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). $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)