The Stacks project

99.4 Descending properties

This section is the analogue of Limits, Section 32.4.

Situation 99.4.1. Let $Y = \mathop{\mathrm{lim}}\nolimits _{i \in I} Y_ i$ be a limit of a directed system of algebraic spaces with affine transition morphisms. We assume that $X_ i$ is quasi-compact and quasi-separated for all $i \in I$. We also choose an element $0 \in I$.

Lemma 99.4.2. In Situation 99.4.1 assume that $\mathcal{X}_0 \to Y_0$ is a morphism from algebraic stack to $Y_0$. Assume $\mathcal{X}_0$ is quasi-compact and quasi-separated. If $Y \times _{Y_0} \mathcal{X}_0 \to Y$ is separated, then $Y_ i \times _{Y_0} \mathcal{X}_0 \to Y_ i$ is separated for all sufficiently large $i \in I$.

Proof. Write $\mathcal{X} = Y \times _{Y_0} \mathcal{X}_0$ and $\mathcal{X}_ i = Y_ i \times _{Y_0} \mathcal{X}_0$. Choose an affine scheme $U_0$ and a surjective smooth morphism $U_0 \to \mathcal{X}_0$. Set $U = Y \times _{Y_0} U_0$ and $U_ i = Y_ i \times _{Y_0} U_0$. Then $U$ and $U_ i$ are affine and $U \to \mathcal{X}$ and $U_ i \to \mathcal{X}_ i$ are smooth and surjective. Set $R_0 = U_0 \times _{\mathcal{X}_0} U_0$. Set $R = Y \times _{Y_0} R_0$ and $R_ i = Y_ i \times _{Y_0} R_0$. Then $R = U \times _\mathcal {X} U$ and $R_ i = U_ i \times _{\mathcal{X}_ i} U_ i$.

With this notation note that $\mathcal{X} \to Y$ is separated implies that $R \to U \times _ Y U$ is proper as the base change of $\mathcal{X} \to \mathcal{X} \times _ Y \mathcal{X}$ by $U \times _ Y U \to \mathcal{X} \times _ Y \mathcal{X}$. Conversely, we see that $\mathcal{X}_ i \to Y_ i$ is separated if $R_ i \to U_ i \times _{Y_ i} U_ i$ is proper because $U_ i \times _{Y_ i} U_ i \to \mathcal{X}_ i \times _{Y_ i} \mathcal{X}_ i$ is surjective and smooth, see Properties of Stacks, Lemma 97.3.3. Observe that $R_0 \to U_0 \times _{Y_0} U_0$ is locally of finite type and that $R_0$ is quasi-compact and quasi-separated. By Limits of Spaces, Lemma 67.6.13 we see that $R_ i \to U_ i \times _{Y_ i} U_ i$ is proper for large enough $i$ which finishes the proof. $\square$

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