## 67.25 Points of finite type

Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. A finite type point $x \in |X|$ is a point which can be represented by a morphism $\mathop{\mathrm{Spec}}(k) \to X$ 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 67.25.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$. The following are equivalent:

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

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

**Proof.**
Let $u \in U$ and $U \to X$ be as in (2). Then $\mathop{\mathrm{Spec}}(\kappa (u)) \to U$ is of finite type, and $U \to X$ is representable and locally of finite type (by the general principle Spaces, Lemma 65.5.8 and Morphisms, Lemmas 29.36.11 and 29.21.8). Hence we see (1) holds by Lemma 67.23.2.

Conversely, assume $\mathop{\mathrm{Spec}}(k) \to X$ is locally of finite type and represents $x$. Let $U \to X$ be a surjective étale morphism where $U$ is a scheme. By assumption $U \times _ X \mathop{\mathrm{Spec}}(k) \to U$ is locally of finite type. Pick a finite type point $v$ of $U \times _ X \mathop{\mathrm{Spec}}(k)$ (there exists at least one, see Morphisms, Lemma 29.16.4). By Morphisms, Lemma 29.16.5 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 67.25.2. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. We say a point $x \in |X|$ is a *finite type point*^{1} if the equivalent conditions of Lemma 67.25.1 are satisfied. We denote $X_{\text{ft-pts}}$ the set of finite type points of $X$.

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

Lemma 67.25.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. We have

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

where $U_0$ is the set of closed points of $U$. Here we may let $U$ range over all schemes étale over $X$ or over all affine schemes étale over $X$.

**Proof.**
Immediate from Lemma 67.25.1.
$\square$

Lemma 67.25.4. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. If $f$ is locally of finite type, then $f(X_{\text{ft-pts}}) \subset Y_{\text{ft-pts}}$.

**Proof.**
Take $x \in X_{\text{ft-pts}}$. Represent $x$ by a locally finite type morphism $x : \mathop{\mathrm{Spec}}(k) \to X$. Then $f \circ x$ is locally of finite type by Lemma 67.23.2. Hence $f(x) \in Y_{\text{ft-pts}}$.
$\square$

Lemma 67.25.5. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. If $f$ is locally of finite type and surjective, then $f(X_{\text{ft-pts}}) = Y_{\text{ft-pts}}$.

**Proof.**
We have $f(X_{\text{ft-pts}}) \subset Y_{\text{ft-pts}}$ by Lemma 67.25.4. Let $y \in |Y|$ be a finite type point. Represent $y$ by a morphism $\mathop{\mathrm{Spec}}(k) \to Y$ which is locally of finite type. As $f$ is surjective the algebraic space $X_ k = \mathop{\mathrm{Spec}}(k) \times _ Y X$ is nonempty, therefore has a finite type point $x \in |X_ k|$ by Lemma 67.25.3. Now $X_ k \to X$ is a morphism which is locally of finite type as a base change of $\mathop{\mathrm{Spec}}(k) \to Y$ (Lemma 67.23.3). Hence the image of $x$ in $X$ is a finite type point by Lemma 67.25.4 which maps to $y$ by construction.
$\square$

Lemma 67.25.6. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. For any locally closed subset $T \subset |X|$ we have

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

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

**Proof.**
Let $i : Z \to X$ be the reduced induce subspace structure on $T$, see Remark 67.12.5. Any immersion is locally of finite type, see Lemma 67.23.7. Hence by Lemma 67.25.4 we see $Z_{\text{ft-pts}} \subset X_{\text{ft-pts}} \cap T$. Finally, any nonempty affine scheme $U$ with an étale morphism towards $Z$ has at least one closed point. Hence $Z$ has at least one finite type point by Lemma 67.25.3. The lemma follows.
$\square$

Here is another, more technical, characterization of a finite type point on an algebraic space.

Lemma 67.25.7. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$. The following are equivalent:

$x$ is a finite type point,

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

there exists an algebraic space $Z$ and a morphism $f : Z \to X$ with the following properties:

there is a surjective étale morphism $z : \mathop{\mathrm{Spec}}(k) \to Z$ where $k$ is a field,

$f$ is locally of finite type,

$f$ is a monomorphism, and

$x = f(z)$.

**Proof.**
Assume $x$ is a finite type point. Choose an affine scheme $U$, a closed point $u \in U$, and an étale morphism $\varphi : U \to X$ with $\varphi (u) = x$, see Lemma 67.25.3. Set $u = \mathop{\mathrm{Spec}}(\kappa (u))$ as usual. The projection morphisms $u \times _ X u \to u$ are the compositions

\[ u \times _ X u \to u \times _ X U \to u \times _ X X = u \]

where the first arrow is a closed immersion (a base change of $u \to U$) and the second arrow is étale (a base change of the étale morphism $U \to X$). Hence $u \times _ X U$ is a disjoint union of spectra of finite separable extensions of $k$ (see Morphisms, Lemma 29.36.7) and therefore the closed subscheme $u \times _ X u$ is a disjoint union of finite separable extension of $k$, i.e., $u \times _ X u \to u$ is étale. By Spaces, Theorem 65.10.5 we see that $Z = u/u \times _ X u$ is an algebraic space. By construction the diagram

\[ \xymatrix{ u \ar[d] \ar[r] & U \ar[d] \\ Z \ar[r] & X } \]

is commutative with étale vertical arrows. Hence $Z \to X$ is locally of finite type (see Lemma 67.23.4). By construction the morphism $Z \to X$ is a monomorphism and the image of $z$ is $x$. Thus (3) holds.

It is clear that (3) implies (2). If (2) holds then $x$ is a finite type point of $X$ by Lemma 67.25.4 (and Lemma 67.25.6 to see that $Z_{\text{ft-pts}}$ is nonempty, i.e., the unique point of $Z$ is a finite type point of $Z$).
$\square$

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