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Tag 01P1

27.6. Jacobson schemes

Recall that a space is said to be Jacobson if the closed points are dense in every closed subset, see Topology, Section 5.18.

Definition 27.6.1. A scheme $S$ is said to be Jacobson if its underlying topological space is Jacobson.

Recall that a ring $R$ is Jacobson if every radical ideal of $R$ is the intersection of maximal ideals, see Algebra, Definition 10.34.1.

Lemma 27.6.2. An affine scheme $\mathop{\rm Spec}(R)$ is Jacobson if and only if the ring $R$ is Jacobson.

Proof. This is Algebra, Lemma 10.34.4. $\square$

Here is the standard result characterizing Jacobson schemes. Intuitively it claims that Jacobson $\Leftrightarrow$ locally Jacobson.

Lemma 27.6.3. Let $X$ be a scheme. The following are equivalent:

  1. The scheme $X$ is Jacobson.
  2. The scheme $X$ is ''locally Jacobson'' in the sense of Definition 27.4.2.
  3. For every affine open $U \subset X$ the ring $\mathcal{O}_X(U)$ is Jacobson.
  4. There exists an affine open covering $X = \bigcup U_i$ such that each $\mathcal{O}_X(U_i)$ is Jacobson.
  5. There exists an open covering $X = \bigcup X_j$ such that each open subscheme $X_j$ is Jacobson.

Moreover, if $X$ is Jacobson then every open subscheme is Jacobson.

Proof. The final assertion of the lemma holds by Topology, Lemma 5.18.5. The equivalence of (5) and (1) is Topology, Lemma 5.18.4. Hence, using Lemma 27.6.2, we see that (1) $\Leftrightarrow$ (2). To finish proving the lemma it suffices to show that ''Jacobson'' is a local property of rings, see Lemma 27.4.3. Any localization of a Jacobson ring at an element is Jacobson, see Algebra, Lemma 10.34.14. Suppose $R$ is a ring, $f_1, \ldots, f_n \in R$ generate the unit ideal and each $R_{f_i}$ is Jacobson. Then we see that $\mathop{\rm Spec}(R) = \bigcup D(f_i)$ is a union of open subsets which are all Jacobson, and hence $\mathop{\rm Spec}(R)$ is Jacobson by Topology, Lemma 5.18.4 again. This proves the second property of Definition 27.4.1. $\square$

Many schemes used commonly in algebraic geometry are Jacobson, see Morphisms, Lemma 28.15.10. We mention here the following interesting case.

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 642–746 (see updates for more information).

    \section{Jacobson schemes}
    \label{section-jacobson}
    
    \noindent
    Recall that a space is said to be {\it Jacobson} if the closed points are
    dense in every closed subset, see
    Topology, Section \ref{topology-section-space-jacobson}.
    
    \begin{definition}
    \label{definition-jacobson}
    A scheme $S$ is said to be {\it Jacobson} if its underlying topological
    space is Jacobson.
    \end{definition}
    
    \noindent
    Recall that a ring $R$ is Jacobson if every radical ideal of $R$
    is the intersection of maximal ideals, see
    Algebra, Definition \ref{algebra-definition-ring-jacobson}.
    
    \begin{lemma}
    \label{lemma-affine-jacobson}
    An affine scheme $\Spec(R)$ is Jacobson if and only if
    the ring $R$ is Jacobson.
    \end{lemma}
    
    \begin{proof}
    This is Algebra, Lemma \ref{algebra-lemma-jacobson}.
    \end{proof}
    
    \noindent
    Here is the standard result characterizing Jacobson schemes.
    Intuitively it claims that Jacobson $\Leftrightarrow$ locally Jacobson.
    
    \begin{lemma}
    \label{lemma-locally-jacobson}
    Let $X$ be a scheme. The following are equivalent:
    \begin{enumerate}
    \item The scheme $X$ is Jacobson.
    \item The scheme $X$ is ``locally Jacobson'' in the sense of
    Definition \ref{definition-locally-P}.
    \item For every affine open $U \subset X$ the ring $\mathcal{O}_X(U)$
    is Jacobson.
    \item There exists an affine open covering $X = \bigcup U_i$ such that
    each $\mathcal{O}_X(U_i)$ is Jacobson.
    \item There exists an open covering $X = \bigcup X_j$
    such that each open subscheme $X_j$ is Jacobson.
    \end{enumerate}
    Moreover, if $X$ is Jacobson then every open subscheme
    is Jacobson.
    \end{lemma}
    
    \begin{proof}
    The final assertion of the lemma holds by
    Topology, Lemma \ref{topology-lemma-jacobson-inherited}.
    The equivalence of (5) and (1) is
    Topology, Lemma \ref{topology-lemma-jacobson-local}.
    Hence, using Lemma \ref{lemma-affine-jacobson},
    we see that (1) $\Leftrightarrow$ (2).
    To finish proving the lemma it suffices to show that ``Jacobson''
    is a local property of rings, see Lemma \ref{lemma-locally-P}.
    Any localization of a Jacobson ring at an element is Jacobson, see
    Algebra, Lemma \ref{algebra-lemma-Jacobson-invert-element}.
    Suppose $R$ is a ring, $f_1, \ldots, f_n \in R$ generate the unit
    ideal and each $R_{f_i}$ is Jacobson. Then we see that
    $\Spec(R) = \bigcup D(f_i)$ is a union of open subsets
    which are all Jacobson, and hence $\Spec(R)$ is Jacobson
    by Topology, Lemma \ref{topology-lemma-jacobson-local} again.
    This proves the second property of Definition \ref{definition-property-local}.
    \end{proof}
    
    \noindent
    Many schemes used commonly in algebraic geometry are Jacobson, see
    Morphisms, Lemma \ref{morphisms-lemma-ubiquity-Jacobson-schemes}.
    We mention here the following interesting case.
    
    \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}

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