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The Stacks project

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

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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  • 3 comment(s) on Section 10.39: Flat modules and flat ring maps

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