Lemma 69.6.13. Assumptions and notation as in Situation 69.6.1. If

1. $f$ is proper, and

2. $f_0$ is locally of finite type,

then there exists an $i$ such that $f_ i$ is proper.

Proof. Choose an affine scheme $V_0$ and a surjective étale morphism $V_0 \to Y_0$. Set $V_ i = Y_ i \times _{Y_0} V_0$ and $V = Y \times _{Y_0} V_0$. It suffices to prove that the base change of $f_ i$ to $V_ i$ is proper, see Morphisms of Spaces, Lemma 66.40.2. Thus we may assume $Y_0$ is affine.

By Lemma 69.6.9 we see that $f_ i$ is separated for some $i \geq 0$. Replacing $0$ by $i$ we may assume that $f_0$ is separated. Observe that $f_0$ is quasi-compact. Thus $f_0$ is separated and of finite type. By Cohomology of Spaces, Lemma 68.18.1 we can choose a diagram

$\xymatrix{ X_0 \ar[rd] & X_0' \ar[d] \ar[l]^\pi \ar[r] & \mathbf{P}^ n_{Y_0} \ar[dl] \\ & Y_0 & }$

where $X_0' \to \mathbf{P}^ n_{Y_0}$ is an immersion, and $\pi : X_0' \to X_0$ is proper and surjective. Introduce $X' = X_0' \times _{Y_0} Y$ and $X_ i' = X_0' \times _{Y_0} Y_ i$. By Morphisms of Spaces, Lemmas 66.40.4 and 66.40.3 we see that $X' \to Y$ is proper. Hence $X' \to \mathbf{P}^ n_ Y$ is a closed immersion (Morphisms of Spaces, Lemma 66.40.6). By Morphisms of Spaces, Lemma 66.40.7 it suffices to prove that $X'_ i \to Y_ i$ is proper for some $i$. By Lemma 69.6.8 we find that $X'_ i \to \mathbf{P}^ n_{Y_ i}$ is a closed immersion for $i$ large enough. Then $X'_ i \to Y_ i$ is proper and we win. $\square$

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