The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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. $\square$

Comments (2)

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 is finite over . In other words, you apply Lemma 10.98.10 to the ring map and the module viewed as a finite module over . This makes sense because here is an ideal of and is a finite module over the bigger ring .

Similarly, the application of 10.98.7 is OK too I think. Namely, here we apply that lemma with and with module (what I mean is that the module called in Lemma 10.98.7 should be taken to be in the current situation). This is OK because we just showed that .


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?

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