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 111.18.5. 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.

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