Lemma 10.99.15 (Critère de platitude par fibres; Noetherian case). Let $R$, $S$, $S'$ be Noetherian local rings and let $R \to S \to S'$ be local ring homomorphisms. Let $\mathfrak m \subset R$ be the maximal ideal. Let $M$ be an $S'$-module. Assume

The module $M$ is finite over $S'$.

The module $M$ is not zero.

The module $M/\mathfrak m M$ is a flat $S/\mathfrak m S$-module.

The module $M$ is a flat $R$-module.

Then $S$ is flat over $R$ and $M$ is a flat $S$-module.

**Proof.**
Set $I = \mathfrak mS \subset S$. Then we see that $M/IM$ is a flat $S/I$-module because of (3). Since $\mathfrak m \otimes _ R S' \to I \otimes _ S S'$ is surjective we see that also $\mathfrak m \otimes _ R M \to I \otimes _ S M$ is surjective. Consider

\[ \mathfrak m \otimes _ R M \to I \otimes _ S M \to M. \]

As $M$ is flat over $R$ the composition is injective and so both arrows are injective. In particular $\text{Tor}_1^ S(S/I, M) = 0$ see Remark 10.75.9. By Lemma 10.99.10 we conclude that $M$ is flat over $S$. Note that since $M/\mathfrak m_{S'}M$ is not zero by Nakayama's Lemma 10.20.1 we see that actually $M$ is faithfully flat over $S$ by Lemma 10.39.15 (since it forces $M/\mathfrak m_ SM \not= 0$).

Consider the exact sequence $0 \to \mathfrak m \to R \to \kappa \to 0$. This gives an exact sequence $0 \to \text{Tor}_1^ R(\kappa , S) \to \mathfrak m \otimes _ R S \to I \to 0$. Since $M$ is flat over $S$ this gives an exact sequence $0 \to \text{Tor}_1^ R(\kappa , S)\otimes _ S M \to \mathfrak m \otimes _ R M \to I \otimes _ S M \to 0$. By the above this implies that $\text{Tor}_1^ R(\kappa , S)\otimes _ S M = 0$. Since $M$ is faithfully flat over $S$ this implies that $\text{Tor}_1^ R(\kappa , S) = 0$ and we conclude that $S$ is flat over $R$ by Lemma 10.99.7.
$\square$

## Comments (2)

Comment #475 by Bas Edixhoven on

Comment #488 by Johan on