Lemma 27.6.4. Let $R$ be a Noetherian local ring with maximal ideal $\mathfrak m$. In this case the scheme $S = \mathop{\mathrm{Spec}}(R) \setminus \{ \mathfrak m\}$ is Jacobson.

Proof. Since $\mathop{\mathrm{Spec}}(R)$ is a Noetherian scheme, $S$ is a Noetherian scheme (Lemma 27.5.6). Hence $S$ is a sober, Noetherian topological space (use Schemes, Lemma 25.11.1). Assume $S$ is not Jacobson to get a contradiction. By Topology, Lemma 5.18.3 there exists some non-closed point $\xi \in S$ such that $\{ \xi \}$ is locally closed. This corresponds to a prime $\mathfrak p \subset R$ such that (1) there exists a prime $\mathfrak q$, $\mathfrak p \subset \mathfrak q \subset \mathfrak m$ with both inclusions strict, and (2) $\{ \mathfrak p\}$ is open in $\mathop{\mathrm{Spec}}(R/\mathfrak p)$. This is impossible by Algebra, Lemma 10.60.1. $\square$

Comment #2435 by Matthew Emerton on

A minor comment: In the first sentence of the proof, having both Since and hence is redundant.

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