Definition 28.11.1. Let $S$ be a scheme. We say $S$ is *catenary* if the underlying topological space of $S$ is catenary.

## 28.11 Catenary schemes

Recall that a topological space $X$ is called *catenary* if for every pair of irreducible closed subsets $T \subset T'$ there exist a maximal chain of irreducible closed subsets

and every such chain has the same length. See Topology, Definition 5.11.4.

Recall that a ring $A$ is called *catenary* if for any pair of prime ideals $\mathfrak p \subset \mathfrak q$ there exists a maximal chain of primes

and all of these have the same length. See Algebra, Definition 10.105.1.

Lemma 28.11.2. Let $S$ be a scheme. The following are equivalent

$S$ is catenary,

there exists an open covering of $S$ all of whose members are catenary schemes,

for every affine open $\mathop{\mathrm{Spec}}(R) = U \subset S$ the ring $R$ is catenary, and

there exists an affine open covering $S = \bigcup U_ i$ such that each $U_ i$ is the spectrum of a catenary ring.

Moreover, in this case any locally closed subscheme of $S$ is catenary as well.

**Proof.**
Combine Topology, Lemma 5.11.5, and Algebra, Lemma 10.105.2.
$\square$

Lemma 28.11.3. Let $S$ be a locally Noetherian scheme. The following are equivalent:

$S$ is catenary, and

locally in the Zariski topology there exists a dimension function on $S$ (see Topology, Definition 5.20.1).

**Proof.**
This follows from Topology, Lemmas 5.11.5, 5.20.2, and 5.20.4, Schemes, Lemma 26.11.1 and finally Lemma 28.5.5.
$\square$

It turns out that a scheme is catenary if and only if its local rings are catenary.

Lemma 28.11.4. Let $X$ be a scheme. The following are equivalent

$X$ is catenary, and

for any $x \in X$ the local ring $\mathcal{O}_{X, x}$ is catenary.

**Proof.**
Assume $X$ is catenary. Let $x \in X$. By Lemma 28.11.2 we may replace $X$ by an affine open neighbourhood of $x$, and then $\Gamma (X, \mathcal{O}_ X)$ is a catenary ring. By Algebra, Lemma 10.105.4 any localization of a catenary ring is catenary. Whence $\mathcal{O}_{X, x}$ is catenary.

Conversely assume all local rings of $X$ are catenary. Let $Y \subset Y'$ be an inclusion of irreducible closed subsets of $X$. Let $\xi \in Y$ be the generic point. Let $\mathfrak p \subset \mathcal{O}_{X, \xi }$ be the prime corresponding to the generic point of $Y'$, see Schemes, Lemma 26.13.2. By that same lemma the irreducible closed subsets of $X$ in between $Y$ and $Y'$ correspond to primes $\mathfrak q \subset \mathcal{O}_{X, \xi }$ with $\mathfrak p \subset \mathfrak q \subset \mathfrak m_{\xi }$. Hence we see all maximal chains of these are finite and have the same length as $\mathcal{O}_{X, \xi }$ is a catenary ring. $\square$

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