## 5.11 Codimension and catenary spaces

We only define the codimension of irreducible closed subsets.

Definition 5.11.1. Let $X$ be a topological space. Let $Y \subset X$ be an irreducible closed subset. The codimension of $Y$ in $X$ is the supremum of the lengths $e$ of chains

$Y = Y_0 \subset Y_1 \subset \ldots \subset Y_ e \subset X$

of irreducible closed subsets in $X$ starting with $Y$. We will denote this $\text{codim}(Y, X)$.

The codimension is an element of $\{ 0, 1, 2, \ldots \} \cup \{ \infty \}$. If $\text{codim}(Y, X) < \infty$, then every chain can be extended to a maximal chain (but these do not all have to have the same length).

Lemma 5.11.2. Let $X$ be a topological space. Let $Y \subset X$ be an irreducible closed subset. Let $U \subset X$ be an open subset such that $Y \cap U$ is nonempty. Then

$\text{codim}(Y, X) = \text{codim}(Y \cap U, U)$

Proof. The rule $T \mapsto \overline{T}$ defines a bijective inclusion preserving map between the closed irreducible subsets of $U$ and the closed irreducible subsets of $X$ which meet $U$. Using this the lemma easily follows. Details omitted. $\square$

Example 5.11.3. Let $X = [0, 1]$ be the unit interval with the following topology: The sets $[0, 1]$, $(1 - 1/n, 1]$ for $n \in \mathbf{N}$, and $\emptyset$ are open. So the closed sets are $\emptyset$, $\{ 0\}$, $[0, 1 - 1/n]$ for $n > 1$ and $[0, 1]$. This is clearly a Noetherian topological space. But the irreducible closed subset $Y = \{ 0\}$ has infinite codimension $\text{codim}(Y, X) = \infty$. To see this we just remark that all the closed sets $[0, 1 - 1/n]$ are irreducible.

Definition 5.11.4. Let $X$ be a topological space. We say $X$ is catenary if for every pair of irreducible closed subsets $T \subset T'$ we have $\text{codim}(T, T') < \infty$ and every maximal chain of irreducible closed subsets

$T = T_0 \subset T_1 \subset \ldots \subset T_ e = T'$

has the same length (equal to the codimension).

Lemma 5.11.5. Let $X$ be a topological space. The following are equivalent:

1. $X$ is catenary,

2. $X$ has an open covering by catenary spaces.

Moreover, in this case any locally closed subspace of $X$ is catenary.

Proof. Suppose that $X$ is catenary and that $U \subset X$ is an open subset. The rule $T \mapsto \overline{T}$ defines a bijective inclusion preserving map between the closed irreducible subsets of $U$ and the closed irreducible subsets of $X$ which meet $U$. Using this the lemma easily follows. Details omitted. $\square$

Lemma 5.11.6. Let $X$ be a topological space. The following are equivalent:

1. $X$ is catenary, and

2. for every pair of irreducible closed subsets $Y \subset Y'$ we have $\text{codim}(Y, Y') < \infty$ and for every triple $Y \subset Y' \subset Y''$ of irreducible closed subsets we have

$\text{codim}(Y, Y'') = \text{codim}(Y, Y') + \text{codim}(Y', Y'').$

Proof. Omitted. $\square$

Comment #5067 by Mario Kummer on

I suggest that the definition of codimension is extended to arbitrary closed subsets $Y\subset X$ as the infimum of the codimension of all closed irreducible subsets of $Y$, as this is a quite common notion.

Comment #5283 by on

@#5067: No because this often leads to problems later on. For example, another reasonable choice would be to use the supremum (after all in the definition the supremum is used too).

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