The Stacks project

Proof. It is immediate from the definitions that if $M^\bullet$ is $m$-pseudo-coherent, so is $E$. To prove the converse, assume $E$ is $m$-pseudo-coherent. As $X = \mathop{\mathrm{Spec}}(A)$ is quasi-compact with a basis for the topology given by standard opens, we can find a standard open covering $X = D(f_1) \cup \ldots \cup D(f_ n)$ and strictly perfect complexes $\mathcal{E}_ i^\bullet$ on $D(f_ i)$ and maps $\alpha _ i : \mathcal{E}_ i^\bullet \to E|_{U_ i}$ inducing isomorphisms on $H^ j$ for $j > m$ and surjections on $H^ m$. By Cohomology, Lemma 20.46.8 after refining the open covering we may assume $\alpha _ i$ is given by a map of complexes $\mathcal{E}_ i^\bullet \to \widetilde{M^\bullet }|_{U_ i}$ for each $i$. By Modules, Lemma 17.14.6 the terms $\mathcal{E}_ i^ n$ are finite locally free modules. Hence after refining the open covering we may assume each $\mathcal{E}_ i^ n$ is a finite free $\mathcal{O}_{U_ i}$-module. From the definition it follows that $M^\bullet _{f_ i}$ is an $m$-pseudo-coherent complex of $A_{f_ i}$-modules. We conclude by applying More on Algebra, Lemma 15.64.14.

The case “pseudo-coherent” follows from the fact that $E$ is pseudo-coherent if and only if $E$ is $m$-pseudo-coherent for all $m$ (by definition) and the same is true for $M^\bullet$ by More on Algebra, Lemma 15.64.5. $\square$