Lemma 38.3.3. Assumptions and notation as in Lemma 38.3.2. If $f$ is locally of finite presentation then $\pi$ is of finite presentation. In this case the following are equivalent

1. $\mathcal{F}$ is an $\mathcal{O}_ X$-module of finite presentation in a neighbourhood of $x$,

2. $\mathcal{G}$ is an $\mathcal{O}_{Z'}$-module of finite presentation in a neighbourhood of $z'$, and

3. $\pi _*\mathcal{G}$ is an $\mathcal{O}_{Y'}$-module of finite presentation in a neighbourhood of $y'$.

Still assuming $f$ locally of finite presentation the following are equivalent to each other

1. $\mathcal{F}_ x$ is an $\mathcal{O}_{X, x}$-module of finite presentation,

2. $\mathcal{G}_{z'}$ is an $\mathcal{O}_{Z', z'}$-module of finite presentation, and

3. $(\pi _*\mathcal{G})_{y'}$ is an $\mathcal{O}_{Y', y'}$-module of finite presentation.

Proof. Assume $f$ locally of finite presentation. Then $Z' \to S$ is locally of finite presentation as a composition of such, see Morphisms, Lemma 29.21.3. Note that $Y' \to S$ is also locally of finite presentation as a composition of a smooth and an étale morphism. Hence Morphisms, Lemma 29.21.11 implies $\pi$ is locally of finite presentation. Since $\pi$ is finite we conclude that it is also separated and quasi-compact, hence $\pi$ is actually of finite presentation.

To prove the equivalence of (1), (2), and (3) we also consider: (4) $g^*\mathcal{F}$ is a $\mathcal{O}_{X'}$-module of finite presentation in a neighbourhood of $x'$. The pullback of a module of finite presentation is of finite presentation, see Modules, Lemma 17.11.4. Hence (1) $\Rightarrow$ (4). The étale morphism $g$ is open, see Morphisms, Lemma 29.36.13. Hence for any open neighbourhood $U' \subset X'$ of $x'$, the image $g(U')$ is an open neighbourhood of $x$ and the map $\{ U' \to g(U')\}$ is an étale covering. Thus (4) $\Rightarrow$ (1) by Descent, Lemma 35.7.3. Using Descent, Lemma 35.7.10 and some easy topological arguments (see More on Morphisms, Lemma 37.47.4) we see that (4) $\Leftrightarrow$ (2) $\Leftrightarrow$ (3).

To prove the equivalence of (a), (b), (c) consider the ring maps

$\mathcal{O}_{X, x} \to \mathcal{O}_{X', x'} \to \mathcal{O}_{Z', z'} \leftarrow \mathcal{O}_{Y', y'}$

The first ring map is faithfully flat. Hence $\mathcal{F}_ x$ is of finite presentation over $\mathcal{O}_{X, x}$ if and only if $g^*\mathcal{F}_{x'}$ is of finite presentation over $\mathcal{O}_{X', x'}$, see Algebra, Lemma 10.83.2. The second ring map is surjective (hence finite) and finitely presented by assumption, hence $g^*\mathcal{F}_{x'}$ is of finite presentation over $\mathcal{O}_{X', x'}$ if and only if $\mathcal{G}_{z'}$ is of finite presentation over $\mathcal{O}_{Z', z'}$, see Algebra, Lemma 10.36.23. Because $\pi$ is finite, of finite presentation, and $\pi ^{-1}(\{ y'\} ) = \{ x'\}$ the ring homomorphism $\mathcal{O}_{Y', y'} \leftarrow \mathcal{O}_{Z', z'}$ is finite and of finite presentation, see More on Morphisms, Lemma 37.47.4. Hence $\mathcal{G}_{z'}$ is of finite presentation over $\mathcal{O}_{Z', z'}$ if and only if $\pi _*\mathcal{G}_{y'}$ is of finite presentation over $\mathcal{O}_{Y', y'}$, see Algebra, Lemma 10.36.23. $\square$

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