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 \input{preamble} % OK, start here. % \begin{document} \title{Decent Algebraic Spaces} \maketitle \phantomsection \label{section-phantom} \tableofcontents \section{Introduction} \label{section-introduction} \noindent In this chapter we study local'' properties of general algebraic spaces, i.e., those algebraic spaces which aren't quasi-separated. Quasi-separated algebraic spaces are studied in \cite{Kn}. It turns out that essentially new phenomena happen, especially regarding points and specializations of points, on more general algebraic spaces. On the other hand, for most basic results on algebraic spaces, one needn't worry about these phenomena, which is why we have decided to have this material in a separate chapter following the standard development of the theory. \section{Conventions} \label{section-conventions} \noindent The standing assumption is that all schemes are contained in a big fppf site $\Sch_{fppf}$. And all rings $A$ considered have the property that $\Spec(A)$ is (isomorphic) to an object of this big site. \medskip\noindent Let $S$ be a scheme and let $X$ be an algebraic space over $S$. In this chapter and the following we will write $X \times_S X$ for the product of $X$ with itself (in the category of algebraic spaces over $S$), instead of $X \times X$. \section{Universally bounded fibres} \label{section-universally-bounded} \noindent We briefly discuss what it means for a morphism from a scheme to an algebraic space to have universally bounded fibres. Please refer to Morphisms, Section \ref{morphisms-section-universally-bounded} for similar definitions and results on morphisms of schemes. \begin{definition} \label{definition-universally-bounded} Let $S$ be a scheme. Let $X$ be an algebraic space over $S$, and let $U$ be a scheme over $S$. Let $f : U \to X$ be a morphism over $S$. We say the {\it fibres of $f$ are universally bounded}\footnote{This is probably nonstandard notation.} if there exists an integer $n$ such that for all fields $k$ and all morphisms $\Spec(k) \to X$ the fibre product $\Spec(k) \times_X U$ is a finite scheme over $k$ whose degree over $k$ is $\leq n$. \end{definition} \noindent This definition makes sense because the fibre product $\Spec(k) \times_Y X$ is a scheme. Moreover, if $Y$ is a scheme we recover the notion of Morphisms, Definition \ref{morphisms-definition-universally-bounded} by virtue of Morphisms, Lemma \ref{morphisms-lemma-characterize-universally-bounded}. \begin{lemma} \label{lemma-composition-universally-bounded} Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $V \to U$ be a morphism of schemes over $S$, and let $U \to X$ be a morphism from $U$ to $X$. If the fibres of $V \to U$ and $U \to X$ are universally bounded, then so are the fibres of $V \to X$. \end{lemma} \begin{proof} Let $n$ be an integer which works for $V \to U$, and let $m$ be an integer which works for $U \to X$ in Definition \ref{definition-universally-bounded}. Let $\Spec(k) \to X$ be a morphism, where $k$ is a field. Consider the morphisms $$\Spec(k) \times_X V \longrightarrow \Spec(k) \times_X U \longrightarrow \Spec(k).$$ By assumption the scheme $\Spec(k) \times_X U$ is finite of degree at most $m$ over $k$, and $n$ is an integer which bounds the degree of the fibres of the first morphism. Hence by Morphisms, Lemma \ref{morphisms-lemma-composition-universally-bounded} we conclude that $\Spec(k) \times_X V$ is finite over $k$ of degree at most $nm$. \end{proof} \begin{lemma} \label{lemma-base-change-universally-bounded} Let $S$ be a scheme. Let $Y \to X$ be a representable morphism of algebraic spaces over $S$. Let $U \to X$ be a morphism from a scheme to $X$. If the fibres of $U \to X$ are universally bounded, then the fibres of $U \times_X Y \to Y$ are universally bounded. \end{lemma} \begin{proof} This is clear from the definition, and properties of fibre products. (Note that $U \times_X Y$ is a scheme as we assumed $Y \to X$ representable, so the definition applies.) \end{proof} \begin{lemma} \label{lemma-descent-universally-bounded} Let $S$ be a scheme. Let $g : Y \to X$ be a representable morphism of algebraic spaces over $S$. Let $f : U \to X$ be a morphism from a scheme towards $X$. Let $f' : U \times_X Y \to Y$ be the base change of $f$. If $$\Im(|f| : |U| \to |X|) \subset \Im(|g| : |Y| \to |X|)$$ and $f'$ has universally bounded fibres, then $f$ has universally bounded fibres. \end{lemma} \begin{proof} Let $n \geq 0$ be an integer bounding the degrees of the fibre products $\Spec(k) \times_Y (U \times_X Y)$ as in Definition \ref{definition-universally-bounded} for the morphism $f'$. We claim that $n$ works for $f$ also. Namely, suppose that $x : \Spec(k) \to X$ is a morphism from the spectrum of a field. Then either $\Spec(k) \times_X U$ is empty (and there is nothing to prove), or $x$ is in the image of $|f|$. By Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-cartesian} and the assumption of the lemma we see that this means there exists a field extension $k \subset k'$ and a commutative diagram $$\xymatrix{ \Spec(k') \ar[r] \ar[d] & Y \ar[d] \\ \Spec(k) \ar[r] & X }$$ Hence we see that $$\Spec(k') \times_Y (U \times_X Y) = \Spec(k') \times_{\Spec(k)} (\Spec(k) \times_X U)$$ Since the scheme $\Spec(k') \times_Y (U \times_X Y)$ is assumed finite of degree $\leq n$ over $k'$ it follows that also $\Spec(k) \times_X U$ is finite of degree $\leq n$ over $k$ as desired. (Some details omitted.) \end{proof} \begin{lemma} \label{lemma-universally-bounded-permanence} Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Consider a commutative diagram $$\xymatrix{ U \ar[rd]_g \ar[rr]_f & & V \ar[ld]^h \\ & X & }$$ where $U$ and $V$ are schemes. If $g$ has universally bounded fibres, and $f$ is surjective and flat, then also $h$ has universally bounded fibres. \end{lemma} \begin{proof} Assume $g$ has universally bounded fibres, and $f$ is surjective and flat. Say $n \geq 0$ is an integer which bounds the degrees of the schemes $\Spec(k) \times_X U$ as in Definition \ref{definition-universally-bounded}. We claim $n$ also works for $h$. Let $\Spec(k) \to X$ be a morphism from the spectrum of a field to $X$. Consider the morphism of schemes $$\Spec(k) \times_X V \longrightarrow \Spec(k) \times_X U$$ It is flat and surjective. By assumption the scheme on the left is finite of degree $\leq n$ over $\Spec(k)$. It follows from Morphisms, Lemma \ref{morphisms-lemma-universally-bounded-permanence} that the degree of the scheme on the right is also bounded by $n$ as desired. \end{proof} \begin{lemma} \label{lemma-universally-bounded-finite-fibres} Let $S$ be a scheme. Let $X$ be an algebraic space over $S$, and let $U$ be a scheme over $S$. Let $\varphi : U \to X$ be a morphism over $S$. If the fibres of $\varphi$ are universally bounded, then there exists an integer $n$ such that each fibre of $|U| \to |X|$ has at most $n$ elements. \end{lemma} \begin{proof} The integer $n$ of Definition \ref{definition-universally-bounded} works. Namely, pick $x \in |X|$. Represent $x$ by a morphism $x : \Spec(k) \to X$. Then we get a commutative diagram $$\xymatrix{ \Spec(k) \times_X U \ar[r] \ar[d] & U \ar[d] \\ \Spec(k) \ar[r]^x & X }$$ which shows (via Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-cartesian}) that the inverse image of $x$ in $|U|$ is the image of the top horizontal arrow. Since $\Spec(k) \times_X U$ is finite of degree $\leq n$ over $k$ it has at most $n$ points. \end{proof} \section{Finiteness conditions and points} \label{section-points-monomorphisms} \noindent In this section we elaborate on the question of when points can be represented by monomorphisms from spectra of fields into the space. \begin{remark} \label{remark-recall} Before we give the proof of the next lemma let us recall some facts about \'etale morphisms of schemes: \begin{enumerate} \item An \'etale morphism is flat and hence generalizations lift along an \'etale morphism (Morphisms, Lemmas \ref{morphisms-lemma-etale-flat} and \ref{morphisms-lemma-generalizations-lift-flat}). \item An \'etale morphism is unramified, an unramified morphism is locally quasi-finite, hence fibres are discrete (Morphisms, Lemmas \ref{morphisms-lemma-flat-unramified-etale}, \ref{morphisms-lemma-unramified-quasi-finite}, and \ref{morphisms-lemma-quasi-finite-at-point-characterize}). \item A quasi-compact \'etale morphism is quasi-finite and in particular has finite fibres (Morphisms, Lemmas \ref{morphisms-lemma-quasi-finite-locally-quasi-compact} and \ref{morphisms-lemma-quasi-finite}). \item An \'etale scheme over a field $k$ is a disjoint union of spectra of finite separable field extension of $k$ (Morphisms, Lemma \ref{morphisms-lemma-etale-over-field}). \end{enumerate} For a general discussion of \'etale morphisms, please see \'Etale Morphisms, Section \ref{etale-section-etale-morphisms}. \end{remark} \begin{lemma} \label{lemma-U-finite-above-x} Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$. The following are equivalent: \begin{enumerate} \item there exists a family of schemes $U_i$ and \'etale morphisms $\varphi_i : U_i \to X$ such that $\coprod \varphi_i : \coprod U_i \to X$ is surjective, and such that for each $i$ the fibre of $|U_i| \to |X|$ over $x$ is finite, and \item for every affine scheme $U$ and \'etale morphism $\varphi : U \to X$ the fibre of $|U| \to |X|$ over $x$ is finite. \end{enumerate} \end{lemma} \begin{proof} The implication (2) $\Rightarrow$ (1) is trivial. Let $\varphi_i : U_i \to X$ be a family of \'etale morphisms as in (1). Let $\varphi : U \to X$ be an \'etale morphism from an affine scheme towards $X$. Consider the fibre product diagrams $$\xymatrix{ U \times_X U_i \ar[r]_-{p_i} \ar[d]_{q_i} & U_i \ar[d]^{\varphi_i} \\ U \ar[r]^\varphi & X } \quad \quad \xymatrix{ \coprod U \times_X U_i \ar[r]_-{\coprod p_i} \ar[d]_{\coprod q_i} & \coprod U_i \ar[d]^{\coprod \varphi_i} \\ U \ar[r]^\varphi & X }$$ Since $q_i$ is \'etale it is open (see Remark \ref{remark-recall}). Moreover, the morphism $\coprod q_i$ is surjective. Hence there exist finitely many indices $i_1, \ldots, i_n$ and a quasi-compact opens $W_{i_j} \subset U \times_X U_{i_j}$ which surject onto $U$. The morphism $p_i$ is \'etale, hence locally quasi-finite (see remark on \'etale morphisms above). Thus we may apply Morphisms, Lemma \ref{morphisms-lemma-locally-quasi-finite-qc-source-universally-bounded} to see the fibres of $p_{i_j}|_{W_{i_j}} : W_{i_j} \to U_i$ are finite. Hence by Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-cartesian} and the assumption on $\varphi_i$ we conclude that the fibre of $\varphi$ over $x$ is finite. In other words (2) holds. \end{proof} \begin{lemma} \label{lemma-R-finite-above-x} Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$. The following are equivalent: \begin{enumerate} \item there exists a scheme $U$, an \'etale morphism $\varphi : U \to X$, and points $u, u' \in U$ mapping to $x$ such that setting $R = U \times_X U$ the fibre of $$|R| \to |U| \times_{|X|} |U|$$ over $(u, u')$ is finite, \item for every scheme $U$, \'etale morphism $\varphi : U \to X$ and any points $u, u' \in U$ mapping to $x$ setting $R = U \times_X U$ the fibre of $$|R| \to |U| \times_{|X|} |U|$$ over $(u, u')$ is finite, \item there exists a morphism $\Spec(k) \to X$ with $k$ a field in the equivalence class of $x$ such that the projections $\Spec(k) \times_X \Spec(k) \to \Spec(k)$ are \'etale and quasi-compact, and \item there exists a monomorphism $\Spec(k) \to X$ with $k$ a field in the equivalence class of $x$. \end{enumerate} \end{lemma} \begin{proof} Assume (1), i.e., let $\varphi : U \to X$ be an \'etale morphism from a scheme towards $X$, and let $u, u'$ be points of $U$ lying over $x$ such that the fibre of $|R| \to |U| \times_{|X|} |U|$ over $(u, u')$ is a finite set. In this proof we think of a point $u = \Spec(\kappa(u))$ as a scheme. Note that $u \to U$, $u' \to U$ are monomorphisms (see Schemes, Lemma \ref{schemes-lemma-injective-points-surjective-stalks}), hence $u \times_X u' \to R = U \times_X U$ is a monomorphism. In this language the assumption really means that $u \times_X u'$ is a scheme whose underlying topological space has finitely many points. Let $\psi : W \to X$ be an \'etale morphism from a scheme towards $X$. Let $w, w' \in W$ be points of $W$ mapping to $x$. We have to show that $w \times_X w'$ is a scheme whose underlying topological space has finitely many points. Consider the fibre product diagram $$\xymatrix{ W \times_X U \ar[r]_p \ar[d]_q & U \ar[d]^\varphi \\ W \ar[r]^\psi & X }$$ As $x$ is the image of $u$ and $u'$ we may pick points $\tilde w, \tilde w'$ in $W \times_X U$ with $q(\tilde w) = w$, $q(\tilde w') = w'$, $u = p(\tilde w)$ and $u' = p(\tilde w')$, see Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-cartesian}. As $p$, $q$ are \'etale the field extensions $\kappa(w) \subset \kappa(\tilde w) \supset \kappa(u)$ and $\kappa(w') \subset \kappa(\tilde w') \supset \kappa(u')$ are finite separable, see Remark \ref{remark-recall}. Then we get a commutative diagram $$\xymatrix{ w \times_X w' \ar[d] & \tilde w \times_X \tilde w' \ar[l] \ar[d] \ar[r] & u \times_X u' \ar[d] \\ w \times_X w' & \tilde w \times_S \tilde w' \ar[l] \ar[r] & u \times_S u' }$$ where the squares are fibre product squares. The lower horizontal morphisms are \'etale and quasi-compact, as any scheme of the form $\Spec(k) \times_S \Spec(k')$ is affine, and by our observations about the field extensions above. Thus we see that the top horizontal arrows are \'etale and quasi-compact and hence have finite fibres. We have seen above that $|u \times_X u'|$ is finite, so we conclude that $|w \times_X w'|$ is finite. In other words, (2) holds. \medskip\noindent Assume (2). Let $U \to X$ be an \'etale morphism from a scheme $U$ such that $x$ is in the image of $|U| \to |X|$. Let $u \in U$ be a point mapping to $x$. Then we have seen in the previous paragraph that $u = \Spec(\kappa(u)) \to X$ has the property that $u \times_X u$ has a finite underlying topological space. On the other hand, the projection maps $u \times_X u \to u$ are the composition $$u \times_X u \longrightarrow u \times_X U \longrightarrow u \times_X X = u,$$ i.e., the composition of a monomorphism (the base change of the monomorphism $u \to U$) by an \'etale morphism (the base change of the \'etale morphism $U \to X$). Hence $u \times_X U$ is a disjoint union of spectra of fields finite separable over $\kappa(u)$ (see Remark \ref{remark-recall}). Since $u \times_X u$ is finite the image of it in $u \times_X U$ is a finite disjoint union of spectra of fields finite separable over $\kappa(u)$. By Schemes, Lemma \ref{schemes-lemma-mono-towards-spec-field} we conclude that $u \times_X u$ is a finite disjoint union of spectra of fields finite separable over $\kappa(u)$. In other words, we see that $u \times_X u \to u$ is quasi-compact and \'etale. This means that (3) holds. \medskip\noindent Let us prove that (3) implies (4). Let $\Spec(k) \to X$ be a morphism from the spectrum of a field into $X$, in the equivalence class of $x$ such that the two projections $t, s : R = \Spec(k) \times_X \Spec(k) \to \Spec(k)$ are quasi-compact and \'etale. This means in particular that $R$ is an \'etale equivalence relation on $\Spec(k)$. By Spaces, Theorem \ref{spaces-theorem-presentation} we know that the quotient sheaf $X' = \Spec(k)/R$ is an algebraic space. By Groupoids, Lemma \ref{groupoids-lemma-quotient-groupoid-restrict} the map $X' \to X$ is a monomorphism. Since $s, t$ are quasi-compact, we see that $R$ is quasi-compact and hence Properties of Spaces, Lemma \ref{spaces-properties-lemma-point-like-spaces} applies to $X'$, and we see that $X' = \Spec(k')$ for some field $k'$. Hence we get a factorization $$\Spec(k) \longrightarrow \Spec(k') \longrightarrow X$$ which shows that $\Spec(k') \to X$ is a monomorphism mapping to $x \in |X|$. In other words (4) holds. \medskip\noindent Finally, we prove that (4) implies (1). Let $\Spec(k) \to X$ be a monomorphism with $k$ a field in the equivalence class of $x$. Let $U \to X$ be a surjective \'etale morphism from a scheme $U$ to $X$. Let $u \in U$ be a point over $x$. Since $\Spec(k) \times_X u$ is nonempty, and since $\Spec(k) \times_X u \to u$ is a monomorphism we conclude that $\Spec(k) \times_X u = u$ (see Schemes, Lemma \ref{schemes-lemma-mono-towards-spec-field}). Hence $u \to U \to X$ factors through $\Spec(k) \to X$, here is a picture $$\xymatrix{ u \ar[r] \ar[d] & U \ar[d] \\ \Spec(k) \ar[r] & X }$$ Since the right vertical arrow is \'etale this implies that $k \subset \kappa(u)$ is a finite separable extension. Hence we conclude that $$u \times_X u = u \times_{\Spec(k)} u$$ is a finite scheme, and we win by the discussion of the meaning of property (1) in the first paragraph of this proof. \end{proof} \begin{lemma} \label{lemma-weak-UR-finite-above-x} Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$. Let $U$ be a scheme and let $\varphi : U \to X$ be an \'etale morphism. The following are equivalent: \begin{enumerate} \item $x$ is in the image of $|U| \to |X|$, and setting $R = U \times_X U$ the fibres of both $$|U| \longrightarrow |X| \quad\text{and}\quad |R| \longrightarrow |X|$$ over $x$ are finite, \item there exists a monomorphism $\Spec(k) \to X$ with $k$ a field in the equivalence class of $x$, and the fibre product $\Spec(k) \times_X U$ is a finite nonempty scheme over $k$. \end{enumerate} \end{lemma} \begin{proof} Assume (1). This clearly implies the first condition of Lemma \ref{lemma-R-finite-above-x} and hence we obtain a monomorphism $\Spec(k) \to X$ in the class of $x$. Taking the fibre product we see that $\Spec(k) \times_X U \to \Spec(k)$ is a scheme \'etale over $\Spec(k)$ with finitely many points, hence a finite nonempty scheme over $k$, i.e., (2) holds. \medskip\noindent Assume (2). By assumption $x$ is in the image of $|U| \to |X|$. The finiteness of the fibre of $|U| \to |X|$ over $x$ is clear since this fibre is equal to $|\Spec(k) \times_X U|$ by Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-cartesian}. The finiteness of the fibre of $|R| \to |X|$ above $x$ is also clear since it is equal to the set underlying the scheme $$(\Spec(k) \times_X U) \times_{\Spec(k)} (\Spec(k) \times_X U)$$ which is finite over $k$. Thus (1) holds. \end{proof} \begin{lemma} \label{lemma-UR-finite-above-x} Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$. The following are equivalent: \begin{enumerate} \item for every affine scheme $U$, any \'etale morphism $\varphi : U \to X$ setting $R = U \times_X U$ the fibres of both $$|U| \longrightarrow |X| \quad\text{and}\quad |R| \longrightarrow |X|$$ over $x$ are finite, \item there exist schemes $U_i$ and \'etale morphisms $U_i \to X$ such that $\coprod U_i \to X$ is surjective and for each $i$, setting $R_i = U_i \times_X U_i$ the fibres of both $$|U_i| \longrightarrow |X| \quad\text{and}\quad |R_i| \longrightarrow |X|$$ over $x$ are finite, \item there exists a monomorphism $\Spec(k) \to X$ with $k$ a field in the equivalence class of $x$, and for any affine scheme $U$ and \'etale morphism $U \to X$ the fibre product $\Spec(k) \times_X U$ is a finite scheme over $k$, and \item there exists a quasi-compact monomorphism $\Spec(k) \to X$ with $k$ a field in the equivalence class of $x$. \end{enumerate} \end{lemma} \begin{proof} The equivalence of (1) and (3) follows on applying Lemma \ref{lemma-weak-UR-finite-above-x} to every \'etale morphism $U \to X$ with $U$ affine. It is clear that (3) implies (2). Assume $U_i \to X$ and $R_i$ are as in (2). We conclude from Lemma \ref{lemma-U-finite-above-x} that for any affine scheme $U$ and \'etale morphism $U \to X$ the fibre of $|U| \to |X|$ over $x$ is finite. Say this fibre is $\{u_1, \ldots, u_n\}$. Then, as Lemma \ref{lemma-R-finite-above-x} (1) applies to $U_i \to X$ for some $i$ such that $x$ is in the image of $|U_i| \to |X|$, we see that the fibre of $|R = U \times_X U| \to |U| \times_{|X|} |U|$ is finite over $(u_a, u_b)$, $a, b \in \{1, \ldots, n\}$. Hence the fibre of $|R| \to |X|$ over $x$ is finite. In this way we see that (1) holds. At this point we know that (1), (2), and (3) are equivalent. \medskip\noindent If (4) holds, then for any affine scheme $U$ and \'etale morphism $U \to X$ the scheme $\Spec(k) \times_X U$ is on the one hand \'etale over $k$ (hence a disjoint union of spectra of finite separable extensions of $k$ by Remark \ref{remark-recall}) and on the other hand quasi-compact over $U$ (hence quasi-compact). Thus we see that (3) holds. Conversely, if $U_i \to X$ is as in (2) and $\Spec(k) \to X$ is a monomorphism as in (3), then $$\coprod \Spec(k) \times_X U_i \longrightarrow \coprod U_i$$ is quasi-compact (because over each $U_i$ we see that $\Spec(k) \times_X U_i$ is a finite disjoint union spectra of fields). Thus $\Spec(k) \to X$ is quasi-compact by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-quasi-compact-local}. \end{proof} \begin{lemma} \label{lemma-U-universally-bounded} Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. The following are equivalent: \begin{enumerate} \item there exist schemes $U_i$ and \'etale morphisms $U_i \to X$ such that $\coprod U_i \to X$ is surjective and each $U_i \to X$ has universally bounded fibres, and \item for every affine scheme $U$ and \'etale morphism $\varphi : U \to X$ the fibres of $U \to X$ are universally bounded. \end{enumerate} \end{lemma} \begin{proof} The implication (2) $\Rightarrow$ (1) is trivial. Assume (1). Let $(\varphi_i : U_i \to X)_{i \in I}$ be a collection of \'etale morphisms from schemes towards $X$, covering $X$, such that each $\varphi_i$ has universally bounded fibres. Let $\psi : U \to X$ be an \'etale morphism from an affine scheme towards $X$. For each $i$ consider the fibre product diagram $$\xymatrix{ U \times_X U_i \ar[r]_{p_i} \ar[d]_{q_i} & U_i \ar[d]^{\varphi_i} \\ U \ar[r]^\psi & X }$$ Since $q_i$ is \'etale it is open (see Remark \ref{remark-recall}). Moreover, we have $U = \bigcup \Im(q_i)$, since the family $(\varphi_i)_{i \in I}$ is surjective. Since $U$ is affine, hence quasi-compact we can finite finitely many $i_1, \ldots, i_n \in I$ and quasi-compact opens $W_j \subset U \times_X U_{i_j}$ such that $U = \bigcup p_{i_j}(W_j)$. The morphism $p_{i_j}$ is \'etale, hence locally quasi-finite (see remark on \'etale morphisms above). Thus we may apply Morphisms, Lemma \ref{morphisms-lemma-locally-quasi-finite-qc-source-universally-bounded} to see the fibres of $p_{i_j}|_{W_j} : W_j \to U_{i_j}$ are universally bounded. Hence by Lemma \ref{lemma-composition-universally-bounded} we see that the fibres of $W_j \to X$ are universally bounded. Thus also $\coprod_{j = 1, \ldots, n} W_j \to X$ has universally bounded fibres. Since $\coprod_{j = 1, \ldots, n} W_j \to X$ factors through the surjective \'etale map $\coprod q_{i_j}|_{W_j} : \coprod_{j = 1, \ldots, n} W_j \to U$ we see that the fibres of $U \to X$ are universally bounded by Lemma \ref{lemma-universally-bounded-permanence}. In other words (2) holds. \end{proof} \begin{lemma} \label{lemma-characterize-very-reasonable} Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. The following are equivalent: \begin{enumerate} \item there exists a Zariski covering $X = \bigcup X_i$ and for each $i$ a scheme $U_i$ and a quasi-compact surjective \'etale morphism $U_i \to X_i$, and \item there exist schemes $U_i$ and \'etale morphisms $U_i \to X$ such that the projections $U_i \times_X U_i \to U_i$ are quasi-compact and $\coprod U_i \to X$ is surjective. \end{enumerate} \end{lemma} \begin{proof} If (1) holds then the morphisms $U_i \to X_i \to X$ are \'etale (combine Morphisms, Lemma \ref{morphisms-lemma-composition-etale} and Spaces, Lemmas \ref{spaces-lemma-composition-representable-transformations-property} and \ref{spaces-lemma-morphism-schemes-gives-representable-transformation-property} ). Moreover, as $U_i \times_X U_i = U_i \times_{X_i} U_i$, both projections $U_i \times_X U_i \to U_i$ are quasi-compact. \medskip\noindent If (2) holds then let $X_i \subset X$ be the open subspace corresponding to the image of the open map $|U_i| \to |X|$, see Properties of Spaces, Lemma \ref{spaces-properties-lemma-etale-image-open}. The morphisms $U_i \to X_i$ are surjective. Hence $U_i \to X_i$ is surjective \'etale, and the projections $U_i \times_{X_i} U_i \to U_i$ are quasi-compact, because $U_i \times_{X_i} U_i = U_i \times_X U_i$. Thus by Spaces, Lemma \ref{spaces-lemma-representable-morphisms-spaces-property} the morphisms $U_i \to X_i$ are quasi-compact. \end{proof} \section{Conditions on algebraic spaces} \label{section-conditions} \noindent In this section we discuss the relationship between various natural conditions on algebraic spaces we have seen above. Please read Section \ref{section-reasonable-decent} to get a feeling for the meaning of these conditions. \begin{lemma} \label{lemma-bounded-fibres} Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Consider the following conditions on $X$: \begin{enumerate} \item[$(\alpha)$] For every $x \in |X|$, the equivalent conditions of Lemma \ref{lemma-U-finite-above-x} hold. \item[$(\beta)$] For every $x \in |X|$, the equivalent conditions of Lemma \ref{lemma-R-finite-above-x} hold. \item[$(\gamma)$] For every $x \in |X|$, the equivalent conditions of Lemma \ref{lemma-UR-finite-above-x} hold. \item[$(\delta)$] The equivalent conditions of Lemma \ref{lemma-U-universally-bounded} hold. \item[$(\epsilon)$] The equivalent conditions of Lemma \ref{lemma-characterize-very-reasonable} hold. \item[$(\zeta)$] The space $X$ is Zariski locally quasi-separated. \item[$(\eta)$] The space $X$ is quasi-separated \item[$(\theta)$] The space $X$ is representable, i.e., $X$ is a scheme. \item[$(\iota)$] The space $X$ is a quasi-separated scheme. \end{enumerate} We have $$\xymatrix{ & (\theta) \ar@{=>}[rd] & & & & \\ (\iota) \ar@{=>}[ru] \ar@{=>}[rd] & & (\zeta) \ar@{=>}[r] & (\epsilon) \ar@{=>}[r] & (\delta) \ar@{=>}[r] & (\gamma) \ar@{<=>}[r] & (\alpha) + (\beta) \\ & (\eta) \ar@{=>}[ru] & & & & }$$ \end{lemma} \begin{proof} The implication $(\gamma) \Leftrightarrow (\alpha) + (\beta)$ is immediate. The implications in the diamond on the left are clear from the definitions. \medskip\noindent Assume $(\zeta)$, i.e., that $X$ is Zariski locally quasi-separated. Then $(\epsilon)$ holds by Properties of Spaces, Lemma \ref{spaces-properties-lemma-quasi-separated-quasi-compact-pieces}. \medskip\noindent Assume $(\epsilon)$. By Lemma \ref{lemma-characterize-very-reasonable} there exists a Zariski open covering $X = \bigcup X_i$ such that for each $i$ there exists a scheme $U_i$ and a quasi-compact surjective \'etale morphism $U_i \to X_i$. Choose an $i$ and an affine open subscheme $W \subset U_i$. It suffices to show that $W \to X$ has universally bounded fibres, since then the family of all these morphisms $W \to X$ covers $X$. To do this we consider the diagram $$\xymatrix{ W \times_X U_i \ar[r]_-p \ar[d]_q & U_i \ar[d] \\ W \ar[r] & X }$$ Since $W \to X$ factors through $X_i$ we see that $W \times_X U_i = W \times_{X_i} U_i$, and hence $q$ is quasi-compact. Since $W$ is affine this implies that the scheme $W \times_X U_i$ is quasi-compact. Thus we may apply Morphisms, Lemma \ref{morphisms-lemma-locally-quasi-finite-qc-source-universally-bounded} and we conclude that $p$ has universally bounded fibres. From Lemma \ref{lemma-descent-universally-bounded} we conclude that $W \to X$ has universally bounded fibres as well. \medskip\noindent Assume $(\delta)$. Let $U$ be an affine scheme, and let $U \to X$ be an \'etale morphism. By assumption the fibres of the morphism $U \to X$ are universally bounded. Thus also the fibres of both projections $R = U \times_X U \to U$ are universally bounded, see Lemma \ref{lemma-base-change-universally-bounded}. And by Lemma \ref{lemma-composition-universally-bounded} also the fibres of $R \to X$ are universally bounded. Hence for any $x \in X$ the fibres of $|U| \to |X|$ and $|R| \to |X|$ over $x$ are finite, see Lemma \ref{lemma-universally-bounded-finite-fibres}. In other words, the equivalent conditions of Lemma \ref{lemma-UR-finite-above-x} hold. This proves that $(\delta) \Rightarrow (\gamma)$. \end{proof} \begin{lemma} \label{lemma-properties-local} Let $S$ be a scheme. Let $\mathcal{P}$ be one of the properties $(\alpha)$, $(\beta)$, $(\gamma)$, $(\delta)$, $(\epsilon)$, $(\zeta)$, or $(\theta)$ of algebraic spaces listed in Lemma \ref{lemma-bounded-fibres}. Then if $X$ is an algebraic space over $S$, and $X = \bigcup X_i$ is a Zariski open covering such that each $X_i$ has $\mathcal{P}$, then $X$ has $\mathcal{P}$. \end{lemma} \begin{proof} Let $X$ be an algebraic space over $S$, and let $X = \bigcup X_i$ is a Zariski open covering such that each $X_i$ has $\mathcal{P}$. \medskip\noindent The case $\mathcal{P} = (\alpha)$. The condition $(\alpha)$ for $X_i$ means that for every $x \in |X_i|$ and every affine scheme $U$, and \'etale morphism $\varphi : U \to X_i$ the fibre of $\varphi : |U| \to |X_i|$ over $x$ is finite. Consider $x \in X$, an affine scheme $U$ and an \'etale morphism $U \to X$. Since $X = \bigcup X_i$ is a Zariski open covering there exits a finite affine open covering $U = U_1 \cup \ldots \cup U_n$ such that each $U_j \to X$ factors through some $X_{i_j}$. By assumption the fibres of $|U_j | \to |X_{i_j}|$ over $x$ are finite for $j = 1, \ldots, n$. Clearly this means that the fibre of $|U| \to |X|$ over $x$ is finite. This proves the result for $(\alpha)$. \medskip\noindent The case $\mathcal{P} = (\beta)$. The condition $(\beta)$ for $X_i$ means that every $x \in |X_i|$ is represented by a monomorphism from the spectrum of a field towards $X_i$. Hence the same follows for $X$ as $X_i \to X$ is a monomorphism and $X = \bigcup X_i$. \medskip\noindent The case $\mathcal{P} = (\gamma)$. Note that $(\gamma) = (\alpha) + (\beta)$ by Lemma \ref{lemma-bounded-fibres} hence the lemma for $(\gamma)$ follows from the cases treated above. \medskip\noindent The case $\mathcal{P} = (\delta)$. The condition $(\delta)$ for $X_i$ means there exist schemes $U_{ij}$ and \'etale morphisms $U_{ij} \to X_i$ with universally bounded fibres which cover $X_i$. These schemes also give an \'etale surjective morphism $\coprod U_{ij} \to X$ and $U_{ij} \to X$ still has universally bounded fibres. \medskip\noindent The case $\mathcal{P} = (\epsilon)$. The condition $(\epsilon)$ for $X_i$ means we can find a set $J_i$ and morphisms $\varphi_{ij} : U_{ij} \to X_i$ such that each $\varphi_{ij}$ is \'etale, both projections $U_{ij} \times_{X_i} U_{ij} \to U_{ij}$ are quasi-compact, and $\coprod_{j \in J_i} U_{ij} \to X_i$ is surjective. In this case the compositions $U_{ij} \to X_i \to X$ are \'etale (combine Morphisms, Lemmas \ref{morphisms-lemma-composition-etale} and \ref{morphisms-lemma-open-immersion-etale} and Spaces, Lemmas \ref{spaces-lemma-composition-representable-transformations-property} and \ref{spaces-lemma-morphism-schemes-gives-representable-transformation-property} ). Since $X_i \subset X$ is a subspace we see that $U_{ij} \times_{X_i} U_{ij} = U_{ij} \times_X