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spaces-properties.tex
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\input{preamble}
% OK, start here.
%
\begin{document}
\title{Properties of Algebraic Spaces}
\maketitle
\phantomsection
\label{section-phantom}
\tableofcontents
\section{Introduction}
\label{section-introduction}
\noindent
Please see
Spaces, Section \ref{spaces-section-introduction}
for a brief introduction to algebraic spaces, and please read
some of that chapter for our basic definitions and conventions
concerning algebraic spaces. In this chapter we start introducing
some basic notions and properties of algebraic spaces. A fundamental
reference for the case of quasi-separated algebraic spaces is
\cite{Kn}.
\medskip\noindent
The discussion is somewhat awkward at times since we made the design
decision to first talk about properties of algebraic spaces by
themselves, and only later about properties of morphisms of algebraic
spaces. We make an exception for this rule regarding
{\it \'etale morphisms} of algebraic spaces, which we introduce in
Section \ref{section-etale-morphisms}.
But until that section whenever we say a morphism has a certain property, it
automatically means the source of the morphism is a scheme (or perhaps the
morphism is representable).
\medskip\noindent
Some of the material in the chapter (especially regarding points)
will be improved upon in the chapter on decent algebraic spaces.
\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 $X \times X$. The reason is that we
want to avoid confusion when changing base schemes, as in
Spaces, Section \ref{spaces-section-change-base-scheme}.
\section{Separation axioms}
\label{section-separation}
\noindent
In this section we collect all the ``absolute'' separation conditions
of algebraic spaces. Since in our language any algebraic space is an
algebraic space over some definite base scheme, any absolute property
of $X$ over $S$ corresponds to a conditions imposed on $X$ viewed
as an algebraic space over $\Spec(\mathbf{Z})$. Here is the
precise formulation.
\begin{definition}
\label{definition-separated}
(Compare Spaces, Definition \ref{spaces-definition-separated}.)
Consider a big fppf site
$\Sch_{fppf} = (\Sch/\Spec(\mathbf{Z}))_{fppf}$.
Let $X$ be an algebraic space over
$\Spec(\mathbf{Z})$. Let $\Delta : X \to X \times X$
be the diagonal morphism.
\begin{enumerate}
\item We say $X$ is {\it separated} if $\Delta$ is a closed immersion.
\item We say $X$ is {\it locally separated}\footnote{In the
literature this often refers to quasi-separated and locally
separated algebraic spaces.} if $\Delta$ is an
immersion.
\item We say $X$ is {\it quasi-separated} if $\Delta$ is quasi-compact.
\item We say $X$ is {\it Zariski locally quasi-separated}\footnote{
This notion was suggested by B.\ Conrad.} if there
exists a Zariski covering $X = \bigcup_{i \in I} X_i$ (see Spaces,
Definition \ref{spaces-definition-Zariski-open-covering}) such that
each $X_i$ is quasi-separated.
\end{enumerate}
Let $S$ is a scheme contained in $\Sch_{fppf}$, and let
$X$ be an algebraic space over $S$. Then we say $X$ is {\it separated},
{\it locally separated}, {\it quasi-separated}, or
{\it Zariski locally quasi-separated}
if $X$ viewed as an algebraic space over $\Spec(\mathbf{Z})$ (see
Spaces, Definition \ref{spaces-definition-base-change})
has the corresponding property.
\end{definition}
\noindent
It is true that an algebraic space $X$ over $S$ which is separated
(in the absolute sense above) is separated over $S$ (and similarly
for the other absolute separation properties above). This will be discussed
in great detail in
Morphisms of Spaces, Section \ref{spaces-morphisms-section-separation-axioms}.
We will see in
Lemma \ref{lemma-quasi-separated-quasi-compact-pieces}
that being Zariski locally separated is independent of the base scheme
(hence equivalent to the absolute notion).
\begin{lemma}
\label{lemma-trivial-implications}
Let $S$ be a scheme.
Let $X$ be an algebraic space over $S$.
We have the following implications among the separation axioms
of Definition \ref{definition-separated}:
\begin{enumerate}
\item separated implies all the others,
\item quasi-separated implies Zariski locally quasi-separated.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\section{Points of algebraic spaces}
\label{section-points}
\noindent
As is clear from
Spaces, Example \ref{spaces-example-affine-line-translation}
a point of an algebraic space should not be defined as a monomorphism
from the spectrum of a field.
Instead we define them as equivalence classes of morphisms of spectra
of fields as equivalence classes of morphisms from spectra of fields.
exactly as explained in
Schemes, Section \ref{schemes-section-points}.
\medskip\noindent
Let $S$ be a scheme.
Let $F$ be a presheaf on $(\Sch/S)_{fppf}$.
