Lemma 101.18.7. Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$. The following are equivalent:

1. $x$ is a finite type point,

2. there exists an algebraic stack $\mathcal{Z}$ whose underlying topological space $|\mathcal{Z}|$ is a singleton, and a morphism $f : \mathcal{Z} \to \mathcal{X}$ which is locally of finite type such that $\{ x\} = |f|(|\mathcal{Z}|)$, and

3. the residual gerbe $\mathcal{Z}_ x$ of $\mathcal{X}$ at $x$ exists and the inclusion morphism $\mathcal{Z}_ x \to \mathcal{X}$ is locally of finite type.

Proof. (All of the morphisms occurring in this paragraph are representable by algebraic spaces, hence the conventions and results of Properties of Stacks, Section 100.3 are applicable.) Assume $x$ is a finite type point. Choose an affine scheme $U$, a closed point $u \in U$, and a smooth morphism $\varphi : U \to \mathcal{X}$ with $\varphi (u) = x$, see Lemma 101.18.3. Set $u = \mathop{\mathrm{Spec}}(\kappa (u))$ as usual. Set $R = u \times _\mathcal {X} u$ so that we obtain a groupoid in algebraic spaces $(u, R, s, t, c)$, see Algebraic Stacks, Lemma 94.16.1. The projection morphisms $R \to u$ are the compositions

$R = u \times _\mathcal {X} u \to u \times _\mathcal {X} U \to u \times _\mathcal {X} X = u$

where the first arrow is of finite type (a base change of the closed immersion of schemes $u \to U$) and the second arrow is smooth (a base change of the smooth morphism $U \to \mathcal{X}$). Hence $s, t : R \to u$ are locally of finite type (as compositions, see Morphisms of Spaces, Lemma 67.23.2). Since $u$ is the spectrum of a field, it follows that $s, t$ are flat and locally of finite presentation (by Morphisms of Spaces, Lemma 67.28.7). We see that $\mathcal{Z} = [u/R]$ is an algebraic stack by Criteria for Representability, Theorem 97.17.2. By Algebraic Stacks, Lemma 94.16.1 we obtain a canonical morphism

$f : \mathcal{Z} \longrightarrow \mathcal{X}$

which is fully faithful. Hence this morphism is representable by algebraic spaces, see Algebraic Stacks, Lemma 94.15.2 and a monomorphism, see Properties of Stacks, Lemma 100.8.4. It follows that the residual gerbe $\mathcal{Z}_ x \subset \mathcal{X}$ of $\mathcal{X}$ at $x$ exists and that $f$ factors through an equivalence $\mathcal{Z} \to \mathcal{Z}_ x$, see Properties of Stacks, Lemma 100.11.12. By construction the diagram

$\xymatrix{ u \ar[d] \ar[r] & U \ar[d] \\ \mathcal{Z} \ar[r]^ f & \mathcal{X} }$

is commutative. By Criteria for Representability, Lemma 97.17.1 the left vertical arrow is surjective, flat, and locally of finite presentation. Consider

$\xymatrix{ u \times _\mathcal {X} U \ar[d] \ar[r] & \mathcal{Z} \times _\mathcal {X} U \ar[r] \ar[d] & U \ar[d] \\ u \ar[r] & \mathcal{Z} \ar[r]^ f & \mathcal{X} }$

As $u \to \mathcal{X}$ is locally of finite type, we see that the base change $u \times _\mathcal {X} U \to U$ is locally of finite type. Moreover, $u \times _\mathcal {X} U \to \mathcal{Z} \times _\mathcal {X} U$ is surjective, flat, and locally of finite presentation as a base change of $u \to \mathcal{Z}$. Thus $\{ u \times _\mathcal {X} U \to \mathcal{Z} \times _\mathcal {X} U\}$ is an fppf covering of algebraic spaces, and we conclude that $\mathcal{Z} \times _\mathcal {X} U \to U$ is locally of finite type by Descent on Spaces, Lemma 74.16.1. By definition this means that $f$ is locally of finite type (because the vertical arrow $\mathcal{Z} \times _\mathcal {X} U \to \mathcal{Z}$ is smooth as a base change of $U \to \mathcal{X}$ and surjective as $\mathcal{Z}$ has only one point). Since $\mathcal{Z} = \mathcal{Z}_ x$ we see that (3) holds.

It is clear that (3) implies (2). If (2) holds then $x$ is a finite type point of $\mathcal{X}$ by Lemma 101.18.4 and Lemma 101.18.6 to see that $\mathcal{Z}_{\text{ft-pts}}$ is nonempty, i.e., the unique point of $\mathcal{Z}$ is a finite type point of $\mathcal{Z}$. $\square$

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