U_{ij}$, and hence the condition on fibre products is preserved. And clearly $\coprod_{i, j} U_{ij} \to X$ is surjective. Hence $X$ satisfies $(\epsilon)$. \medskip\noindent The case $\mathcal{P} = (\zeta)$. The condition $(\zeta)$ for $X_i$ means that $X_i$ is Zariski locally quasi-separated. It is immediately clear that this means $X$ is Zariski locally quasi-separated. \medskip\noindent For $(\theta)$, see Properties of Spaces, Lemma \ref{spaces-properties-lemma-subscheme}. \end{proof} \begin{lemma} \label{lemma-representable-properties} Let $S$ be a scheme. Let $\mathcal{P}$ be one of the properties $(\beta)$, $(\gamma)$, $(\delta)$, $(\epsilon)$, or $(\theta)$ of algebraic spaces listed in Lemma \ref{lemma-bounded-fibres}. Let $X$, $Y$ be algebraic spaces over $S$. Let $X \to Y$ be a representable morphism. If $Y$ has property $\mathcal{P}$, so does $X$. \end{lemma} \begin{proof} Assume $f : X \to Y$ is a representable morphism of algebraic spaces, and assume that $Y$ has $\mathcal{P}$. Let $x \in |X|$, and set $y = f(x) \in |Y|$. \medskip\noindent The case $\mathcal{P} = (\beta)$. Condition $(\beta)$ for $Y$ means there exists a monomorphism $\Spec(k) \to Y$ representing $y$. The fibre product $X_y = \Spec(k) \times_Y X$ is a scheme, and $x$ corresponds to a point of $X_y$, i.e., to a monomorphism $\Spec(k') \to X_y$. As $X_y \to X$ is a monomorphism also we see that $x$ is represented by the monomorphism $\Spec(k') \to X_y \to X$. In other words $(\beta)$ holds for $X$. \medskip\noindent The case $\mathcal{P} = (\gamma)$. Since $(\gamma) \Rightarrow (\beta)$ we have seen in the preceding paragraph that $y$ and $x$ can be represented by monomorphisms as in the following diagram $$\xymatrix{ \Spec(k') \ar[r]_-x \ar[d] & X \ar[d] \\ \Spec(k) \ar[r]^-y & Y }$$ Also, by definition of property $(\gamma)$ via Lemma \ref{lemma-UR-finite-above-x} (2) there exist schemes $V_i$ and \'etale morphisms $V_i \to Y$ such that $\coprod V_i \to Y$ is surjective and for each $i$, setting $R_i = V_i \times_Y V_i$ the fibres of both $$|V_i| \longrightarrow |Y| \quad\text{and}\quad |R_i| \longrightarrow |Y|$$ over $y$ are finite. This means that the schemes $(V_i)_y$ and $(R_i)_y$ are finite schemes over $y = \Spec(k)$. As $X \to Y$ is representable, the fibre products $U_i = V_i \times_Y X$ are schemes. The morphisms $U_i \to X$ are \'etale, and $\coprod U_i \to X$ is surjective. Finally, for each $i$ we have $$(U_i)_x = (V_i \times_Y X)_x = (V_i)_y \times_{\Spec(k)} \Spec(k')$$ and $$(U_i \times_X U_i)_x = \left((V_i \times_Y X) \times_X (V_i \times_Y X)\right)_x = (R_i)_y \times_{\Spec(k)} \Spec(k')$$ hence these are finite over $k'$ as base changes of the finite schemes $(V_i)_y$ and $(R_i)_y$. This implies that $(\gamma)$ holds for $X$, again via the second condition of Lemma \ref{lemma-UR-finite-above-x}. \medskip\noindent The case $\mathcal{P} = (\delta)$. Let $V \to Y$ be an \'etale morphism with $V$ an affine scheme. Since $Y$ has property $(\delta)$ this morphism has universally bounded fibres. By Lemma \ref{lemma-base-change-universally-bounded} the base change $V \times_Y X \to X$ also has universally bounded fibres. Hence the first part of Lemma \ref{lemma-U-universally-bounded} applies and we see that $Y$ also has property $(\delta)$. \medskip\noindent The case $\mathcal{P} = (\epsilon)$. We will repeatedly use Spaces, Lemma \ref{spaces-lemma-base-change-representable-transformations-property}. Let $V_i \to Y$ be as in Lemma \ref{lemma-characterize-very-reasonable} (2). Set $U_i = X \times_Y V_i$. The morphisms $U_i \to X$ are \'etale, and $\coprod U_i \to X$ is surjective. Because $U_i \times_X U_i = X \times_Y (V_i \times_Y V_i)$ we see that the projections $U_i \times_Y U_i \to U_i$ are base changes of the projections $V_i \times_Y V_i \to V_i$, and so quasi-compact as well. Hence $X$ satisfies Lemma \ref{lemma-characterize-very-reasonable} (2). \medskip\noindent The case $\mathcal{P} = (\theta)$. In this case the result is Categories, Lemma \ref{categories-lemma-representable-over-representable}. \end{proof} \section{Reasonable and decent algebraic spaces} \label{section-reasonable-decent} \noindent In Lemma \ref{lemma-bounded-fibres} we have seen a number of conditions on algebraic spaces related to the behaviour of \'etale morphisms from affine schemes into $X$ and related to the existence of special \'etale coverings of $X$ by schemes. We tabulate the different types of conditions here: $$\boxed{ \begin{matrix} (\alpha) & \text{fibres of \'etale morphisms from affines are finite} \\ (\beta) & \text{points come from monomorphisms of spectra of fields} \\ (\gamma) & \text{points come from quasi-compact monomorphisms of spectra of fields} \\ (\delta) & \text{fibres of \'etale morphisms from affines are universally bounded} \\ (\epsilon) & \text{cover by \'etale morphisms from schemes quasi-compact onto their image} \end{matrix} }$$ \medskip\noindent The conditions in the following definition are not exactly conditions on the diagonal of $X$, but they are in some sense separation conditions on $X$. \begin{definition} \label{definition-very-reasonable} Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. \begin{enumerate} \item We say $X$ is {\it decent} if for every point $x \in X$ the equivalent conditions of Lemma \ref{lemma-UR-finite-above-x} hold, in other words property $(\gamma)$ of Lemma \ref{lemma-bounded-fibres} holds. \item We say $X$ is {\it reasonable} if the equivalent conditions of Lemma \ref{lemma-U-universally-bounded} hold, in other words property $(\delta)$ of Lemma \ref{lemma-bounded-fibres} holds. \item We say $X$ is {\it very reasonable} if the equivalent conditions of Lemma \ref{lemma-characterize-very-reasonable} hold, i.e., property $(\epsilon)$ of Lemma \ref{lemma-bounded-fibres} holds. \end{enumerate} \end{definition} \noindent We have the following implications among these conditions on algebraic spaces: $$\xymatrix{ \text{representable} \ar@{=>}[rd] & & & \\ & \text{very reasonable} \ar@{=>}[r] & \text{reasonable} \ar@{=>}[r] & \text{decent} \\ \text{quasi-separated} \ar@{=>}[ru] & & & }$$ The notion of a very reasonable algebraic space is obsolete. It was introduced because the assumption was needed to prove some results which are now proven for the class of decent spaces. The class of decent spaces is the largest class of spaces $X$ where one has a good relationship between the topology of $|X|$ and properties of $X$ itself. \begin{example} \label{example-not-decent} The algebraic space $\mathbf{A}^1_{\mathbf{Q}}/\mathbf{Z}$ constructed in Spaces, Example \ref{spaces-example-affine-line-translation} is not decent as its generic point'' cannot be represented by a monomorphism from the spectrum of a field. \end{example} \begin{remark} \label{remark-reasonable} Reasonable algebraic spaces are technically easier to work with than very reasonable algebraic spaces. For example, if $X \to Y$ is a quasi-compact \'etale surjective morphism of algebraic spaces and $X$ is reasonable, then so is $Y$, see Lemma \ref{lemma-descent-conditions} but we don't know if this is true for the property very reasonable''. Below we give another technical property enjoyed by reasonable algebraic spaces. \end{remark} \begin{lemma} \label{lemma-fun-property-reasonable} Let $S$ be a scheme. Let $X$ be a quasi-compact reasonable algebraic space. Then there exists a directed system of quasi-compact and quasi-separated algebraic spaces $X_i$ such that $X = \colim_i X_i$ (colimit in the category of sheaves). \end{lemma} \begin{proof} We sketch the proof. By Properties of Spaces, Lemma \ref{spaces-properties-lemma-quasi-compact-affine-cover} we have $X = U/R$ with $U$ affine. In this case, reasonable means $U \to X$ is universally bounded. Hence there exists an integer $N$ such that the fibres'' of $U \to X$ have degree at most $N$, see Definition \ref{definition-universally-bounded}. Denote $s, t : R \to U$ and $c : R \times_{s, U, t} R \to R$ the groupoid structural maps. \medskip\noindent Claim: for every quasi-compact open $A \subset R$ there exists an open $R' \subset R$ such that \begin{enumerate} \item $A \subset R'$, \item $R'$ is quasi-compact, and \item $(U, R', s|_{R'}, t|_{R'}, c|_{R' \times_{s, U, t} R'})$ is a groupoid scheme. \end{enumerate} Note that $e : U \to R$ is open as it is a section of the \'etale morphism $s : R \to U$, see \'Etale Morphisms, Proposition \ref{etale-proposition-properties-sections}. Moreover $U$ is affine hence quasi-compact. Hence we may replace $A$ by $A \cup e(U) \subset R$, and assume that $A$ contains $e(U)$. Next, we define inductively $A^1 = A$, and $$A^n = c(A^{n - 1} \times_{s, U, t} A) \subset R$$ for $n \geq 2$. Arguing inductively, we see that $A^n$ is quasi-compact for all $n \geq 2$, as the image of the quasi-compact fibre product $A^{n - 1} \times_{s, U, t} A$. If $k$ is an algebraically closed field over $S$, and we consider $k$-points then $$A^n(k) = \left\{(u, u') \in U(k) : \begin{matrix} \text{there exist } u = u_1, u_2, \ldots, u_n \in U(k)\text{ with} \\ (u_i , u_{i + 1}) \in A \text{ for all }i = 1, \ldots, n - 1. \end{matrix} \right\}$$ But as the fibres of $U(k) \to X(k)$ have size at most $N$ we see that if $n > N$ then we get a repeat in the sequence above, and we can shorten it proving $A^N = A^n$ for all $n \geq N$. This implies that $R' = A^N$ gives a groupoid scheme $(U, R', s|_{R'}, t|_{R'}, c|_{R' \times_{s, U, t} R'})$, proving the claim above. \medskip\noindent Consider the map of sheaves on $(\Sch/S)_{fppf}$ $$\colim_{R' \subset R} U/R' \longrightarrow U/R$$ where $R' \subset R$ runs over the quasi-compact open subschemes of $R$ which give \'etale equivalence relations as above. Each of the quotients $U/R'$ is an algebraic space (see Spaces, Theorem \ref{spaces-theorem-presentation}). Since $R'$ is quasi-compact, and $U$ affine the morphism $R' \to U \times_{\Spec(\mathbf{Z})} U$ is quasi-compact, and hence $U/R'$ is quasi-separated. Finally, if $T$ is a quasi-compact scheme, then $$\colim_{R' \subset R} U(T)/R'(T) \longrightarrow U(T)/R(T)$$ is a bijection, since every morphism from $T$ into $R$ ends up in one of the open subrelations $R'$ by the claim above. This clearly implies that the colimit of the sheaves $U/R'$ is $U/R$. In other words the algebraic space $X = U/R$ is the colimit of the quasi-separated algebraic spaces $U/R'$. \end{proof} \begin{lemma} \label{lemma-representable-named-properties} Let $S$ be a scheme. Let $X$, $Y$ be algebraic spaces over $S$. Let $X \to Y$ be a representable morphism. If $Y$ is decent (resp.\ reasonable), then so is $X$. \end{lemma} \begin{proof} Translation of Lemma \ref{lemma-representable-properties}. \end{proof} \begin{lemma} \label{lemma-etale-named-properties} Let $S$ be a scheme. Let $X \to Y$ be an \'etale morphism of algebraic spaces over $S$. If $Y$ is decent, resp.\ reasonable, then so is $X$. \end{lemma} \begin{proof} Let $U$ be an affine scheme and $U \to X$ an \'etale morphism. Set $R = U \times_X U$ and $R' = U \times_Y U$. Note that $R \to R'$ is a monomorphism. \medskip\noindent Let $x \in |X|$. To show that $X$ is decent, we have to show that the fibres of $|U| \to |X|$ and $|R| \to |X|$ over $x$ are finite. But if $Y$ is decent, then the fibres of $|U| \to |Y|$ and $|R'| \to |Y|$ are finite. Hence the result for decent''. \medskip\noindent To show that $X$ is reasonable, we have to show that the fibres of $U \to X$ are universally bounded. However, if $Y$ is reasonable, then the fibres of $U \to Y$ are universally bounded, which immediately implies the same thing for the fibres of $U \to X$. Hence the result for reasonable''. \end{proof} \section{Points and specializations} \label{section-specializations} \noindent There exists an \'etale morphism of algebraic spaces $f : X \to Y$ and a nontrivial specialization between points in a fibre of $|f| : |X| \to |Y|$, see Examples, Lemma \ref{examples-lemma-specializations-fibre-etale}. If the source of the morphism is a scheme we can avoid this by imposing condition ($\alpha$) on $Y$. \begin{lemma} \label{lemma-no-specializations-map-to-same-point} Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $U \to X$ be an \'etale morphism from a scheme to $X$. Assume $u, u' \in |U|$ map to the same point $x$ of $|X|$, and $u' \leadsto u$. If the pair $(X, x)$ satisfies the equivalent conditions of Lemma \ref{lemma-U-finite-above-x} then $u = u'$. \end{lemma} \begin{proof} Assume the pair $(X, x)$ satisfies the equivalent conditions for Lemma \ref{lemma-U-finite-above-x}. Let $U$ be a scheme, $U \to X$ \'etale, and let $u, u' \in |U|$ map to $x$ of $|X|$, and $u' \leadsto u$. We may and do replace $U$ by an affine neighbourhood of $u$. Let $t, s : R = U \times_X U \to U$ be the \'etale projection maps. \medskip\noindent Pick a point $r \in R$ with $t(r) = u$ and $s(r) = u'$. This is possible by Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-presentation}. Because generalizations lift along the \'etale morphism $t$ (Remark \ref{remark-recall}) we can find a specialization $r' \leadsto r$ with $t(r') = u'$. Set $u'' = s(r')$. Then $u'' \leadsto u'$. Thus we may repeat and find $r'' \leadsto r'$ with $t(r'') = u''$. Set $u''' = s(r'')$, and so on. Here is a picture: $$\xymatrix{ & r'' \ar[rd]^s \ar[ld]_t \ar@{~>}[d] & \\ u'' \ar@{~>}[d] & r' \ar[rd]^s \ar[ld]_t \ar@{~>}[d] & u''' \ar@{~>}[d] \\ u' \ar@{~>}[d] & r \ar[rd]^s \ar[ld]_t & u'' \ar@{~>}[d] \\ u & & u' }$$ In Remark \ref{remark-recall} we have seen that there are no specializations among points in the fibres of the \'etale morphism $s$. Hence if $u^{(n + 1)} = u^{(n)}$ for some $n$, then also $r^{(n)} = r^{(n - 1)}$ and hence also (by taking $t$) $u^{(n)} = u^{(n - 1)}$. This then forces the whole tower to collapse, in particular $u = u'$. Thus we see that if $u \not = u'$, then all the specializations are strict and $\{u, u', u'', \ldots\}$ is an infinite set of points in $U$ which map to the point $x$ in $|X|$. As we chose $U$ affine this contradicts the second part of Lemma \ref{lemma-U-finite-above-x}, as desired. \end{proof} \begin{lemma} \label{lemma-specialization} Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x, x' \in |X|$ and assume $x' \leadsto x$, i.e., $x$ is a specialization of $x'$. Assume the pair $(X, x')$ satisfies the equivalent conditions of Lemma \ref{lemma-UR-finite-above-x}. Then for every \'etale morphism $\varphi : U \to X$ from a scheme $U$ and any $u \in U$ with $\varphi(u) = x$, exists a point $u'\in U$, $u' \leadsto u$ with $\varphi(u') = x'$. \end{lemma} \begin{proof} We may replace $U$ by an affine open neighbourhood of $u$. Hence we may assume that $U$ is affine. As $x$ is in the image of the open map $|U| \to |X|$, so is $x'$. Thus we may replace $X$ by the Zariski open subspace corresponding to the image of $|U| \to |X|$, see Properties of Spaces, Lemma \ref{spaces-properties-lemma-etale-image-open}. In other words we may assume that $U \to X$ is surjective and \'etale. Let $s, t : R = U \times_X U \to U$ be the projections. By our assumption that $(X, x')$ satisfies the equivalent conditions of Lemma \ref{lemma-UR-finite-above-x} we see that the fibres of $|U| \to |X|$ and $|R| \to |X|$ over $x'$ are finite. Say $\{u'_1, \ldots, u'_n\} \subset U$ and $\{r'_1, \ldots, r'_m\} \subset R$ form the complete inverse image of $\{x'\}$. Consider the closed sets $$T = \overline{\{u'_1\}} \cup \ldots \cup \overline{\{u'_n\}} \subset |U|, \quad T' = \overline{\{r'_1\}} \cup \ldots \cup \overline{\{r'_m\}} \subset |R|.