Let $K$ is a field. Consider a morphism
$$
\Spec(K) \longrightarrow F.
$$
By the Yoneda Lemma this is given by an
element $p \in F(\Spec(K))$. We say that two such
pairs $(\Spec(K), p)$ and $(\Spec(L), q)$
are {\it equivalent} if there exists
a third field $\Omega$ and a commutative diagram
$$
\xymatrix{
\Spec(\Omega) \ar[r] \ar[d] &
\Spec(L) \ar[d]^q \\
\Spec(K) \ar[r]^p &
F.
}
$$
In other words, there are field extensions
$K \to \Omega$ and $L \to \Omega$ such that
$p$ and $q$ map to the same element
of $F(\Spec(\Omega))$. We omit the verification that this
defines an equivalence relation.
\begin{definition}
\label{definition-points}
Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.
A {\it point} of $X$ is an equivalence class of morphisms
from spectra of fields into $X$.
The set of points of $X$ is denoted $|X|$.
\end{definition}
\noindent
Note that if $f : X \to Y$ is a morphism of algebraic spaces
over $S$, then there is an induced map $|f| : |X| \to |Y|$ which
maps a representative $x : \Spec(K) \to X$ to the representative
$f \circ x : \Spec(K) \to Y$.
\begin{lemma}
\label{lemma-scheme-points}
Let $S$ be a scheme. Let $X$ be a scheme over $S$.
The points of $X$ as a scheme are in canonical 1-1 correspondence
with the points of $X$ as an algebraic space.
\end{lemma}
\begin{proof}
This is Schemes, Lemma \ref{schemes-lemma-characterize-points}.
\end{proof}
\begin{lemma}
\label{lemma-points-cartesian}
Let $S$ be a scheme. Let
$$
\xymatrix{
Z \times_Y X \ar[r] \ar[d] & X \ar[d] \\
Z \ar[r] & Y
}
$$
be a cartesian diagram of algebraic spaces. Then the map of sets
of points
$$
|Z \times_Y X|
\longrightarrow
|Z| \times_{|Y|} |X|
$$
is surjective.
\end{lemma}
\begin{proof}
Namely, suppose given fields $K$, $L$ and morphisms
$\Spec(K) \to X$, $\Spec(L) \to Z$, then the
assumption that they agree as elements of $|Y|$ means that
there is a common extension $K \subset M$ and $L \subset M$
such that
$\Spec(M) \to \Spec(K) \to X \to Y$ and
$\Spec(M) \to \Spec(L) \to Z \to Y$ agree.
And this is exactly the condition which says you get a
morphism $\Spec(M) \to Z \times_Y X$.
\end{proof}
\begin{lemma}
\label{lemma-characterize-surjective}
Let $S$ be a scheme.
Let $X$ be an algebraic space over $S$.
Let $f : T \to X$ be a morphism from a scheme to $X$.
The following are equivalent
\begin{enumerate}
\item $f : T \to X$ is surjective (according to
Spaces, Definition \ref{spaces-definition-relative-representable-property}),
and
\item $|f| : |T| \to |X|$ is surjective.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume (1). Let $x : \Spec(K) \to X$ be a morphism
from the spectrum of a field into $X$. By assumption the morphism of
schemes $\Spec(K) \times_X T \to \Spec(K)$ is surjective.
Hence there exists a field extension $K \subset K'$ and a morphism
$\Spec(K') \to \Spec(K) \times_X T$ such that the left
square in the diagram
$$
\xymatrix{
\Spec(K') \ar[r] \ar[d] &
\Spec(K) \times_X T \ar[d] \ar[r] &
T \ar[d] \\
\Spec(K) \ar@{=}[r] &
\Spec(K) \ar[r]^-x & X
}
$$
is commutative. This shows that $|f| : |T| \to |X|$ is surjective.
\medskip\noindent
Assume (2). Let $Z \to X$ be a morphism where $Z$ is
a scheme. We have to show that the morphism of schemes $Z \times_X T \to T$
is surjective, i.e., that $|Z \times_X T| \to |Z|$ is surjective.
This follows from (2) and
Lemma \ref{lemma-points-cartesian}.
\end{proof}
\begin{lemma}
\label{lemma-points-presentation}
Let $S$ be a scheme.
Let $X$ be an algebraic space over $S$.
Let $X = U/R$ be a presentation of $X$, see
Spaces, Definition \ref{spaces-definition-presentation}.
Then the image of $|R| \to |U| \times |U|$ is an equivalence relation
and $|X|$ is the quotient of $|U|$ by this equivalence relation.
