5.10 Krull dimension

Definition 5.10.1. Let $X$ be a topological space.

1. A chain of irreducible closed subsets of $X$ is a sequence $Z_0 \subset Z_1 \subset \ldots \subset Z_ n \subset X$ with $Z_ i$ closed irreducible and $Z_ i \not= Z_{i + 1}$ for $i = 0, \ldots , n - 1$.

2. The length of a chain $Z_0 \subset Z_1 \subset \ldots \subset Z_ n \subset X$ of irreducible closed subsets of $X$ is the integer $n$.

3. The dimension or more precisely the Krull dimension $\dim (X)$ of $X$ is the element of $\{ -\infty , 0, 1, 2, 3, \ldots , \infty \}$ defined by the formula:

$\dim (X) = \sup \{ \text{lengths of chains of irreducible closed subsets}\}$

Thus $\dim (X) = -\infty$ if and only if $X$ is the empty space.

4. Let $x \in X$. The Krull dimension of $X$ at $x$ is defined as

$\dim _ x(X) = \min \{ \dim (U), x\in U\subset X\text{ open}\}$

the minimum of $\dim (U)$ where $U$ runs over the open neighbourhoods of $x$ in $X$.

Note that if $U' \subset U \subset X$ are open then $\dim (U') \leq \dim (U)$. Hence if $\dim _ x(X) = d$ then $x$ has a fundamental system of open neighbourhoods $U$ with $\dim (U) = \dim _ x(X)$.

Lemma 5.10.2. Let $X$ be a topological space. Then $\dim (X) = \sup \dim _ x(X)$ where the supremum runs over the points $x$ of $X$.

Proof. It is clear that $\dim (X) \geq \dim _ x(X)$ for all $x \in X$ (see discussion following Definition 5.10.1). Thus an inequality in one direction. For the converse, let $n \geq 0$ and suppose that $\dim (X) \geq n$. Then we can find a chain of irreducible closed subsets $Z_0 \subset Z_1 \subset \ldots \subset Z_ n \subset X$. Pick $x \in Z_0$. Then we see that every open neighbourhood $U$ of $x$ has a chain of irreducible closed subsets $Z_0 \cap U \subset Z_1 \cap U \subset \ldots Z_ n \cap U \subset U$. In this way we see that $\dim _ x(X) \geq n$ which proves the other inequality. $\square$

Example 5.10.3. The Krull dimension of the usual Euclidean space $\mathbf{R}^ n$ is $0$.

Example 5.10.4. Let $X = \{ s, \eta \}$ with open sets given by $\{ \emptyset , \{ \eta \} , \{ s, \eta \} \}$. In this case a maximal chain of irreducible closed subsets is $\{ s\} \subset \{ s, \eta \}$. Hence $\dim (X) = 1$. It is easy to generalize this example to get a $(n + 1)$-element topological space of Krull dimension $n$.

Definition 5.10.5. Let $X$ be a topological space. We say that $X$ is equidimensional if every irreducible component of $X$ has the same dimension.

Comment #5026 by James A. Myer on

Is Example 5.10.3. meant to say "The Krull dimension of the usual Euclidean space $\mathbb{R}^n$ is $n$."?

Comment #5114 by on

@James A. Myer: No. Every closed set with $\geq 2$ elements is reducible.

Comment #5455 by James A. Myer on

Oh yeah, sorry: I thought "the usual Euclidean space $\mathbb{R}^n$" meant the scheme $\mathbb{A}^n_{\mathbb{R}}=\text{Spec}(\mathbb{R}[x_1,...,x_n])$, in which case there is a chain of irreducible closed subsets given by $Z_0=\mathbb{V}((x_1,...,x_n))\subset Z_1=\mathbb{V}((x_1,...,x_{n-1}))\subset...\subset Z_{n-1}=\mathbb{V}((x_1))\subset Z_n=\mathbb{V}((0))\subset\mathbb{A}^n_{\mathbb{R}}$ of length $n$, but of course it makes more sense to interpret "the usual Euclidean space $\mathbb{R}^n$" as $\mathbb{R}^n$ equipped with the Euclidean topology.

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