The Stacks project

Lemma 28.5.13. Let $S$ be a Noetherian scheme. Let $T \subset S$ be an infinite dense subset. Then there exist a countable subset $E \subset T$ which is dense in $S$.

Proof. Let $T'$ be the set of points $s \in S$ such that $\overline{\{ s\} } \cap T$ contains a countable subset whose closure is $\overline{\{ s\} }$. Since a finite set is countable we have $T \subset T'$. For $s \in T'$ choose such a countable subset $E_ s \subset \overline{\{ s\} } \cap T$. Let $E' = \{ s_1, s_2, s_3, \ldots \} \subset T'$ be a countable subset. Then the closure of $E'$ in $S$ is the closure of the countable subset $\bigcup _ n E_{s_ n}$ of $T$. It follows that if $Z$ is an irreducible component of the closure of $E'$, then the generic point of $Z$ is in $T'$.

Denote $T'_0 \subset T'$ the subset of $t \in T'$ such that there is no nontrivial specialization $t' \leadsto t$ with $t' \in T'$ as in Lemma 28.5.12 whose results we will use without further mention. If $T'_0$ is infinite, then we choose a countable subset $E' \subset T'_0$. By the argument in the first paragraph, the generic points of the irreducible components of the closure of $E'$ are in $T'$. However, since one of these points specializes to infinitely many distinct elements of $E' \subset T'_0$ this is a contradiction. Thus $T'_0$ is finite, say $T'_0 = \{ s_1, \ldots , s_ m\} $. Then it follows that $S$, which is the closure of $T$, is contained in the closure of $\{ s_1, \ldots , s_ m\} $, which in turn is contained in the closure of the countable subset $E_{s_1} \cup \ldots \cup E_{s_ m} \subset T$ as desired. $\square$


Comments (0)

There are also:

  • 2 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Your email address will not be published.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 0G2F. Beware of the difference between the letter 'O' and the digit '0'.