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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