The Stacks project

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 $x \in Z_0$. For every open neighbourhood $U$ of $x$ we get a chain of irreducible closed subsets

\[ Z_0 \cap U \subset Z_1 \cap U \subset \ldots \subset Z_ n \cap U \]

in $U$. Namely, the sets $U \cap Z_ i$ are irreducible closed in $U$ and the inclusions are strict (details omitted; hint: the closure of $U \cap Z_ i$ is $Z_ i$). 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.


Comments (3)

Comment #5026 by James A. Myer on

Is Example 5.10.3. meant to say "The Krull dimension of the usual Euclidean space is ."?

Comment #5114 by on

@James A. Myer: No. Every closed set with elements is reducible.

Comment #5455 by James A. Myer on

Oh yeah, sorry: I thought "the usual Euclidean space " meant the scheme , in which case there is a chain of irreducible closed subsets given by of length , but of course it makes more sense to interpret "the usual Euclidean space " as equipped with the Euclidean topology.


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