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24.6. Jacobson schemes

Recall that a space is said to be Jacobson if the closed points are dense in every closed subset, see Topology, Section 5.13.

Definition 24.6.1. 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 9.32.1.

Lemma 24.6.2. An affine scheme $\mathop{\rm Spec}(R)$ is Jacobson if and only if the ring $R$ is Jacobson.

Proof. This is Algebra, Lemma 9.32.4. $\square$

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

Lemma 24.6.3. 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 24.4.2.
  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 5.13.5. The equivalence of (5) and (1) is Topology, Lemma 5.13.4. Hence, using Lemma 24.6.2, 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 24.4.3. Any localization of a Jacobson ring at an element is Jacobson, see Algebra, Lemma 9.32.14. 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 5.13.4 again. This proves the second property of Definition 24.4.1. $\square$

Many schemes used commonly in algebraic geometry are Jacobson, see Morphisms, Lemma 25.17.10. We mention here the following interesting case.

Lemma 24.6.4. 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 24.5.7). Hence $S$ is a sober, Noetherian topological space (use Schemes, Lemma 22.11.1). Assume $S$ is not Jacobson to get a contradiction. By Topology, Lemma 5.13.3 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 9.60.1. $\square$

\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, 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
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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