$$ Trivially we have $s(T') \subset T$. Because $R$ is an equivalence relation we also have $t(T') = s(T')$ as the set $\{r_j'\}$ is invariant under the inverse of $R$ by construction. Let $w \in T$ be any point. Then $u'_i \leadsto w$ for some $i$. Choose $r \in R$ with $s(r) = w$. Since generalizations lift along $s : R \to U$, see Remark \ref{remark-recall}, we can find $r' \leadsto r$ with $s(r') = u_i'$. Then $r' = r'_j$ for some $j$ and we conclude that $w \in s(T')$. Hence $T = s(T') = t(T')$ is an $|R|$-invariant closed set in $|U|$. This means $T$ is the inverse image of a closed (!) subset $T'' = \varphi(T)$ of $|X|$, see Properties of Spaces, Lemmas \ref{spaces-properties-lemma-points-presentation} and \ref{spaces-properties-lemma-topology-points}. Hence $T'' = \overline{\{x'\}}$. Thus $T$ contains some point $u_1$ mapping to $x$ as $x \in T''$. I.e., we see that for some $i$ there exists a specialization $u'_i \leadsto u_1$ which maps to the given specialization $x' \leadsto x$. \medskip\noindent To finish the proof, choose a point $r \in R$ such that $s(r) = u$ and $t(r) = u_1$ (using Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-cartesian}). As generalizations lift along $t$, and $u'_i \leadsto u_1$ we can find a specialization $r' \leadsto r$ such that $t(r') = u'_i$. Set $u' = s(r')$. Then $u' \leadsto u$ and $\varphi(u') = x'$ as desired. \end{proof} \begin{lemma} \label{lemma-generalizations-lift-flat} Let $S$ be a scheme. Let $f : Y \to X$ be a flat morphism of algebraic spaces over $S$. Let $x, x' \in |X|$ and assume $x' \leadsto x$, i.e., $x$ is a specialization of $x'$. Assume the pair $(X, x')$ satisfies the equivalent conditions of Lemma \ref{lemma-UR-finite-above-x} (for example if $X$ is decent, $X$ is quasi-separated, or $X$ is representable). Then for every $y \in |Y|$ with $f(y) = x$, there exists a point $y' \in |Y|$, $y' \leadsto y$ with $f(y') = x'$. \end{lemma} \begin{proof} (The parenthetical statement holds by the definition of decent spaces and the implications between the different separation conditions mentioned in Section \ref{section-reasonable-decent}.) Choose a scheme $V$ and a surjective \'etale morphism $V \to Y$. Choose $v \in V$ mapping to $y$. Then we see that it suffices to prove the lemma for $V \to X$. Thus we may assume $Y$ is a scheme. Choose a scheme $U$ and a surjective \'etale morphism $U \to X$. Choose $u \in U$ mapping to $x$. By Lemma \ref{lemma-specialization} we may choose $u' \leadsto u$ mapping to $x'$. By Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-cartesian} we may choose $z \in U \times_X Y$ mapping to $y$ and $u$. Thus we reduce to the case of the flat morphism of schemes $U \times_X Y \to U$ which is Morphisms, Lemma \ref{morphisms-lemma-generalizations-lift-flat}. \end{proof} \section{Stratifying algebraic spaces by schemes} \label{section-stratifications} \noindent In this section we prove that a quasi-compact and quasi-separated algebraic space has a finite stratification by locally closed subspaces each of which is a scheme and such that the glueing of the parts is by elementary distinguished squares. We first prove a slightly weaker result for reasonable algebraic spaces. \begin{lemma} \label{lemma-quasi-compact-reasonable-stratification} Let $S$ be a scheme. Let $W \to X$ be a morphism of a scheme $W$ to an algebraic space $X$ which is flat, locally of finite presentation, separated, locally quasi-finite with universally bounded fibres. There exist reduced closed subspaces $$\emptyset = Z_{-1} \subset Z_0 \subset Z_1 \subset Z_2 \subset \ldots \subset Z_n = X$$ such that with $X_r = Z_r \setminus Z_{r - 1}$ the stratification $X = \coprod_{r = 0, \ldots, n} X_r$ is characterized by the following universal property: Given $g : T \to X$ the projection $W \times_X T \to T$ is finite locally free of degree $r$ if and only if $g(|T|) \subset |X_r|$. \end{lemma} \begin{proof} Let $n$ be an integer bounding the degrees of the fibres of $W \to X$. Choose a scheme $U$ and a surjective \'etale morphism $U \to X$. Apply More on Morphisms, Lemma \ref{more-morphisms-lemma-stratify-flat-fp-lqf-universally-bounded} to $W \times_X U \to U$. We obtain closed subsets $$\emptyset = Y_{-1} \subset Y_0 \subset Y_1 \subset Y_2 \subset \ldots \subset Y_n = U$$ characterized by the property stated in the lemma for the morphism $W \times_X U \to U$. Clearly, the formation of these closed subsets commutes with base change. Setting $R = U \times_X U$ with projection maps $s, t : R \to U$ we conclude that $$s^{-1}(Y_r) = t^{-1}(Y_r)$$ as closed subsets of $R$. In other words the closed subsets $Y_r \subset U$ are $R$-invariant. This means that $|Y_r|$ is the inverse image of a closed subset $Z_r \subset |X|$. Denote $Z_r \subset X$ also the reduced induced algebraic space structure, see Properties of Spaces, Definition \ref{spaces-properties-definition-reduced-induced-space}. \medskip\noindent Let $g : T \to X$ be a morphism of algebraic spaces. Choose a scheme $V$ and a surjective \'etale morphism $V \to T$. To prove the final assertion of the lemma it suffices to prove the assertion for the composition $V \to X$ (by our definition of finite locally free morphisms, see Morphisms of Spaces, Section \ref{spaces-morphisms-section-finite-locally-free}). Similarly, the morphism of schemes $W \times_X V \to V$ is finite locally free of degree $r$ if and only if the morphism of schemes $$W \times_X (U \times_X V) \longrightarrow U \times_X V$$ is finite locally free of degree $r$ (see Descent, Lemma \ref{descent-lemma-descending-property-finite-locally-free}). By construction this happens if and only if $|U \times_X V| \to |U|$ maps into $|Y_r|$, which is true if and only if $|V| \to |X|$ maps into $|Z_r|$. \end{proof} \begin{lemma} \label{lemma-stratify-flat-fp-lqf} Let $S$ be a scheme. Let $W \to X$ be a morphism of a scheme $W$ to an algebraic space $X$ which is flat, locally of finite presentation, separated, and locally quasi-finite. Then there exist open subspaces $$X = X_0 \supset X_1 \supset X_2 \supset \ldots$$ such that a morphism $\Spec(k) \to X$ factors through $X_d$ if and only if $W \times_X \Spec(k)$ has degree $\geq d$ over $k$. \end{lemma} \begin{proof} Choose a scheme $U$ and a surjective \'etale morphism $U \to X$. Apply More on Morphisms, Lemma \ref{more-morphisms-lemma-stratify-flat-fp-lqf} to $W \times_X U \to U$. We obtain open subschemes $$U = U_0 \supset U_1 \supset U_2 \supset \ldots$$ characterized by the property stated in the lemma for the morphism $W \times_X U \to U$. Clearly, the formation of these closed subsets commutes with base change. Setting $R = U \times_X U$ with projection maps $s, t : R \to U$ we conclude that $$s^{-1}(U_d) = t^{-1}(U_d)$$ as open subschemes of $R$. In other words the open subschemes $U_d \subset U$ are $R$-invariant. This means that $U_d$ is the inverse image of an open subspace $X_d \subset X$ (Properties of Spaces, Lemma \ref{spaces-properties-lemma-subspaces-presentation}). \end{proof} \begin{lemma} \label{lemma-filter-quasi-compact} Let $S$ be a scheme. Let $X$ be a quasi-compact algebraic space over $S$. There exist open subspaces $$\ldots \subset U_4 \subset U_3 \subset U_2 \subset U_1 = X$$ with the following properties: \begin{enumerate} \item setting $T_p = U_p \setminus U_{p + 1}$ (with reduced induced subspace structure) there exists a separated scheme $V_p$ and a surjective \'etale morphism $f_p : V_p \to U_p$ such that $f_p^{-1}(T_p) \to T_p$ is an isomorphism, \item if $x \in |X|$ can be represented by a quasi-compact morphism $\Spec(k) \to X$ from a field, then $x \in T_p$ for some $p$. \end{enumerate} \end{lemma} \begin{proof} By Properties of Spaces, Lemma \ref{spaces-properties-lemma-quasi-compact-affine-cover} we can choose an affine scheme $U$ and a surjective \'etale morphism $U \to X$. For $p \geq 0$ set $$W_p = U \times_X \ldots \times_X U \setminus \text{all diagonals}$$ where the fibre product has $p$ factors. Since $U$ is separated, the morphism $U \to X$ is separated and all fibre products $U \times_X \ldots \times_X U$ are separated schemes. Since $U \to X$ is separated the diagonal $U \to U \times_X U$ is a closed immersion. Since $U \to X$ is \'etale the diagonal $U \to U \times_X U$ is an open immersion, see Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-etale-unramified} and \ref{spaces-morphisms-lemma-diagonal-unramified-morphism}. Similarly, all the diagonal morphisms are open and closed immersions and $W_p$ is an open and closed subscheme of $U \times_X \ldots \times_X U$. Moreover, the morphism $$U \times_X \ldots \times_X U \longrightarrow U \times_{\Spec(\mathbf{Z})} \ldots \times_{\Spec(\mathbf{Z})} U$$ is locally quasi-finite and separated (Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-fibre-product-after-map}) and its target is an affine scheme. Hence every finite set of points of $U \times_X \ldots \times_X U$ is contained in an affine open, see More on Morphisms, Lemma \ref{more-morphisms-lemma-separated-locally-quasi-finite-over-affine}. Therefore, the same is true for $W_p$. There is a free action of the symmetric group $S_p$ on $W_p$ over $X$ (because we threw out the fix point locus from $U \times_X \ldots \times_X U$). By the above and Properties of Spaces, Proposition \ref{spaces-properties-proposition-finite-flat-equivalence-global} the quotient $V_p = W_p/S_p$ is a scheme. Since the action of $S_p$ on $W_p$ was over $X$, there is a morphism $V_p \to X$. Since $W_p \to X$ is \'etale and since $W_p \to V_p$ is surjective \'etale, it follows that also $V_p \to X$ is \'etale, see Properties of Spaces, Lemma \ref{spaces-properties-lemma-etale-local}. Observe that $V_p$ is a separated scheme by Properties of Spaces, Lemma \ref{spaces-properties-lemma-quotient-separated}. \medskip\noindent We let $U_p \subset X$ be the open subspace which is the image of $V_p \to X$. By construction a morphism $\Spec(k) \to X$ with $k$ algebraically closed, factors through $U_p$ if and only if $U \times_X \Spec(k)$ has $\geq p$ points; as usual observe that $U \times_X \Spec(k)$ is scheme theoretically a disjoint union of (possibly infinitely many) copies of $\Spec(k)$, see Remark \ref{remark-recall}. It follows that the $U_p$ give a filtration of $X$ as stated in the lemma. Moreover, our morphism $\Spec(k) \to X$ factors through $T_p$ if and only if $U \times_X \Spec(k)$ has exactly $p$ points. In this case we see that $V_p \times_X \Spec(k)$ has exactly one point. Set $Z_p = f_p^{-1}(T_p) \subset V_p$. This is a closed subscheme of $V_p$. Then $Z_p \to T_p$ is an \'etale morphism between algebraic spaces which induces a bijection on $k$-valued points for any algebraically closed field $k$. To be sure this implies that $Z_p \to T_p$ is universally injective, whence an open immersion by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-etale-universally-injective-open} hence an isomorphism and (1) has been proved. \medskip\noindent Let $x : \Spec(k) \to X$ be a quasi-compact morphism where $k$ is a field. Then the composition $\Spec(\overline{k}) \to \Spec(k) \to X$ is quasi-compact as well (Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-composition-quasi-compact}). In this case the scheme $U \times_X \Spec(\overline{k})$ is quasi-compact. In view of the fact (seen above) that it is a disjoint union of copies of $\Spec(\overline{k})$ we find that it has finitely many points. If the number of points is $p$, then we see that indeed $x \in T_p$ and the proof is finished. \end{proof} \begin{lemma} \label{lemma-filter-reasonable} Let $S$ be a scheme. Let $X$ be a quasi-compact, reasonable algebraic space over $S$. There exist an integer $n$ and open subspaces $$\emptyset = U_{n + 1} \subset U_n \subset U_{n - 1} \subset \ldots \subset U_1 = X$$ with the following property: setting $T_p = U_p \setminus U_{p + 1}$ (with reduced induced subspace structure) there exists a separated scheme $V_p$ and a surjective \'etale morphism $f_p : V_p \to U_p$ such that $f_p^{-1}(T_p) \to T_p$ is an isomorphism. \end{lemma} \begin{proof} The proof of this lemma is identical to the proof of Lemma \ref{lemma-filter-quasi-compact}. Let $n$ be an integer bounding the degrees of the fibres of $U \to X$ which exists as $X$ is reasonable, see Definition \ref{definition-very-reasonable}. Then we see that $U_{n + 1} = \emptyset$ and the proof is complete. \end{proof} \begin{lemma} \label{lemma-stratify-reasonable} Let $S$ be a scheme. Let $X$ be a quasi-compact, reasonable algebraic space over $S$. There exist an integer $n$ and open subspaces $$\emptyset = U_{n + 1} \subset U_n \subset U_{n - 1} \subset \ldots \subset U_1 = X$$ such that each $T_p = U_p \setminus U_{p + 1}$ (with reduced induced subspace structure) is a scheme. \end{lemma} \begin{proof} Immediate consequence of Lemma \ref{lemma-filter-reasonable}. \end{proof} \noindent The following result is almost identical to \cite[Proposition 5.7.8]{GruRay}. \begin{lemma} \label{lemma-filter-quasi-compact-quasi-separated} \begin{reference} This result is almost identical to \cite[Proposition 5.7.8]{GruRay}. \end{reference} Let $X$ be a quasi-compact and quasi-separated algebraic space over $\Spec(\mathbf{Z})$. There exist an integer $n$ and open subspaces $$\emptyset = U_{n + 1} \subset U_n \subset U_{n - 1} \subset \ldots \subset U_1 = X$$ with the following property: setting $T_p = U_p \setminus U_{p + 1}$ (with reduced induced subspace structure) there exists a quasi-compact separated scheme $V_p$ and a surjective \'etale morphism $f_p : V_p \to U_p$ such that $f_p^{-1}(T_p) \to T_p$ is an isomorphism. \end{lemma} \begin{proof} The proof of this lemma is identical to the proof of Lemma \ref{lemma-filter-quasi-compact}. Observe that a quasi-separated space is reasonable, see Lemma \ref{lemma-bounded-fibres} and Definition \ref{definition-very-reasonable}. Hence we find that $U_{n + 1} = \emptyset$ as in Lemma \ref{lemma-filter-reasonable}. At the end of the argument we add that since $X$ is quasi-separated the schemes $U \times_X \ldots \times_X U$ are all quasi-compact. Hence the schemes $W_p$ are quasi-compact. Hence the quotients $V_p = W_p/S_p$ by the symmetric group $S_p$ are quasi-compact schemes. \end{proof} \section{Integral cover by a scheme} \label{section-integral-cover} \noindent Here we prove that given any quasi-compact and quasi-separated algebraic space $X$, there is a scheme $Y$ and a surjective, integral morphism $Y \to X$. After we develop some theory about limits of algebraic spaces, we will prove that one can do this with a finite morphism, see Limits of Spaces, Section \ref{spaces-limits-section-finite-cover}. \begin{lemma} \label{lemma-there-is-a-scheme-integral-over} Let $S$ be a scheme. Let $X$ be a quasi-compact and quasi-separated algebraic space over $S$. \begin{enumerate} \item There exists a surjective integral morphism $Y \to X$ where $Y$ is a scheme, \item given a surjective \'etale morphism $U \to X$ we may choose $Y \to X$ such that for every $y \in Y$ there is an open neighbourhood $V \subset Y$ such that $V \to X$ factors through $U$. \end{enumerate} \end{lemma} \begin{proof} Part (1) is the special case of part (2) where $U = X$. Choose a surjective \'etale morphism $U' \to U$ where $U'$ is a scheme. It is clear that we may replace $U$ by $U'$ and hence we may assume $U$ is a scheme. Since $X$ is quasi-compact, there exist finitely many affine opens $U_i \subset U$ such that $U' = \coprod U_i \to X$ is surjective. After replacing $U$ by $U'$ again, we see that we may assume $U$ is affine. Since $X$ is quasi-separated, hence