The Stacks project

Proof. One direction of the lemma is trivial. For the other, say $\mathcal{E}_ j \to \mathcal{I}_ j$ is a surjection with $\mathcal{E}_ j$ finite locally free. In the next paragraph, we reduce to the Noetherian case; we suggest the reader skip it.

The first observation is that $U_ j \cap U_{j'}$ is quasi-compact as the complement of the zero scheme of the quasi-coherent finite type ideal $\mathcal{I}_{j'}|{U_ j}$ on the affine scheme $U_ j$, see Properties, Lemma 28.24.1. Hence $X$ is quasi-compact and quasi-separated, see Schemes, Lemma 26.21.6. By Limits, Proposition 32.5.4 we can write $X = \mathop{\mathrm{lim}}\nolimits X_ i$ as the limit of a direct system of Noetherian schemes with affine transition morphisms. For each $j$ we can find an $i$ and a finite locally free $\mathcal{O}_{X_ i}$-module $\mathcal{E}_{i, j}$ pulling back to $\mathcal{E}_ j$, see Limits, Lemma 32.10.3. After increasing $i$ we may assume that the composition $\mathcal{E}_ j \to \mathcal{I}_ j \to \mathcal{O}_ X$ is the pullback of a map $\mathcal{E}_{i, j} \to \mathcal{O}_{X_ i}$, see Limits, Lemma 32.10.2. Denote $\mathcal{I}_{i, j} \subset \mathcal{O}_{X_ i}$ the image of this map; this is a quasi-coherent ideal sheaf on the Noetherian scheme $X_ i$ whose pullback to $X$ is $\mathcal{I}_ j$. Denoting $U_{i, j} \subset X_ i$ the complementary opens, we may assume these are affine for all $i, j$, see Limits, Lemma 32.4.13. If we can prove the lemma for the opens $U_{i, j}$ and the ideal sheaves $\mathcal{I}_{i, j}$ on $X_ i$ then $X$, being affine over $X_ i$, will have the resolution property by Lemma 36.36.4. In this way we reduce to the case of a Noetherian scheme.

Assume $X$ is Noetherian. For every coherent module $\mathcal{F}$ we can choose a finite list of sections $s_{jk} \in \mathcal{F}(U_ j)$, $k = 1, \ldots , e_ j$ which generate the restriction of $\mathcal{F}$ to $U_ j$. By Cohomology of Schemes, Lemma 30.10.5 we can extend $s_{jk}$ to a map $s'_{jk} : \mathcal{I}_ i^{n_{jk}} \to \mathcal{F}$ for some $n_{jk} \geq 1$. Then we can consider the compositions

to conclude. $\square$