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Tag 02IM

Chapter 27: Properties of Schemes > Section 27.6: Jacobson schemes

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, $S$ 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,
    $S$ 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
    get a contradiction. By
    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}

    Comments (2)

    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 #2479 by Johan (site) on April 13, 2017 a 10:16 pm UTC

    OK, thanks, fixed here.

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