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 10.147.10. Let $R \to S$ be a ring map. Let $\mathfrak q$ be a prime of $S$ lying over $\mathfrak p \subset R$. Assume that $R \to S$ is of finite type and unramified at $\mathfrak q$. Then there exist

  1. an étale ring map $R \to R'$,

  2. a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$.

  3. a product decomposition

    \[ R' \otimes _ R S = A \times B \]

with the following properties

  1. $R' \to A$ is surjective, and

  2. $\mathfrak p'A$ is a prime of $A$ lying over $\mathfrak p'$ and over $\mathfrak q$.

Proof. We may replace $(R \to S, \mathfrak p, \mathfrak q)$ with any base change $(R' \to R'\otimes _ R S, \mathfrak p', \mathfrak q')$ by an étale ring map $R \to R'$ with a prime $\mathfrak p'$ lying over $\mathfrak p$, and a choice of $\mathfrak q'$ lying over both $\mathfrak q$ and $\mathfrak p'$. Note also that given $R \to R'$ and $\mathfrak p'$ a suitable $\mathfrak q'$ can always be found.

The assumption that $R \to S$ is of finite type means that we may apply Lemma 10.141.23. Thus we may assume that $S = A_1 \times \ldots \times A_ n \times B$, that each $R \to A_ i$ is finite with exactly one prime $\mathfrak r_ i$ lying over $\mathfrak p$ such that $\kappa (\mathfrak p) \subset \kappa (\mathfrak r_ i)$ is purely inseparable and that $R \to B$ is not quasi-finite at any prime lying over $\mathfrak p$. Then clearly $\mathfrak q = \mathfrak r_ i$ for some $i$, since an unramified morphism is quasi-finite (see Lemma 10.147.6). Say $\mathfrak q = \mathfrak r_1$. By Lemma 10.147.5 we see that $\kappa (\mathfrak p) \subset \kappa (\mathfrak r_1)$ is separable hence the trivial field extension, and that $\mathfrak p(A_1)_{\mathfrak r_1}$ is the maximal ideal. Also, by Lemma 10.40.11 (which applies to $R \to A_1$ because a finite ring map satisfies going up by Lemma 10.35.22) we have $(A_1)_{\mathfrak r_1} = (A_1)_{\mathfrak p}$. It follows from Nakayama's Lemma 10.19.1 that the map of local rings $R_{\mathfrak p} \to (A_1)_{\mathfrak p} = (A_1)_{\mathfrak r_1}$ is surjective. Since $A_1$ is finite over $R$ we see that there exists a $f \in R$, $f \not\in \mathfrak p$ such that $R_ f \to (A_1)_ f$ is surjective. After replacing $R$ by $R_ f$ we win. $\square$


Comments (2)

Comment #3260 by Dario Weißmann on

Typo: by a etale ring map -> by an etale ring map.


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