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 10.77.7. Let $R$ be a local ring with maximal ideal $\mathfrak m$ and infinite residue field. Let $R \to S$ be a ring map. Let $M$ be an $S$-module and let $N \subset M$ be an $R$-submodule. Assume

  1. $S$ is semi-local and $\mathfrak mS$ is contained in the Jacobson radical of $S$,

  2. $M$ is a finite free $S$-module, and

  3. $N$ generates $M$ as an $S$-module.

Then $N$ contains an $S$-basis of $M$.

Proof. Assume $M$ is free of rank $n$. Let $I \subset S$ be the Jacobson radical. By Nakayama's Lemma 10.19.1 a sequence of elements $m_1, \ldots , m_ n$ is a basis for $M$ if and only if $\overline{m}_ i \in M/IM$ generate $M/IM$. Hence we may replace $M$ by $M/IM$, $N$ by $N/(N \cap IM)$, $R$ by $R/\mathfrak m$, and $S$ by $S/IS$. In this case we see that $S$ is a finite product of fields $S = k_1 \times \ldots \times k_ r$ and $M = k_1^{\oplus n} \times \ldots \times k_ r^{\oplus n}$. The fact that $N \subset M$ generates $M$ as an $S$-module means that there exist $x_ j \in N$ such that a linear combination $\sum a_ j x_ j$ with $a_ j \in S$ has a nonzero component in each factor $k_ i^{\oplus n}$. Because $R = k$ is an infinite field, this means that also some linear combination $y = \sum c_ j x_ j$ with $c_ j \in k$ has a nonzero component in each factor. Hence $y \in N$ generates a free direct summand $Sy \subset M$. By induction on $n$ the result holds for $M/Sy$ and the submodule $\overline{N} = N/(N \cap Sy)$. In other words there exist $\overline{y}_2, \ldots , \overline{y}_ n$ in $\overline{N}$ which (freely) generate $M/Sy$. Then $y, y_2, \ldots , y_ n$ (freely) generate $M$ and we win. $\square$


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