reasonable, there exists an integer $d$ bounding the degree of the geometric fibres of $U \to X$ (see Lemma \ref{lemma-bounded-fibres}). We will prove the lemma by induction on $d$ for all quasi-compact and separated schemes $U$ mapping surjective and \'etale onto $X$. If $d = 1$, then $U = X$ and the result holds with $Y = U$. Assume $d > 1$. \medskip\noindent We apply Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-quasi-finite-separated-quasi-affine} and we obtain a factorization $$\xymatrix{ U \ar[rr]_j \ar[rd] & & Y \ar[ld]^\pi \\ & X }$$ with $\pi$ integral and $j$ a quasi-compact open immersion. We may and do assume that $j(U)$ is scheme theoretically dense in $Y$. Note that $$U \times_X Y = U \amalg W$$ where the first summand is the image of $U \to U \times_X Y$ (which is closed by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-semi-diagonal} and open because it is \'etale as a morphism between algebraic spaces \'etale over $Y$) and the second summand is the (open and closed) complement. The image $V \subset Y$ of $W$ is an open subspace containing $Y \setminus U$. \medskip\noindent The \'etale morphism $W \to Y$ has geometric fibres of cardinality $< d$. Namely, this is clear for geometric points of $U \subset Y$ by inspection. Since $|U| \subset |Y|$ is dense, it holds for all geometric points of $Y$ by Lemma \ref{lemma-quasi-compact-reasonable-stratification} (the degree of the fibres of a quasi-compact \'etale morphism does not go up under specialization). Thus we may apply the induction hypothesis to $W \to V$ and find a surjective integral morphism $Z \to V$ with $Z$ a scheme, which Zariski locally factors through $W$. Choose a factorization $Z \to Z' \to Y$ with $Z' \to Y$ integral and $Z \to Z'$ open immersion (Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-quasi-finite-separated-quasi-affine}). After replacing $Z'$ by the scheme theoretic closure of $Z$ in $Z'$ we may assume that $Z$ is scheme theoretically dense in $Z'$. After doing this we have $Z' \times_Y V = Z$. Finally, let $T \subset Y$ be the induced closed subspace structure on $Y \setminus V$. Consider the morphism $$Z' \amalg T \longrightarrow X$$ This is a surjective integral morphism by construction. Since $T \subset U$ it is clear that the morphism $T \to X$ factors through $U$. On the other hand, let $z \in Z'$ be a point. If $z \not \in Z$, then $z$ maps to a point of $Y \setminus V \subset U$ and we find a neighbourhood of $z$ on which the morphism factors through $U$. If $z \in Z$, then we have a neighbourhood $V \subset Z$ which factors through $W \subset U \times_X Y$ and hence through $U$. \end{proof} \section{Schematic locus} \label{section-schematic} \noindent In this section we prove that a decent algebraic space has a dense open subspace which is a scheme. We first prove this for reasonable algebraic spaces. \begin{proposition} \label{proposition-reasonable-open-dense-scheme} Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. If $X$ is reasonable, then there exists a dense open subspace of $X$ which is a scheme. \end{proposition} \begin{proof} By Properties of Spaces, Lemma \ref{spaces-properties-lemma-subscheme} the question is local on $X$. Hence we may assume there exists an affine scheme $U$ and a surjective \'etale morphism $U \to X$ (Properties of Spaces, Lemma \ref{spaces-properties-lemma-cover-by-union-affines}). Let $n$ be an integer bounding the degrees of the fibres of $U \to X$ which exists as $X$ is reasonable, see Definition \ref{definition-very-reasonable}. We will argue by induction on $n$ that whenever \begin{enumerate} \item $U \to X$ is a surjective \'etale morphism whose fibres have degree $\leq n$, and \item $U$ is isomorphic to a locally closed subscheme of an affine scheme \end{enumerate} then the schematic locus is dense in $X$. \medskip\noindent Let $X_n \subset X$ be the open subspace which is the complement of the closed subspace $Z_{n - 1} \subset X$ constructed in Lemma \ref{lemma-quasi-compact-reasonable-stratification} using the morphism $U \to X$. Let $U_n \subset U$ be the inverse image of $X_n$. Then $U_n \to X_n$ is finite locally free of degree $n$. Hence $X_n$ is a scheme by Properties of Spaces, Proposition \ref{spaces-properties-proposition-finite-flat-equivalence-global} (and the fact that any finite set of points of $U_n$ is contained in an affine open of $U_n$, see Properties, Lemma \ref{properties-lemma-ample-finite-set-in-affine}). \medskip\noindent Let $X' \subset X$ be the open subspace such that $|X'|$ is the interior of $|Z_{n - 1}|$ in $|X|$ (see Topology, Definition \ref{topology-definition-nowhere-dense}). Let $U' \subset U$ be the inverse image. Then $U' \to X'$ is surjective \'etale and has degrees of fibres bounded by $n - 1$. By induction we see that the schematic locus of $X'$ is an open dense $X'' \subset X'$. By elementary topology we see that $X'' \cup X_n \subset X$ is open and dense and we win. \end{proof} \begin{theorem}[David Rydh] \label{theorem-decent-open-dense-scheme} Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. If $X$ is decent, then there exists a dense open subspace of $X$ which is a scheme. \end{theorem} \begin{proof} Assume $X$ is a decent algebraic space for which the theorem is false. By Properties of Spaces, Lemma \ref{spaces-properties-lemma-subscheme} there exists a largest open subspace $X' \subset X$ which is a scheme. Since $X'$ is not dense in $X$, there exists an open subspace $X'' \subset X$ such that $|X''| \cap |X'| = \emptyset$. Replacing $X$ by $X''$ we get a nonempty decent algebraic space $X$ which does not contain {\it any} open subspace which is a scheme. \medskip\noindent Choose a nonempty affine scheme $U$ and an \'etale morphism $U \to X$. We may and do replace $X$ by the open subscheme corresponding to the image of $|U| \to |X|$. Consider the sequence of open subspaces $$X = X_0 \supset X_1 \supset X_2 \ldots$$ constructed in Lemma \ref{lemma-stratify-flat-fp-lqf} for the morphism $U \to X$. Note that $X_0 = X_1$ as $U \to X$ is surjective. Let $U = U_0 = U_1 \supset U_2 \ldots$ be the induced sequence of open subschemes of $U$. \medskip\noindent Choose a nonempty open affine $V_1 \subset U_1$ (for example $V_1 = U_1$). By induction we will construct a sequence of nonempty affine opens $V_1 \supset V_2 \supset \ldots$ with $V_n \subset U_n$. Namely, having constructed $V_1, \ldots, V_{n - 1}$ we can always choose $V_n$ unless $V_{n - 1} \cap U_n = \emptyset$. But if $V_{n - 1} \cap U_n = \emptyset$, then the open subspace $X' \subset X$ with $|X'| = \Im(|V_{n - 1}| \to |X|)$ is contained in $|X| \setminus |X_n|$. Hence $V_{n - 1} \to X'$ is an \'etale morphism whose fibres have degree bounded by $n - 1$. In other words, $X'$ is reasonable (by definition), hence $X'$ contains a nonempty open subscheme by Proposition \ref{proposition-reasonable-open-dense-scheme}. This is a contradiction which shows that we can pick $V_n$. \medskip\noindent By Limits, Lemma \ref{limits-lemma-limit-nonempty} the limit $V_\infty = \lim V_n$ is a nonempty scheme. Pick a morphism $\Spec(k) \to V_\infty$. The composition $\Spec(k) \to V_\infty \to U \to X$ has image contained in all $X_d$ by construction. In other words, the fibred $U \times_X \Spec(k)$ has infinite degree which contradicts the definition of a decent space. This contradiction finishes the proof of the theorem. \end{proof} \begin{lemma} \label{lemma-when-quotient-scheme-at-point} Let $S$ be a scheme. Let $X \to Y$ be a surjective finite locally free morphism of algebraic spaces over $S$. For $y \in |Y|$ the following are equivalent \begin{enumerate} \item $y$ is in the schematic locus of $Y$, and \item there exists an affine open $U \subset X$ containing the preimage of $y$. \end{enumerate} \end{lemma} \begin{proof} If $y \in Y$ is in the schematic locus, then it has an affine open neighbourhood $V \subset Y$ and the inverse image $U$ of $V$ in $X$ is an open finite over $V$, hence affine. Thus (1) implies (2). \medskip\noindent Conversely, assume that $U \subset X$ as in (2) is given. Set $R = X \times_Y X$ and denote the projections $s, t : R \to X$. Consider $Z = R \setminus s^{-1}(U) \cap t^{-1}(U)$. This is a closed subset of $R$. The image $t(Z)$ is a closed subset of $X$ which can loosely be described as the set of points of $X$ which are $R$-equivalent to a point of $X \setminus U$. Hence $U' = X \setminus t(Z)$ is an $R$-invariant, open subspace of $X$ contained in $U$ which contains the fibre of $X \to Y$ over $y$. Since $X \to Y$ is open (Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-fppf-open}) the image of $U'$ is an open subspace $V' \subset Y$. Since $U'$ is $R$-invariant and $R = X \times_Y X$, we see that $U'$ is the inverse image of $V'$ (use Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-cartesian}). After replacing $Y$ by $V'$ and $X$ by $U'$ we see that we may assume $X$ is a scheme isomorphic to an open subscheme of an affine scheme. \medskip\noindent Assume $X$ is a scheme isomorphic to an open subscheme of an affine scheme. In this case the fppf quotient sheaf $X/R$ is a scheme, see Properties of Spaces, Proposition \ref{spaces-properties-proposition-finite-flat-equivalence-global}. Since $Y$ is a sheaf in the fppf topology, obtain a canonical map $X/R \to Y$ factoring $X \to Y$. Since $X \to Y$ is surjective finite locally free, it is surjective as a map of sheaves (Spaces, Lemma \ref{spaces-lemma-surjective-flat-locally-finite-presentation}). We conclude that $X/R \to Y$ is surjective as a map of sheaves. On the other hand, since $R = X \times_Y X$ as sheaves we conclude that $X/R \to Y$ is injective as a map of sheaves. Hence $X/R \to Y$ is an isomorphism and we see that $Y$ is representable. \end{proof} \noindent At this point we have several different ways for proving the following lemma. \begin{lemma} \label{lemma-finite-etale-cover-dense-open-scheme} Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. If there exists a finite, \'etale, surjective morphism $U \to X$ where $U$ is a scheme, then there exists a dense open subspace of $X$ which is a scheme. \end{lemma} \begin{proof}[First proof] The morphism $U \to X$ is finite locally free. Hence there is a decomposition of $X$ into open and closed subspaces $X_d \subset X$ such that $U \times_X X_d \to X_d$ is finite locally free of degree $d$. Thus we may assume $U \to X$ is finite locally free of degree $d$. In this case, let $U_i \subset U$, $i \in I$ be the set of affine opens. For each $i$ the morphism $U_i \to X$ is \'etale and has universally bounded fibres (namely, bounded by $d$). In other words, $X$ is reasonable and the result follows from Proposition \ref{proposition-reasonable-open-dense-scheme}. \end{proof} \begin{proof}[Second proof] The question is local on $X$ (Properties of Spaces, Lemma \ref{spaces-properties-lemma-subscheme}), hence may assume $X$ is quasi-compact. Then $U$ is quasi-compact. Then there exists a dense open subscheme $W \subset U$ which is separated (Properties, Lemma \ref{properties-lemma-quasi-compact-dense-open-separated}). Set $Z = U \setminus W$. Let $R = U \times_X U$ and $s, t : R \to U$ the projections. Then $t^{-1}(Z)$ is nowhere dense in $R$ (Topology, Lemma \ref{topology-lemma-open-inverse-image-closed-nowhere-dense}) and hence $\Delta = s(t^{-1}(Z))$ is an $R$-invariant closed nowhere dense subset of $U$ (Morphisms, Lemma \ref{morphisms-lemma-image-nowhere-dense-finite}). Let $u \in U \setminus \Delta$ be a generic point of an irreducible component. Since these points are dense in $U \setminus \Delta$ and since $\Delta$ is nowhere dense, it suffices to show that the image $x \in X$ of $u$ is in the schematic locus of $X$. Observe that $t(s^{-1}(\{u\})) \subset W$ is a finite set of generic points of irreducible components of $W$ (compare with Properties of Spaces, Lemma \ref{spaces-properties-lemma-codimension-0-points}). By Properties, Lemma \ref{properties-lemma-maximal-points-affine} we can find an affine open $V \subset W$ such that $t(s^{-1}(\{u\})) \subset V$. Since $t(s^{-1}(\{u\}))$ is the fibre of $|U| \to |X|$ over $x$, we conclude by Lemma \ref{lemma-when-quotient-scheme-at-point}. \end{proof} \begin{proof}[Third proof] (This proof is essentially the same as the second proof, but uses fewer references.) Assume $X$ is an algebraic space, $U$ a scheme, and $U \to X$ is a finite \'etale surjective morphism. Write $R = U \times_X U$ and denote $s, t : R \to U$ the projections as usual. Note that $s, t$ are surjective, finite and \'etale. Claim: The union of the $R$-invariant affine opens of $U$ is topologically dense in $U$. \medskip\noindent Proof of the claim. Let $W \subset U$ be an affine open. Set $W' = t(s^{-1}(W)) \subset U$. Since $s^{-1}(W)$ is affine (hence quasi-compact) we see that $W' \subset U$ is a quasi-compact open. By Properties, Lemma \ref{properties-lemma-quasi-compact-dense-open-separated} there exists a dense open $W'' \subset W'$ which is a separated scheme. Set $\Delta' = W' \setminus W''$. This is a nowhere dense closed subset of $W''$. Since $t|_{s^{-1}(W)} : s^{-1}(W) \to W'$ is open (because it is \'etale) we see that the inverse image $(t|_{s^{-1}(W)})^{-1}(\Delta') \subset s^{-1}(W)$ is a nowhere dense closed subset (see Topology, Lemma \ref{topology-lemma-open-inverse-image-closed-nowhere-dense}). Hence, by Morphisms, Lemma \ref{morphisms-lemma-image-nowhere-dense-finite} we see that $$\Delta = s\left((t|_{s^{-1}(W)})^{-1}(\Delta')\right)$$ is a nowhere dense closed subset of $W$. Pick any point $\eta \in W$, $\eta \not \in \Delta$ which is a generic point of an irreducible component of $W$ (and hence of $U$). By our choices above the finite set $t(s^{-1}(\{\eta\})) = \{\eta_1, \ldots, \eta_n\}$ is contained in the separated scheme $W''$. Note that the fibres of $s$ is are finite discrete spaces, and that generalizations lift along the \'etale morphism $t$, see Morphisms, Lemmas \ref{morphisms-lemma-etale-flat} and \ref{morphisms-lemma-generalizations-lift-flat}. In this way we see that each $\eta_i$ is a generic point of an irreducible component of $W''$. Thus, by Properties, Lemma \ref{properties-lemma-maximal-points-affine} we can find an affine open $V \subset W''$ such that $\{\eta_1, \ldots, \eta_n\} \subset V$. By Groupoids, Lemma \ref{groupoids-lemma-find-invariant-affine} this implies that $\eta$ is contained in an $R$-invariant affine open subscheme of $U$. The claim follows as $W$ was chosen as an arbitrary affine open of $U$ and because the set of generic points of irreducible components of $W \setminus \Delta$ is dense in $W$. \medskip\noindent Using the claim we can finish the proof. Namely, if $W \subset U$ is an $R$-invariant affine open, then the restriction $R_W$ of $R$ to $W$ equals $R_W = s^{-1}(W) = t^{-1}(W)$ (see Groupoids, Definition \ref{groupoids-definition-invariant-open} and discussion following it). In particular the maps $R_W \to W$ are finite \'etale also. It follows in particular that $R_W$ is affine. Thus we see that $W/R_W$ is a scheme, by Groupoids, Proposition \ref{groupoids-proposition-finite-flat-equivalence}. On the other hand, $W/R_W$ is an open subspace of $X$ by Spaces, Lemma \ref{spaces-lemma-finding-opens}. Hence having a dense collection of points contained in $R$-invariant affine open of $U$ certainly implies that the schematic locus of $X$ (see Properties of Spaces, Lemma \ref{spaces-properties-lemma-subscheme}) is open dense in $X$. \end{proof} \section{Points on spaces} \label{section-points} \noindent In this section we prove some properties of points