Lemma 101.22.1. Let $\mathcal{X}$ be an algebraic stack. There exist open substacks

such that $\mathcal{X}''$ is DM, $\mathcal{X}'$ is quasi-DM, and such that these are the largest open substacks with these properties.

Every algebraic stack has a largest open substack which is a Deligne-Mumford stack; this is more or less clear but we also write out the proof below. Of course this substack may be empty, for example if $X = [\mathop{\mathrm{Spec}}(\mathbf{Z})/\mathbf{G}_{m, \mathbf{Z}}]$. Below we will characterize the points of the DM locus.

Lemma 101.22.1. Let $\mathcal{X}$ be an algebraic stack. There exist open substacks

\[ \mathcal{X}'' \subset \mathcal{X}' \subset \mathcal{X} \]

such that $\mathcal{X}''$ is DM, $\mathcal{X}'$ is quasi-DM, and such that these are the largest open substacks with these properties.

**Proof.**
All we are really saying here is that if $\mathcal{U} \subset \mathcal{X}$ and $\mathcal{V} \subset \mathcal{X}$ are open substacks which are DM, then the open substack $\mathcal{W} \subset \mathcal{X}$ with $|\mathcal{W}| = |\mathcal{U}| \cup |\mathcal{V}|$ is DM as well. (Similarly for quasi-DM.) Although this is a cheat, let us use Theorem 101.21.6 to prove this. By that theorem we can choose schemes $U$ and $V$ and surjective étale morphisms $U \to \mathcal{U}$ and $V \to \mathcal{V}$. Then of course $U \amalg V \to \mathcal{W}$ is surjective and étale. The quasi-DM case is proven by exactly the same method using Theorem 101.21.3.
$\square$

Lemma 101.22.2. Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$ correspond to $x : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$. Let $G_ x/k$ be the automorphism group algebraic space of $x$. Then

$x$ is in the DM locus of $\mathcal{X}$ if and only if $G_ x \to \mathop{\mathrm{Spec}}(k)$ is unramified, and

$x$ is in the quasi-DM locus of $\mathcal{X}$ if and only if $G_ x \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite.

**Proof.**
Proof of (2). Choose a scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$. Consider the fibre product

\[ \xymatrix{ G \ar[r] \ar[d] & \mathcal{I}_\mathcal {X} \ar[d] \\ U \ar[r] & \mathcal{X} } \]

Recall that $G$ is the automorphism group algebraic space of $U \to \mathcal{X}$. By Groupoids in Spaces, Lemma 78.6.3 there is a maximal open subscheme $U' \subset U$ such that $G_{U'} \to U'$ is locally quasi-finite. Moreover, formation of $U'$ commutes with arbitrary base change. In particular the two inverse images of $U'$ in $R = U \times _\mathcal {X} U$ are the same open subspace of $R$ (since after all the two maps $R \to \mathcal{X}$ are isomorphic and hence have isomorphic automorphism group spaces). Hence $U'$ is the inverse image of an open substack $\mathcal{X}' \subset \mathcal{X}$ by Properties of Stacks, Lemma 100.9.11 and we have a cartesian diagram

\[ \xymatrix{ G_{U'} \ar[r] \ar[d] & \mathcal{I}_{\mathcal{X}'} \ar[d] \\ U' \ar[r] & \mathcal{X}' } \]

Thus the morphism $\mathcal{I}_{\mathcal{X}'} \to \mathcal{X}'$ is locally quasi-finite and we conclude that $\mathcal{X}'$ is quasi-DM by Lemma 101.6.1 part (5). On the other hand, if $\mathcal{W} \subset \mathcal{X}$ is an open substack which is quasi-DM, then the inverse image $W \subset U$ of $\mathcal{W}$ must be contained in $U'$ by our construction of $U'$ since $\mathcal{I}_\mathcal {W} = \mathcal{W} \times _\mathcal {X} \mathcal{I}_\mathcal {X}$ is locally quasi-finite over $\mathcal{W}$. Thus $\mathcal{X}'$ is the quasi-DM locus. Finally, choose a field extension $K/k$ and a $2$-commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(k) \ar[d]^ x \\ U \ar[r] & \mathcal{X} } \]

Then we find an isomorphism $G_ x \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(K) \cong G \times _ U \mathop{\mathrm{Spec}}(K)$ of group algebraic spaces over $K$. Hence $G_ x$ is locally quasi-finite over $k$ if and only if $\mathop{\mathrm{Spec}}(K) \to U$ maps into $U'$ (use the commutation of formation of $U'$ and Groupoids in Spaces, Lemma 78.6.3 applied to $\mathop{\mathrm{Spec}}(K) \to \mathop{\mathrm{Spec}}(k)$ and $G_ x$ to see this). This finishes the proof of (2). The proof of (1) is exactly the same. $\square$

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