# The Stacks Project

## Tag 02IM

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

Proof. Since $\mathop{\rm Spec}(R)$ is a Noetherian scheme, hence $S$ 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 $\mathop{\rm Spec}(R/\mathfrak p)$. This is impossible by Algebra, Lemma 10.60.1. $\square$

The code snippet corresponding to this tag is a part of the file properties.tex and is located in lines 717–722 (see updates for more information).

\begin{lemma}
\label{lemma-complement-closed-point-Jacobson}
Let $R$ be a Noetherian local ring with maximal ideal $\mathfrak m$.
In this case the scheme $S = \Spec(R) \setminus \{\mathfrak m\}$
is Jacobson.
\end{lemma}

\begin{proof}
Since $\Spec(R)$ is a Noetherian scheme, hence
$S$ is a Noetherian scheme (Lemma \ref{lemma-locally-closed-in-Noetherian}).
Hence $S$ is a sober, Noetherian topological space (use
Schemes, Lemma \ref{schemes-lemma-scheme-sober}).
Assume $S$ is not Jacobson to
Topology, Lemma \ref{topology-lemma-non-jacobson-Noetherian-characterize}
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
$\Spec(R/\mathfrak p)$. This is impossible by Algebra,
Lemma \ref{algebra-lemma-Noetherian-local-domain-dim-2-infinite-opens}.
\end{proof}

Comment #2434 by Matthew Emerton on February 18, 2017 a 1:38 am UTC

A couple of minor typos: In the first sentence of the proof, having both Since and hence is redundant. In point (2) of the second-to-last sentence, I think $\{\mathfrak p\}$ should actually be $\{\mathfrak q\}.$

Comment #2435 by Matthew Emerton on February 18, 2017 a 3:46 am UTC

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

Comment #2436 by Matthew Emerton on February 18, 2017 a 3:47 am UTC

(Sorry for the near duplicates; the first comment demonstrates a mistake on my part, rather than yours!)

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