66.3 Separation axioms
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 $\mathop{\mathrm{Spec}}(\mathbf{Z})$. Here is the precise formulation.
Definition 66.3.1. (Compare Spaces, Definition 65.13.2.) Consider a big fppf site $\mathit{Sch}_{fppf} = (\mathit{Sch}/\mathop{\mathrm{Spec}}(\mathbf{Z}))_{fppf}$. Let $X$ be an algebraic space over $\mathop{\mathrm{Spec}}(\mathbf{Z})$. Let $\Delta : X \to X \times X$ be the diagonal morphism.
We say $X$ is separated if $\Delta $ is a closed immersion.
We say $X$ is locally separated1 if $\Delta $ is an immersion.
We say $X$ is quasi-separated if $\Delta $ is quasi-compact.
We say $X$ is Zariski locally quasi-separated2 if there exists a Zariski covering $X = \bigcup _{i \in I} X_ i$ (see Spaces, Definition 65.12.5) such that each $X_ i$ is quasi-separated.
Let $S$ is a scheme contained in $\mathit{Sch}_{fppf}$, and let $X$ be an algebraic space over $S$. Then we say $X$ is separated, locally separated, quasi-separated, or Zariski locally quasi-separated if $X$ viewed as an algebraic space over $\mathop{\mathrm{Spec}}(\mathbf{Z})$ (see Spaces, Definition 65.16.2) has the corresponding property.
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 67.4. We will see in Lemma 66.6.6 that being Zariski locally separated is independent of the base scheme (hence equivalent to the absolute notion).
Lemma 66.3.2. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. We have the following implications among the separation axioms of Definition 66.3.1:
separated implies all the others,
quasi-separated implies Zariski locally quasi-separated.
Proof.
Omitted.
$\square$
Lemma 66.3.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. The following are equivalent
$X$ is a quasi-separated algebraic space,
for $U \to X$, $V \to X$ with $U$, $V$ quasi-compact schemes the fibre product $U \times _ X V$ is quasi-compact,
for $U \to X$, $V \to X$ with $U$, $V$ affine the fibre product $U \times _ X V$ is quasi-compact.
Proof.
Using Spaces, Lemma 65.16.3 we see that we may assume $S = \mathop{\mathrm{Spec}}(\mathbf{Z})$. Since $U \times _ X V = X \times _{X \times X} (U \times V)$ and since $U \times V$ is quasi-compact if $U$ and $V$ are so, we see that (1) implies (2). It is clear that (2) implies (3). Assume (3). Choose a scheme $W$ and a surjective étale morphism $W \to X$. Then $W \times W \to X \times X$ is surjective étale. Hence it suffices to show that
\[ j : W \times _ X W = X \times _{(X \times X)} (W \times W) \to W \times W \]
is quasi-compact, see Spaces, Lemma 65.5.6. If $U \subset W$ and $V \subset W$ are affine opens, then $j^{-1}(U \times V) = U \times _ X V$ is quasi-compact by assumption. Since the affine opens $U \times V$ form an affine open covering of $W \times W$ (Schemes, Lemma 26.17.4) we conclude by Schemes, Lemma 26.19.2.
$\square$
Lemma 66.3.4. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. The following are equivalent
$X$ is a separated algebraic space,
for $U \to X$, $V \to X$ with $U$, $V$ affine the fibre product $U \times _ X V$ is affine and
\[ \mathcal{O}(U) \otimes _\mathbf {Z} \mathcal{O}(V) \longrightarrow \mathcal{O}(U \times _ X V) \]
is surjective.
Proof.
Using Spaces, Lemma 65.16.3 we see that we may assume $S = \mathop{\mathrm{Spec}}(\mathbf{Z})$. Since $U \times _ X V = X \times _{X \times X} (U \times V)$ and since $U \times V$ is affine if $U$ and $V$ are so, we see that (1) implies (2). Assume (2). Choose a scheme $W$ and a surjective étale morphism $W \to X$. Then $W \times W \to X \times X$ is surjective étale. Hence it suffices to show that
\[ j : W \times _ X W = X \times _{(X \times X)} (W \times W) \to W \times W \]
is a closed immersion, see Spaces, Lemma 65.5.6. If $U \subset W$ and $V \subset W$ are affine opens, then $j^{-1}(U \times V) = U \times _ X V$ is affine by assumption and the map $U \times _ X V \to U \times V$ is a closed immersion because the corresponding ring map is surjective. Since the affine opens $U \times V$ form an affine open covering of $W \times W$ (Schemes, Lemma 26.17.4) we conclude by Morphisms, Lemma 29.2.1.
$\square$
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