Lemma 100.32.4. Let $\mathcal{X}$ be an algebraic stack. Assume $\mathcal{X}$ is quasi-DM with separated diagonal (equivalently $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is locally quasi-finite and separated). Let $x \in |\mathcal{X}|$. Assume the automorphism group of $\mathcal{X}$ at $x$ is finite (Remark 100.19.3). Then there exists a morphism of algebraic stacks

$g : \mathcal{U} \longrightarrow \mathcal{X}$

with the following properties

1. there exists a point $u \in |\mathcal{U}|$ mapping to $x$ and $g$ induces an isomorphism between automorphism groups at $u$ and $x$ (Remark 100.19.5),

2. $\mathcal{U} \to \mathcal{X}$ is representable by algebraic spaces and étale,

3. $\mathcal{U} = [U/R]$ where $(U, R, s, t, c)$ is a groupoid scheme with $U$, $R$ affine, and $s, t$ finite, flat, and locally of finite presentation.

Proof. Observe that $G_ x$ is a group scheme by Lemma 100.19.1. The first part of the proof is exactly the same as the first part of the proof of Lemma 100.32.3. Thus we may assume $\mathcal{X} = [U/R]$ where $(U, R, s, t, c)$ and $u \in U$ mapping to $x$ satisfy all the assumptions of More on Groupoids in Spaces, Lemma 78.15.13. Our assumption on $G_ x$ implies that $G_ u$ is finite over $u$. Hence all the assumptions of More on Groupoids in Spaces, Lemma 78.15.12 are satisfied. Hence we can find an elementary étale neighbourhood $(U', u') \to (U, u)$ such that the restriction $R'$ of $R$ to $U'$ is split over $u$. Note that $R' = U' \times _\mathcal {X} U'$ (small detail omitted; hint: transitivity of fibre products). Replacing $(U, R, s, t, c)$ by $(U', R', s', t', c')$ and shrinking $\mathcal{X}$ as above, we may assume that $(U, R, s, t, c)$ has a splitting over $u$. Let $P \subset R$ be a splitting of $R$ over $u$. Apply Lemma 100.32.2 to see that

$\mathcal{U} = [U/P] \longrightarrow [U/R] = \mathcal{X}$

is representable by algebraic spaces and étale. By construction $G_ u$ is contained in $P$, hence this morphism defines an isomorphism on automorphism groups at $u$ as desired. $\square$

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