, j, \alpha ) \leq (i', j', \alpha ')$ if and only if $i \leq i'$, $j \leq j'$, and the diagram

commutes. It follows from the above that $K$ is a directed partially ordered set,

and we win. $\square$

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