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.

There are also:

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

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00HM. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 10.39.13 as pdf