## 66.14 Decent spaces

In this section we collect some useful facts on decent spaces.

Lemma 66.14.1. Any locally Noetherian decent algebraic space is quasi-separated.

Proof. Namely, let $X$ be an algebraic space (over some base scheme, for example over $\mathbf{Z}$) which is decent and locally Noetherian. Let $U \to X$ and $V \to X$ be étale morphisms with $U$ and $V$ affine schemes. We have to show that $W = U \times _ X V$ is quasi-compact (Properties of Spaces, Lemma 64.3.3). Since $X$ is locally Noetherian, the schemes $U$, $V$ are Noetherian and $W$ is locally Noetherian. Since $X$ is decent, the fibres of the morphism $W \to U$ are finite. Namely, we can represent any $x \in |X|$ by a quasi-compact monomorphism $\mathop{\mathrm{Spec}}(k) \to X$. Then $U_ k$ and $V_ k$ are finite disjoint unions of spectra of finite separable extensions of $k$ (Remark 66.4.1) and we see that $W_ k = U_ k \times _{\mathop{\mathrm{Spec}}(k)} V_ k$ is finite. Let $n$ be the maximum degree of a fibre of $W \to U$ at a generic point of an irreducible component of $U$. Consider the stratification

$U = U_0 \supset U_1 \supset U_2 \supset \ldots$

associated to $W \to U$ in More on Morphisms, Lemma 37.40.5. By our choice of $n$ above we conclude that $U_{n + 1}$ is empty. Hence we see that the fibres of $W \to U$ are universally bounded. Then we can apply More on Morphisms, Lemma 37.40.3 to find a stratification

$\emptyset = Z_{-1} \subset Z_0 \subset Z_1 \subset Z_2 \subset \ldots \subset Z_ n = U$

by closed subsets such that with $S_ r = Z_ r \setminus Z_{r - 1}$ the morphism $W \times _ U S_ r \to S_ r$ is finite locally free. Since $U$ is Noetherian, the schemes $S_ r$ are Noetherian, whence the schemes $W \times _ U S_ r$ are Noetherian, whence $W = \coprod W \times _ U S_ r$ is quasi-compact as desired. $\square$

Lemma 66.14.2. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$.

1. If $|X|$ is a singleton then $X$ is a scheme.

2. If $|X|$ is a singleton and $X$ is reduced, then $X \cong \mathop{\mathrm{Spec}}(k)$ for some field $k$.

Proof. Assume $|X|$ is a singleton. It follows immediately from Theorem 66.10.2 that $X$ is a scheme, but we can also argue directly as follows. Choose an affine scheme $U$ and a surjective étale morphism $U \to X$. Set $R = U \times _ X U$. Then $U$ and $R$ have finitely many points by Lemma 66.4.5 (and the definition of a decent space). All of these points are closed in $U$ and $R$ by Lemma 66.12.1. It follows that $U$ and $R$ are affine schemes. We may shrink $U$ to a singleton space. Then $U$ is the spectrum of a henselian local ring, see Algebra, Lemma 10.152.10. The projections $R \to U$ are étale, hence finite étale because $U$ is the spectrum of a $0$-dimensional henselian local ring, see Algebra, Lemma 10.152.3. It follows that $X$ is a scheme by Groupoids, Proposition 39.23.9.

Part (2) follows from (1) and the fact that a reduced singleton scheme is the spectrum of a field. $\square$

Remark 66.14.3. We will see in Limits of Spaces, Lemma 68.15.3 that an algebraic space whose reduction is a scheme is a scheme.

Lemma 66.14.4. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Consider a commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(k) \ar[rr] \ar[rd] & & X \ar[ld] \\ & S }$

Assume that the image point $s \in S$ of $\mathop{\mathrm{Spec}}(k) \to S$ is a closed point and that $\kappa (s) \subset k$ is algebraic. Then the image $x$ of $\mathop{\mathrm{Spec}}(k) \to X$ is a closed point of $|X|$.

