Lemma 10.101.8 (06A5): Critère de platitude par fibres: Nilpotent case

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.101: Flatness criteria over Artinian rings
Lemma 10.101.8: Critère de platitude par fibres: Nilpotent case (cite)

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Lemma 10.101.8 (Critère de platitude par fibres: Nilpotent case). Let

$\xymatrix{ S \ar[rr] & & S' \\ & R \ar[lu] \ar[ru] }$

be a commutative diagram in the category of rings. Let $I \subset R$ be a nilpotent ideal and $M$ an $S'$ -module. Assume

The module $M/IM$ is a flat $S/IS$ -module.
The module $M$ is a flat $R$ -module.

Then $M$ is a flat $S$ -module and $S_{\mathfrak q}$ is flat over $R$ for every $\mathfrak q \subset S$ such that $M \otimes _ S \kappa (\mathfrak q)$ is nonzero.

Proof. As $M$ is flat over $R$ tensoring with the short exact sequence $0 \to I \to R \to R/I \to 0$ gives a short exact sequence

$0 \to I \otimes _ R M \to M \to M/IM \to 0.$

Note that $I \otimes _ R M \to IS \otimes _ S M$ is surjective. Combined with the above this means both maps in

$I \otimes _ R M \to IS \otimes _ S M \to M$

are injective. Hence $\text{Tor}_1^ S(IS, M) = 0$ (see Remark 10.75.9) and we conclude that $M$ is a flat $S$ -module by Lemma 10.99.8. To finish we need to show that $S_{\mathfrak q}$ is flat over $R$ for any prime $\mathfrak q \subset S$ such that $M \otimes _ S \kappa (\mathfrak q)$ is nonzero. This follows from Lemma 10.39.15 and 10.39.10. $\square$

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