The Stacks project

Proof. The implication (1) $\Rightarrow$ (2) is a special case of Lemma 10.89.6. Assume (2). By Theorem 10.81.4 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 10.88.8, 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

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

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

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 $Q_ j\subset Q$. $\square$

Proof. Combining Lemma 10.90.5 and Proposition 10.90.6 we see that $M$ is flat over $R$. We show that $M$ satisfies the condition of Lemma 10.91.2. Let $F$ be a free finite $R$-module. If $F'$ is any submodule of $F$ then it is finitely presented since $R$ is Noetherian. So by Proposition 10.89.3 we have a commutative diagram

by which we can identify the map $F' \otimes _ R M \to F \otimes _ R M$ with $(F')^ A \to F^ A$. Hence if $x \in F \otimes _ R M$ corresponds to $(x_\alpha ) \in F^ A$, then the submodule of $F'$ of $F$ generated by the $x_\alpha$ is the smallest submodule of $F$ such that $x \in F' \otimes _ R M$. $\square$

Proof. As an $R$-module, we have $M = R^ A$ for a (countable) set $A$. Hence this lemma is a special case of Lemma 10.91.3. $\square$