\end{lemma}
\begin{proof}
The assumption means that $U$ is a scheme, $p : U \to X$ is a surjective,
\'etale morphism, $R = U \times_X U$ is a scheme and defines an \'etale
equivalence relation on $U$ such that $X = U/R$ as sheaves. By
Lemma \ref{lemma-characterize-surjective}
we see that $|U| \to |X|$ is surjective. By
Lemma \ref{lemma-points-cartesian}
the map
$$
|R| \longrightarrow |U| \times_{|X|} |U|
$$
is surjective. Hence the image of $|R| \to |U| \times |U|$ is
exactly the set of pairs $(u_1, u_2) \in |U| \times |U|$
such that $u_1$ and $u_2$ have the same image in $|X|$.
Combining these two statements we get the result of the lemma.
\end{proof}
\begin{lemma}
\label{lemma-topology-points}
Let $S$ be a scheme. There exists a unique topology on the set of points
of algebraic spaces over $S$ with the following properties:
\begin{enumerate}
\item for every morphism of algebraic spaces $X \to Y$ over $S$
the map $|X| \to |Y|$ is continuous, and
\item for every \'etale morphism $U \to X$ with $U$ a scheme
the map of topological spaces $|U| \to |X|$ is continuous and open.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $X$ be an algebraic space over $S$. Let $p : U \to X$ be a
surjective \'etale morphism where $U$ is a scheme over $S$.
We define $W \subset |X|$ is open if and only if $|p|^{-1}(W)$
is an open subset of $|U|$. This is a topology on $|X|$.
\medskip\noindent
Let us prove that the topology is independent of the choice of
the presentation. To do this it suffices to show that if $U'$ is a scheme,
and $U' \to X$ is an \'etale morphism, then the map $|U'| \to |X|$
(with topology on $|X|$ defined using $U \to X$ as above)
is open and continuous; which in addition will prove that (2) holds.
Set $U'' = U \times_X U'$, so that we have the commutative diagram
$$
\xymatrix{
U'' \ar[r] \ar[d] & U' \ar[d] \\
U \ar[r] & X
}
$$
As $U \to X$ and $U' \to X$ are \'etale we see that
both $U'' \to U$ and $U'' \to U'$ are \'etale morphisms of schemes.
Moreover, $U'' \to U'$ is surjective. Hence
we get a commutative diagram of maps of sets
$$
\xymatrix{
|U''| \ar[r] \ar[d] & |U'| \ar[d] \\
|U| \ar[r] & |X|
}
$$
The lower horizontal arrow is surjective (see
Lemma \ref{lemma-characterize-surjective}
or
Lemma \ref{lemma-points-presentation})
and continuous by definition of the topology on $|X|$.
The top horizontal arrow is surjective, continuous, and open by
Morphisms, Lemma \ref{morphisms-lemma-etale-open}.
The left vertical arrow is continuous and open (by
Morphisms, Lemma \ref{morphisms-lemma-etale-open}
again.) Hence it follows formally that the right vertical
arrow is continuous and open.
\medskip\noindent
To finish the proof we prove (1).
Let $a : X \to Y$ be a morphism of algebraic spaces. According to
Spaces, Lemma \ref{spaces-lemma-lift-morphism-presentations}
we can find a diagram
$$
\xymatrix{
U \ar[d]_p \ar[r]_\alpha & V \ar[d]^q \\
X \ar[r]^a & Y
}
$$
where $U$ and $V$ are schemes, and $p$ and $q$ are surjective and \'etale.
This gives rise to the diagram
$$
\xymatrix{
|U| \ar[d]_p \ar[r]_\alpha & |V| \ar[d]^q \\
|X| \ar[r]^a & |Y|
}
$$
where all but the lower horizontal arrows are known to be continuous and
the two vertical arrows are surjective and open. It follows that the
lower horizontal arrow is continuous as desired.
\end{proof}
\begin{definition}
\label{definition-topological-space}
Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.
The underlying {\it topological space} of $X$ is the set of points
$|X|$ endowed with the topology constructed in
Lemma \ref{lemma-topology-points}.
\end{definition}
\noindent
It turns out that this topological space carries the same information
as the small Zariski site $X_{Zar}$ of
Spaces, Definition \ref{spaces-definition-small-Zariski-site}.
\begin{lemma}
\label{lemma-open-subspaces}
Let $S$ be a scheme.
Let $X$ be an algebraic space over $S$.
\begin{enumerate}
\item The rule $X' \mapsto |X'|$ defines an inclusion preserving
bijection between open subspaces $X'$ (see
Spaces, Definition \ref{spaces-definition-immersion})
of $X$, and opens of the topological space $|X|$.
\item A family $\{X_i \subset X\}_{i \in I}$ of open subspaces of $X$
is a Zariski covering (see
Spaces, Definition \ref{spaces-definition-Zariski-open-covering})
if and only if $|X| = \bigcup |X_i|$.
\end{enumerate}
In other words, the small Zariski site $X_{Zar}$ of $X$ is canonically
identified with a site associated to the topological space $|X|$ (see
Sites, Example \ref{sites-example-site-topological}).
