
Lemma 10.74.8. Let $R$ be a ring. Let $M$ be an $R$-module. The following are equivalent:

1. The module $M$ is flat over $R$.

2. For all $i>0$ the functor $\text{Tor}_ i^ R(M, -)$ is zero.

3. The functor $\text{Tor}_1^ R(M, -)$ is zero.

4. For all ideals $I \subset R$ we have $\text{Tor}_1^ R(M, R/I) = 0$.

5. For all finitely generated ideals $I \subset R$ we have $\text{Tor}_1^ R(M, R/I) = 0$.

Proof. Suppose $M$ is flat. Let $N$ be an $R$-module. Let $F_\bullet$ be a free resolution of $N$. Then $F_\bullet \otimes _ R M$ is a resolution of $N \otimes _ R M$, by flatness of $M$. Hence all higher Tor groups vanish.

It now suffices to show that the last condition implies that $M$ is flat. Let $I \subset R$ be an ideal. Consider the short exact sequence $0 \to I \to R \to R/I \to 0$. Apply Lemma 10.74.2. We get an exact sequence

$\text{Tor}_1^ R(M, R/I) \to M \otimes _ R I \to M \otimes _ R R \to M \otimes _ R R/I \to 0$

Since obviously $M \otimes _ R R = M$ we conclude that the last hypothesis implies that $M \otimes _ R I \to M$ is injective for every finitely generated ideal $I$. Thus $M$ is flat by Lemma 10.38.5. $\square$

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