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 10.53.5. Let $\varphi : R \to S$ be essentially of finite type with $R$ and $S$ local (but not necessarily $\varphi $ local). Then there exists an $n$ and a maximal ideal $\mathfrak m \subset R[x_1, \ldots , x_ n]$ lying over $\mathfrak m_ R$ such that $S$ is a localization of a quotient of $R[x_1, \ldots , x_ n]_\mathfrak m$.

Proof. We can write $S$ as a localization of a quotient of $R[x_1, \ldots , x_ n]$. Hence it suffices to prove the lemma in case $S = R[x_1, \ldots , x_ n]_\mathfrak q$ for some prime $\mathfrak q \subset R[x_1, \ldots , x_ n]$. If $\mathfrak q + \mathfrak m_ R R[x_1, \ldots , x_ n] \not= R[x_1, \ldots , x_ n]$ then we can find a maximal ideal $\mathfrak m$ as in the statement of the lemma with $\mathfrak q \subset \mathfrak m$ and the result is clear.

Choose a valuation ring $A \subset \kappa (\mathfrak q)$ which dominates the image of $R \to \kappa (\mathfrak q)$ (Lemma 10.49.2). If the image $\lambda _ i \in \kappa (\mathfrak q)$ of $x_ i$ is contained in $A$, then $\mathfrak q$ is contained in the inverse image of $\mathfrak m_ A$ via $R[x_1, \ldots , x_ n] \to A$ which means we are back in the preceding case. Hence there exists an $i$ such that $\lambda _ i^{-1} \in A$ and such that $\lambda _ j/\lambda _ i \in A$ for all $j = 1, \ldots , n$ (because the value group of $A$ is totally ordered, see Lemma 10.49.12). Then we consider the map

\[ R[y_0, y_1, \ldots , \hat{y_ i}, \ldots , y_ n] \to R[x_1, \ldots , x_ n]_\mathfrak q,\quad y_0 \mapsto 1/x_ i,\quad y_ j \mapsto x_ j/x_ i \]

Let $\mathfrak q' \subset R[y_0, \ldots , \hat{y_ i}, \ldots , y_ n]$ be the inverse image of $\mathfrak q$. Since $y_0 \not\in \mathfrak q'$ it is easy to see that the displayed arrow defines an isomorphism on localizations. On the other hand, the result of the first paragraph applies to $R[y_0, \ldots , \hat{y_ i}, \ldots , y_ n]$ because $y_ j$ maps to an element of $A$. This finishes the proof. $\square$


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