on decent algebraic spaces. \begin{lemma} \label{lemma-decent-points-monomorphism} Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Consider the map $$\{\Spec(k) \to X \text{ monomorphism}\} \longrightarrow |X|$$ This map is always injective. If $X$ is decent then this map is a bijection. \end{lemma} \begin{proof} We have seen in Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-monomorphism} that the map is an injection in general. By Lemma \ref{lemma-bounded-fibres} it is surjective when $X$ is decent (actually one can say this is part of the definition of being decent). \end{proof} \noindent The following lemma tells us that the henselian local ring of a point on a decent algebraic space is defined. \begin{lemma} \label{lemma-decent-space-elementary-etale-neighbourhood} Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. For every point $x \in |X|$ there exists an \'etale morphism $$(U, u) \longrightarrow (X, x)$$ where $U$ is an affine scheme, $u$ is the only point of $U$ lying over $x$, and the induced morphism $\Spec(\kappa(u)) \to X$ is a monomorphism. \end{lemma} \begin{proof} We may assume that $X$ is quasi-compact by replacing $X$ with a quasi-compact open containing $x$. Recall that $x$ can be represented by a quasi-compact (mono)morphism from the spectrum a field (by definition of decent spaces). Thus the lemma follows from Lemma \ref{lemma-filter-quasi-compact}. \end{proof} \begin{definition} \label{definition-elemenary-etale-neighbourhood} Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in X$ be a point. An {\it elementary \'etale neighbourhood} is an \'etale morphism $(U, u) \to (X, x)$ where $U$ is a scheme, $u \in U$ is a point mapping to $x$, and $\Spec(\kappa(u)) \to X$ is a monomorphism. A {\it morphism of elementary \'etale neighbourhoods} $(U, u) \to (U', u')$ is defined as a morphism $U \to U'$ over $X$ mapping $u$ to $u'$. \end{definition} \noindent If $X$ is not decent then the category of elementary \'etale neighbourhoods may be empty. \begin{lemma} \label{lemma-elementary-etale-neighbourhoods} Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Let $x$ be a point of $X$. The category of elementary \'etale neighborhoods of $(X, x)$ is cofiltered (see Categories, Definition \ref{categories-definition-codirected}). \end{lemma} \begin{proof} The category is nonempty by Lemma \ref{lemma-decent-space-elementary-etale-neighbourhood}. Suppose that we have two elementary \'etale neighbourhoods $(U_i, u_i) \to (X, x)$. Then consider $U = U_1 \times_X U_2$. Since $\Spec(\kappa(u_i)) \to X$, $i = 1, 2$ are both monomorphisms in the class of $x$, we see that $$u = \Spec(\kappa(u_1)) \times_X \Spec(\kappa(u_2))$$ is the spectrum of a field $\kappa(u)$ such that the induced maps $\kappa(u_i) \to \kappa(u)$ are isomorphisms. Then $u \to U$ is a point of $U$ and we see that $(U, u) \to (X, x)$ is an elementary \'etale neighbourhood dominating $(U_i, u_i)$. If $a, b : (U_1, u_1) \to (U_2, u_2)$ are two morphisms between our elementary \'etale neighbourhoods, then we consider the scheme $$U = U_1 \times_{(a, b), (U_2 \times_X U_2), \Delta} U_2$$ Using Properties of Spaces, Lemma \ref{spaces-properties-lemma-etale-permanence} we see that $U \to X$ is \'etale. Moreover, in exactly the same manner as before we see that $U$ has a point $u$ such that $(U, u) \to (X, x)$ is an elementary \'etale neighbourhood. Finally, $U \to U_1$ equalizes $a$ and $b$ and the proof is finished. \end{proof} \begin{definition} \label{definition-henselian-local-ring} Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Let $x \in |X|$. The {\it henselian local ring of $X$ at $x$}, is $$\mathcal{O}_{X, x}^h = \colim \Gamma(U, \mathcal{O}_U)$$ where the colimit is over the elementary \'etale neighbourhoods $(U, u) \to (X, x)$. \end{definition} \noindent To be sure, the henselian local ring of $X$ at $x$ is equal to the henselization $\mathcal{O}_{U, u}^h$ of the local ring $\mathcal{O}_{U, u}$ of any elementary \'etale neighbourhood. This follows from the definition, Lemma \ref{lemma-elementary-etale-neighbourhoods}, and More on Morphisms, Lemma \ref{more-morphisms-lemma-describe-henselization}. \medskip\noindent The following lemma shows that specialization of points behaves well on decent algebraic spaces. Spaces, Example \ref{spaces-example-infinite-product} shows that this is {\bf not} true in general. \begin{lemma} \label{lemma-decent-no-specializations-map-to-same-point} Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Let $U \to X$ be an \'etale morphism from a scheme to $X$. If $u, u' \in |U|$ map to the same point of $|X|$, and $u' \leadsto u$, then $u = u'$. \end{lemma} \begin{proof} Combine Lemmas \ref{lemma-bounded-fibres} and \ref{lemma-no-specializations-map-to-same-point}. \end{proof} \begin{lemma} \label{lemma-decent-specialization} Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Let $x, x' \in |X|$ and assume $x' \leadsto x$, i.e., $x$ is a specialization of $x'$. Then for every \'etale morphism $\varphi : U \to X$ from a scheme $U$ and any $u \in U$ with $\varphi(u) = x$, exists a point $u'\in U$, $u' \leadsto u$ with $\varphi(u') = x'$. \end{lemma} \begin{proof} Combine Lemmas \ref{lemma-bounded-fibres} and \ref{lemma-specialization}. \end{proof} \begin{lemma} \label{lemma-kolmogorov} Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Then $|X|$ is Kolmogorov (see Topology, Definition \ref{topology-definition-generic-point}). \end{lemma} \begin{proof} Let $x_1, x_2 \in |X|$ with $x_1 \leadsto x_2$ and $x_2 \leadsto x_1$. We have to show that $x_1 = x_2$. Pick a scheme $U$ and an \'etale morphism $U \to X$ such that $x_1, x_2$ are both in the image of $|U| \to |X|$. By Lemma \ref{lemma-decent-specialization} we can find a specialization $u_1 \leadsto u_2$ in $U$ mapping to $x_1 \leadsto x_2$. By Lemma \ref{lemma-decent-specialization} we can find $u_2' \leadsto u_1$ mapping to $x_2 \leadsto x_1$. This means that $u_2' \leadsto u_2$ is a specialization between points of $U$ mapping to the same point of $X$, namely $x_2$. This is not possible, unless $u_2' = u_2$, see Lemma \ref{lemma-decent-no-specializations-map-to-same-point}. Hence also $u_1 = u_2$ as desired. \end{proof} \begin{proposition} \label{proposition-reasonable-sober} Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Then the topological space $|X|$ is sober (see Topology, Definition \ref{topology-definition-generic-point}). \end{proposition} \begin{proof} We have seen in Lemma \ref{lemma-kolmogorov} that $|X|$ is Kolmogorov. Hence it remains to show that every irreducible closed subset $T \subset |X|$ has a generic point. By Properties of Spaces, Lemma \ref{spaces-properties-lemma-reduced-closed-subspace} there exists a closed subspace $Z \subset X$ with $|Z| = |T|$. By definition this means that $Z \to X$ is a representable morphism of algebraic spaces. Hence $Z$ is a decent algebraic space by Lemma \ref{lemma-representable-properties}. By Theorem \ref{theorem-decent-open-dense-scheme} we see that there exists an open dense subspace $Z' \subset Z$ which is a scheme. This means that $|Z'| \subset T$ is open dense. Hence the topological space $|Z'|$ is irreducible, which means that $Z'$ is an irreducible scheme. By Schemes, Lemma \ref{schemes-lemma-scheme-sober} we conclude that $|Z'|$ is the closure of a single point $\eta \in T$ and hence also $T = \overline{\{\eta\}}$, and we win. \end{proof} \noindent For decent algebraic spaces dimension works as expected. \begin{lemma} \label{lemma-dimension-decent-space} Let $S$ be a scheme. Dimension as defined in Properties of Spaces, Section \ref{spaces-properties-section-dimension} behaves well on decent algebraic spaces $X$ over $S$. \begin{enumerate} \item If $x \in |X|$, then $\dim_x(|X|) = \dim_x(X)$, and \item $\dim(|X|) = \dim(X)$. \end{enumerate} \end{lemma} \begin{proof} Proof of (1). Choose a scheme $U$ with a point $u \in U$ and an \'etale morphism $h : U \to X$ mapping $u$ to $x$. By definition the dimension of $X$ at $x$ is $\dim_u(|U|)$. Thus we may pick $U$ such that $\dim_x(X) = \dim(|U|)$. Let $d$ be an integer. If $\dim(U) \geq d$, then there exists a sequence of nontrivial specializations $u_d \leadsto \ldots \leadsto u_0$ in $U$. Taking the image we find a corresponding sequence $h(u_d) \leadsto \ldots \leadsto h(u_0)$ each of which is nontrivial by Lemma \ref{lemma-decent-no-specializations-map-to-same-point}. Hence we see that the image of $|U|$ in $|X|$ has dimension at least $d$. Conversely, suppose that $x_d \leadsto \ldots \leadsto x_0$ is a sequence of specializations in $|X|$ with $x_0$ in the image of $|U| \to |X|$. Then we can lift this to a sequence of specializations in $U$ by Lemma \ref{lemma-decent-specialization}. \medskip\noindent Part (2) is an immediate consequence of part (1), Topology, Lemma \ref{topology-lemma-dimension-supremum-local-dimensions}, and Properties of Spaces, Section \ref{spaces-properties-section-dimension}. \end{proof} \begin{lemma} \label{lemma-dimension-local-ring-quasi-finite} Let $S$ be a scheme. Let $X \to Y$ be a locally quasi-finite morphism of algebraic spaces over $S$. Let $x \in |X|$ with image $y \in |Y|$. Then the dimension of the local ring of $Y$ at $y$ is $\geq$ to the dimension of the local ring of $X$ at $x$. \end{lemma} \begin{proof} The definition of the dimension of the local ring of a point on an algebraic space is given in Properties of Spaces, Definition \ref{spaces-properties-definition-dimension-local-ring}. Choose an \'etale morphism $(V, v) \to (Y, y)$ where $V$ is a scheme. Choose an \'etale morphism $U \to V \times_Y X$ and a point $u \in U$ mapping to $x \in |X|$ and $v \in V$. Then $U \to V$ is locally quasi-finite and we have to prove that $$\dim(\mathcal{O}_{V, v}) \geq \dim(\mathcal{O}_{U, u})$$ This is Algebra, Lemma \ref{algebra-lemma-dimension-inequality-quasi-finite}. \end{proof} \noindent The following lemma is a tiny bit stronger than Properties of Spaces, Lemma \ref{spaces-properties-lemma-point-like-spaces}. We will improve this lemma in Lemma \ref{lemma-when-field}. \begin{lemma} \label{lemma-decent-point-like-spaces} Let $S$ be a scheme. Let $k$ be a field. Let $X$ be an algebraic space over $S$ and assume that there exists a surjective \'etale morphism $\Spec(k) \to X$. If $X$ is decent, then $X \cong \Spec(k')$ where $k' \subset k$ is a finite separable extension. \end{lemma} \begin{proof} The assumption implies that $|X| = \{x\}$ is a singleton. Since $X$ is decent we can find a quasi-compact monomorphism $\Spec(k') \to X$ whose image is $x$. Then the projection $U = \Spec(k') \times_X \Spec(k) \to \Spec(k)$ is a monomorphism, whence $U = \Spec(k)$, see Schemes, Lemma \ref{schemes-lemma-mono-towards-spec-field}. Hence the projection $\Spec(k) = U \to \Spec(k')$ is \'etale and we win. \end{proof} \section{Reduced singleton spaces} \label{section-singleton} \noindent A {\it singleton} space is an algebraic space $X$ such that $|X|$ is a singleton. It turns out that these can be more interesting than just being the spectrum of a field, see Spaces, Example \ref{spaces-example-Qbar}. We develop a tiny bit of machinery to be able to talk about these. \begin{lemma} \label{lemma-flat-cover-by-field} Let $S$ be a scheme. Let $Z$ be an algebraic space over $S$. Let $k$ be a field and let $\Spec(k) \to Z$ be surjective and flat. Then any morphism $\Spec(k') \to Z$ where $k'$ is a field is surjective and flat. \end{lemma} \begin{proof} Consider the fibre square $$\xymatrix{ T \ar[d] \ar[r] & \Spec(k) \ar[d] \\ \Spec(k') \ar[r] & Z }$$ Note that $T \to \Spec(k')$ is flat and surjective hence $T$ is not empty. On the other hand $T \to \Spec(k)$ is flat as $k$ is a field. Hence $T \to Z$ is flat and surjective. It follows from Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-flat-permanence} that $\Spec(k') \to Z$ is flat. It is surjective as by assumption $|Z|$ is a singleton. \end{proof} \begin{lemma} \label{lemma-unique-point} Let $S$ be a scheme. Let $Z$ be an algebraic space over $S$. The following are equivalent \begin{enumerate} \item $Z$ is reduced and $|Z|$ is a singleton, \item there exists a surjective flat morphism $\Spec(k) \to Z$ where $k$ is a field, and \item there exists a locally of finite type, surjective, flat morphism $\Spec(k) \to Z$ where $k$ is a field. \end{enumerate} \end{lemma} \begin{proof} Assume (1). Let $W$ be a scheme and let $W \to Z$ be a surjective \'etale morphism. Then $W$ is a reduced scheme. Let $\eta \in W$ be a generic point of an irreducible component of $W$. Since $W$ is reduced we have $\mathcal{O}_{W, \eta} = \kappa(\eta)$. It follows that the canonical morphism $\eta = \Spec(\kappa(\eta)) \to W$ is flat. We see that the composition $\eta \to Z$ is flat (see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-composition-flat}). It is also surjective as $|Z|$ is a singleton. In other words (2) holds. \medskip\noindent Assume (2). Let $W$ be a scheme and let $W \to Z$ be a surjective \'etale morphism. Choose a field $k$ and a surjective flat morphism $\Spec(k) \to Z$. Then $W \times_Z \Spec(k)$ is a scheme \'etale over $k$. Hence $W \times_Z \Spec(k)$ is a disjoint union of spectra of fields (see Remark \ref{remark-recall}), in particular reduced. Since $W \times_Z \Spec(k) \to W$ is surjective and flat we conclude that $W$ is reduced (Descent, Lemma \ref{descent-lemma-descend-reduced}). In other words (1) holds. \medskip\noindent It is clear that (3) implies (2). Finally, assume (2). Pick a nonempty affine scheme $W$ and an \'etale morphism $W \to Z$. Pick a closed point $w \in W$ and set $k = \kappa(w)$. The composition $$\Spec(k) \xrightarrow{w} W \longrightarrow Z$$ is locally of finite type by Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-composition-finite-type} and \ref{spaces-morphisms-lemma-etale-locally-finite-type}. It is also flat and surjective by Lemma \ref{lemma-flat-cover-by-field}. Hence (3) holds. \end{proof} \noindent The following lemma singles out a slightly better class of singleton algebraic spaces than the preceding lemma. \begin{lemma} \label{lemma-unique-point-better} Let $S$ be a scheme. Let $Z$ be an algebraic space over $S$. The following are equivalent \begin{enumerate} \item $Z$ is reduced, locally Noetherian, and $|Z|$ is a singleton, and \item there exists a locally finitely presented, surjective, flat morphism $\Spec(k) \to Z$ where $k$ is a field. \end{enumerate} \end{lemma} \begin{proof} Assume (2) holds. By Lemma \ref{lemma-unique-point} we see that $Z$ is reduced and $|Z|$ is a singleton. Let $W$ be a scheme and let $W \to Z$ be a surjective \'etale morphism. Choose a field $k$ and a locally finitely presented, surjective, flat morphism $\Spec(k) \to Z$. Then $W \times_Z \Spec(k)$ is a scheme \'etale over $k$, hence a disjoint union of spectra of fields (see Remark \ref{remark-recall}), hence locally Noetherian. Since $W \times_Z \Spec(k) \to W$ is flat, surjective, and locally of finite presentation, we see that $\{W \times_Z \Spec(k) \to W\}$ is an fppf covering and we conclude that $W$ is locally Noetherian (Descent, Lemma \ref{descent-lemma-Noetherian-local-fppf}). In other words (1) holds. \medskip\noindent Assume (1). Pick a nonempty affine scheme $W$ and an \'etale morphism $W \to Z$. Pick a closed point $w \in W$ and set $k = \kappa(w)$. Because $W$ is locally Noetherian the morphism $w : \Spec(k) \to W$ is of finite presentation, see Morphisms, Lemma \ref{morphisms-lemma-closed-immersion-finite-presentation}. Hence the composition $$\Spec(k) \xrightarrow{w} W \longrightarrow Z$$ is locally of finite presentation by Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-composition-finite-presentation} and \ref{spaces-morphisms-lemma-etale-locally-finite-presentation}. It is also flat and surjective by Lemma \ref{lemma-flat-cover-by-field}. Hence (2) holds. \end{proof} \begin{lemma} \label{lemma-monomorphism-into-point} Let $S$ be a scheme. Let $Z' \to Z$ be a monomorphism of algebraic spaces over $S$. Assume there exists a field $k$ and a locally finitely presented, surjective, flat morphism $\Spec(k) \to Z$. Then either $Z'$ is empty or $Z' = Z$. \end{lemma} \begin{proof} We may assume that $Z'$ is nonempty. In this case the fibre product $T = Z' \times_Z \Spec(k)$ is nonempty, see Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-cartesian}. Now $T$ is an algebraic space and the projection $T \to \Spec(k)$ is a monomorphism. Hence $T = \Spec(k)$, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-monomorphism-toward-field}. We conclude that $\Spec(k) \to Z$ factors through $Z'$. But as $\Spec(k) \to Z$ is surjective, flat and locally of finite presentation, we see that $\Spec(k) \to Z$ is surjective as a map of sheaves on $(\Sch/S)_{fppf}$ (see Spaces, Remark \ref{spaces-remark-warning}) and we conclude that $Z' = Z$. \end{proof} \noindent The following lemma says that to each point of an algebraic space we can associate a canonical reduced, locally Noetherian singleton algebraic space. \begin{lemma} \label{lemma-find-singleton-from-point} Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$. Then there exists a unique monomorphism $Z \to X$ of algebraic spaces over $S$ such that $Z$ is an algebraic space which satisfies the equivalent conditions of Lemma \ref{lemma-unique-point-better} and such that the image of $|Z| \to |X|$ is $\{x\}$. \end{lemma} \begin{proof} Choose a scheme $U$ and a surjective \'etale morphism $U \to X$. Set $R = U \times_X U$ so that $X = U/R$ is a presentation (see Spaces, Section \ref{spaces-section-presentations}). Set $$U' = \coprod\nolimits_{u \in U\text{ lying over }x} \Spec(\kappa(u)).