The Stacks project

Lemma 106.13.8. Let $\mathcal{X}$ be an algebraic stack. Assume $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is finite. Then there exist a set $I$ and for $i \in I$ a morphism of algebraic stacks

\[ g_ i : \mathcal{X}_ i \longrightarrow \mathcal{X} \]

with the following properties

  1. $|\mathcal{X}| = \bigcup |g_ i|(|\mathcal{X}_ i|)$,

  2. $\mathcal{X}_ i$ is well-nigh affine,

  3. $\mathcal{I}_{\mathcal{X}_ i} \to \mathcal{X}_ i \times _\mathcal {X} \mathcal{I}_\mathcal {X}$ is an isomorphism, and

  4. $g_ i : \mathcal{X}_ i \to \mathcal{X}$ is representable by algebraic spaces, separated, and ├ętale,

Proof. For any $x \in |\mathcal{X}|$ we can choose $g : \mathcal{U} \to \mathcal{X}$, $\mathcal{U} = [U/R]$, and $u$ as in Morphisms of Stacks, Lemma 101.32.4. Then by Morphisms of Stacks, Lemma 101.45.4 we see that there exists an open substack $\mathcal{U}' \subset \mathcal{U}$ containing $u$ such that $\mathcal{I}_{\mathcal{U}'} \to \mathcal{U}' \times _\mathcal {X} \mathcal{I}_\mathcal {X}$ is an isomorphism. Let $U' \subset U$ be the $R$-invariant open corresponding to the open substack $\mathcal{U}'$. Let $u' \in U'$ be a point of $U'$ mapping to $u$. Observe that $t(s^{-1}(\{ u'\} ))$ is finite as $s : R \to U$ is finite. By Properties, Lemma 28.29.5 and Groupoids, Lemma 39.24.1 we can find an $R$-invariant affine open $U'' \subset U'$ containing $u'$. Let $R''$ be the restriction of $R$ to $U''$. Then $\mathcal{U}'' = [U''/R'']$ is an open substack of $\mathcal{U}'$ containing $u$, is well-nigh affine, $\mathcal{I}_{\mathcal{U}''} \to \mathcal{U}'' \times _\mathcal {X} \mathcal{I}_\mathcal {X}$ is an isomorphism, and $\mathcal{U}'' \to \mathcal{X}$ and is representable by algebraic spaces and ├ętale. Finally, $\mathcal{U}'' \to \mathcal{X}$ is separated as $\mathcal{U}''$ is separated (Lemma 106.13.2) the diagonal of $\mathcal{X}$ is separated (Morphisms of Stacks, Lemma 101.6.1) and separatedness follows from Morphisms of Stacks, Lemma 101.4.12. Since the point $x \in |\mathcal{X}|$ is arbitrary the proof is complete. $\square$

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