The Stacks project

Lemma 28.5.12. Let $S$ be a Noetherian scheme. Let $T \subset S$ be a subset. Let $T_0 \subset T$ be the set of $t \in T$ such that there is no nontrivial specialization $t' \leadsto t$ with $t' \in T'$. Then (a) there are no specializations among the points of $T_0$, (b) every point of $T$ is a specialization of a point of $T_0$, and (c) the closures of $T$ and $T_0$ are the same.

Proof. Recall that $\dim (\mathcal{O}_{S, s}) < \infty $ for any $s \in S$, see Algebra, Proposition 10.60.9. Let $t \in T$. If $t' \leadsto t$, then by dimension theory $\dim (\mathcal{O}_{S, t'}) \leq \dim (\mathcal{O}_{S, t})$ with equality if and only if $t' = t$. Thus if we pick $t' \leadsto t$ with $\dim (\mathcal{O}_{T, t'})$ minimal, then $t' \in T_0$. In other words, every $t \in T$ is the specialization of an element of $T_0$. $\square$


Comments (0)

There are also:

  • 4 comment(s) on Section 28.5: Noetherian schemes

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Your email address will not be published.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 0G2R. Beware of the difference between the letter 'O' and the digit '0'.