The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.34.4. A ring $R$ is Jacobson if and only if $\mathop{\mathrm{Spec}}(R)$ is Jacobson, see Topology, Definition 5.18.1.

Proof. Suppose $R$ is Jacobson. Let $Z \subset \mathop{\mathrm{Spec}}(R)$ be a closed subset. We have to show that the set of closed points in $Z$ is dense in $Z$. Let $U \subset \mathop{\mathrm{Spec}}(R)$ be an open such that $U \cap Z$ is nonempty. We have to show $Z \cap U$ contains a closed point of $\mathop{\mathrm{Spec}}(R)$. We may assume $U = D(f)$ as standard opens form a basis for the topology on $\mathop{\mathrm{Spec}}(R)$. According to Lemma 10.16.2 we may assume that $Z = V(I)$, where $I$ is a radical ideal. We see also that $f \not\in I$. By assumption, there exists a maximal ideal $\mathfrak m \subset R$ such that $I \subset \mathfrak m$ but $f \not\in \mathfrak m$. Hence $\mathfrak m \in D(f) \cap V(I) = U \cap Z$ as desired.

Conversely, suppose that $\mathop{\mathrm{Spec}}(R)$ is Jacobson. Let $I \subset R$ be a radical ideal. Let $J = \cap _{I \subset \mathfrak m} \mathfrak m$ be the intersection of the maximal ideals containing $I$. Clearly $J$ is a radical ideal, $V(J) \subset V(I)$, and $V(J)$ is the smallest closed subset of $V(I)$ containing all the closed points of $V(I)$. By assumption we see that $V(J) = V(I)$. But Lemma 10.16.2 shows there is a bijection between Zariski closed sets and radical ideals, hence $I = J$ as desired. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00G3. Beware of the difference between the letter 'O' and the digit '0'.