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 10.30.9. Let $R$ be a ring and $\mathfrak p \subset R$ be a prime. There exists an $f \in R$, $f \not\in \mathfrak p$ such that $R_ f \to R_\mathfrak p$ is injective in each of the following cases

  1. $R$ is a domain,

  2. $R$ is Noetherian, or

  3. $R$ is reduced and has finitely many minimal primes.

Proof. If $R$ is a domain, then $R \subset R_\mathfrak p$, hence $f = 1$ works. If $R$ is Noetherian, then the kernel $I$ of $R \to R_\mathfrak p$ is a finitely generated ideal and we can find $f \in R$, $f \not\in \mathfrak p$ such that $IR_ f = 0$. For this $f$ the map $R_ f \to R_\mathfrak p$ is injective and $f$ works. If $R$ is reduced with finitely many minimal primes $\mathfrak p_1, \ldots , \mathfrak p_ n$, then we can choose $f \in \bigcap _{\mathfrak p_ i \not\subset \mathfrak p} \mathfrak p_ i$, $f \not\in \mathfrak p$. Indeed, if $\mathfrak {p}_ i\not\subset \mathfrak {p}$ then there exist $f_ i \in \mathfrak {p}_ i$, $f_ i \not\in \mathfrak {p}$ and $f = \prod f_ i$ works. For this $f$ we have $R_ f \subset R_\mathfrak p$ because the minimal primes of $R_ f$ correspond to minimal primes of $R_\mathfrak p$ and we can apply Lemma 10.24.2 (some details omitted). $\square$


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