\end{lemma}
\begin{proof}
In order to prove (1) let us construct the inverse of the rule.
Namely, suppose that $W \subset |X|$ is open. Choose a presentation
$X = U/R$ corresponding to the surjective \'etale map
$p : U \to X$ and \'etale maps $s, t : R \to U$.
By construction we see that $|p|^{-1}(W)$ is an
open of $U$. Denote $W' \subset U$ the corresponding open subscheme.
It is clear that $R' = s^{-1}(W') = t^{-1}(W')$ is a Zariski open
of $R$ which defines an \'etale equivalence relation on $W'$.
By Spaces, Lemma \ref{spaces-lemma-finding-opens} the morphism
$X' = W'/R' \to X$ is an open immersion. Hence $X'$ is an algebraic space
by Spaces, Lemma \ref{spaces-lemma-representable-over-space}.
By construction $|X'| = W$, i.e., $X'$ is a subspace of $X$
corresponding to $W$. Thus (1) is proved.
\medskip\noindent
To prove (2), note that if $\{X_i \subset X\}_{i \in I}$ is a collection
of open subspaces, then it is a Zariski covering if and only if the
$U = \bigcup U \times_X X_i$ is an open covering. This follows from
the definition of a Zariski covering and the fact that the morphism
$U \to X$ is surjective as a map of presheaves on $(\Sch/S)_{fppf}$.
On the other hand, we see that $|X| = \bigcup |X_i|$ if and only if
$U = \bigcup U \times_X X_i$ by Lemma \ref{lemma-points-presentation}
(and the fact that the projections $U \times_X X_i \to X_i$ are surjective
and \'etale). Thus the equivalence of (2) follows.
\end{proof}
\begin{lemma}
\label{lemma-factor-through-open-subspace}
Let $S$ be a scheme.
Let $X$, $Y$ be algebraic spaces over $S$.
Let $X' \subset X$ be an open subspace.
Let $f : Y \to X$ be a morphism of algebraic spaces over $S$.
Then $f$ factors through $X'$ if and only if $|f| : |Y| \to |X|$
factors through $|X'| \subset |X|$.
\end{lemma}
\begin{proof}
By Spaces, Lemma \ref{spaces-lemma-base-change-immersions}
we see that $Y' = Y \times_X X' \to Y$ is an open immersion.
If $|f|(|Y|) \subset |X'|$, then clearly $|Y'| = |Y|$. Hence $Y' = Y$ by
Lemma \ref{lemma-open-subspaces}.
\end{proof}
\begin{lemma}
\label{lemma-etale-image-open}
Let $S$ be a scheme. Let $X$ be an algebraic spaces over $S$.
Let $U$ be a scheme and let $f : U \to X$ be an \'etale morphism.
Let $X' \subset X$ be the open subspace corresponding to
the open $|f|(|U|) \subset |X|$ via
Lemma \ref{lemma-open-subspaces}.
Then $f$ factors through a surjective \'etale morphism $f' : U \to X'$.
Moreover, if $R = U \times_X U$, then $R = U \times_{X'} U$ and $X'$ has
the presentation $X' = U/R$.
\end{lemma}
\begin{proof}
The existence of the factorization follows from
Lemma \ref{lemma-factor-through-open-subspace}.
The morphism $f'$ is surjective according to
Lemma \ref{lemma-characterize-surjective}.
To see $f'$ is \'etale, suppose that $T \to X'$ is a morphism
where $T$ is a scheme. Then $T \times_X U = T \times_{X'} U$
as $X" \to X$ is a monomorphism of sheaves. Thus the projection
$T \times_{X'} U \to T$ is \'etale as we assumed $f$ \'etale.
We have $U \times_X U = U \times_{X'} U$ as $X' \to X$ is a monomorphism.
Then $X' = U/R$ follows from
Spaces, Lemma \ref{spaces-lemma-space-presentation}.
\end{proof}
\begin{lemma}
\label{lemma-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 injective.
\end{lemma}
\begin{proof}
Suppose that $\varphi_i : \Spec(k_i) \to X$ are monomorphisms
for $i = 1, 2$. If $\varphi_1$ and $\varphi_2$ define the same point
of $|X|$, then we see that the scheme
$$
Y = \Spec(k_1) \times_{\varphi_1, X, \varphi_2} \Spec(k_2)
$$
is nonempty. Since the base change of a monomorphism is a monomorphism
this means that the projection morphisms $Y \to \Spec(k_i)$
are monomorphisms. Hence $\Spec(k_1) = Y = \Spec(k_2)$
as schemes over $X$, see
Schemes, Lemma \ref{schemes-lemma-mono-towards-spec-field}.
We conclude that $\varphi_1 = \varphi_2$, which proves the lemma.
