Lemma 10.39.13 (00HM)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.39: Flat modules and flat ring maps
Lemma 10.39.13 (cite)

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Lemma 10.39.13. Suppose that $0 \to M' \to M \to M'' \to 0$ is a short exact sequence of $R$-modules. If $M'$ and $M''$ are flat so is $M$. If $M$ and $M''$ are flat so is $M'$.

Proof. We will use the criterion that a module $N$ is flat if for every ideal $I \subset R$ the map $N \otimes _ R I \to N$ is injective, see Lemma 10.39.5. Consider an ideal $I \subset R$. Consider the diagram

\[ \begin{matrix} 0 & \to & M' & \to & M & \to & M'' & \to & 0 \\ & & \uparrow & & \uparrow & & \uparrow & & \\ & & M'\otimes _ R I & \to & M \otimes _ R I & \to & M''\otimes _ R I & \to & 0 \end{matrix} \]

with exact rows. This immediately proves the first assertion. The second follows because if $M''$ is flat then the lower left horizontal arrow is injective by Lemma 10.39.12. $\square$

Comments (1)

Comment #9876 by YoyoPan on November 28, 2024 at 05:42

Sorry but I didnt really understand the first proof... the flat module M is right exact functor that be like _*M, and it seems not use M' and M'' are flat.

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