The Stacks project

Lemma 10.75.8. 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 10.75.2. 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 10.39.5. $\square$

Comments (2)

Comment #5459 by Pavel on

what is this in the end of the proof? there is a long exact sequence of tors with three tensor products being the last three nonzero terms of it, AND if they all the tors vanish we are left with short exact sequence of products, hence M is flat, it's just simple as that. You don't even have to consider some ideals of the ring, it doesn't change anything. Link to lemma at the very end leads to some immense amount of logic manipulations as well

Comment #5677 by on

Dear Pavel, I do not see anything wrong with the proof as given. (I do agree that the proof of Lemma 10.39.5 is too long.) Note that the last statement only works for finitely generated ideal which is sort of the whole point of Lemma 10.39.5.

There are also:

  • 2 comment(s) on Section 10.75: Tor groups and flatness

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