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*}

Proposition 10.34.19. Let $R$ be a Jacobson ring. Let $R \to S$ be a ring map of finite type. Then

  1. The ring $S$ is Jacobson.

  2. The map $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ transforms closed points to closed points.

  3. For $\mathfrak m' \subset S$ maximal lying over $\mathfrak m \subset R$ the field extension $\kappa (\mathfrak m')/\kappa (\mathfrak m)$ is finite.

Proof. Let $\mathfrak m' \subset S$ be a maximal ideal and $R \cap \mathfrak m' = \mathfrak m$. Then $R/\mathfrak m \to S/\mathfrak m'$ satisfies the conditions of Lemma 10.34.18 by Lemma 10.34.17. Hence $R/\mathfrak m$ is a field and $\mathfrak m$ a maximal ideal and the induced residue field extension is finite. This proves (2) and (3).

If $S$ is not Jacobson, then by Lemma 10.34.5 there exists a non-maximal prime ideal $\mathfrak q$ of $S$ and an $g \in S$, $g \not\in \mathfrak q$ such that $(S/\mathfrak q)_ g$ is a field. To arrive at a contradiction we show that $\mathfrak q$ is a maximal ideal. Let $\mathfrak p = \mathfrak q \cap R$. Then $R/\mathfrak p \to (S/\mathfrak q)_ g$ satisfies the conditions of Lemma 10.34.18 by Lemma 10.34.17. Hence $R/\mathfrak p$ is a field and the field extension $\kappa (\mathfrak p) \to (S/\mathfrak q)_ g = \kappa (\mathfrak q)$ is finite, thus algebraic. Then $\mathfrak q$ is a maximal ideal of $S$ by Lemma 10.34.9. Contradiction. $\square$


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