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 10.60.8. 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 $T_0$. $\square$

There are also:

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

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).