
Lemma 10.90.3. Let $R$ be a Noetherian ring and $A$ a set. Then $M = R^ A$ is a flat and Mittag-Leffler $R$-module.

Proof. Combining Lemma 10.89.5 and Proposition 10.89.6 we see that $M$ is flat over $R$. We show that $M$ satisfies the condition of Lemma 10.90.2. 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 10.88.3 we have a commutative diagram

$\xymatrix{ F' \otimes _ R M \ar[r] \ar[d]^{\cong } & F \otimes _ R M \ar[d]^{\cong } \\ (F')^ A \ar[r] & F^ A }$

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 $x \in F' \otimes _ R M$. $\square$

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).