Lemma 37.60.13. The property $\mathcal{P}(f) =$“$f$ is pseudo-coherent” is fppf local on the source.

**Proof.**
Let $f : X \to S$ be a morphism of schemes. Let $\{ g_ i : X_ i \to X\} $ be an fppf covering such that each composition $f \circ g_ i$ is pseudo-coherent. According to Lemma 37.48.2 there exist

a Zariski open covering $X = \bigcup U_ j$,

surjective finite locally free morphisms $W_ j \to U_ j$,

Zariski open coverings $W_ j = \bigcup _ k W_{j, k}$,

surjective finite locally free morphisms $T_{j, k} \to W_{j, k}$

such that the fppf covering $\{ h_{j, k} : T_{j, k} \to X\} $ refines the given covering $\{ X_ i \to X\} $. Denote $\psi _{j, k} : T_{j, k} \to X_{\alpha (j, k)}$ the morphisms that witness the fact that $\{ T_{j, k} \to X\} $ refines the given covering $\{ X_ i \to X\} $. Note that $T_{j, k} \to X$ is a flat, locally finitely presented morphism, so both $X_ i$ and $T_{j, k}$ are pseudo-coherent over $X$ by Lemma 37.60.6. Hence $\psi _{j, k} : T_{j, k} \to X_ i$ is pseudo-coherent, see Lemma 37.60.7. Hence $T_{j, k} \to S$ is pseudo coherent as the composition of $\psi _{j, k}$ and $f \circ g_{\alpha (j, k)}$, see Lemma 37.60.4. Thus we see we have reduced the lemma to the case of a Zariski open covering (which is OK) and the case of a covering given by a single surjective finite locally free morphism which we deal with in the following paragraph.

Assume that $X' \to X \to S$ is a sequence of morphisms of schemes with $X' \to X$ surjective finite locally free and $X' \to Y$ pseudo-coherent. Our goal is to show that $X \to S$ is pseudo-coherent. Note that by Descent, Lemma 35.14.3 the morphism $X \to S$ is locally of finite presentation. It is clear that the problem reduces to the case that $X'$, $X$ and $S$ are affine and $X' \to X$ is free of some rank $r > 0$. The corresponding algebra problem is the following: Suppose $R \to A \to A'$ are ring maps such that $R \to A'$ is pseudo-coherent, $R \to A$ is of finite presentation, and $A' \cong A^{\oplus r}$ as an $A$-module. Goal: Show $R \to A$ is pseudo-coherent. The assumption that $R \to A'$ is pseudo-coherent means that $A'$ as an $A'$-module is pseudo-coherent relative to $R$. By More on Algebra, Lemma 15.81.5 this implies that $A'$ as an $A$-module is pseudo-coherent relative to $R$. Since $A' \cong A^{\oplus r}$ as an $A$-module we see that $A$ as an $A$-module is pseudo-coherent relative to $R$, see More on Algebra, Lemma 15.81.8. This by definition means that $R \to A$ is pseudo-coherent and we win. $\square$

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