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 10.60.1. Let $R$ be a Noetherian local domain of dimension $\geq 2$. A nonempty open subset $U \subset \mathop{\mathrm{Spec}}(R)$ is infinite.

Proof. To get a contradiction, assume that $U \subset \mathop{\mathrm{Spec}}(R)$ is finite. In this case $(0) \in U$ and $\{ (0)\} $ is an open subset of $U$ (because the complement of $\{ (0)\} $ is the union of the closures of the other points). Thus we may assume $U = \{ (0)\} $. Let $\mathfrak m \subset R$ be the maximal ideal. We can find an $x \in \mathfrak m$, $x \not= 0$ such that $V(x) \cup U = \mathop{\mathrm{Spec}}(R)$. In other words we see that $D(x) = \{ (0)\} $. In particular we see that $\dim (R/xR) = \dim (R) - 1 \geq 1$, see Lemma 10.59.12. Let $\overline{y}_2, \ldots , \overline{y}_{\dim (R)} \in R/xR$ generate an ideal of definition of $R/xR$, see Proposition 10.59.8. Choose lifts $y_2, \ldots , y_{\dim (R)} \in R$, so that $x, y_2, \ldots , y_{\dim (R)}$ generate an ideal of definition in $R$. This implies that $\dim (R/(y_2)) = \dim (R) - 1$ and $\dim (R/(y_2, x)) = \dim (R) - 2$, see Lemma 10.59.13. Hence there exists a prime $\mathfrak p$ containing $y_2$ but not $x$. This contradicts the fact that $D(x) = \{ (0)\} $. $\square$


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