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 10.88.6. Let $M$ be a flat Mittag-Leffler module over $R$. Let $F$ be an $R$-module and let $x \in F \otimes _ R M$. Then there exists a smallest submodule $F' \subset F$ such that $x \in F' \otimes _ R M$.

Proof. Since $M$ is flat we have $F' \otimes _ R M \subset F \otimes _ R M$ if $F' \subset F$ is a submodule, hence the statement makes sense. Let $I = \{ F' \subset F \mid x \in F' \otimes _ R M\} $ and for $i \in I$ denote $F_ i \subset F$ the corresponding submodule. Then $x$ maps to zero under the map

\[ F \otimes _ R M \longrightarrow \prod (F/F_ i \otimes _ R M) \]

whence by Proposition 10.88.5 $x$ maps to zero under the map

\[ F \otimes _ R M \longrightarrow \left(\prod F/F_ i\right) \otimes _ R M \]

Since $M$ is flat the kernel of this arrow is $(\bigcap F_ i) \otimes _ R M$ which proves the lemma. $\square$


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