Proposition 10.89.2. Let $M$ be an $R$-module. The following are equivalent:

1. $M$ is finitely generated.

2. For every family $(Q_{\alpha })_{\alpha \in A}$ of $R$-modules, the canonical map $M \otimes _ R \left( \prod _{\alpha } Q_{\alpha } \right) \to \prod _{\alpha } (M \otimes _ R Q_{\alpha })$ is surjective.

3. For every $R$-module $Q$ and every set $A$, the canonical map $M \otimes _ R Q^{A} \to (M \otimes _ R Q)^{A}$ is surjective.

4. For every set $A$, the canonical map $M \otimes _ R R^{A} \to M^{A}$ is surjective.

Proof. First we prove (1) implies (2). Choose a surjection $R^ n \to M$ and consider the commutative diagram

$\xymatrix{ R^ n \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r]^{\cong } \ar[d] & \prod _{\alpha } (R^ n \otimes _ R Q_{\alpha }) \ar[d] \\ M \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] & \prod _{\alpha } ( M \otimes _ R Q_{\alpha }). }$

The top arrow is an isomorphism and the vertical arrows are surjections. We conclude that the bottom arrow is a surjection.

Obviously (2) implies (3) implies (4), so it remains to prove (4) implies (1). In fact for (1) to hold it suffices that the element $d = (x)_{x \in M}$ of $M^ M$ is in the image of the map $f: M \otimes _ R R^{M} \to M^ M$. In this case $d = \sum _{i = 1}^{n} f(x_ i \otimes a_ i)$ for some $x_ i \in M$ and $a_ i \in R^ M$. If for $x \in M$ we write $p_ x: M^ M \to M$ for the projection onto the $x$-th factor, then

$x = p_ x(d) = \sum \nolimits _{i = 1}^{n} p_ x(f(x_ i \otimes a_ i)) = \sum \nolimits _{i=1}^{n} p_ x(a_ i) x_ i.$

Thus $x_1, \ldots , x_ n$ generate $M$. $\square$

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