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 10.125.10. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $S \to S'$ be an $R$-algebra map. Assume that

  1. $I$ is locally nilpotent,

  2. $S/IS \to S'/IS'$ is an isomorphism,

  3. $S$ is of finite type over $R$,

  4. $S'$ of finite presentation over $R$, and

  5. $S'$ is flat over $R$.

Then $S \to S'$ is an isomorphism.

Proof. By Lemma 10.125.8 the map $S \to S'$ is surjective. As $I$ is locally nilpotent, so are the ideals $IS$ and $IS'$ (Lemma 10.31.3). Hence every prime ideal $\mathfrak q$ of $S$ contains $IS$ and (trivially) $S_\mathfrak q/IS_\mathfrak q \cong S'_\mathfrak q/IS'_\mathfrak q$. Thus Lemma 10.125.9 applies and we see that $S_\mathfrak q \to S'_\mathfrak q$ is an isomorphism for every prime $\mathfrak q \subset S$. It follows that $S \to S'$ is injective for example by Lemma 10.22.1. $\square$


Comments (1)

Comment #213 by Rex on

Typo: "is an ismorphism"


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