The Stacks project

Proof. The map $\mathcal{O}' \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{I}, \mathcal{I})$ factors through $\mathcal{O}$ as $\mathcal{I} \cdot \mathcal{I} = 0$ by assumption. Hence $\mathcal{I}$ is a sheaf of $\mathcal{O}$-modules and this defines the $R$-module structure on $M$. The boundary map is additive hence it suffices to prove the Leibniz rule. Let $f \in R$. Choose an open covering $\mathcal{U} : X = \bigcup U_ i$ such that there exist $f_ i \in \mathcal{O}'(U_ i)$ lifting $f|_{U_ i} \in \mathcal{O}(U_ i)$. Observe that $f_ i - f_ j$ is an element of $\mathcal{I}(U_ i \cap U_ j)$. Then $\partial (f)$ corresponds to the Čech cohomology class of the $1$-cocycle $\alpha $ with $\alpha _{i_0i_1} = f_{i_0} - f_{i_1}$. (Observe that by Lemma 20.11.3 the first Čech cohomology group with respect to $\mathcal{U}$ is a submodule of $M$.) Next, let $g \in R$ be a second element and assume (after possibly refining the open covering) that $g_ i \in \mathcal{O}'(U_ i)$ lifts $g|_{U_ i} \in \mathcal{O}(U_ i)$. Then we see that $\partial (g)$ is given by the cocycle $\beta $ with $\beta _{i_0i_1} = g_{i_0} - g_{i_1}$. Since $f_ ig_ i \in \mathcal{O}'(U_ i)$ lifts $fg|_{U_ i}$ we see that $\partial (fg)$ is given by the cocycle $\gamma $ with

by our definition of the $\mathcal{O}$-module structure on $\mathcal{I}$. This proves the Leibniz rule and the proof is complete. $\square$