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 10.60.4. Noetherian Jacobson rings.

  1. Any Noetherian domain $R$ of dimension $1$ with infinitely many primes is Jacobson.

  2. Any Noetherian ring such that every prime $\mathfrak p$ is either maximal or contained in infinitely many prime ideals is Jacobson.

Proof. Part (1) is a reformulation of Lemma 10.34.6.

Let $R$ be a Noetherian ring such that every non-maximal prime $\mathfrak p$ is contained in infinitely many prime ideals. Assume $\mathop{\mathrm{Spec}}(R)$ is not Jacobson to get a contradiction. By Lemmas 10.25.1 and 10.30.5 we see that $\mathop{\mathrm{Spec}}(R)$ is a sober, Noetherian topological space. By Topology, Lemma 5.18.3 we see that there exists a non-maximal ideal $\mathfrak p \subset R$ such that $\{ \mathfrak p\} $ is a locally closed subset of $\mathop{\mathrm{Spec}}(R)$. In other words, $\mathfrak p$ is not maximal and $\{ \mathfrak p\} $ is an open subset of $V(\mathfrak p)$. Consider a prime $\mathfrak q \subset R$ with $\mathfrak p \subset \mathfrak q$. Recall that the topology on the spectrum of $(R/\mathfrak p)_{\mathfrak q} = R_{\mathfrak q}/\mathfrak pR_{\mathfrak q}$ is induced from that of $\mathop{\mathrm{Spec}}(R)$, see Lemmas 10.16.5 and 10.16.7. Hence we see that $\{ (0)\} $ is a locally closed subset of $\mathop{\mathrm{Spec}}((R/\mathfrak p)_{\mathfrak q})$. By Lemma 10.60.1 we conclude that $\dim ((R/\mathfrak p)_{\mathfrak q}) = 1$. Since this holds for every $\mathfrak q \supset \mathfrak p$ we conclude that $\dim (R/\mathfrak p) = 1$. At this point we use the assumption that $\mathfrak p$ is contained in infinitely many primes to see that $\mathop{\mathrm{Spec}}(R/\mathfrak p)$ is infinite. Hence by part (1) of the lemma we see that $V(\mathfrak p) \cong \mathop{\mathrm{Spec}}(R/\mathfrak p)$ is the closure of its closed points. This is the desired contradiction since it means that $\{ \mathfrak p\} \subset V(\mathfrak p)$ cannot be open. $\square$


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 00KX. Beware of the difference between the letter 'O' and the digit '0'.