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 10.114.7. Let $R \to S$ be an injective finite type map of domains. Then there exists an integer $d$ and a factorization

\[ R \to R[y_1, \ldots , y_ d] \to S' \to S \]

by injective maps such that $S'$ is finite over $R[y_1, \ldots , y_ d]$ and such that $S'_ f \cong S_ f$ for some nonzero $f \in R$.

Proof. Pick $x_1, \ldots , x_ n \in S$ which generate $S$ over $R$. Let $K$ be the fraction field of $R$ and $S_ K = S \otimes _ R K$. By Lemma 10.114.4 we can find $y_1, \ldots , y_ d \in S$ such that $K[y_1, \ldots , y_ d] \to S_ K$ is a finite injective map. Note that $y_ i \in S$ because we may pick the $y_ j$ in the $\mathbf{Z}$-algebra generated by $x_1, \ldots , x_ n$. As a finite ring map is integral (see Lemma 10.35.3) we can find monic $P_ i \in K[y_1, \ldots , y_ d][T]$ such that $P_ i(x_ i) = 0$ in $S_ K$. Let $f \in R$ be a nonzero element such that $fP_ i \in R[y_1, \ldots , y_ d][T]$ for all $i$. Set $x_ i' = fx_ i$ and let $S' \subset S$ be the subalgebra generated by $y_1, \ldots , y_ d$ and $x'_1, \ldots , x'_ n$. Note that $x'_ i$ is integral over $R[y_1, \ldots , y_ d]$ as we have $Q_ i(x_ i') = 0$ where $Q_ i = f^{\deg _ T(P_ i)}P_ i(T/f)$ which is a monic polynomial in $T$ with coefficients in $R[y_1, \ldots , y_ d]$ by our choice of $f$. Hence $R[y_1, \ldots , y_ n] \subset S'$ is finite by Lemma 10.35.5. By construction $S'_ f \cong S_ f$ and we win. $\square$


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