decent-spaces.tex

\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 talk 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 point.
\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 specializations 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 an algebraic space.
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}







\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 distinguihsed 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-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}
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 $U \to X$. 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}.
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}.

\medskip\noindent
We let $U_p \subset X$ be the open subspace which is the
image of $V_p \to X$. By construction 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. It follows that
the $U_p$ give a filtration of $X$ as stated in the lemma.
Moreover, $\Spec(k) \to X$ 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 we win.
\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-reasonable}.
Observe that a quasi-separated space is reasonable, see
Lemma \ref{lemma-bounded-fibres} and
Definition \ref{definition-very-reasonable}.
At the end of the argument we add that since $X$ is quasi-separated
the schemes $V \times_X \ldots \times_X V$ are all quasi-compact.
Hence the schemes $W_p$ are quasi-compact. Hence the schemes
$V_p = W_p/S_p$ are quasi-compact.
\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}
















\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 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}

\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 an algebraic space. 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) and the definitions.
\end{proof}

\begin{lemma}
\label{lemma-decent-generic-points}
Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$.
Let $x \in |X|$. The following are equivalent
\begin{enumerate}
\item $x$ is a generic point of an irreducible component of $|X|$,
\item for any \'etale morphism $(Y, y) \to (X, x)$ of pointed algebraic
spaces, $y$ is a generic point of an irreducible component of $|Y|$,
\item for some \'etale morphism $(Y, y) \to (X, x)$ of pointed algebraic
spaces, $y$ is a generic point of an irreducible component of $|Y|$,
\item the dimension of the local ring of $X$ at $x$ is zero
(Properties of Spaces, Definition
\ref{spaces-properties-definition-dimension-local-ring}).
\end{enumerate}
\end{lemma}

\begin{proof}
Observe that any $Y$ as in (2) and (3) is decent by
Lemma \ref{lemma-etale-named-properties}.
Thus it suffices to prove the equivalence of (1) and (4)
as then the equivalence with (2) and (3) follows since the dimension
of the local ring of $Y$ at $y$ is equal to the dimension
of the local ring of $X$ at $x$.
Let $f : U \to X$ be an \'etale morphism from an affine scheme and let
$u \in U$ be a point mapping to $x$.

\medskip\noindent
Assume (1). Let $u' \leadsto u$ be a specialization in $U$.
Then $f(u') = f(u) = x$. By
Lemma \ref{lemma-decent-no-specializations-map-to-same-point}
we see that $u' = u$. Hence $u$ is a generic point of an irreducible component
of $U$. Thus $\dim(\mathcal{O}_{U, u}) = 0$ and we see that (4) holds.

\medskip\noindent
Assume (4). The point $x$ is contained in an irreducible component
$T \subset |X|$. Since $|X|$ is sober
(Proposition \ref{proposition-reasonable-sober})
we $T$ has a generic point $x'$. Of course $x' \leadsto x$.
Then we can lift this specialization to $u' \leadsto u$ in $U$
(Lemma \ref{lemma-decent-specialization}). This contradicts the assumption
that $\dim(\mathcal{O}_{U, u}) = 0$ unless $u' = u$, i.e., $x' = x$.
\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}






\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-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 $k \to \kappa(\mathfrak p)$ 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 base scheme. Let $X$ be a locally separated algebraic space
over $S$. 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 $e : U \to R$ be the
diagonal map. Using that $e$ is a morphism between \'etale schemes
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 an immersion and $j|_{e(U)}$ is a closed
immersion we conclude that
$\overline{j(W)} \cap \Delta(U) = \emptyset$ 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}

\noindent
The following lemma is a special case of the more general
Lemma \ref{lemma-re-characterize-universally-closed}.

\begin{lemma}
\label{lemma-characterize-universally-closed-quasi-separated}
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces
over $S$. Assume $f$ is quasi-compact and quasi-separated. Then
$f$ is universally closed if and only if the existence part of the
valuative criterion holds (Morphisms of Spaces, Definition
\ref{spaces-morphisms-definition-valuative-criterion}).
\end{lemma}

\begin{proof}
This is a combination of
Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-quasi-compact-existence-universally-closed}
and
Proposition \ref{proposition-characterize-universally-closed}.
Namely, the implication in one direction is given by
Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-quasi-compact-existence-universally-closed}.
For the converse, assume $f$ is quasi-separated, quasi-compact, and
universally closed and assume given a diagram
$$
\xymatrix{
\Spec(K) \ar[r] \ar[d] & X \ar[d] \\
\Spec(A) \ar[r] & Y
}
$$
as in Morphisms of Spaces,
Definition \ref{spaces-morphisms-definition-valuative-criterion}.
A formal argument shows that the existence of the
desired diagram
$$
\xymatrix{
\Spec(K') \ar[r] \ar[d] & \Spec(K) \ar[r] & X \ar[d] \\
\Spec(A') \ar[r] \ar[rru] & \Spec(A) \ar[r] & Y
}
$$
can be reduced to the case of the morphism $X_A \to \Spec(A)$.
In this case the algebraic space $X_A$ is quasi-separated, hence
decent (property $(\gamma)$ of Lemma \ref{lemma-bounded-fibres}).
Hence the existence of $A \subset A'$ and the arrow
$\Spec(A') \to X_A$ follows from
Proposition \ref{proposition-characterize-universally-closed}.
\end{proof}