$$ The canonical morphism $U' \to U$ is a monomorphism. Let $$R' = U' \times_X U' = R \times_{(U \times_S U)} (U' \times_S U').$$ Because $U' \to U$ is a monomorphism we see that the projections $s', t' : R' \to U'$ factor as a monomorphism followed by an \'etale morphism. Hence, as $U'$ is a disjoint union of spectra of fields, using Remark \ref{remark-recall}, and using Schemes, Lemma \ref{schemes-lemma-mono-towards-spec-field} we conclude that $R'$ is a disjoint union of spectra of fields and that the morphisms $s', t' : R' \to U'$ are \'etale. Hence $Z = U'/R'$ is an algebraic space by Spaces, Theorem \ref{spaces-theorem-presentation}. As $R'$ is the restriction of $R$ by $U' \to U$ we see $Z \to X$ is a monomorphism by Groupoids, Lemma \ref{groupoids-lemma-quotient-groupoid-restrict}. Since $Z \to X$ is a monomorphism we see that $|Z| \to |X|$ is injective, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-monomorphism-injective-points}. By Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-cartesian} we see that $$|U'| = |Z \times_X U'| \to |Z| \times_{|X|} |U'|$$ is surjective which implies (by our choice of $U'$) that $|Z| \to |X|$ has image $\{x\}$. We conclude that $|Z|$ is a singleton. Finally, by construction $U'$ is locally Noetherian and reduced, i.e., we see that $Z$ satisfies the equivalent conditions of Lemma \ref{lemma-unique-point-better}. \medskip\noindent Let us prove uniqueness of $Z \to X$. Suppose that $Z' \to X$ is a second such monomorphism of algebraic spaces. Then the projections $$Z' \longleftarrow Z' \times_X Z \longrightarrow Z$$ are monomorphisms. The algebraic space in the middle is nonempty by Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-cartesian}. Hence the two projections are isomorphisms by Lemma \ref{lemma-monomorphism-into-point} and we win. \end{proof} \noindent We introduce the following terminology which foreshadows the residual gerbes we will introduce later, see Properties of Stacks, Definition \ref{stacks-properties-definition-residual-gerbe}. \begin{definition} \label{definition-residual-space} Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$. The {\it residual space of $X$ at $x$}\footnote{This is nonstandard notation.} is the monomorphism $Z_x \to X$ constructed in Lemma \ref{lemma-find-singleton-from-point}. \end{definition} \noindent In particular we know that $Z_x$ is a locally Noetherian, reduced, singleton algebraic space and that there exists a field and a surjective, flat, locally finitely presented morphism $$\Spec(k) \longrightarrow Z_x.$$ It turns out that $Z_x$ is a regular algebraic space as follows from the following lemma. \begin{lemma} \label{lemma-residual-space-regular} A reduced, locally Noetherian singleton algebraic space $Z$ is regular. \end{lemma} \begin{proof} Let $Z$ be a reduced, locally Noetherian singleton algebraic space over a scheme $S$. Let $W \to Z$ be a surjective \'etale morphism where $W$ is a scheme. Let $k$ be a field and let $\Spec(k) \to Z$ be surjective, flat, and locally of finite presentation (see Lemma \ref{lemma-unique-point-better}). The scheme $T = W \times_Z \Spec(k)$ is \'etale over $k$ in particular regular, see Remark \ref{remark-recall}. Since $T \to W$ is locally of finite presentation, flat, and surjective it follows that $W$ is regular, see Descent, Lemma \ref{descent-lemma-descend-regular}. By definition this means that $Z$ is regular. \end{proof} \section{Decent spaces} \label{section-decent} \noindent In this section we collect some useful facts on decent spaces. \begin{lemma} \label{lemma-locally-Noetherian-decent-quasi-separated} Any locally Noetherian decent algebraic space is quasi-separated. \end{lemma} \begin{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 \'etale morphisms with $U$ and $V$ affine schemes. We have to show that $W = U \times_X V$ is quasi-compact (Properties of Spaces, Lemma \ref{spaces-properties-lemma-characterize-quasi-separated}). 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 $\Spec(k) \to X$. Then $U_k$ and $V_k$ are finite disjoint unions of spectra of finite separable extensions of $k$ (Remark \ref{remark-recall}) and we see that $W_k = U_k \times_{\Spec(k)} V_k$ 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 \ref{more-morphisms-lemma-stratify-flat-fp-lqf}. 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 \ref{more-morphisms-lemma-stratify-flat-fp-lqf-universally-bounded} 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. \end{proof} \begin{lemma} \label{lemma-when-field} Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. \begin{enumerate} \item If $|X|$ is a singleton then $X$ is a scheme. \item If $|X|$ is a singleton and $X$ is reduced, then $X \cong \Spec(k)$ for some field $k$. \end{enumerate} \end{lemma} \begin{proof} Assume $|X|$ is a singleton. It follows immediately from Theorem \ref{theorem-decent-open-dense-scheme} that $X$ is a scheme, but we can also argue directly as follows. Choose an affine scheme $U$ and a surjective \'etale morphism $U \to X$. Set $R = U \times_X U$. Then $U$ and $R$ have finitely many points by Lemma \ref{lemma-UR-finite-above-x} (and the definition of a decent space). All of these points are closed in $U$ and $R$ by Lemma \ref{lemma-decent-no-specializations-map-to-same-point}. 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 \ref{algebra-lemma-local-dimension-zero-henselian}. The projections $R \to U$ are \'etale, hence finite \'etale because $U$ is the spectrum of a $0$-dimensional henselian local ring, see Algebra, Lemma \ref{algebra-lemma-characterize-henselian}. It follows that $X$ is a scheme by Groupoids, Proposition \ref{groupoids-proposition-finite-flat-equivalence}. \medskip\noindent Part (2) follows from (1) and the fact that a reduced singleton scheme is the spectrum of a field. \end{proof} \begin{remark} \label{remark-one-point-decent-scheme} We will see in Limits of Spaces, Lemma \ref{spaces-limits-lemma-reduction-scheme} that an algebraic space whose reduction is a scheme is a scheme. \end{remark} \begin{lemma} \label{lemma-algebraic-residue-field-extension-closed-point} Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Consider a commutative diagram $$\xymatrix{ \Spec(k) \ar[rr] \ar[rd] & & X \ar[ld] \\ & S }$$ Assume that the image point $s \in S$ of $\Spec(k) \to S$ is a closed point and that $\kappa(s) \subset k$ is algebraic. Then the image $x$ of $\Spec(k) \to X$ is a closed point of $|X|$. \end{lemma} \begin{proof} Suppose that $x \leadsto x'$ for some $x' \in |X|$. Choose an \'etale 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 \ref{lemma-decent-specialization}. Then $u$ is the image of a point $w$ of the scheme $W = \Spec(k) \times_X U$. Since the projection $W \to \Spec(k)$ is \'etale 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 \ref{morphisms-lemma-algebraic-residue-field-extension-closed-point-fibre}. Thus $u = u'$, whence $x = x'$. \end{proof} \begin{lemma} \label{lemma-finite-residue-field-extension-finite} Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Consider a commutative diagram $$\xymatrix{ \Spec(k) \ar[rr] \ar[rd] & & X \ar[ld] \\ & S }$$ Assume that the image point $s \in S$ of $\Spec(k) \to S$ is a closed point and that $\kappa(s) \subset k$ is finite. Then $\Spec(k) \to X$ is finite morphism. If $\kappa(s) = k$ then $\Spec(k) \to X$ is closed immersion. \end{lemma} \begin{proof} By Lemma \ref{lemma-algebraic-residue-field-extension-closed-point} the image point $x \in |X|$ is closed. Let $Z \subset X$ be the reduced closed subspace with $|Z| = \{x\}$ (Properties of Spaces, Lemma \ref{spaces-properties-lemma-reduced-closed-subspace}). Note that $Z$ is a decent algebraic space by Lemma \ref{lemma-representable-named-properties}. By Lemma \ref{lemma-when-field} we see that $Z = \Spec(k')$ for some field $k'$. Of course $k \supset k' \supset \kappa(s)$. Then $\Spec(k) \to Z$ is a finite morphism of schemes and $Z \to X$ is a finite morphism as it is a closed immersion. Hence $\Spec(k) \to X$ is finite (Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-composition-integral}). If $k = \kappa(s)$, then $\Spec(k) = Z$ and $\Spec(k) \to X$ is a closed immersion. \end{proof} \begin{lemma} \label{lemma-decent-space-closed-point} 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 $i : \Spec(k) \to X$ from the spectrum of a field. \end{lemma} \begin{proof} We know that $x$ can be represented by a quasi-compact monomorphism $i : \Spec(k) \to X$ where $k$ is a field (Definition \ref{definition-very-reasonable}). Let $U \to X$ be an \'etale 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 \ref{lemma-decent-no-specializations-map-to-same-point}). As $i$ is a monomorphism, so is $U_k = U \times_X \Spec(k) \to U$. In particular, the map $|U_k| \to F$ is injective. Since $U_k$ is quasi-compact and \'etale over a field, we see that $U_k$ is a finite disjoint union of spectra of fields (Remark \ref{remark-recall}). Say $U_k = \Spec(k_1) \amalg \ldots \amalg \Spec(k_r)$. Since $\Spec(k_i) \to U$ 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 $\Spec(k_i) \to U$ 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 \ref{spaces-morphisms-lemma-closed-immersion-local}). This finishes the proof. \end{proof} \section{Locally separated spaces} \label{section-locally-separated} \noindent It turns out that a locally separated algebraic space is decent. \begin{lemma} \label{lemma-infinite-number} Let $A$ be a ring. Let $k$ be a field. Let $\mathfrak p_n$, $n \geq 1$ be a sequence of pairwise distinct primes of $A$. Moreover, for each $n$ let $k \to \kappa(\mathfrak p_n)$ be an embedding. Then the closure of the image of $$\coprod\nolimits_{n \not = m} \Spec(\kappa(\mathfrak p_n) \otimes_k \kappa(\mathfrak p_m)) \longrightarrow \Spec(A \otimes A)$$ meets the diagonal. \end{lemma} \begin{proof} Set $k_n = \kappa(\mathfrak p_n)$. We may assume that $A = \prod k_n$. Denote $x_n = \Spec(k_n)$ the open and closed point corresponding to $A \to k_n$. Then $\Spec(A) = Z \amalg \{x_n\}$ where $Z$ is a nonempty closed subset. Namely, $Z = V(e_n; n \geq 1)$ where $e_n$ is the idempotent of $A$ corresponding to the factor $k_n$ and $Z$ is nonempty as the ideal generated by the $e_n$ is not equal to $A$. We will show that the closure of the image contains $\Delta(Z)$. The kernel of the map $$(\prod k_n) \otimes_k (\prod k_m) \longrightarrow \prod\nolimits_{n \not = m} k_n \otimes_k k_m$$ is the ideal generated by $e_n \otimes e_n$, $n \geq 1$. Hence the closure of the image of the map on spectra is $V(e_n \otimes e_n; n \geq 1)$ whose intersection with $\Delta(\Spec(A))$ is $\Delta(Z)$. Thus it suffices to show that $$\coprod\nolimits_{n \not = m} \Spec(k_n \otimes_k k_m) \longrightarrow \Spec(\prod\nolimits_{n \not = m} k_n \otimes_k k_m)$$ has dense image. This follows as the family of ring maps $\prod_{n \not = m} k_n \otimes_k k_m \to k_n \otimes_k k_m$ is jointly injective. \end{proof} \begin{lemma}[David Rydh] \label{lemma-locally-separated-decent} A locally separated algebraic space is decent. \end{lemma} \begin{proof} Let $S$ be a scheme and let $X$ be a locally separated algebraic space over $S$. We may assume $S = \Spec(\mathbf{Z})$, see Properties of Spaces, Definition \ref{spaces-properties-definition-separated}. Unadorned fibre products will be over $\mathbf{Z}$. Let $x \in |X|$. Choose a scheme $U$, an \'etale morphism $U \to X$, and a point $u \in U$ mapping to $x$ in $|X|$. As usual we identify $u = \Spec(\kappa(u))$. As $X$ is locally separated the morphism $$u \times_X u \to u \times u$$ is an immersion (Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-fibre-product-after-map}). Hence More on Groupoids, Lemma \ref{more-groupoids-lemma-locally-closed-image-is-closed} tells us that it is a closed immersion (use Schemes, Lemma \ref{schemes-lemma-immersion-when-closed}). As $u \times_X u \to u \times_X U$ is a monomorphism (base change of $u \to U$) and as $u \times_X U \to u$ is \'etale we conclude that $u \times_X u$ is a disjoint union of spectra of fields (see Remark \ref{remark-recall} and Schemes, Lemma \ref{schemes-lemma-mono-towards-spec-field}). Since it is also closed in the affine scheme $u \times u$ we conclude $u \times_X u$ is a finite disjoint union of spectra of fields. Thus $x$ can be represented by a monomorphism $\Spec(k) \to X$ where $k$ is a field, see Lemma \ref{lemma-R-finite-above-x}. \medskip\noindent Next, let $U = \Spec(A)$ be an affine scheme and let $U \to X$ be an \'etale morphism. To finish the proof it suffices to show that $F = U \times_X \Spec(k)$ is finite. Write $F = \coprod_{i \in I} \Spec(k_i)$ as the disjoint union of finite separable extensions of $k$. We have to show that $I$ is finite. Set $R = U \times_X U$. As $X$ is locally separated, the morphism $j : R \to U \times U$ is an immersion. Let $U' \subset U \times U$ be an open such that $j$ factors through a closed immersion $j' : R \to U'$. Let $e : U \to R$ be the diagonal map. Using that $e$ is a morphism between schemes \'etale over $U$ such that $\Delta = j \circ e$ is a closed immersion, we conclude that $R = e(U) \amalg W$ for some open and closed subscheme $W \subset R$. Since $j'$ is a closed immersion we conclude that $j'(W) \subset U'$ is closed and disjoint from $j'(e(U))$. Therefore $\overline{j(W)} \cap \Delta(U) = \emptyset$ in $U \times U$. Note that $W$ contains $\Spec(k_i \otimes_k k_{i'})$ for all $i \not = i'$, $i, i' \in I$. By Lemma \ref{lemma-infinite-number} we conclude that $I$ is finite as desired. \end{proof} \section{Valuative