The Stacks project

Lemma 18.28.15. Let $\mathcal{C}$ be a site. Let $\mathcal{O}' \to \mathcal{O}$ be a surjection of sheaves of rings whose kernel $\mathcal{I}$ is an ideal of square zero. Let $\mathcal{F}'$ be an $\mathcal{O}'$-module and set $\mathcal{F} = \mathcal{F}'/\mathcal{I}\mathcal{F}'$. The following are equivalent

  1. $\mathcal{F}'$ is a flat $\mathcal{O}'$-module, and

  2. $\mathcal{F}$ is a flat $\mathcal{O}$-module and $\mathcal{I} \otimes _\mathcal {O} \mathcal{F} \to \mathcal{F}'$ is injective.

Proof. If (1) holds, then $\mathcal{F} = \mathcal{F}' \otimes _{\mathcal{O}'} \mathcal{O}$ is flat over $\mathcal{O}$ by Lemma 18.28.13 and we see the map $\mathcal{I} \otimes _\mathcal {O} \mathcal{F} \to \mathcal{F}'$ is injective by applying $- \otimes _{\mathcal{O}'} \mathcal{F}'$ to the exact sequence $0 \to \mathcal{I} \to \mathcal{O}' \to \mathcal{O} \to 0$, see Lemma 18.28.9. Assume (2). In the rest of the proof we will use without further mention that $\mathcal{K} \otimes _{\mathcal{O}'} \mathcal{F}' = \mathcal{K} \otimes _\mathcal {O} \mathcal{F}$ for any $\mathcal{O}'$-module $\mathcal{K}$ annihilated by $\mathcal{I}$. Let $\alpha : \mathcal{G}' \to \mathcal{H}'$ be an injective map of $\mathcal{O}'$-modules. Let $\mathcal{G} \subset \mathcal{G}'$, resp. $\mathcal{H} \subset \mathcal{H}'$ be the subsheaf of sections annihilated by $\mathcal{I}$. Consider the diagram

\[ \xymatrix{ \mathcal{G} \otimes _{\mathcal{O}'} \mathcal{F}' \ar[r] \ar[d] & \mathcal{G}' \otimes _{\mathcal{O}'} \mathcal{F}' \ar[r] \ar[d] & \mathcal{G}'/\mathcal{G} \otimes _{\mathcal{O}'} \mathcal{F}' \ar[r] \ar[d] & 0 \\ \mathcal{H} \otimes _{\mathcal{O}'} \mathcal{F}' \ar[r] & \mathcal{H}' \otimes _{\mathcal{O}'} \mathcal{F}' \ar[r] & \mathcal{H}'/\mathcal{H} \otimes _{\mathcal{O}'} \mathcal{F}' \ar[r] & 0 } \]

Note that $\mathcal{G}'/\mathcal{G}$ and $\mathcal{H}'/\mathcal{H}$ are annihilated by $\mathcal{I}$ and that $\mathcal{G}'/\mathcal{G} \to \mathcal{H}'/\mathcal{H}$ is injective. Thus the right vertical arrow is injective as $\mathcal{F}$ is flat over $\mathcal{O}$. The same is true for the left vertical arrow. Hence the middle vertical arrow is injective and $\mathcal{F}'$ is flat. $\square$

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