The Stacks project

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

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

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

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. $\square$