Proposition 10.89.6. 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 10.89.6. 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 10.38.5. 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 10.88.3. 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 10.88.2 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 10.88.3. $\square$

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## Comments (2)

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