Lemma 51.2.4. Let $S \subset A$ be a multiplicative set of a ring $A$. Let $M$ be an $A$-module with $S^{-1}M = 0$. Then $\mathop{\mathrm{colim}}\nolimits _{f \in S} H^0_{V(f)}(M) = M$ and $\mathop{\mathrm{colim}}\nolimits _{f \in S} H^1_{V(f)}(M) = 0$.

Proof. The statement on $H^0$ follows directly from the definitions. To see the statement on $H^1$ observe that $R\Gamma _{V(f)}$ and $H^1_{V(f)}$ commute with colimits. Hence we may assume $M$ is annihilated by some $f \in S$. Then $H^1_{V(ff')}(M) = 0$ for all $f' \in S$ (for example by Lemma 51.2.3). $\square$

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