The Stacks project

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. $\square$

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