Definition 66.25.1. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. We say $X$ is *catenary* if $|X|$ is catenary (Topology, Definition 5.11.4).

## 66.25 Catenary algebraic spaces

This section extends the material in Properties, Section 28.11 and Morphisms, Section 29.17 to algebraic spaces.

If $X$ is representable, then this is equivalent to the corresponding notion for the scheme representing $X$.

Lemma 66.25.2. Let $S$ be a locally Noetherian and universally catenary scheme. Let $\delta : S \to \mathbf{Z}$ be a dimension function. Let $X$ be a decent algebraic space over $S$ such that the structure morphism $X \to S$ is locally of finite type. Let $\delta _ X : |X| \to \mathbf{Z}$ be the map sending $x$ to $\delta (f(x))$ plus the transcendence degree of $x/f(x)$. Then $\delta _ X$ is a dimension function on $|X|$.

**Proof.**
Let $\varphi : U \to X$ be a surjective étale morphism where $U$ is a scheme. Then the similarly defined function $\delta _ U$ is a dimension function on $U$ by Morphisms, Lemma 29.51.3. On the other hand, by the definition of relative transcendence degree in (Morphisms of Spaces, Definition 65.33.1) we see that $\delta _ U(u) = \delta _ X(\varphi (u))$.

Let $x \leadsto x'$ be a specialization of points in $|X|$. by Lemma 66.12.2 we can find a specialization $u \leadsto u'$ of points of $U$ with $\varphi (u) = x$ and $\varphi (u') = x'$. Moreover, we see that $x = x'$ if and only if $u = u'$, see Lemma 66.12.1. Thus the fact that $\delta _ U$ is a dimension function implies that $\delta _ X$ is a dimension function, see Topology, Definition 5.20.1. $\square$

Lemma 66.25.3. Let $S$ be a locally Noetherian and universally catenary scheme. Let $X$ be an algebraic space over $S$ such that $X$ is decent and such that the structure morphism $X \to S$ is locally of finite type. Then $X$ is catenary.

**Proof.**
The question is local on $S$ (use Topology, Lemma 5.11.5). Thus we may assume that $S$ has a dimension function, see Topology, Lemma 5.20.4. Then we conclude that $|X|$ has a dimension function by Lemma 66.25.2. Since $|X|$ is sober (Proposition 66.12.4) we conclude that $|X|$ is catenary by Topology, Lemma 5.20.2.
$\square$

By Lemma 66.25.3 the following definition is compatible with the already existing notion for representable algebraic spaces.

Definition 66.25.4. Let $S$ be a scheme. Let $X$ be a decent and locally Noetherian algebraic space over $S$. We say $X$ is *universally catenary* if for every morphism $Y \to X$ of algebraic spaces which is locally of finite type and with $Y$ decent, the algebraic space $Y$ is catenary.

If $X$ is an algebraic space, then the condition “$X$ is decent and locally Noetherian” is equivalent to “$X$ is quasi-separated and locally Noetherian”. This is Lemma 66.14.1. Thus another way to understand the definition above is that $X$ is universally catenary if and only if $Y$ is catenary for all morphisms $Y \to X$ which are quasi-separated and locally of finite type.

Lemma 66.25.5. Let $S$ be a scheme. Let $X$ be a decent, locally Noetherian, and universally catenary algebraic space over $S$. Then any decent algebraic space locally of finite type over $X$ is universally catenary.

**Proof.**
This is formal from the definitions and the fact that compositions of morphisms locally of finite type are locally of finite type (Morphisms of Spaces, Lemma 65.23.2).
$\square$

Lemma 66.25.6. Let $S$ be a scheme. Let $f : Y \to X$ be a surjective finite morphism of decent and locally Noetherian algebraic spaces. Let $\delta : |X| \to \mathbf{Z}$ be a function. If $\delta \circ |f|$ is a dimension function, then $\delta $ is a dimension function.

**Proof.**
Let $x \mapsto x'$, $x \not= x'$ be a specialization in $|X|$. Choose $y \in |Y|$ with $|f|(y) = x$. Since $|f|$ is closed (Morphisms of Spaces, Lemma 65.45.9) we find a specialization $y \leadsto y'$ with $|f|(y') = x'$. Thus we conclude that $\delta (x) = \delta (|f|(y)) > \delta (|f|(y')) = \delta (x')$ (see Topology, Definition 5.20.1). If $x \leadsto x'$ is an immediate specialization, then $y \leadsto y'$ is an immediate specialization too: namely if $y \leadsto y'' \leadsto y'$, then $|f|(y'')$ must be either $x$ or $x'$ and there are no nontrivial specializations between points of fibres of $|f|$ by Lemma 66.18.10.
$\square$

The discussion will be continued in More on Morphisms of Spaces, Section 74.32.

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