The Stacks project

Lemma 5.20.3. Let $X$ be a topological space. Let $\delta $, $\delta '$ be two dimension functions on $X$. If $X$ is locally Noetherian and sober then $\delta - \delta '$ is locally constant on $X$.

Proof. Let $x \in X$ be a point. We will show that $\delta - \delta '$ is constant in a neighbourhood of $x$. We may replace $X$ by an open neighbourhood of $x$ in $X$ which is Noetherian. Hence we may assume $X$ is Noetherian and sober. Let $Z_1, \ldots , Z_ r$ be the irreducible components of $X$ passing through $x$. (There are finitely many as $X$ is Noetherian, see Lemma 5.9.2.) Let $\xi _ i \in Z_ i$ be the generic point. Note $Z_1 \cup \ldots \cup Z_ r$ is a neighbourhood of $x$ in $X$ (not necessarily closed). We claim that $\delta - \delta '$ is constant on $Z_1 \cup \ldots \cup Z_ r$. Namely, if $y \in Z_ i$, then

\[ \delta (x) - \delta (y) = \delta (x) - \delta (\xi _ i) + \delta (\xi _ i) - \delta (y) = - \text{codim}(\overline{\{ x\} }, Z_ i) + \text{codim}(\overline{\{ y\} }, Z_ i) \]

by Lemma 5.20.2. Similarly for $\delta '$. Whence the result. $\square$

Comments (0)

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Your email address will not be published.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 02IB. Beware of the difference between the letter 'O' and the digit '0'.