\begin{lemma}
\label{lemma-characterize-universally-closed-separated}
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces
over $S$. Assume $f$ is quasi-compact and separated. Then the following
are equivalent
\begin{enumerate}
\item $f$ is universally closed,
\item the existence part of the valuative criterion holds
as in Morphisms of Spaces, Definition
\ref{spaces-morphisms-definition-valuative-criterion}, and
\item given any commutative solid diagram
$$
\xymatrix{
\Spec(K) \ar[r] \ar[d] & X \ar[d] \\
\Spec(A) \ar[r] \ar@{-->}[ru] & Y
}
$$
where $A$ is a valuation ring with field of fractions $K$, there exists
a dotted arrow, i.e., $f$ satisfies the existence part of the valuative
criterion as in
Schemes, Definition \ref{schemes-definition-valuative-criterion}.
\end{enumerate}
\end{lemma}

\begin{proof}
Since $f$ is separated parts (2) and (3) are equivalent by
Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-usual-enough}.
The equivalence of (3) and (1) follows from
Lemma \ref{lemma-characterize-universally-closed-quasi-separated}.
\end{proof}

\begin{lemma}
\label{lemma-characterize-separated-and-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 quasi-separated. Then the
following are equivalent
\begin{enumerate}
\item $f$ is separated and universally closed,
\item the valuative criterion holds as in Morphisms of Spaces, Definition
\ref{spaces-morphisms-definition-valuative-criterion},
\item given any commutative solid diagram
$$
\xymatrix{
\Spec(K) \ar[r] \ar[d] & X \ar[d] \\
\Spec(A) \ar[r] \ar@{-->}[ru] & Y
}
$$
where $A$ is a valuation ring with field of fractions $K$, there exists
a unique dotted arrow, i.e., $f$ satisfies the valuative
criterion as in
Schemes, Definition \ref{schemes-definition-valuative-criterion}.
\end{enumerate}
\end{lemma}

\begin{proof}
Since $f$ is quasi-separated, the uniqueness
part of the valutative criterion implies $f$ is separated
(Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-valuative-criterion-separatedness}).
Conversely, if $f$ is separated, then it satisfies the
uniqueness part of the valuative criterion
(Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-separated-implies-valuative}).
Having said this, we see that in each of the three cases the
morphism $f$ is separated and satisfies the uniqueness part
of the valuative criterion. In this case the lemma is a formal
consequence of
Lemma \ref{lemma-characterize-universally-closed-separated}.
\end{proof}

\begin{lemma}[Valuative criterion for properness]
\label{lemma-characterize-proper}
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces
over $S$. Assume $f$ is of finite type and quasi-separated. Then the
following are equivalent
\begin{enumerate}
\item $f$ is proper,
\item the valuative criterion holds as in Morphisms of Spaces, Definition
\ref{spaces-morphisms-definition-valuative-criterion},
\item given any commutative solid diagram
$$
\xymatrix{
\Spec(K) \ar[r] \ar[d] & X \ar[d] \\
\Spec(A) \ar[r] \ar@{-->}[ru] & Y
}
$$
where $A$ is a valuation ring with field of fractions $K$, there exists
a unique dotted arrow, i.e., $f$ satisfies the valuative
criterion as in
Schemes, Definition \ref{schemes-definition-valuative-criterion}.
\end{enumerate}
\end{lemma}

\begin{proof}
Formal consequence of
Lemma \ref{lemma-characterize-separated-and-universally-closed}
and the definitions.
\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
subschemes $W_i \subset Z \times_Y V$ such that $Z = \bigcup p(W_i)$.
Consider the commutative diagram
$$
\xymatrix{
V \times_Y X \ar[d] &
(\coprod W_i) \times_Y X \ar[l] \ar[d] \ar[r] &
Z \times_Y X \ar[d] \\
V &
\coprod W_i \ar[l] \ar[r] &
Z
}
$$
We know $V \times_Y X$ has property $\mathcal{P}$. By
Lemma \ref{lemma-representable-properties}
we see that $(\coprod W_i) \times_Y X$ has property $\mathcal{P}$.
Note that the morphism $(\coprod W_i) \times_Y X \to Z \times_Y X$
is \'etale and quasi-compact as the base change of $\coprod W_i \to Z$.
Hence by Lemma \ref{lemma-descent-conditions}
we conclude that $Z \times_Y X$ has property $\mathcal{P}$.
\end{proof}

\begin{remark}
\label{remark-very-reasonable}
An informal description of the properties $(\beta)$, decent, reasonable,
very reasonable was given in Section \ref{section-reasonable-decent}.
A morphism has one of these properties if (very) loosely speaking the
fibres of the morphism have the corresponding properties.
Being decent is useful to prove things about specializations of
points on $|X|$. Being reasonable is a bit stronger and technically
quite easy to work with.
\end{remark}

