Lemma 10.98.15 (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

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

2. The module $M$ is not zero.

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

4. 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 10.74.9. By Lemma 10.98.10 we conclude that $M$ is flat over $S$. Note that since $M/\mathfrak m_{S'}M$ is not zero by Nakayama's Lemma 10.19.1 we see that actually $M$ is faithfully flat over $S$ by Lemma 10.38.15 (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 10.98.7. $\square$

Comment #475 by Bas Edixhoven on

The reference to Lemma 10.93.9 is not appropriate because that lemma assumes that M is a finite S-module.

A similar problem occurs at the end of the proof. There one concludes that S is flat over R but the argument uses that S is finite over R.

Comment #488 by on

Actually, it is appropriate because $M$ is finite over $S'$. In other words, you apply Lemma 10.98.10 to the ring map $S \to S'$ and the module $M$ viewed as a finite module over $S'$. This makes sense because here $I \subset S$ is an ideal of $S$ and $M$ is a finite module over the bigger ring $S'$.

Similarly, the application of 10.98.7 is OK too I think. Namely, here we apply that lemma with $R \to S$ and with module $M = S$ (what I mean is that the module called $M$ in Lemma 10.98.7 should be taken to be $S$ in the current situation). This is OK because we just showed that $\text{Tor}_1^R(\kappa, S) = 0$.

OK?

Anyway, it is very confusing. But I am not sure how to write the proof in such a way that it is less confusing. Do you have a suggestion?

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).