Lemma 65.25.7. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$. The following are equivalent:

1. $x$ is a finite type point,

2. 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

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

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

2. $f$ is locally of finite type,

3. $f$ is a monomorphism, and

4. $x = f(z)$.

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 65.25.3. 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 63.10.5 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 65.23.4). 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 65.25.4 (and Lemma 65.25.6 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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