Section 109.18 (027N): Catenary rings

Definition 109.18.1. 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 109.18.2. Show that the notion of catenary defined in Algebra, Definition 10.104.1 agrees with the notion of Definition 109.18.1 for Noetherian rings.

Exercise 109.18.3. Show that a Noetherian local domain of dimension $2$ is catenary.

Exercise 109.18.4. Let $k$ be a field. Show that a finite type $k$-algebra is catenary.

Exercise 109.18.5. 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 succintly explain why there cannot be a dimension function.

The Stacks project

109.18 Catenary rings

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