\noindent
Here is a lemma we promised earlier which uses decent morphisms.

\begin{lemma}
\label{lemma-re-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 decent.
(For example if $f$ is representable, or quasi-separated, see
Lemma \ref{lemma-properties-trivial-implications}.)
Then $f$ is universally closed if and only if the
existence part of the valuative criterion holds.
\end{lemma}

\begin{proof}
In
Morphisms of Spaces,
Lemma \ref{spaces-morphisms-lemma-quasi-compact-existence-universally-closed}
we proved that any quasi-compact morphism which satisfies the existence
part of the valuative criterion is universally closed.
To prove the other, assume that $f$ is universally closed.
In the proof of
Proposition \ref{proposition-characterize-universally-closed}
we have seen that it suffices to show, for any valuation ring $A$,
and any morphism $\Spec(A) \to Y$, that the base change
$f_A : X_A \to \Spec(A)$ satisfies the existence part of the valuative
criterion. By definition the algebraic space $X_A$ has property $(\gamma)$
and hence
Proposition \ref{proposition-characterize-universally-closed}
applies to the morphism $f_A$ and we win.
\end{proof}





\section{Points of fibres}
\label{section-points-fibres}

\noindent
Let $S$ be a scheme. Consider a cartesian diagram
\begin{equation}
\label{equation-points-fibres}
\xymatrix{
W \ar[r]_q \ar[d]_p & Z \ar[d]^g \\
X \ar[r]^f & Y
}
\end{equation}
of algebraic spaces over $S$. Let $x \in |X|$ and $z \in |Z|$
be points mapping to the same point $y \in |Y|$. We may ask:
When is the set
\begin{equation}
\label{equation-fibre}
F_{x, z} = \{ w \in |W| \text{ such that }p(w) = x\text{ and }q(w) = z\}
\end{equation}
finite?

\begin{example}
\label{example-schemes}
If $X, Y, Z$ are schemes, then the set $F_{x, z}$
is equal to the spectrum of $\kappa(x) \otimes_{\kappa(y)} \kappa(z)$
(Schemes, Lemma \ref{schemes-lemma-points-fibre-product}). Thus we
obtain a finite set if either $\kappa(y) \subset \kappa(x)$ is finite or if
$\kappa(y) \subset \kappa(z)$ is finite. In particular, this is always
the case if $g$ is quasi-finite at $z$ (Morphisms, Lemma
\ref{morphisms-lemma-residue-field-quasi-finite}).
\end{example}

\begin{example}
\label{example-not-finite}
Let $K$ be a characteristic $0$ field endowed with an automorphism
$\sigma$ of infinite order. Set $Y = \Spec(K)/\mathbf{Z}$ and
$X = \mathbf{A}^1_K/\mathbf{Z}$ where $\mathbf{Z}$ acts on $K$ via $\sigma$
and on $\mathbf{A}^1_K = \Spec(K[t])$ via $t \mapsto t + 1$.
Let $Z = \Spec(K)$. Then $W = \mathbf{A}^1_K$. Picture
$$
\xymatrix{
\mathbf{A}^1_K \ar[r]_q \ar[d]_p & \Spec(K) \ar[d]^g \\
\mathbf{A}^1_K/\mathbf{Z} \ar[r]^f & \Spec(K)/\mathbf{Z}
}
$$
Take $x$ corresponding to $t = 0$ and $z$ the unique point of $\Spec(K)$.
Then we see that $F_{x, z} = \mathbf{Z}$ as a set.
\end{example}

\begin{lemma}
\label{lemma-surjective-on-fibres}
In the situation of (\ref{equation-points-fibres}) if $Z' \to Z$
is a morphism and $z' \in |Z'|$ maps to $z$, then the induced map
$F_{x, z'} \to F_{x, z}$ is surjective.
\end{lemma}

\begin{proof}
Set $W' = X \times_Y Z' = W \times_Z Z'$. Then
$|W'| \to |W| \times_{|Z|} |Z'|$ is surjective by
Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-cartesian}.
Hence the surjectivity of $F_{x, z'} \to F_{x, z}$.
\end{proof}

\begin{lemma}
\label{lemma-qf-and-qc-finite-fibre}
In diagram (\ref{equation-points-fibres}) the set (\ref{equation-fibre})
is finite if $f$ is of finite type and $f$ is quasi-finite at $x$.
\end{lemma}

\begin{proof}
The morphism $p$ is quasi-finite at every $w \in F_{x, z}$, see
Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-base-change-quasi-finite-locus}.
Hence the lemma follows from
Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-quasi-finite-at-a-finite-number-of-points}.
\end{proof}

\begin{lemma}
\label{lemma-decent-finite-fibre}
In diagram (\ref{equation-points-fibres}) the set (\ref{equation-fibre})
is finite if $y$ can be represented by a monomorphism $\Spec(k) \to Y$
where $k$ is a field and $g$ is quasi-finite at $z$.
(Special case: $Y$ is decent and $g$ is \'etale.)
\end{lemma}

