## 98.18 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 98.18.1. Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$. The following are equivalent:

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

2. 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 64.39.8 and 64.28.5). Hence we see (1) holds by Lemma 98.17.2.

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 64.25.3). By Morphisms of Spaces, Lemma 64.25.4 the image $u \in U$ of $v$ is a finite type point of $U$. Hence by Morphisms, Lemma 29.15.4 after shrinking $U$ we may assume that $u$ is a closed point of $U$, i.e., (2) holds. $\square$

Definition 98.18.2. Let $\mathcal{X}$ be an algebraic stack. We say a point $x \in |\mathcal{X}|$ is a finite type point1 if the equivalent conditions of Lemma 98.18.1 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 98.18.3. Let $\mathcal{X}$ be an algebraic stack. We have

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

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 98.18.1. $\square$

Lemma 98.18.4. 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 98.17.2. Hence $f(x) \in \mathcal{Y}_{\text{ft-pts}}$. $\square$

Lemma 98.18.5. 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 98.18.4. 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 98.18.3. 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 98.17.3). Hence the image of $x$ in $\mathcal{X}$ is a finite type point by Lemma 98.18.4 which maps to $y$ by construction. $\square$

Lemma 98.18.6. 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 97.10.5. An immersion is locally of finite type, see Lemma 98.17.4. Hence by Lemma 98.18.4 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 98.18.3. 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 98.18.7. 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 97.3 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 98.18.3. 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 91.16.1. 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 64.23.2). 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 64.28.7). We see that $\mathcal{Z} = [u/R]$ is an algebraic stack by Criteria for Representability, Theorem 94.17.2. By Algebraic Stacks, Lemma 91.16.1 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 91.15.2 and a monomorphism, see Properties of Stacks, Lemma 97.8.4. 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 97.11.11. 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 94.17.1 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 71.15.1. 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 98.18.4 and Lemma 98.18.6 to see that $\mathcal{Z}_{\text{ft-pts}}$ is nonempty, i.e., the unique point of $\mathcal{Z}$ is a finite type point of $\mathcal{Z}$. $\square$

 This is a slight abuse of language as it would perhaps be more correct to say “locally finite type point”.

Comment #4881 by on

There seems to be a small typo in the subscript of the union in Lemma 06FZ, where $X$ should be replaced by $\mathcal X$.

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