\end{proof}
\noindent
We will see in
Decent Spaces,
Lemma \ref{decent-spaces-lemma-decent-points-monomorphism}
that this map is a bijection when $X$ is decent.
\section{Quasi-compact spaces}
\label{section-quasi-compact}
\begin{definition}
\label{definition-quasi-compact}
Let $S$ be a scheme.
Let $X$ be an algebraic space over $S$.
We say $X$ is {\it quasi-compact} if there exists a surjective
\'etale morphism $U \to X$ with $U$ quasi-compact.
\end{definition}
\begin{lemma}
\label{lemma-quasi-compact-space}
Let $S$ be a scheme.
Let $X$ be an algebraic space over $S$.
Then $X$ is quasi-compact if and only if $|X|$ is quasi-compact.
\end{lemma}
\begin{proof}
Choose a scheme $U$ and an \'etale surjective morphism $U \to X$.
We will use Lemma \ref{lemma-characterize-surjective}.
If $U$ is quasi-compact, then since $|U| \to |X|$ is surjective
we conclude that $|X|$ is quasi-compact.
If $|X|$ is quasi-compact, then since $|U| \to |X|$ is open
we see that there exists a quasi-compact open $U' \subset U$
such that $|U'| \to |X|$ is surjective (and still \'etale).
Hence we win.
\end{proof}
\begin{lemma}
\label{lemma-finite-disjoint-quasi-compact}
A finite disjoint union of quasi-compact algebraic spaces is
a quasi-compact algebraic space.
\end{lemma}
\begin{proof}
This is clear from
Lemma \ref{lemma-quasi-compact-space}
and the corresponding topological fact.
\end{proof}
\begin{example}
\label{example-quasi-compact-not-very-reasonable}
The space $\mathbf{A}^1_{\mathbf{Q}}/\mathbf{Z}$ is a quasi-compact
algebraic space.
\end{example}
\begin{lemma}
\label{lemma-space-locally-quasi-compact}
Let $S$ be a scheme.
Let $X$ be an algebraic space over $S$.
Every point of $|X|$ has a fundamental system of open
quasi-compact neighbourhoods.
In particular $|X|$ is locally quasi-compact in the sense of
Topology, Definition \ref{topology-definition-locally-quasi-compact}.
\end{lemma}
\begin{proof}
This follows formally from the fact that there exists a scheme
$U$ and a surjective, open, continuous map
$U \to |X|$ of topological spaces. To be a bit more precise, if
$u \in U$ maps to $x \in |X|$, then the images of the affine
neighbourhoods of $u$ will give a fundamental system of quasi-compact
open neighbourhoods of $x$.
\end{proof}
\section{Special coverings}
\label{section-special-coverings}
\noindent
In this section we collect some straightforward lemmas on the existence
of \'etale surjective coverings of algebraic spaces.
\begin{lemma}
\label{lemma-cover-by-union-affines}
Let $S$ be a scheme.
Let $X$ be an algebraic space over $S$.
There exists a surjective \'etale morphism $U \to X$ where
$U$ is a disjoint union of affine schemes.
We may in addition assume each of these affines
maps into an affine open of $S$.
\end{lemma}
\begin{proof}
Let $V \to X$ be a surjective \'etale morphism.
Let $V = \bigcup_{i \in I} V_i$ be a Zariski open covering
such that each $V_i$ maps into an affine open of $S$.
Then set $U = \coprod_{i \in I} V_i$ with induced morphism
$U \to V \to X$. This is \'etale and surjective as a composition
of \'etale and surjective representable
transformations of functors (via the general principle
Spaces, Lemma
\ref{spaces-lemma-composition-representable-transformations-property}
and
Morphisms, Lemmas \ref{morphisms-lemma-composition-surjective} and
\ref{morphisms-lemma-composition-etale}).
\end{proof}
\begin{lemma}
\label{lemma-union-of-quasi-compact}
Let $S$ be a scheme.
Let $X$ be an algebraic space over $S$.
There exists a Zariski covering $X = \bigcup X_i$
such that each algebraic space $X_i$ has a surjective
\'etale covering by an affine scheme. We may in addition assume
each $X_i$ maps into an affine open of $S$.
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-cover-by-union-affines} we can find a surjective
\'etale morphism $U = \coprod U_i \to X$, with $U_i$ affine and mapping
into an affine open of $S$. Let $X_i \subset X$ be the open subspace
of $X$ such that $U_i \to X$ factors through an \'etale surjective morphism
$U_i \to X_i$, see
Lemma \ref{lemma-etale-image-open}.
Since $U = \bigcup U_i$ we see that $X = \bigcup X_i$.
As $U_i \to X_i$ is surjective it follows that $X_i \to S$ maps into
an affine open of $S$.
\end{proof}
\begin{lemma}
\label{lemma-quasi-compact-affine-cover}
Let $S$ be a scheme.
