## 100.18 Points of finite type

Let $\mathcal{X}$ be an algebraic stack. A finite type point $x \in |\mathcal{X}|$ is a point which can be represented by a morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ which is locally of finite type. Finite type points are a suitable replacement of closed points for algebraic spaces and algebraic stacks. There are always “enough of them” for example.

Lemma 100.18.1. Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$. The following are equivalent:

There exists a morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ which is locally of finite type and represents $x$.

There exists a scheme $U$, a closed point $u \in U$, and a smooth morphism $\varphi : U \to \mathcal{X}$ such that $\varphi (u) = x$.

**Proof.**
Let $u \in U$ and $U \to \mathcal{X}$ be as in (2). Then $\mathop{\mathrm{Spec}}(\kappa (u)) \to U$ is of finite type, and $U \to \mathcal{X}$ is representable and locally of finite type (by Morphisms of Spaces, Lemmas 66.39.8 and 66.28.5). Hence we see (1) holds by Lemma 100.17.2.

Conversely, assume $\mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ is locally of finite type and represents $x$. Let $U \to \mathcal{X}$ be a surjective smooth morphism where $U$ is a scheme. By assumption $U \times _\mathcal {X} \mathop{\mathrm{Spec}}(k) \to U$ is a morphism of algebraic spaces which is locally of finite type. Pick a finite type point $v$ of $U \times _\mathcal {X} \mathop{\mathrm{Spec}}(k)$ (there exists at least one, see Morphisms of Spaces, Lemma 66.25.3). By Morphisms of Spaces, Lemma 66.25.4 the image $u \in U$ of $v$ is a finite type point of $U$. Hence by Morphisms, Lemma 29.16.4 after shrinking $U$ we may assume that $u$ is a closed point of $U$, i.e., (2) holds.
$\square$

Definition 100.18.2. Let $\mathcal{X}$ be an algebraic stack. We say a point $x \in |\mathcal{X}|$ is a *finite type point*^{1} if the equivalent conditions of Lemma 100.18.1 are satisfied. We denote $\mathcal{X}_{\text{ft-pts}}$ the set of finite type points of $\mathcal{X}$.

We can describe the set of finite type points as follows.

Lemma 100.18.3. Let $\mathcal{X}$ be an algebraic stack. We have

\[ \mathcal{X}_{\text{ft-pts}} = \bigcup \nolimits _{\varphi : U \to \mathcal{X}\text{ smooth}} |\varphi |(U_0) \]

where $U_0$ is the set of closed points of $U$. Here we may let $U$ range over all schemes smooth over $\mathcal{X}$ or over all affine schemes smooth over $\mathcal{X}$.

**Proof.**
Immediate from Lemma 100.18.1.
$\square$

Lemma 100.18.4. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. If $f$ is locally of finite type, then $f(\mathcal{X}_{\text{ft-pts}}) \subset \mathcal{Y}_{\text{ft-pts}}$.

**Proof.**
Take $x \in \mathcal{X}_{\text{ft-pts}}$. Represent $x$ by a locally finite type morphism $x : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$. Then $f \circ x$ is locally of finite type by Lemma 100.17.2. Hence $f(x) \in \mathcal{Y}_{\text{ft-pts}}$.
$\square$

Lemma 100.18.5. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. If $f$ is locally of finite type and surjective, then $f(\mathcal{X}_{\text{ft-pts}}) = \mathcal{Y}_{\text{ft-pts}}$.

**Proof.**
We have $f(\mathcal{X}_{\text{ft-pts}}) \subset \mathcal{Y}_{\text{ft-pts}}$ by Lemma 100.18.4. Let $y \in |\mathcal{Y}|$ be a finite type point. Represent $y$ by a morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{Y}$ which is locally of finite type. As $f$ is surjective the algebraic stack $\mathcal{X}_ k = \mathop{\mathrm{Spec}}(k) \times _\mathcal {Y} \mathcal{X}$ is nonempty, therefore has a finite type point $x \in |\mathcal{X}_ k|$ by Lemma 100.18.3. Now $\mathcal{X}_ k \to \mathcal{X}$ is a morphism which is locally of finite type as a base change of $\mathop{\mathrm{Spec}}(k) \to \mathcal{Y}$ (Lemma 100.17.3). Hence the image of $x$ in $\mathcal{X}$ is a finite type point by Lemma 100.18.4 which maps to $y$ by construction.
$\square$

Lemma 100.18.6. Let $\mathcal{X}$ be an algebraic stack. For any locally closed subset $T \subset |\mathcal{X}|$ we have

\[ T \not= \emptyset \Rightarrow T \cap \mathcal{X}_{\text{ft-pts}} \not= \emptyset . \]

In particular, for any closed subset $T \subset |\mathcal{X}|$ we see that $T \cap \mathcal{X}_{\text{ft-pts}}$ is dense in $T$.

**Proof.**
Let $i : \mathcal{Z} \to \mathcal{X}$ be the reduced induced substack structure on $T$, see Properties of Stacks, Remark 99.10.5. An immersion is locally of finite type, see Lemma 100.17.4. Hence by Lemma 100.18.4 we see $\mathcal{Z}_{\text{ft-pts}} \subset \mathcal{X}_{\text{ft-pts}} \cap T$. Finally, any nonempty affine scheme $U$ with a smooth morphism towards $\mathcal{Z}$ has at least one closed point, hence $\mathcal{Z}$ has at least one finite type point by Lemma 100.18.3. The lemma follows.
$\square$

Here is another, more technical, characterization of a finite type point on an algebraic stack. It tells us in particular that the residual gerbe of $\mathcal{X}$ at $x$ exists whenever $x$ is a finite type point!

Lemma 100.18.7. Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$. The following are equivalent:

$x$ is a finite type point,

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

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 99.3 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 100.18.3. 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 93.16.1. 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 66.23.2). 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 66.28.7). We see that $\mathcal{Z} = [u/R]$ is an algebraic stack by Criteria for Representability, Theorem 96.17.2. By Algebraic Stacks, Lemma 93.16.1 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 93.15.2 and a monomorphism, see Properties of Stacks, Lemma 99.8.4. 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 99.11.12. 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 96.17.1 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 73.16.1. 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 100.18.4 and Lemma 100.18.6 to see that $\mathcal{Z}_{\text{ft-pts}}$ is nonempty, i.e., the unique point of $\mathcal{Z}$ is a finite type point of $\mathcal{Z}$.
$\square$

## Comments (2)

Comment #4881 by Olivier de Gaay Fortman on

Comment #5161 by Johan on