Proof. Suppose that $x \leadsto x'$ for some $x' \in |X|$. Choose an étale morphism $U \to X$ where $U$ is a scheme and a point $u' \in U'$ mapping to $x'$. Choose a specialization $u \leadsto u'$ in $U$ with $u$ mapping to $x$ in $X$, see Lemma 66.12.2. Then $u$ is the image of a point $w$ of the scheme $W = \mathop{\mathrm{Spec}}(k) \times _ X U$. Since the projection $W \to \mathop{\mathrm{Spec}}(k)$ is étale we see that $\kappa (w) \supset k$ is finite. Hence $\kappa (w) \supset \kappa (s)$ is algebraic. Hence $\kappa (u) \supset \kappa (s)$ is algebraic. Thus $u$ is a closed point of $U$ by Morphisms, Lemma 29.20.2. Thus $u = u'$, whence $x = x'$. $\square$

Lemma 66.14.5. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Consider a commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(k) \ar[rr] \ar[rd] & & X \ar[ld] \\ & S }$

Assume that the image point $s \in S$ of $\mathop{\mathrm{Spec}}(k) \to S$ is a closed point and that $\kappa (s) \subset k$ is finite. Then $\mathop{\mathrm{Spec}}(k) \to X$ is finite morphism. If $\kappa (s) = k$ then $\mathop{\mathrm{Spec}}(k) \to X$ is closed immersion.

Proof. By Lemma 66.14.4 the image point $x \in |X|$ is closed. Let $Z \subset X$ be the reduced closed subspace with $|Z| = \{ x\}$ (Properties of Spaces, Lemma 64.12.3). Note that $Z$ is a decent algebraic space by Lemma 66.6.5. By Lemma 66.14.2 we see that $Z = \mathop{\mathrm{Spec}}(k')$ for some field $k'$. Of course $k \supset k' \supset \kappa (s)$. Then $\mathop{\mathrm{Spec}}(k) \to Z$ is a finite morphism of schemes and $Z \to X$ is a finite morphism as it is a closed immersion. Hence $\mathop{\mathrm{Spec}}(k) \to X$ is finite (Morphisms of Spaces, Lemma 65.45.4). If $k = \kappa (s)$, then $\mathop{\mathrm{Spec}}(k) = Z$ and $\mathop{\mathrm{Spec}}(k) \to X$ is a closed immersion. $\square$

Lemma 66.14.6. Let $S$ be a scheme. Suppose $X$ is a decent algebraic space over $S$. Let $x \in |X|$ be a closed point. Then $x$ can be represented by a closed immersion $i : \mathop{\mathrm{Spec}}(k) \to X$ from the spectrum of a field.

Proof. We know that $x$ can be represented by a quasi-compact monomorphism $i : \mathop{\mathrm{Spec}}(k) \to X$ where $k$ is a field (Definition 66.6.1). Let $U \to X$ be an étale morphism where $U$ is an affine scheme. As $x$ is closed and $X$ decent, the fibre $F$ of $|U| \to |X|$ over $x$ consists of closed points (Lemma 66.12.1). As $i$ is a monomorphism, so is $U_ k = U \times _ X \mathop{\mathrm{Spec}}(k) \to U$. In particular, the map $|U_ k| \to F$ is injective. Since $U_ k$ is quasi-compact and étale over a field, we see that $U_ k$ is a finite disjoint union of spectra of fields (Remark 66.4.1). Say $U_ k = \mathop{\mathrm{Spec}}(k_1) \amalg \ldots \amalg \mathop{\mathrm{Spec}}(k_ r)$. Since $\mathop{\mathrm{Spec}}(k_ i) \to U$ is a monomorphism, we see that its image $u_ i$ has residue field $\kappa (u_ i) = k_ i$. Since $u_ i \in F$ is a closed point we conclude the morphism $\mathop{\mathrm{Spec}}(k_ i) \to U$ is a closed immersion. As the $u_ i$ are pairwise distinct, $U_ k \to U$ is a closed immersion. Hence $i$ is a closed immersion (Morphisms of Spaces, Lemma 65.12.1). This finishes the proof. $\square$

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