Let $X$ be an algebraic space over $S$.
Then $X$ is quasi-compact if and only if
there exists an \'etale surjective morphism $U \to X$
with $U$ an affine scheme.
\end{lemma}
\begin{proof}
If there exists an \'etale surjective morphism $U \to X$ with $U$
affine then $X$ is quasi-compact by Definition \ref{definition-quasi-compact}.
Conversely, if $X$ is quasi-compact, then $|X|$ is quasi-compact.
Let $U = \coprod_{i \in I} U_i$ be a disjoint union of affine schemes
with an \'etale and surjective map $\varphi : U \to X$
(Lemma \ref{lemma-cover-by-union-affines}).
Then $|X| = \bigcup \varphi(|U_i|)$ and
by quasi-compactness there is a finite subset $i_1, \ldots, i_n$
such that $|X| = \bigcup \varphi(|U_{i_j}|)$. Hence
$U_{i_1} \cup \ldots \cup U_{i_n}$ is an affine scheme with a
finite surjective morphism towards $X$.
\end{proof}
\noindent
The following lemma will be obsoleted by the discussion of
separated morphisms in the chapter on morphisms of algebraic spaces.
\begin{lemma}
\label{lemma-separated-cover}
Let $S$ be a scheme.
Let $X$ be an algebraic space over $S$.
Let $U$ be a separated scheme and $U \to X$ \'etale.
Then $U \to X$ is separated, and $R = U \times_X U$ is a separated scheme.
\end{lemma}
\begin{proof}
Let $X' \subset X$ be the open subscheme such that $U \to X$ factors
through an \'etale surjection $U \to X'$, see
Lemma \ref{lemma-etale-image-open}.
If $U \to X'$ is separated, then so is $U \to X$, see
Spaces, Lemma
\ref{spaces-lemma-composition-representable-transformations-property}
(as the open immersion $X' \to X$ is separated by
Spaces, Lemma
\ref{spaces-lemma-representable-transformations-property-implication}
and
Schemes, Lemma \ref{schemes-lemma-immersions-monomorphisms}).
Moreover, since $U \times_{X'} U = U \times_X U$ it suffices
to prove the result after replacing $X$ by $X'$, i.e., we may
assume $U \to X$ surjective.
Consider the commutative diagram
$$
\xymatrix{
R = U \times_X U \ar[r] \ar[d] & U \ar[d] \\
U \ar[r] & X
}
$$
In the proof of
Spaces, Lemma \ref{spaces-lemma-properties-diagonal}
we have seen that $j : R \to U \times_S U$ is separated.
The morphism of schemes $U \to S$ is separated as $U$ is a separated
scheme, see
Schemes, Lemma \ref{schemes-lemma-compose-after-separated}.
Hence $U \times_S U \to U$ is separated as a base change, see
Schemes, Lemma \ref{schemes-lemma-separated-permanence}.
Hence the scheme $U \times_S U$ is separated (by the same lemma).
Since $j$ is separated we see in the same way that $R$ is separated.
Hence $R \to U$ is a separated morphism (by
Schemes, Lemma \ref{schemes-lemma-compose-after-separated}
again). Thus by
Spaces, Lemma \ref{spaces-lemma-representable-morphisms-spaces-property}
and the diagram above we conclude that $U \to X$ is separated.
\end{proof}
\begin{lemma}
\label{lemma-quasi-separated}
Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.
If there exists a quasi-separated scheme $U$ and a surjective
\'etale morphism $U \to X$ such that either of the projections
$U \times_X U \to U$ is quasi-compact, then $X$ is quasi-separated.
\end{lemma}
\begin{proof}
We may think of $X$ as an algebraic space over $\mathbf{Z}$.
Consider the cartesian diagram
$$
\xymatrix{
U \times_X U \ar[r] \ar[d]_j & X \ar[d]^\Delta \\
U \times U \ar[r] & X \times X
}
$$
Since $U$ is quasi-separated the projection $U \times U \to U$ is
quasi-separated (as a base change of a quasi-separated morphism
of schemes, see Schemes, Lemma \ref{schemes-lemma-separated-permanence}).
Hence the assumption in the lemma implies $j$ is quasi-compact by
Schemes, Lemma \ref{schemes-lemma-quasi-compact-permanence}.
By Spaces, Lemma \ref{spaces-lemma-representable-morphisms-spaces-property}
we see that $\Delta$ is quasi-compact as desired.
\end{proof}
\begin{lemma}
\label{lemma-quasi-separated-quasi-compact-pieces}
Let $S$ be a scheme.
Let $X$ be an algebraic space over $S$.
