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 10.90.2. Let $M$ be a flat $R$-module. The following are equivalent

  1. $M$ is Mittag-Leffler, and

  2. if $F$ is a finite free $R$-module and $x \in F \otimes _ R M$, then there exists a smallest submodule $F'$ of $F$ such that $x \in F' \otimes _ R M$.

Proof. The implication (1) $\Rightarrow $ (2) is a special case of Lemma 10.88.6. Assume (2). By Theorem 10.80.4 we can write $M$ as the colimit $M = \mathop{\mathrm{colim}}\nolimits _{i \in I} M_ i$ of a directed system $(M_ i, f_{ij})$ of finite free $R$-modules. By Remark 10.87.8, it suffices to show that the inverse system $(\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R), \mathop{\mathrm{Hom}}\nolimits _ R(f_{ij}, R))$ is Mittag-Leffler. In other words, fix $i \in I$ and for $j \geq i$ let $Q_ j$ be the image of $\mathop{\mathrm{Hom}}\nolimits _ R(M_ j, R) \to \mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R)$; we must show that the $Q_ j$ stabilize.

Since $M_ i$ is free and finite, we can make the identification $\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, M_ j) = \mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R) \otimes _ R M_ j$ for all $j$. Using the fact that the $M_ j$ are free, it follows that for $j \geq i$, $Q_ j$ is the smallest submodule of $\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R)$ such that $f_{ij} \in Q_ j \otimes _ R M_ j$. Under the identification $\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, M) = \mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R) \otimes _ R M$, the canonical map $f_ i: M_ i \to M$ is in $\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R) \otimes _ R M$. By the assumption on $M$, there exists a smallest submodule $Q$ of $\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R)$ such that $f_ i \in Q \otimes _ R M$. We are going to show that the $Q_ j$ stabilize to $Q$.

For $j \geq i$ we have a commutative diagram

\[ \xymatrix{ Q_ j \otimes _ R M_ j \ar[r] \ar[d] & \mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R) \otimes _ R M_ j \ar[d] \\ Q_ j \otimes _ R M \ar[r] & \mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R) \otimes _ R M. } \]

Since $f_{ij} \in Q_ j \otimes _ R M_ j$ maps to $f_ i \in \mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R) \otimes _ R M$, it follows that $f_ i \in Q_ j \otimes _ R M$. Hence, by the choice of $Q$, we have $Q \subset Q_ j$ for all $j \geq i$.

Since the $Q_ j$ are decreasing and $Q \subset Q_ j$ for all $j \geq i$, to show that the $Q_ j$ stabilize to $Q$ it suffices to find a $j \geq i$ such that $Q_ j \subset Q$. As an element of

\[ \mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R) \otimes _ R M = \mathop{\mathrm{colim}}\nolimits _{j \in J} (\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R) \otimes _ R M_ j), \]

$f_ i$ is the colimit of $f_{ij}$ for $j \geq i$, and $f_ i$ also lies in the submodule

\[ \mathop{\mathrm{colim}}\nolimits _{j \in J} (Q \otimes _ R M_ j) \subset \mathop{\mathrm{colim}}\nolimits _{j \in J} (\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R) \otimes _ R M_ j). \]

It follows that for some $j \geq i$, $f_{ij}$ lies in $Q \otimes _ R M_ j$. Since $Q_ j$ is the smallest submodule of $\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R)$ with $f_{ij} \in Q_ j \otimes _ R M_ j$, we conclude $Q_ j\subset Q$. $\square$


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