Proposition 10.90.6. Let R be a ring. The following are equivalent
R is coherent,
any product of flat R-modules is flat, and
for every set A the module R^ A is flat.
This is [Theorem 2.1, Chase].
Proposition 10.90.6. Let R be a ring. The following are equivalent
R is coherent,
any product of flat R-modules is flat, and
for every set A the module R^ A is flat.
Proof. Assume R coherent, and let Q_\alpha , \alpha \in A be a set of flat R-modules. We have to show that I \otimes _ R \prod _\alpha Q_\alpha \to \prod Q_\alpha is injective for every finitely generated ideal I of R, see Lemma 10.39.5. Since R is coherent I is an R-module of finite presentation. Hence I \otimes _ R \prod _\alpha Q_\alpha = \prod I \otimes _ R Q_\alpha by Proposition 10.89.3. The desired injectivity follows as I \otimes _ R Q_\alpha \to Q_\alpha is injective by flatness of Q_\alpha .
The implication (2) \Rightarrow (3) is trivial.
Assume that the R-module R^ A is flat for every set A. Let I be a finitely generated ideal in R. Then I \otimes _ R R^ A \to R^ A is injective by assumption. By Proposition 10.89.2 and the finiteness of I the image is equal to I^ A. Hence I \otimes _ R R^ A = I^ A for every set A and we conclude that I is finitely presented by Proposition 10.89.3. \square
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