The following are equivalent
\begin{enumerate}
\item $X$ is Zariski locally quasi-separated over $S$,
\item $X$ is Zariski locally quasi-separated,
\item there exists a Zariski open covering $X = \bigcup X_i$
such that for each $i$ there exists an affine scheme
$U_i$ and a quasi-compact surjective \'etale
morphism $U_i \to X_i$, and
\item there exists a Zariski open covering $X = \bigcup X_i$
such that for each $i$ there exists an affine scheme
$U_i$ which maps into an affine open of $S$ and a quasi-compact
surjective \'etale morphism $U_i \to X_i$.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume $U_i \to X_i \subset X$ are as in (3). To prove (4)
choose for each $i$ a finite affine open covering $U_i =
U_{i1} \cup \ldots \cup U_{in_i}$ such that each $U_{ij}$ maps
into an affine open of $S$. The compositions
$U_{ij} \to U_i \to X_i$ are \'etale and quasi-compact (see
Spaces, Lemma
\ref{spaces-lemma-composition-representable-transformations-property}).
Let $X_{ij} \subset X_i$ be the open subspace corresponding to
the image of $|U_{ij}| \to |X_i|$, see
Lemma \ref{lemma-etale-image-open}.
Note that $U_{ij} \to X_{ij}$ is quasi-compact as $X_{ij} \subset X_i$
is a monomorphism and as $U_{ij} \to X$ is quasi-compact.
Then $X = \bigcup X_{ij}$ is a covering as in (4).
The implication (4) $\Rightarrow$ (3) is immediate.
\medskip\noindent
Assume (4). To show that $X$ is Zariski locally quasi-separated over $S$
it suffices to show that $X_i$ is quasi-separated over $S$.
Hence we may assume there exists an affine scheme $U$ mapping into
an affine open of $S$ and a quasi-compact surjective \'etale
morphism $U \to X$. Consider the fibre product square
$$
\xymatrix{
U \times_X U \ar[r] \ar[d] & U \times_S U \ar[d] \\
X \ar[r]^-{\Delta_{X/S}} & X \times_S X
}
$$
The right vertical arrow is surjective \'etale (see
Spaces, Lemma
\ref{spaces-lemma-product-representable-transformations-property})
and $U \times_S U$ is affine (as $U$ maps into an affine open of $S$, see
Schemes, Section \ref{schemes-section-fibre-products}),
and $U \times_X U$ is quasi-compact
because the projection $U \times_X U \to U$ is quasi-compact as a
base change of $U \to X$. It follows from
Spaces, Lemma \ref{spaces-lemma-representable-morphisms-spaces-property}
that $\Delta_{X/S}$ is quasi-compact as desired.
\medskip\noindent
Assume (1). To prove (3) there is an immediate reduction to the case
where $X$ is quasi-separated over $S$. By
Lemma \ref{lemma-union-of-quasi-compact}
we can find a Zariski open covering $X = \bigcup X_i$ such that each
$X_i$ maps into an affine open of $S$, and such that there exist affine
schemes $U_i$ and surjective \'etale morphisms $U_i \to X_i$.
Since $U_i \to S$ maps into an affine open of $S$ we see that
$U_i \times_S U_i$ is affine, see
Schemes, Section \ref{schemes-section-fibre-products}.
As $X$ is quasi-separated over $S$, the morphisms
$$
R_i = U_i \times_{X_i} U_i = U_i \times_X U_i
\longrightarrow
U_i \times_S U_i
$$
as base changes of $\Delta_{X/S}$ are quasi-compact. Hence we conclude
that $R_i$ is a quasi-compact scheme. This in turn implies that each
projection $R_i \to U_i$ is quasi-compact. Hence, applying
Spaces, Lemma \ref{spaces-lemma-representable-morphisms-spaces-property}
to the covering $U_i \to X_i$ and the morphism $U_i \to X_i$
we conclude that the morphisms $U_i \to X_i$ are quasi-compact as desired.
\medskip\noindent
At this point we see that (1), (3), and (4) are equivalent. Since (3) does
not refer to the base scheme we conclude that these are also equivalent
with (2).
\end{proof}
\section{Properties of Spaces defined by properties of schemes}
\label{section-types-properties}
\noindent
Any \'etale local property of schemes gives rise to a corresponding
property of algebraic spaces via the following lemma.
\begin{lemma}
\label{lemma-type-property}
Let $S$ be a scheme.
Let $X$ be an algebraic space over $S$.
Let $\mathcal{P}$ be a property of schemes which is local in the \'etale
topology, see
Descent, Definition \ref{descent-definition-property-local}.
The following are equivalent
\begin{enumerate}
\item for some scheme $U$ and surjective \'etale morphism $U \to X$
the scheme $U$ has property $\mathcal{P}$, and
\item for every scheme $U$ and every \'etale morphism $U \to X$
the scheme $U$ has property $\mathcal{P}$.