\begin{proof}
By Lemma \ref{lemma-surjective-on-fibres} applied twice
we may replace $Z$ by $Z_k = \Spec(k) \times_Y Z$ and
$X$ by $X_k = \Spec(k) \times_Y X$. We may and do
replace $Y$ by $\Spec(k)$ as well. Note that $Z_k \to \Spec(k)$
is quasi-finite at $z$ by Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-base-change-quasi-finite-locus}.
Choose a scheme $V$, a point $v \in V$, and an \'etale morphism
$V \to Z_k$ mapping $v$ to $z$. Choose a scheme $U$, a point $u \in U$,
and an \'etale morphism $U \to X_k$ mapping $u$ to $x$.
Again by Lemma \ref{lemma-surjective-on-fibres}
it suffices to show $F_{u, v}$ is finite for the diagram
$$
\xymatrix{
U \times_{\Spec(k)} V \ar[r] \ar[d] & V \ar[d] \\
U \ar[r] & \Spec(k)
}
$$
The morphism $V \to \Spec(k)$ is quasi-finite at $v$
(follows from the general discussion in
Morphisms of Spaces, Section \ref{spaces-morphisms-section-local-source-target}
and the definition of being quasi-finite at a point).
At this point the finiteness follows from Example \ref{example-schemes}.
The parenthetical remark of the statement of the lemma follows
from the fact that on decent spaces points are represented by
monomorphisms from fields and from the fact that an \'etale
morphism of algebraic spaces is quasi-finite.
\end{proof}

\begin{lemma}
\label{lemma-topology-fibre}
\begin{slogan}
Fibers of field points of algebraic spaces have the
expected Zariski topologies.
\end{slogan}
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces
over $S$.
Let $y \in |Y|$ and assume that $y$ is represented by a quasi-compact
monomorphism $\Spec(k) \to Y$. Then $|X_k| \to |X|$ is a
homeomorphism onto $f^{-1}(\{y\}) \subset |X|$ with induced topology.
\end{lemma}

\begin{proof}
We will use
Properties of Spaces, Lemma \ref{spaces-properties-lemma-etale-open}
and
Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-monomorphism-injective-points}
without further mention.
Let $V \to Y$ be an \'etale morphism with $V$ affine such that there
exists a $v \in V$ mapping to $y$. Since $\Spec(k) \to Y$ is quasi-compact
there are a finite number of points of $V$ mapping to $y$
(Lemma \ref{lemma-UR-finite-above-x}). After shrinking
$V$ we may assume $v$ is the only one. Choose a scheme $U$ and
a surjective \'etale morphism $U \to X$.
Consider the commutative diagram
$$
\xymatrix{
U \ar[d] & U_V \ar[l] \ar[d] & U_v \ar[l] \ar[d] \\
X \ar[d] & X_V \ar[l] \ar[d] & X_v \ar[l] \ar[d] \\
Y & V \ar[l] & v \ar[l]
}
$$
Since $U_v \to U_V$ identifies $U_v$ with a subset of $U_V$ with
the induced topology (Schemes, Lemma \ref{schemes-lemma-fibre-topological}),
and since $|U_V| \to |X_V|$ and $|U_v| \to |X_v|$ are surjective and open,
we see that $|X_v| \to |X_V|$ is a homeomorphism onto its image (with
induced topology).
On the other hand, the inverse image of $f^{-1}(\{y\})$
under the open map $|X_V| \to |X|$ is equal to $|X_v|$.
We conclude that $|X_v| \to f^{-1}(\{y\})$ is open.
The morphism $X_v \to X$ factors through $X_k$
and $|X_k| \to |X|$ is injective with image $f^{-1}(\{y\})$
by Properties of Spaces, Lemma
\ref{spaces-properties-lemma-points-cartesian}. Using
$|X_v| \to |X_k| \to f^{-1}(\{y\})$ the lemma follows because
$X_v \to X_k$ is surjective.
\end{proof}

\begin{lemma}
\label{lemma-conditions-on-point-on-space-over-field}
Let $X$ be an algebraic space locally of finite type over a field $k$.
Let $x \in |X|$. Consider the conditions
\begin{enumerate}
\item $\dim_x(|X|) = 0$,
\item $x$ is closed in $|X|$ and if $x' \leadsto x$  in $|X|$ then $x' = x$,
\item $x$ is an isolated point of $|X|$,
\item $\dim_x(X) = 0$,
\item $X \to \Spec(k)$ is quasi-finite at $x$.
\end{enumerate}
Then (2), (3), (4), and (5) are equivalent.
If $X$ is decent, then (1) is equivalent to the others.
\end{lemma}

\begin{proof}
Parts (4) and (5) are equivalent for example by
Morphisms of Spaces, Lemmas
\ref{spaces-morphisms-lemma-locally-finite-type-quasi-finite-part} and
\ref{spaces-morphisms-lemma-quasi-finite-at-point}.

