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

  1. $R$ is Noetherian,

  2. $R$ is a domain,

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

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

Then there exists a nonzero $f \in R$ such that $M_ f$ is a free $R_ f$-module.

Proof. Let $K$ be the fraction field of $R$. Set $S_ K = K \otimes _ R S$. This is an algebra of finite type over $K$. We will argue by induction on $d = \dim (S_ K)$ (which is finite for example by Noether normalization, see Section 10.114). Fix $d \geq 0$. Assume we know that the lemma holds in all cases where $\dim (S_ K) < d$.

Suppose given $R \to S$ and $M$ as in the lemma with $\dim (S_ K) = d$. By Lemma 10.61.1 there exists a filtration $0 \subset M_1 \subset M_2 \subset \ldots \subset M_ n = M$ so that $M_ i/M_{i - 1}$ is isomorphic to $S/\mathfrak q$ for some prime $\mathfrak q$ of $S$. Note that $\dim ((S/\mathfrak q)_ K) \leq \dim (S_ K)$. Also, note that an extension of free modules is free (see basic notion 50). Thus we may assume $M = S$ and that $S$ is a domain of finite type over $R$.

If $R \to S$ has a nontrivial kernel, then take a nonzero $f \in R$ in this kernel. In this case $S_ f = 0$ and the lemma holds. (This is really the case $d = -1$ and the start of the induction.) Hence we may assume that $R \to S$ is a finite type extension of Noetherian domains.

Apply Lemma 10.114.7 and replace $R$ by $R_ f$ (with $f$ as in the lemma) to get a factorization

\[ R \subset R[y_1, \ldots , y_ d] \subset S \]

where the second extension is finite. Choose $z_1, \ldots , z_ r \in S$ which form a basis for the fraction field of $S$ over the fraction field of $R[y_1, \ldots , y_ d]$. This gives a short exact sequence

\[ 0 \to R[y_1, \ldots , y_ d]^{\oplus r} \xrightarrow {(z_1, \ldots , z_ r)} S \to N \to 0 \]

By construction $N$ is a finite $R[y_1, \ldots , y_ d]$-module whose support does not contain the generic point $(0)$ of $\mathop{\mathrm{Spec}}(R[y_1, \ldots , y_ d])$. By Lemma 10.39.5 there exists a nonzero $g \in R[y_1, \ldots , y_ d]$ such that $g$ annihilates $N$, so we may view $N$ as a finite module over $S' = R[y_1, \ldots , y_ d]/(g)$. Since $\dim (S'_ K) < d$ by induction there exists a nonzero $f \in R$ such that $N_ f$ is a free $R_ f$-module. Since $(R[y_1, \ldots , y_ d])_ f \cong R_ f[y_1, \ldots , y_ d]$ is free also we conclude by the already mentioned fact that an extension of free modules is free. $\square$


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