\end{enumerate}
If $X$ is representable this is equivalent to $\mathcal{P}(X)$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{definition}
\label{definition-type-property}
Let $\mathcal{P}$ be a property of schemes which is
local in the \'etale topology.
Let $S$ be a scheme.
Let $X$ be an algebraic space over $S$.
We say $X$ {\it has property $\mathcal{P}$}
if any of the equivalent conditions of
Lemma \ref{lemma-type-property}
hold.
\end{definition}
\begin{remark}
\label{remark-list-properties-local-etale-topology}
Here is a list of properties which are local for the \'etale topology
(keep in mind that the fpqc, fppf, syntomic, and smooth topologies are
stronger than the \'etale topology):
\begin{enumerate}
\item locally Noetherian, see
Descent, Lemma \ref{descent-lemma-Noetherian-local-fppf},
\item Jacobson, see
Descent, Lemma \ref{descent-lemma-Jacobson-local-fppf},
\item locally Noetherian and $(S_k)$, see
Descent, Lemma \ref{descent-lemma-Sk-local-syntomic},
\item Cohen-Macaulay, see
Descent, Lemma \ref{descent-lemma-CM-local-syntomic},
\item reduced, see
Descent, Lemma \ref{descent-lemma-reduced-local-smooth},
\item normal, see
Descent, Lemma \ref{descent-lemma-normal-local-smooth},
\item locally Noetherian and $(R_k)$, see
Descent, Lemma \ref{descent-lemma-Rk-local-smooth},
\item regular, see
Descent, Lemma \ref{descent-lemma-regular-local-smooth},
\item Nagata, see
Descent, Lemma \ref{descent-lemma-Nagata-local-smooth}.
\end{enumerate}
\end{remark}
\noindent
Any \'etale local property of germs of schemes gives rise to a corresponding
property of algebraic spaces. Here is the obligatory lemma.
\begin{lemma}
\label{lemma-local-source-target-at-point}
Let $\mathcal{P}$ be a property of germs of schemes which is \'etale local, see
Descent, Definition \ref{descent-definition-local-at-point}.
Let $S$ be a scheme.
Let $X$ be an algebraic space over $S$.
Let $x \in |X|$ be a point of $X$.
Consider \'etale morphisms $a : U \to X$ where $U$ is a scheme.
The following are equivalent
\begin{enumerate}
\item for any $U \to X$ as above and $u \in U$ with $a(u) = x$ we have
$\mathcal{P}(U, u)$, and
\item for some $U \to X$ as above and $u \in U$ with $a(u) = x$ we have
$\mathcal{P}(U, u)$.
\end{enumerate}
If $X$ is representable, then this is equivalent to $\mathcal{P}(X, x)$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{definition}
\label{definition-property-at-point}
Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.
Let $x \in |X|$. Let $\mathcal{P}$ be a property of germs of schemes which is
\'etale local.
We say $X$ {\it has property $\mathcal{P}$ at $x$} if any of the
equivalent conditions of
Lemma \ref{lemma-local-source-target-at-point}
hold.
\end{definition}
\section{Dimension at a point}
\label{section-dimension}
\noindent
We can use
Descent, Lemma \ref{descent-lemma-dimension-at-point-local}
to define the dimension of an algebraic
space $X$ at a point $x$. This will give us a different notion than the
topological one (i.e., the dimension of $|X|$ at $x$).
\begin{definition}
\label{definition-dimension-at-point}
Let $S$ be a scheme.
Let $X$ be an algebraic space over $S$.
Let $x \in |X|$ be a point of $X$.
We define the {\it dimension of $X$ at $x$} to be
the element $\dim_x(X) \in \{0, 1, 2, \ldots, \infty\}$
such that $\dim_x(X) = \dim_u(U)$ for any (equivalently some)
pair $(a : U \to X, u)$ consisting of an \'etale morphism $a : U \to X$
from a scheme to $X$ and a point $u \in U$ with $a(u) = x$.
See
Definition \ref{definition-property-at-point},
Lemma \ref{lemma-local-source-target-at-point}, and
Descent, Lemma \ref{descent-lemma-dimension-at-point-local}.
\end{definition}
\noindent
Warning: It is {\bf not} the case that $\dim_x(X) = \dim_x(|X|)$
in general. A counter example is the algebraic space $X$ of
Spaces, Example \ref{spaces-example-infinite-product}.
Namely, in this example we have $\dim_x(X) = 0$ and
$\dim_x(|X|) = 1$ (this holds for any $x \in |X|$).
In particular, it also means that the dimension of $X$ (as defined
below) is different from the dimension of $|X|$.
\begin{definition}
\label{definition-dimension}
Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.
The {\it dimension} $\dim(X)$ of $X$ is defined by the rule
$$
\dim(X) = \sup\nolimits_{x \in |X|} \dim_x(X)