Section 111.18 (027N): Catenary rings

Definition 111.18.1. A Noetherian ring $A$ is said to be catenary if for any triple of prime ideals ${\mathfrak p}_1 \subset {\mathfrak p}_2 \subset {\mathfrak p}_3$ we have

\[ ht({\mathfrak p}_3 / {\mathfrak p}_1) = ht({\mathfrak p}_3/{\mathfrak p}_2) + ht({\mathfrak p}_2/{\mathfrak p}_1). \]

Here $ht(\mathfrak p/\mathfrak q)$ means the height of $\mathfrak p/\mathfrak q$ in the ring $A/\mathfrak q$. In a formula

\[ ht(\mathfrak p/\mathfrak q) = \dim (A_\mathfrak p/\mathfrak qA_\mathfrak p) = \dim ((A/\mathfrak q)_\mathfrak p) = \dim ((A/\mathfrak q)_{\mathfrak p/\mathfrak q}) \]

A topological space $X$ is catenary, if given $T \subset T' \subset X$ with $T$ and $T'$ closed and irreducible, then there exists a maximal chain of irreducible closed subsets

\[ T = T_0 \subset T_1 \subset \ldots \subset T_ n = T' \]

and every such chain has the same (finite) length.

Exercise 111.18.2. Show that the notion of catenary defined in Algebra, Definition 10.105.1 agrees with the notion of Definition 111.18.1 for Noetherian rings.

Exercise 111.18.3. Show that a Noetherian local domain of dimension $2$ is catenary.

Exercise 111.18.4. Let $k$ be a field. Show that a finite type $k$-algebra is catenary.

Exercise 111.18.5. Give an example of a finite, sober, catenary topological space $X$ which does not have a dimension function $\delta : X \to \mathbf{Z}$. Here $\delta : X \to \mathbf{Z}$ is a dimension function if for $x, y \in X$ we have

$x \leadsto y$ and $x \not= y$ implies $\delta (x) > \delta (y)$,
$x \leadsto y$ and $\delta (x) \geq \delta (y) + 2$ implies there exists a $z \in X$ with $x \leadsto z \leadsto y$ and $\delta (x) > \delta (z) > \delta (y)$.

Describe your space clearly and succinctly explain why there cannot be a dimension function.

The Stacks project

111.18 Catenary rings

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