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

\[ \xymatrix{ R^ m \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d]^{\cong } & R^ n \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d]^{\cong } & M \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & 0 \\ \prod _{\alpha } (R^ m \otimes _ R Q_{\alpha }) \ar[r] & \prod _{\alpha } (R^ n \otimes _ R Q_{\alpha }) \ar[r] & \prod _{\alpha } (M \otimes _ R Q_{\alpha }) \ar[r] & 0. } \]

The first two vertical arrows are isomorphisms and the rows are exact. This implies that the map $M \otimes _ R (\prod _{\alpha } Q_{\alpha }) \to \prod _{\alpha } ( M \otimes _ R Q_{\alpha })$ is surjective and, by a diagram chase, also injective. Hence (2) holds.

Obviously (2) implies (3) implies (4), so it remains to prove (4) implies (1). From Proposition 10.89.2, if (4) holds we already know that $M$ is finitely generated. So we can choose a surjection $F \to M$ where $F$ is free and finite. Let $K$ be the kernel. We must show $K$ is finitely generated. For any set $A$, we have a commutative diagram

\[ \xymatrix{ & K \otimes _ R R^ A \ar[r] \ar[d]_{f_3} & F \otimes _ R R^ A \ar[r] \ar[d]_{f_2}^{\cong } & M \otimes _ R R^ A \ar[r] \ar[d]_{f_1}^{\cong } & 0 \\ 0 \ar[r] & K^ A \ar[r] & F^ A \ar[r] & M^ A \ar[r] & 0 . } \]

The map $f_1$ is an isomorphism by assumption, the map $f_2$ is a isomorphism since $F$ is free and finite, and the rows are exact. A diagram chase shows that $f_3$ is surjective, hence by Proposition 10.89.2 we get that $K$ is finitely generated.
$\square$

## Comments (1)

Comment #8710 by Nancium on

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