\medskip\noindent
Let $U \to X$ be an \'etale morphism where $U$ is an affine scheme and let
$u \in U$ be a point mapping to $x$. Moreover, if $x$ is a closed
point, e.g., in case (2) or (3), then we may and do assume that $u$
is a closed point. Observe that $\dim_u(U) = \dim_x(X)$ by definition
and that this is equal to $\dim(\mathcal{O}_{U, u})$ if $u$ is a closed
point, see Algebra, Lemma
\ref{algebra-lemma-dimension-closed-point-finite-type-field}.

\medskip\noindent
If $\dim_x(X) > 0$ and $u$ is closed, by the arguments above
we can choose a nontrivial
specialization $u' \leadsto u$ in $U$. Then the transcendence degree
of $\kappa(u')$ over $k$ exceeds the transcendence degree of
$\kappa(u)$ over $k$. It follows that the images $x$ and $x'$ in $X$
are distinct, because the transcendence degree of $x/k$ and $x'/k$
are well defined, see Morphisms of Spaces, Definition
\ref{spaces-morphisms-definition-dimension-fibre}.
This applies in particular in cases (2) and (3) and we
conclude that (2) and (3) imply (4).

\medskip\noindent
Conversely, if $X \to \Spec(k)$ is locally quasi-finite at $x$, then
$U \to \Spec(k)$ is locally quasi-finite at $u$, hence $u$ is an
isolated point of $U$
(Morphisms, Lemma \ref{morphisms-lemma-quasi-finite-at-point-characterize}).
It follows that (5) implies (2) and (3) as
$|U| \to |X|$ is continuous and open.

\medskip\noindent
Assume $X$ is decent and (1) holds. Then $\dim_x(X) = \dim_x(|X|)$
by Lemma \ref{lemma-dimension-decent-space} and the proof is complete.
\end{proof}

\begin{lemma}
\label{lemma-conditions-on-space-over-field}
Let $X$ be an algebraic space locally of finite type over a field $k$.
Consider the conditions
\begin{enumerate}
\item $|X|$ is a finite set,
\item $|X|$ is a discrete space,
\item $\dim(|X|) = 0$,
\item $\dim(X) = 0$,
\item $X \to \Spec(k)$ is locally quasi-finite,
\end{enumerate}
Then (2), (3), (4), and (5) are equivalent.
If $X$ is decent, then (1) implies the others.
\end{lemma}

\begin{proof}
Parts (4) and (5) are equivalent for example by
Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-locally-finite-type-quasi-finite-part}.

\medskip\noindent
Let $U \to X$ be a surjective \'etale morphism where $U$ is a scheme.

\medskip\noindent
If $\dim(U) > 0$, then choose a nontrivial specialization
$u \leadsto u'$ in $U$ and the transcendence degree of $\kappa(u)$
over $k$ exceeds the transcendence degree of $\kappa(u')$ over $k$.
It follows that the images $x$ and $x'$ in $X$ are distinct, because
the transcendence degree of $x/k$ and $x'/k$ is well defined, see
Morphisms of Spaces, Definition
\ref{spaces-morphisms-definition-dimension-fibre}.
We conclude that (2) and (3) imply (4).

\medskip\noindent
Conversely, if $X \to \Spec(k)$ is locally quasi-finite, then $U$ is
locally Noetherian
(Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian})
of dimension $0$
(Morphisms, Lemma \ref{morphisms-lemma-locally-quasi-finite-rel-dimension-0})
and hence is a disjoint union of spectra of Artinian local rings
(Properties, Lemma \ref{properties-lemma-locally-Noetherian-dimension-0}).
Hence $U$ is a discrete topological space, and since $|U| \to |X|$
is continuous and open, the same is true for $|X|$.
In other words, (4) implies (2) and (3).

\medskip\noindent
Assume $X$ is decent and (1) holds. Then we may choose $U$ above to
be affine. The fibres of $|U| \to |X|$ are finite (this is a part of the
defining property of decent spaces). Hence $U$ is a finite type scheme
over $k$ with finitely many points. Hence $U$ is quasi-finite over $k$
(Morphisms, Lemma \ref{morphisms-lemma-finite-fibre})
which by definition means that $X \to \Spec(k)$ is locally quasi-finite.
\end{proof}

