The Stacks project

Lemma 73.10.4. Let $S$ be a scheme. Let $\mathcal{P}$ be a property of morphisms of algebraic spaces over $S$. Assume

  1. if $X_ i \to Y_ i$, $i = 1, 2$ have property $\mathcal{P}$ so does $X_1 \amalg X_2 \to Y_1 \amalg Y_2$,

  2. a morphism of algebraic spaces $f : X \to Y$ has property $\mathcal{P}$ if and only if for every affine scheme $Z$ and morphism $Z \to Y$ the base change $Z \times _ Y X \to Z$ of $f$ has property $\mathcal{P}$, and

  3. for any surjective flat morphism of affine schemes $Z' \to Z$ over $S$ and a morphism $f : X \to Z$ from an algebraic space to $Z$ we have

    \[ f' : Z' \times _ Z X \to Z'\text{ has }\mathcal{P} \Rightarrow f\text{ has }\mathcal{P}. \]

Then $\mathcal{P}$ is fpqc local on the base.

Proof. If $\mathcal{P}$ has property (2), then it is automatically stable under any base change. Hence the direct implication in Definition 73.10.1.

Let $\{ Y_ i \to Y\} _{i \in I}$ be an fpqc covering of algebraic spaces over $S$. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume each base change $f_ i : Y_ i \times _ Y X \to Y_ i$ has property $\mathcal{P}$. Our goal is to show that $f$ has $\mathcal{P}$. Let $Z$ be an affine scheme, and let $Z \to Y$ be a morphism. By (2) it suffices to show that the morphism of algebraic spaces $Z \times _ Y X \to Z$ has $\mathcal{P}$. Since $\{ Y_ i \to Y\} _{i \in I}$ is an fpqc covering we know there exists a standard fpqc covering $\{ Z_ j \to Z\} _{j = 1, \ldots , n}$ and morphisms $Z_ j \to Y_{i_ j}$ over $Y$ for suitable indices $i_ j \in I$. Since $f_{i_ j}$ has $\mathcal{P}$ we see that

\[ Z_ j \times _ Y X = Z_ j \times _{Y_{i_ j}} (Y_{i_ j} \times _ Y X) \longrightarrow Z_ j \]

has $\mathcal{P}$ as a base change of $f_{i_ j}$ (see first remark of the proof). Set $Z' = \coprod _{j = 1, \ldots , n} Z_ j$, so that $Z' \to Z$ is a flat and surjective morphism of affine schemes over $S$. By (1) we conclude that $Z' \times _ Y X \to Z'$ has property $\mathcal{P}$. Since this is the base change of the morphism $Z \times _ Y X \to Z$ by the morphism $Z' \to Z$ we conclude that $Z \times _ Y X \to Z$ has property $\mathcal{P}$ as desired. $\square$

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