criterion} \label{section-valuative-criterion-universally-closed} \noindent For a quasi-compact morphism from a decent space the valuative criterion is necessary in order for the morphism to be universally closed. \begin{proposition} \label{proposition-characterize-universally-closed} Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is quasi-compact, and $X$ is decent. Then $f$ is universally closed if and only if the existence part of the valuative criterion holds. \end{proposition} \begin{proof} In Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-quasi-compact-existence-universally-closed} we have seen one of the implications. To prove the other, assume that $f$ is universally closed. Let $$\xymatrix{ \Spec(K) \ar[r] \ar[d] & X \ar[d] \\ \Spec(A) \ar[r] & Y }$$ be a diagram as in Morphisms of Spaces, Definition \ref{spaces-morphisms-definition-valuative-criterion}. Let $X_A = \Spec(A) \times_Y X$, so that we have $$\xymatrix{ \Spec(K) \ar[r] \ar[rd] & X_A \ar[d] \\ & \Spec(A) }$$ By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-quasi-compact} we see that $X_A \to \Spec(A)$ is quasi-compact. Since $X_A \to X$ is representable, we see that $X_A$ is decent also, see Lemma \ref{lemma-representable-properties}. Moreover, as $f$ is universally closed, we see that $X_A \to \Spec(A)$ is universally closed. Hence we may and do replace $X$ by $X_A$ and $Y$ by $\Spec(A)$. \medskip\noindent Let $x' \in |X|$ be the equivalence class of $\Spec(K) \to X$. Let $y \in |Y| = |\Spec(A)|$ be the closed point. Set $y' = f(x')$; it is the generic point of $\Spec(A)$. Since $f$ is universally closed we see that $f(\overline{\{x'\}})$ contains $\overline{\{y'\}}$, and hence contains $y$. Let $x \in \overline{\{x'\}}$ be a point such that $f(x) = y$. Let $U$ be a scheme, and $\varphi : U \to X$ an \'etale morphism such that there exists a $u \in U$ with $\varphi(u) = x$. By Lemma \ref{lemma-specialization} and our assumption that $X$ is decent there exists a specialization $u' \leadsto u$ on $U$ with $\varphi(u') = x'$. This means that there exists a common field extension $K \subset K' \supset \kappa(u')$ such that $$\xymatrix{ \Spec(K') \ar[r] \ar[d] & U \ar[d] \\ \Spec(K) \ar[r] \ar[rd] & X \ar[d] \\ & \Spec(A) }$$ is commutative. This gives the following commutative diagram of rings $$\xymatrix{ K' & \mathcal{O}_{U, u} \ar[l] \\ K \ar[u] & \\ & A \ar[lu] \ar[uu] }$$ By Algebra, Lemma \ref{algebra-lemma-dominate} we can find a valuation ring $A' \subset K'$ dominating the image of $\mathcal{O}_{U, u}$ in $K'$. Since by construction $\mathcal{O}_{U, u}$ dominates $A$ we see that $A'$ dominates $A$ also. Hence we obtain a diagram resembling the second diagram of Morphisms of Spaces, Definition \ref{spaces-morphisms-definition-valuative-criterion} and the proposition is proved. \end{proof} \section{Relative conditions} \label{section-relative-conditions} \noindent This is a (yet another) technical section dealing with conditions on algebraic spaces having to do with points. It is probably a good idea to skip this section. \begin{definition} \label{definition-relative-conditions} Let $S$ be a scheme. We say an algebraic space $X$ over $S$ {\it has property $(\beta)$} if $X$ has the corresponding property of Lemma \ref{lemma-bounded-fibres}. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. \begin{enumerate} \item We say $f$ {\it has property $(\beta)$} if for any scheme $T$ and morphism $T \to Y$ the fibre product $T \times_Y X$ has property $(\beta)$. \item We say $f$ is {\it decent} if for any scheme $T$ and morphism $T \to Y$ the fibre product $T \times_Y X$ is a decent algebraic space. \item We say $f$ is {\it reasonable} if for any scheme $T$ and morphism $T \to Y$ the fibre product $T \times_Y X$ is a reasonable algebraic space. \item We say $f$ is {\it very reasonable} if for any scheme $T$ and morphism $T \to Y$ the fibre product $T \times_Y X$ is a very reasonable algebraic space. \end{enumerate} \end{definition} \noindent We refer to Remark \ref{remark-very-reasonable} for an informal discussion. It will turn out that the class of very reasonable morphisms is not so useful, but that the classes of decent and reasonable morphisms are useful. \begin{lemma} \label{lemma-properties-trivial-implications} Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. We have the following implications among the conditions on $f$: $$\xymatrix{ \text{representable} \ar@{=>}[rd] & & & & \\ & \text{very reasonable} \ar@{=>}[r] & \text{reasonable} \ar@{=>}[r] & \text{decent} \ar@{=>}[r] & (\beta) \\ \text{quasi-separated} \ar@{=>}[ru] & & & & }$$ \end{lemma} \begin{proof} This is clear from the definitions, Lemma \ref{lemma-bounded-fibres} and Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-separated-local}. \end{proof} \noindent Here is another sanity check. \begin{lemma} \label{lemma-property-for-morphism-out-of-property} Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. If $X$ is decent (resp.\ is reasonable, resp.\ has property $(\beta)$ of Lemma \ref{lemma-bounded-fibres}), then $f$ is decent (resp.\ reasonable, resp.\ has property $(\beta)$). \end{lemma} \begin{proof} Let $T$ be a scheme and let $T \to Y$ be a morphism. Then $T \to Y$ is representable, hence the base change $T \times_Y X \to X$ is representable. Hence if $X$ is decent (or reasonable), then so is $T \times_Y X$, see Lemma \ref{lemma-representable-named-properties}. Similarly, for property $(\beta)$, see Lemma \ref{lemma-representable-properties}. \end{proof} \begin{lemma} \label{lemma-base-change-relative-conditions} Having property $(\beta)$, being decent, or being reasonable is preserved under arbitrary base change. \end{lemma} \begin{proof} This is immediate from the definition. \end{proof} \begin{lemma} \label{lemma-property-over-property} Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $\omega \in \{\beta, decent, reasonable\}$. Suppose that $Y$ has property $(\omega)$ and $f : X \to Y$ has $(\omega)$. Then $X$ has $(\omega)$. \end{lemma} \begin{proof} Let us prove the lemma in case $\omega = \beta$. In this case we have to show that any $x \in |X|$ is represented by a monomorphism from the spectrum of a field into $X$. Let $y = f(x) \in |Y|$. By assumption there exists a field $k$ and a monomorphism $\Spec(k) \to Y$ representing $y$. Then $x$ corresponds to a point $x'$ of $\Spec(k) \times_Y X$. By assumption $x'$ is represented by a monomorphism $\Spec(k') \to \Spec(k) \times_Y X$. Clearly the composition $\Spec(k') \to X$ is a monomorphism representing $x$. \medskip\noindent Let us prove the lemma in case $\omega = decent$. Let $x \in |X|$ and $y = f(x) \in |Y|$. By the result of the preceding paragraph we can choose a diagram $$\xymatrix{ \Spec(k') \ar[r]_x \ar[d] & X \ar[d]^f \\ \Spec(k) \ar[r]^y & Y }$$ whose horizontal arrows monomorphisms. As $Y$ is decent the morphism $y$ is quasi-compact. As $f$ is decent the algebraic space $\Spec(k) \times_Y X$ is decent. Hence the monomorphism $\Spec(k') \to \Spec(k) \times_Y X$ is quasi-compact. Then the monomorphism $x : \Spec(k') \to X$ is quasi-compact as a composition of quasi-compact morphisms (use Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-base-change-quasi-compact} and \ref{spaces-morphisms-lemma-composition-quasi-compact}). As the point $x$ was arbitrary this implies $X$ is decent. \medskip\noindent Let us prove the lemma in case $\omega = reasonable$. Choose $V \to Y$ \'etale with $V$ an affine scheme. Choose $U \to V \times_Y X$ \'etale with $U$ an affine scheme. By assumption $V \to Y$ has universally bounded fibres. By Lemma \ref{lemma-base-change-universally-bounded} the morphism $V \times_Y X \to X$ has universally bounded fibres. By assumption on $f$ we see that $U \to V \times_Y X$ has universally bounded fibres. By Lemma \ref{lemma-composition-universally-bounded} the composition $U \to X$ has universally bounded fibres. Hence there exists sufficiently many \'etale morphisms $U \to X$ from schemes with universally bounded fibres, and we conclude that $X$ is reasonable. \end{proof} \begin{lemma} \label{lemma-composition-relative-conditions} Having property $(\beta)$, being decent, or being reasonable is preserved under compositions. \end{lemma} \begin{proof} Let $\omega \in \{\beta, decent, reasonable\}$. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of algebraic spaces over the scheme $S$. Assume $f$ and $g$ both have property $(\omega)$. Then we have to show that for any scheme $T$ and morphism $T \to Z$ the space $T \times_Z X$ has $(\omega)$. By Lemma \ref{lemma-base-change-relative-conditions} this reduces us to the following claim: Suppose that $Y$ is an algebraic space having property $(\omega)$, and that $f : X \to Y$ is a morphism with $(\omega)$. Then $X$ has $(\omega)$. This is the content of Lemma \ref{lemma-property-over-property}. \end{proof} \begin{lemma} \label{lemma-fibre-product-conditions} Let $S$ be a scheme. Let $f : X \to Y$, $g : Z \to Y$ be morphisms of algebraic spaces over $S$. If $X$ and $Z$ are decent (resp.\ reasonable, resp.\ have property $(\beta)$ of Lemma \ref{lemma-bounded-fibres}), then so does $X \times_Y Z$. \end{lemma} \begin{proof} Namely, by Lemma \ref{lemma-property-for-morphism-out-of-property} the morphism $X \to Y$ has the property. Then the base change $X \times_Y Z \to Z$ has the property by Lemma \ref{lemma-base-change-relative-conditions}. And finally this implies $X \times_Y Z$ has the property by Lemma \ref{lemma-property-over-property}. \end{proof} \begin{lemma} \label{lemma-descent-conditions} Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $\mathcal{P} \in \{(\beta), decent, reasonable\}$. Assume \begin{enumerate} \item $f$ is quasi-compact, \item $f$ is \'etale, \item $|f| : |X| \to |Y|$ is surjective, and \item the algebraic space $X$ has property $\mathcal{P}$. \end{enumerate} Then $Y$ has property $\mathcal{P}$. \end{lemma} \begin{proof} Let us prove this in case $\mathcal{P} = (\beta)$. Let $y \in |Y|$ be a point. We have to show that $y$ can be represented by a monomorphism from a field. Choose a point $x \in |X|$ with $f(x) = y$. By assumption we may represent $x$ by a monomorphism $\Spec(k) \to X$, with $k$ a field. By Lemma \ref{lemma-R-finite-above-x} it suffices to show that the projections $\Spec(k) \times_Y \Spec(k) \to \Spec(k)$ are \'etale and quasi-compact. We can factor the first projection as $$\Spec(k) \times_Y \Spec(k) \longrightarrow \Spec(k) \times_Y X \longrightarrow \Spec(k)$$ The first morphism is a monomorphism, and the second is \'etale and quasi-compact. By Properties of Spaces, Lemma \ref{spaces-properties-lemma-etale-over-field-scheme} we see that $\Spec(k) \times_Y X$ is a scheme. Hence it is a finite disjoint union of spectra of finite separable field extensions of $k$. By Schemes, Lemma \ref{schemes-lemma-mono-towards-spec-field} we see that the first arrow identifies $\Spec(k) \times_Y \Spec(k)$ with a finite disjoint union of spectra of finite separable field extensions of $k$. Hence the projection morphism is \'etale and quasi-compact. \medskip\noindent Let us prove this in case $\mathcal{P} = decent$. We have already seen in the first paragraph of the proof that this implies that every $y \in |Y|$ can be represented by a monomorphism $y : \Spec(k) \to Y$. Pick such a $y$. Pick an affine scheme $U$ and an \'etale morphism $U \to X$ such that the image of $|U| \to |Y|$ contains $y$. By Lemma \ref{lemma-UR-finite-above-x} it suffices to show that $U_y$ is a finite scheme over $k$. The fibre product $X_y = \Spec(k) \times_Y X$ is a quasi-compact \'etale algebraic space over $k$. Hence by Properties of Spaces, Lemma \ref{spaces-properties-lemma-etale-over-field-scheme} it is a scheme. So it is a finite disjoint union of spectra of finite separable extensions of $k$. Say $X_y = \{x_1, \ldots, x_n\}$ so $x_i$ is given by $x_i : \Spec(k_i) \to X$ with $[k_i : k] < \infty$. By assumption $X$ is decent, so the schemes $U_{x_i} = \Spec(k_i) \times_X U$ are finite over $k_i$. Finally, we note that $U_y = \coprod U_{x_i}$ as a scheme and we conclude that $U_y$ is finite over $k$ as desired. \medskip\noindent Let us prove this in case $\mathcal{P} = reasonable$. Pick an affine scheme $V$ and an \'etale morphism $V \to Y$. We have the show the fibres of $V \to Y$ are universally bounded. The algebraic space $V \times_Y X$ is quasi-compact. Thus we can find an affine scheme $W$ and a surjective \'etale morphism $W \to V \times_Y X$, see Properties of Spaces, Lemma \ref{spaces-properties-lemma-quasi-compact-affine-cover}. Here is a picture (solid diagram) $$\xymatrix{ W \ar[r] \ar[rd] & V \times_Y X \ar[r] \ar[d] & X \ar[d]_f & \Spec(k) \ar@{..>}[l]^x \ar@{..>}[ld]^y \\ & V \ar[r] & Y }$$ The morphism $W \to X$ is universally bounded by our assumption that the space $X$ is reasonable. Let $n$ be an integer bounding the degrees of the fibres of $W \to X$. We claim that the same integer works for bounding the fibres of $V \to Y$. Namely, suppose $y \in |Y|$ is a point. Then there exists a $x \in |X|$ with $f(x) = y$ (see above). This means we can find a field $k$ and morphisms $x, y$ given as dotted arrows in the diagram above. In particular we get a surjective \'etale morphism $$\Spec(k) \times_{x, X} W \to \Spec(k) \times_{x, X} (V \times_Y X) = \Spec(k) \times_{y, Y} V$$ which shows that the degree of $\Spec(k) \times_{y, Y} V$ over $k$ is less than or equal to the degree of $\Spec(k) \times_{x, X} W$ over $k$, i.e., $\leq n$, and we win. (This last part of the argument is the same as the argument in the proof of Lemma \ref{lemma-descent-universally-bounded}. Unfortunately that lemma is not general enough because it only applies to representable morphisms.) \end{proof} \begin{lemma} \label{lemma-relative-conditions-local} Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $\mathcal{P} \in \{(\beta), decent, reasonable, very\ reasonable\}$. The following are equivalent \begin{enumerate} \item $f$ is $\mathcal{P}$, \item for every affine scheme $Z$ and every morphism $Z \to Y$ the base change $Z \times_Y X \to Z$ of $f$ is $\mathcal{P}$, \item for every affine scheme $Z$ and every morphism $Z \to Y$ the algebraic space $Z \times_Y X$ is $\mathcal{P}$, and \item there exists a Zariski covering $Y = \bigcup Y_i$ such that each morphism $f^{-1}(Y_i) \to Y_i$ has $\mathcal{P}$. \end{enumerate} If $\mathcal{P} \in \{(\beta), decent, reasonable\}$, then this is also equivalent to \begin{enumerate} \item[(5)] there exists a scheme $V$ and a surjective \'etale morphism $V \to Y$ such that the base change $V \times_Y X \to V$ has $\mathcal{P}$. \end{enumerate} \end{lemma} \begin{proof} The implications (1) $\Rightarrow$ (2) $\Rightarrow$ (3) $\Rightarrow$ (4) are trivial. The implication (3) $\Rightarrow$ (1) can be seen as follows. Let $Z \to Y$ be a morphism whose source is a scheme over $S$. Consider the algebraic space $Z \times_Y X$. If we assume (3), then for any affine open $W \subset Z$, the open subspace $W \times_Y X$ of $Z \times_Y X$ has property $\mathcal{P}$. Hence by Lemma \ref{lemma-properties-local} the space $Z \times_Y X$ has property $\mathcal{P}$, i.e., (1) holds. A similar argument (omitted) shows that (4) implies (1). \medskip\noindent The implication (1) $\Rightarrow$ (5) is trivial. Let $V \to Y$ be an \'etale morphism from a scheme as in (5). Let $Z$ be an affine scheme, and let $Z \to Y$ be a morphism. Consider the diagram $$\xymatrix{ Z \times_Y V \ar[r]_q \ar[d]_p & V \ar[d] \\ Z \ar[r] & Y }$$ Since $p$ is \'etale, and hence open, we can choose finitely many affine open