Lemma 10.82.7. Let $0 \to M_1 \to M_2 \to M_3 \to 0$ be a universally exact sequence of $R$-modules, and suppose $M_2$ is flat. Then $M_1$ and $M_3$ are flat.
Proof. Let $0 \to N \to N' \to N'' \to 0$ be a short exact sequence of $R$-modules. Consider the commutative diagram
\[ \xymatrix{ M_1 \otimes _ R N \ar[r] \ar[d] & M_2 \otimes _ R N \ar[r] \ar[d] & M_3 \otimes _ R N \ar[d] \\ M_1 \otimes _ R N' \ar[r] \ar[d] & M_2 \otimes _ R N' \ar[r] \ar[d] & M_3 \otimes _ R N' \ar[d] \\ M_1 \otimes _ R N'' \ar[r] & M_2 \otimes _ R N'' \ar[r] & M_3 \otimes _ R N'' } \]
(we have dropped the $0$'s on the boundary). By assumption the rows give short exact sequences and the arrow $M_2 \otimes N \to M_2 \otimes N'$ is injective. Clearly this implies that $M_1 \otimes N \to M_1 \otimes N'$ is injective and we see that $M_1$ is flat. In particular the left and middle columns give rise to short exact sequences. It follows from a diagram chase that the arrow $M_3 \otimes N \to M_3 \otimes N'$ is injective. Hence $M_3$ is flat. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: