Lemma 10.91.2. 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 10.89.6. 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

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

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

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 059S. Beware of the difference between the letter 'O' and the digit '0'.