\begin{lemma}
\label{lemma-conditions-on-point-in-fibre-and-qf}
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces
over $S$ which is locally of finite type. Let $x \in |X|$ with image
$y \in |Y|$. Let $F = f^{-1}(\{y\})$ with induced topology from $|X|$.
Let $k$ be a field and let $\Spec(k) \to Y$ be in the
equivalence class defining $y$. Set $X_k = \Spec(k) \times_Y X$.
Let $\tilde x \in |X_k|$ map to $x \in |X|$.
Consider the following conditions
\begin{enumerate}
\item
\label{item-fibre-at-x-dim-0}
$\dim_x(F) = 0$,
\item
\label{item-isolated-in-fibre}
$x$ is isolated in $F$,
\item
\label{item-no-specializations-in-fibre}
$x$ is closed in $F$ and if $x' \leadsto x$ in $F$, then $x = x'$,
\item
\label{item-dimension-top-k-fibre}
$\dim_{\tilde x}(|X_k|) = 0$,
\item
\label{item-isolated-in-k-fibre}
$\tilde x$ is isolated in $|X_k|$,
\item
\label{item-no-specializations-in-k-fibre}
$\tilde x$ is closed in $|X_k|$ and if $\tilde x' \leadsto \tilde x$
in $|X_k|$, then $\tilde x = \tilde x'$,
\item
\label{item-k-fibre-at-x-dim-0}
$\dim_{\tilde x}(X_k) = 0$,
\item
\label{item-quasi-finite-at-x}
$f$ is quasi-finite at $x$.
\end{enumerate}
Then we have
$$
\xymatrix{
(\ref{item-dimension-top-k-fibre}) \ar@{=>}[r]_{f\text{ decent}} &
(\ref{item-isolated-in-k-fibre}) \ar@{<=>}[r] &
(\ref{item-no-specializations-in-k-fibre}) \ar@{<=>}[r] &
(\ref{item-k-fibre-at-x-dim-0}) \ar@{<=>}[r] &
(\ref{item-quasi-finite-at-x})
}
$$
If $Y$ is decent, then conditions (\ref{item-isolated-in-fibre}) and
(\ref{item-no-specializations-in-fibre}) are equivalent to each other
and to conditions
(\ref{item-isolated-in-k-fibre}),
(\ref{item-no-specializations-in-k-fibre}),
(\ref{item-k-fibre-at-x-dim-0}), and
(\ref{item-quasi-finite-at-x}).
If $Y$ and $X$ are decent, then all conditions are equivalent.
\end{lemma}

\begin{proof}
By Lemma \ref{lemma-conditions-on-point-on-space-over-field} conditions
(\ref{item-isolated-in-k-fibre}),
(\ref{item-no-specializations-in-k-fibre}), and (\ref{item-k-fibre-at-x-dim-0})
are equivalent to each other and to the condition that
$X_k \to \Spec(k)$ is quasi-finite at $\tilde x$.
Thus by Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-base-change-quasi-finite-locus}
they are also equivalent to (\ref{item-quasi-finite-at-x}).
If $f$ is decent, then $X_k$ is a decent algebraic space and
Lemma \ref{lemma-conditions-on-point-on-space-over-field}
shows that (\ref{item-dimension-top-k-fibre}) implies
(\ref{item-isolated-in-k-fibre}).

\medskip\noindent
If $Y$ is decent, then we can pick a quasi-compact monomorphism
$\Spec(k') \to Y$ in the equivalence class of $y$. In this case
Lemma \ref{lemma-topology-fibre}
tells us that $|X_{k'}| \to F$ is a homeomorphism.
Combined with the arguments given above this implies
the remaining statements of the lemma; details omitted.
\end{proof}

\begin{lemma}
\label{lemma-conditions-on-fibre-and-qf}
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces
over $S$ which is locally of finite type. Let $y \in |Y|$. Let $k$ be a field
and let $\Spec(k) \to Y$ be in the equivalence class defining $y$.
Set $X_k = \Spec(k) \times_Y X$ and let $F = f^{-1}(\{y\})$ with the
induced topology from $|X|$. Consider the following conditions
\begin{enumerate}
\item
\label{item-fibre-finite}
$F$ is finite,
\item
\label{item-fibre-discrete}
$F$ is a discrete topological space,
\item
\label{item-fibre-no-specializations}
$\dim(F) = 0$,
\item
\label{item-k-fibre-finite}
$|X_k|$ is a finite set,
\item
\label{item-k-fibre-discrete}
$|X_k|$ is a discrete space,
\item
\label{item-k-fibre-no-specializations}
$\dim(|X_k|) = 0$,
\item
\label{item-k-fibre-dim-0}
$\dim(X_k) = 0$,
\item
\label{item-quasi-finite-at-points-fibre}
$f$ is quasi-finite at all points of $|X|$ lying over $y$.
\end{enumerate}
Then we have
$$
\xymatrix{
(\ref{item-fibre-finite}) &
(\ref{item-k-fibre-finite}) \ar@{=>}[l] \ar@{=>}[r]_{f\text{ decent}} &
(\ref{item-k-fibre-discrete}) \ar@{<=>}[r] &
(\ref{item-k-fibre-no-specializations}) \ar@{<=>}[r] &
(\ref{item-k-fibre-dim-0}) \ar@{<=>}[r] &
(\ref{item-quasi-finite-at-points-fibre})
}
$$
If $Y$ is decent, then conditions (\ref{item-fibre-discrete}) and
(\ref{item-fibre-no-specializations})
are equivalent to each other and to conditions (\ref{item-k-fibre-discrete}),
(\ref{item-k-fibre-no-specializations}), (\ref{item-k-fibre-dim-0}), and
(\ref{item-quasi-finite-at-points-fibre}).
If $Y$ and $X$ are decent, then (\ref{item-fibre-finite}) implies
all the other conditions.
\end{lemma}

\begin{proof}
By Lemma \ref{lemma-conditions-on-space-over-field}
conditions  (\ref{item-k-fibre-discrete}),
(\ref{item-k-fibre-no-specializations}), and (\ref{item-k-fibre-dim-0})
are equivalent to each other and to the condition that
$X_k \to \Spec(k)$ is locally quasi-finite.
Thus by Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-base-change-quasi-finite-locus}
they are also equivalent to (\ref{item-quasi-finite-at-points-fibre}).
If $f$ is decent, then $X_k$ is a decent algebraic space and
Lemma \ref{lemma-conditions-on-space-over-field}
shows that (\ref{item-k-fibre-finite}) implies (\ref{item-k-fibre-discrete}).

