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.

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0054. Beware of the difference between the letter 'O' and the digit '0'.



View Section 5.10 as pdf