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 10.29.4. Let $R$ be a domain. Let $\varphi : R \to S$ be a ring map. The following are equivalent:

  1. The ring map $R \to S$ is injective.

  2. The image $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ contains a dense set of points.

  3. There exists a prime ideal $\mathfrak q \subset S$ whose inverse image in $R$ is $(0)$.

Proof. Let $K$ be the field of fractions of the domain $R$. Assume that $R \to S$ is injective. Since localization is exact we see that $K \to S \otimes _ R K$ is injective. Hence there is a prime mapping to $(0)$ by Lemma 10.16.9.

Note that $(0)$ is dense in $\mathop{\mathrm{Spec}}(R)$, so that the last condition implies the second.

Suppose the second condition holds. Let $f \in R$, $f \not= 0$. As $R$ is a domain we see that $V(f)$ is a proper closed subset of $R$. By assumption there exists a prime $\mathfrak q$ of $S$ such that $\varphi (f) \not\in \mathfrak q$. Hence $\varphi (f) \not= 0$. Hence $R \to S$ is injective. $\square$


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