Lemma 37.78.5. Let $S = \mathop{\mathrm{lim}}\nolimits _{\lambda \in \Lambda } S_\lambda$ be a limit of a directed system of schemes with affine transition morphisms. Let $0 \in \Lambda$ and let $f_0 : X_0 \to Y_0$ be a morphism of schemes over $S_0$. For $\lambda \geq 0$ let $f_\lambda : X_\lambda \to Y_\lambda$ be the base change of $f_0$ to $S_\lambda$ and let $f : X \to Y$ be the base change of $f_0$ to $S$. If

1. $f$ is completely decomposed,

2. $Y_0$ is quasi-compact and quasi-separated, and

3. $f_0$ is locally of finite presentation,

then there exists an $\lambda \geq 0$ such that $f_\lambda$ is completely decomposed.

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

$Z_ i \xrightarrow {\sigma _ i} X \to X_0$

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. $\square$

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