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 10.117.3. Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Assume

  1. $R$ is a domain,

  2. $R \to S$ is of finite type, and

  3. $M$ is a finite type $S$-module.

Then there exists a nonzero $f \in R$ such that

  1. $M_ f$ and $S_ f$ are free as $R_ f$-modules, and

  2. $S_ f$ is a finitely presented $R_ f$-algebra and $M_ f$ is a finitely presented $S_ f$-module.

Proof. We first prove the lemma for $S = R[x_1, \ldots , x_ n]$, and then we deduce the result in general.

Assume $S = R[x_1, \ldots , x_ n]$. Choose elements $m_1, \ldots , m_ t$ which generate $M$. This gives a short exact sequence

\[ 0 \to N \to S^{\oplus t} \xrightarrow {(m_1, \ldots , m_ t)} M \to 0. \]

Denote $K$ the fraction field of $R$. Denote $S_ K = K \otimes _ R S = K[x_1, \ldots , x_ n]$, and similarly $N_ K = K \otimes _ R N$, $M_ K = K \otimes _ R M$. As $R \to K$ is flat the sequence remains exact after tensoring with $K$. As $S_ K = K[x_1, \ldots , x_ n]$ is a Noetherian ring (see Lemma 10.30.1) we can find finitely many elements $n'_1, \ldots , n'_ s \in N_ K$ which generate it. Choose $n_1, \ldots , n_ r \in N$ such that $n'_ i = \sum a_{ij}n_ j$ for some $a_{ij} \in K$. Set

\[ M' = S^{\oplus t}/\sum \nolimits _{i = 1, \ldots , r} Sn_ i \]

By construction $M'$ is a finitely presented $S$-module, and there is a surjection $M' \to M$ which induces an isomorphism $M'_ K \cong M_ K$. We may apply Lemma 10.117.2 to $R \to S$ and $M'$ and we find an $f \in R$ such that $M'_ f$ is a free $R_ f$-module. Thus $M'_ f \to M_ f$ is a surjection of modules over the domain $R_ f$ where the source is a free module and which becomes an isomorphism upon tensoring with $K$. Thus it is injective as $M'_ f \subset M'_ K$ as it is free over the domain $R_ f$. Hence $M'_ f \to M_ f$ is an isomorphism and the result is proved.

For the general case, choose a surjection $R[x_1, \ldots , x_ n] \to S$. Think of both $S$ and $M$ as finite modules over $R[x_1, \ldots , x_ n]$. By the special case proved above there exists a nonzero $f \in R$ such that both $S_ f$ and $M_ f$ are free as $R_ f$-modules and finitely presented as $R_ f[x_1, \ldots , x_ n]$-modules. Clearly this implies that $S_ f$ is a finitely presented $R_ f$-algebra and that $M_ f$ is a finitely presented $S_ f$-module. $\square$


Comments (2)

Comment #3265 by Samir Canning on

Typo: "the sequence remains flat"--->"the sequence remains exact"


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