Lemma 10.91.3. 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 10.90.5 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
\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
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