The Stacks project

Lemma 21.12.3. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $\mathcal{U} = \{ U_ i \to U\} _{i \in I}$ be a covering of $\mathcal{C}$. Let $\mathcal{I}$ be an injective $\mathcal{O}$-module, i.e., an injective object of $\textit{Mod}(\mathcal{O})$. Then

\[ \check{H}^ p(\mathcal{U}, \mathcal{I}) = \left\{ \begin{matrix} \mathcal{I}(U) & \text{if} & p = 0 \\ 0 & \text{if} & p > 0 \end{matrix} \right. \]

Proof. Lemma 21.9.3 gives the first equality in the following sequence of equalities

\begin{align*} \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{I}) & = \mathop{\mathrm{Mor}}\nolimits _{\textit{PAb}(\mathcal{C})}( \mathbf{Z}_{\mathcal{U}, \bullet }, \mathcal{I}) \\ & = \mathop{\mathrm{Mor}}\nolimits _{\textit{PMod}(\mathbf{Z})}( \mathbf{Z}_{\mathcal{U}, \bullet }, \mathcal{I}) \\ & = \mathop{\mathrm{Mor}}\nolimits _{\textit{PMod}(\mathcal{O})}( \mathbf{Z}_{\mathcal{U}, \bullet } \otimes _{p, \mathbf{Z}} \mathcal{O}, \mathcal{I}) \end{align*}

The third equality by Modules on Sites, Lemma 18.9.2. By Lemma 21.12.1 we see that $\mathcal{I}$ is an injective object in $\textit{PMod}(\mathcal{O})$. Hence $\mathop{\mathrm{Hom}}\nolimits _{\textit{PMod}(\mathcal{O})}(-, \mathcal{I})$ is an exact functor. By Lemma 21.9.5 we see the vanishing of higher Čech cohomology groups. For the zeroth see Lemma 21.8.2. $\square$


Comments (0)


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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 03FC. Beware of the difference between the letter 'O' and the digit '0'.