\medskip\noindent
The map $|X_k| \to F$ is surjective by
Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-cartesian}
and we see
(\ref{item-k-fibre-finite}) $\Rightarrow$ (\ref{item-fibre-finite}).

\medskip\noindent
If $Y$ is decent, then we can pick a quasi-compact monomorphism
$\Spec(k') \to Y$ in the equivalence class of $y$. In this case
Lemma \ref{lemma-topology-fibre}
tells us that $|X_{k'}| \to F$ is a homeomorphism.
Combined with the arguments given above this implies
the remaining statements of the lemma; details omitted.
\end{proof}







\section{Monomorphisms}
\label{section-monomorphisms}

\noindent
Here is another case where monomorphisms are representable.

\begin{lemma}
\label{lemma-monomorphism-toward-disjoint-union-dim-0-rings}
Let $S$ be a scheme. Let $Y$ be a disjoint union of spectra of
zero dimensional local rings over $S$.
Let $f : X \to Y$ be a monomorphism of algebraic spaces over $S$.
Then $f$ is representable, i.e., $X$ is a scheme.
\end{lemma}

\begin{proof}
This immediately reduces to the case $Y = \Spec(A)$ where
$A$ is a zero dimensional local ring, i.e.,
$\Spec(A) = \{\mathfrak m_A\}$
is a singleton. If $X = \emptyset$, then there is nothing to prove.
If not, choose a nonempty affine scheme $U = \Spec(B)$
and an \'etale morphism $U \to X$. As $|X|$ is a singleton (as a
subset of $|Y|$, see
Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-monomorphism-injective-points})
we see that $U \to X$ is surjective. Note that
$U \times_X U = U \times_Y U = \Spec(B \otimes_A B)$.
Thus we see that the ring maps $B \to B \otimes_A B$ are \'etale.
Since
$$
(B \otimes_A B)/\mathfrak m_A(B \otimes_A B)
=
(B/\mathfrak m_AB) \otimes_{A/\mathfrak m_A} (B/\mathfrak m_AB)
$$
we see that
$B/\mathfrak m_AB \to (B \otimes_A B)/\mathfrak m_A(B \otimes_A B)$
is flat and in fact free of rank equal to the dimension of
$B/\mathfrak m_AB$ as a $A/\mathfrak m_A$-vector space. Since
$B \to B \otimes_A B$ is \'etale, this can only happen if this
dimension is finite (see for example
Morphisms, Lemmas \ref{morphisms-lemma-etale-universally-bounded} and
\ref{morphisms-lemma-locally-quasi-finite-qc-source-universally-bounded}).
Every prime of $B$ lies over $\mathfrak m_A$ (the unique prime of $A$).
Hence $\Spec(B) = \Spec(B/\mathfrak m_A)$ as a topological
space, and this space is a finite discrete set as $B/\mathfrak m_A B$
is an Artinian ring, see
Algebra, Lemmas \ref{algebra-lemma-finite-dimensional-algebra} and
\ref{algebra-lemma-artinian-finite-length}.
Hence all prime ideals of $B$ are maximal and
$B = B_1 \times \ldots \times B_n$ is a product of finitely many
local rings of dimension zero, see
Algebra, Lemma \ref{algebra-lemma-product-local}.
Thus $B \to B \otimes_A B$ is finite \'etale as all the local rings
$B_i$ are henselian by
Algebra, Lemma \ref{algebra-lemma-local-dimension-zero-henselian}.
Thus $X$ is an affine scheme by
Groupoids, Proposition \ref{groupoids-proposition-finite-flat-equivalence}.
\end{proof}









\section{Birational morphisms}
\label{section-birational}

\noindent
The following definition of a birational morphism of algebraic spaces
seems to be the closest to our definition
(Morphisms, Definition \ref{morphisms-definition-birational})
of a birational morphism of schemes.

\begin{definition}
\label{definition-birational}
Let $S$ be a scheme. Let $X$ and $Y$ algebraic spaces over $S$.
Assume $X$ and $Y$ are decent and that $|X|$ and $|Y|$ have finitely many
irreducible components. We say a morphism $f : X \to Y$ is
{\it birational} if
\begin{enumerate}
\item $|f|$ induces a bijection between the set of generic points
of irreducible components of $|X|$ and the set of generic points
of the irreducible components of $|Y|$, and
\item for every generic point $x \in |X|$ of an irreducible component
the local ring map $\mathcal{O}_{Y, f(x)} \to \mathcal{O}_{X, x}$
is an isomorphism (see clarification below).
\end{enumerate}
\end{definition}

