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 10.34.6. The ring $\mathbf{Z}$ is a Jacobson ring. More generally, let $R$ be a ring such that

  1. $R$ is a domain,

  2. $R$ is Noetherian,

  3. any nonzero prime ideal is a maximal ideal, and

  4. $R$ has infinitely many maximal ideals.

Then $R$ is a Jacobson ring.

Proof. Let $R$ satisfy (1), (2), (3) and (4). The statement means that $(0) = \bigcap _{\mathfrak m \subset R} \mathfrak m$. Since $R$ has infinitely many maximal ideals it suffices to show that any nonzero $x \in R$ is contained in at most finitely many maximal ideals, in other words that $V(x)$ is finite. By Lemma 10.16.7 we see that $V(x)$ is homeomorphic to $\mathop{\mathrm{Spec}}(R/xR)$. By assumption (3) every prime of $R/xR$ is minimal and hence corresponds to an irreducible component of $\mathop{\mathrm{Spec}}(R)$ (Lemma 10.25.1). As $R/xR$ is Noetherian, the topological space $\mathop{\mathrm{Spec}}(R/xR)$ is Noetherian (Lemma 10.30.5) and has finitely many irreducible components (Topology, Lemma 5.9.2). Thus $V(x)$ is finite as desired. $\square$


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