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 10.40.5. Let $R \to S$ be a ring map. Let $T \subset \mathop{\mathrm{Spec}}(R)$ be the image of $\mathop{\mathrm{Spec}}(S)$. If $T$ is stable under specialization, then $T$ is closed.

Proof. We give two proofs.

First proof. Let $\mathfrak p \subset R$ be a prime ideal such that the corresponding point of $\mathop{\mathrm{Spec}}(R)$ is in the closure of $T$. This means that for every $f \in R$, $f \not\in \mathfrak p$ we have $D(f) \cap T \not= \emptyset $. Note that $D(f) \cap T$ is the image of $\mathop{\mathrm{Spec}}(S_ f)$ in $\mathop{\mathrm{Spec}}(R)$. Hence we conclude that $S_ f \not= 0$. In other words, $1 \not= 0$ in the ring $S_ f$. Since $S_{\mathfrak p}$ is the directed colimit of the rings $S_ f$ we conclude that $1 \not= 0$ in $S_{\mathfrak p}$. In other words, $S_{\mathfrak p} \not= 0$ and considering the image of $\mathop{\mathrm{Spec}}(S_{\mathfrak p}) \to \mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ we see there exists a $\mathfrak p' \in T$ with $\mathfrak p' \subset \mathfrak p$. As we assumed $T$ closed under specialization we conclude $\mathfrak p$ is a point of $T$ as desired.

Second proof. Let $I = \mathop{\mathrm{Ker}}(R \to S)$. We may replace $R$ by $R/I$. In this case the ring map $R \to S$ is injective. By Lemma 10.29.5 all the minimal primes of $R$ are contained in the image $T$. Hence if $T$ is stable under specialization then it contains all primes. $\square$


Comments (3)

Comment #3001 by David Holmes on

"Since S_p is the directed limit of the rings S_f..." Here the "limit" should be "colimit" I think.


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