The Stacks project

Proof. Omitted. $\square$

Proof. Let $f : X \to Y$ and $g : Y' \to Y$ be morphisms of schemes. Let $y' \in Y'$ be a point with image $y = g(y')$ in $Y$. If $x \in X$ is a point such that $f(x) = y$ and $\kappa (x) = \kappa (y)$, then there exists a unique point $x' \in X' = X \times _ Y Y'$ which maps to $y'$ in $Y'$ and to $x$ in $X$ and moreover $\kappa (x') = \kappa (y')$, see Schemes, Lemma 26.17.5. From this fact the lemma follows easily; we omit the details. $\square$

Proof. Let $y \in Y$. By assumption there is a morphism $\sigma : \mathop{\mathrm{Spec}}(\kappa (y)) \to X$ over $Y$. We can write $\mathop{\mathrm{Spec}}(\kappa (y))$ as a directed limit of affine schemes $Z$ over $Y$ such that $Z \to Y$ is an immersion of finite presentation. Namely, choose an affine open $y \in \mathop{\mathrm{Spec}}(A) \subset Y$ and say $y$ corresponds to the prime ideal $\mathfrak p$ of $A$. Then $\kappa (\mathfrak p)$ is the filtered colimit of the rings $(A/I)_ f$ where $I \subset \mathfrak p$ is a finitely generated ideal and $f \in A$, $f \not\in \mathfrak p$. The morphisms $Z = \mathop{\mathrm{Spec}}((A/I)_ f) \to Y$ are immersions of finite presentation; quasi-compactness of $Z \to Y$ follows as $Y$ is quasi-separated, see Schemes, Lemma 26.21.14. By Limits, Proposition 32.6.1 for some such $Z$ there is a morphism $\sigma ' : Z \to X$ over $Y$ agreeing with $\sigma$ on the spectrum of $\kappa (\mathfrak p)$. Since $\sigma '$ is a morphism over $Y$, we obtain a section of the projection $X \times _ Y Z \to Z$

We conclude that $Y$ is the union of the images of immersions $Z \to Y$ of finite presentation such that $X \times _ Y Z \to Z$ has a section. Since the image of $Z \to Y$ is constructible (Morphisms, Lemma 29.22.2) and since $Y$ is compact in the constructible topology (Properties, Lemma 28.2.4 and Topology, Lemma 5.23.2), we see that a finite number of these suffice. $\square$

Proof. Since $Y_0$ is quasi-compact and quasi-separated, the scheme $Y$, which is affine over $Y_0$, is quasi-compact and quasi-separated. Choose $n \geq 0$ and $Z_ i \to Y$, $i = 1, \ldots , n$ as in Lemma 37.78.4. Denote $\sigma _ i : Z_ i \to X$ morphisms over $Y$ which exist by our choice of $Z_ i$. After increasing $0 \in \Lambda$ we may assume there exist morphisms $Z_{i, 0} \to Y_0$ of finite presentation whose base changes to $S$ are the morphisms $Z_ i \to Y$, see Limits, Lemma 32.10.1. By Limits, Lemma 32.8.13 we may assume, after possibly increasing $0$, that $Z_{i, 0} \to Y_0$ is an immersion. Since $\coprod Z_ i \to Y$ is surjective, we may assume, after possibly increasing $0$, that $\coprod Z_{i, 0} \to Y_0$ is surjective, see Limits, Lemma 32.8.15. Observe that $Z_ i = \mathop{\mathrm{lim}}\nolimits _{\lambda \geq 0} Z_{i, \lambda }$ where $Z_{i, \lambda } = Y_\lambda \times _{Y_0} Z_{i, 0}$. Let us view the compositions

as morphisms over $Y_0$. Since $f_0$ is locally of finite presentation by Limits, Proposition 32.6.1 we can find a $\lambda \geq 0$ such that there exist morphisms $\sigma '_{i, \lambda } : Z_{i, \lambda } \to X_0$ over $Y_0$ whose precomposition with $Z_ i \to Z_{i, \lambda }$ are the displayed arrows. Of course, then $\sigma '_{i, \lambda }$ determines a morphism $\sigma _{i, \lambda } : Z_{i, \lambda } \to X_\lambda = X_0 \times _{Y_0} Y_\lambda$ over $Y_\lambda$. Since $\coprod Z_{i, \lambda } \to Y_\lambda$ is surjective we conclude that $X_\lambda \to Y_\lambda$ is completely decomposed. $\square$