\noindent
Clarification: Since $X$ and $Y$ are decent the topological spaces
$|X|$ and $|Y|$ are sober (Proposition \ref{proposition-reasonable-sober}).
Hence condition (1) makes sense. Moreover, because we have assumed
that $|X|$ and $|Y|$ have finitely many irreducible components, we
see that the generic points $x_1, \ldots, x_n \in |X|$,
resp.\ $y_1, \ldots, y_n \in |Y|$ are contained in any dense open
of $|X|$, resp.\ $|Y|$. In particular, they are contained in
the schematic locus of $X$, resp.\ $Y$ by
Theorem \ref{theorem-decent-open-dense-scheme}.
Thus we can define $\mathcal{O}_{X, x_i}$, resp.\ $\mathcal{O}_{Y, y_i}$
to be the local ring of this scheme at $x_i$, resp.\ $y_i$.

\medskip\noindent
Another and perhaps better way to say all of this is that the morphism
$f : X \to Y$ is birational if there exist dense open subspaces
$X' \subset X$ and $Y' \subset Y$ such that
\begin{enumerate}
\item $f(X') \subset Y'$,
\item $X'$ and $Y'$ are representable, and
\item $f|_{X'} : X' \to Y'$ is birational in
the sense of Morphisms, Definition \ref{morphisms-definition-birational}.
\end{enumerate}
However, we do insist that $X$ and $Y$ are decent with finitely many
irreducible components.

\begin{lemma}
\label{lemma-birational-dominant}
Let $S$ be a scheme.
Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which
are decent and have finitely many irreducible components. If $f$ is
birational then $f$ is dominant.
\end{lemma}

\begin{proof}
Follows immediately from the definitions. See
Morphisms of Spaces, Definition \ref{spaces-morphisms-definition-dominant}.
\end{proof}

\begin{lemma}
\label{lemma-birational-etale-localization}
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic
spaces over $S$ which are decent and have finitely many irreducible
components. If $f$ is birational and $V \to Y$ is an \'etale morphism
with $V$ affine, then $X \times_Y V$ is decent with finitely
many irreducible components and $X \times_Y V \to V$ is birational.
\end{lemma}

\begin{proof}
The algebraic space $U = X \times_Y V$ is decent
(Lemma \ref{lemma-etale-named-properties}).
The generic points of $V$ and $U$ are the elements of $|V|$ and $|U|$
which lie over generic points of $|Y|$ and $|X|$
(Lemma \ref{lemma-decent-generic-points}).
Since $Y$ is decent we conclude there are finitely many generic points
on $V$. Let $\xi \in |X|$ be a generic point of an irreducible component.
By the discussion following Definition \ref{definition-birational}
we have a cartesian square
$$
\xymatrix{
\Spec(\mathcal{O}_{X, \xi}) \ar[d] \ar[r] & X \ar[d] \\
\Spec(\mathcal{O}_{Y, f(\xi)}) \ar[r] & Y
}
$$
whose horizontal morphisms are monomorphisms identifying local rings
and where the left vertical arrow is an isomorphism. It follows that
in the diagram
$$
\xymatrix{
\Spec(\mathcal{O}_{X, \xi}) \times_X U \ar[d] \ar[r] & U \ar[d] \\
\Spec(\mathcal{O}_{Y, f(\xi)}) \times_Y V \ar[r] & V
}
$$
the vertical arrow on the left is an isomorphism. The horizonal arrows
have image contained in the schematic locus of $U$ and $V$ and
identify local rings (some details omitted). Since the image of
the horizontal arrows are the points of $|U|$, resp.\ $|V|$
lying over $\xi$, resp.\ $f(\xi)$ we conclude.
\end{proof}

\begin{lemma}
\label{lemma-finite-birational-over-normal}
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic
spaces over $S$. Assume
\begin{enumerate}
\item $X$ and $Y$ are decent and have finitely many irreducible components,
\item $f$ is integral and birational,
\item $Y$ is normal, and
\item $X$ is reduced.
\end{enumerate}
Then $f$ is an isomorphism.
\end{lemma}

\begin{proof}
Let $V \to Y$ be an \'etale morphism with $V$ affine. It suffices to show that
$U = X \times_Y V \to V$ is an isomorphism. By
Lemma \ref{lemma-birational-etale-localization} and its proof
we see that $U$ and $V$ are decent and have finitely many
irreducible components and that $U \to V$ is birational.
By Properties, Lemma
\ref{properties-lemma-normal-locally-finite-nr-irreducibles}
$V$ is a finite disjoint union of integral schemes.
Thus we may assume $V$ is integral. As $f$ is birational, we
see that $U$ is irreducible and reduced, i.e., integral
(note that $U$ is a scheme as $f$ is integral, hence representable).
Thus we may assume that $X$ and $Y$ are integral schemes
and the result follows from the case of schemes, see
Morphisms, Lemma \ref{morphisms-lemma-finite-birational-over-normal}.
\end{proof}








\input{chapters}

\bibliography{my}
\bibliographystyle{amsalpha}

\end{document}