The Stacks project

Lemma 30.2.1. Let $X$ be a scheme. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $\mathcal{U} : U = \bigcup _{i = 1}^ n D(f_ i)$ be a standard open covering of an affine open of $X$. Then $\check{H}^ p(\mathcal{U}, \mathcal{F}) = 0$ for all $p > 0$.

Proof. Write $U = \mathop{\mathrm{Spec}}(A)$ for some ring $A$. In other words, $f_1, \ldots , f_ n$ are elements of $A$ which generate the unit ideal of $A$. Write $\mathcal{F}|_ U = \widetilde{M}$ for some $A$-module $M$. Clearly the Čech complex $\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F})$ is identified with the complex

\[ \prod \nolimits _{i_0} M_{f_{i_0}} \to \prod \nolimits _{i_0i_1} M_{f_{i_0}f_{i_1}} \to \prod \nolimits _{i_0i_1i_2} M_{f_{i_0}f_{i_1}f_{i_2}} \to \ldots \]

We are asked to show that the extended complex
\begin{equation} \label{coherent-equation-extended} 0 \to M \to \prod \nolimits _{i_0} M_{f_{i_0}} \to \prod \nolimits _{i_0i_1} M_{f_{i_0}f_{i_1}} \to \prod \nolimits _{i_0i_1i_2} M_{f_{i_0}f_{i_1}f_{i_2}} \to \ldots \end{equation}

(whose truncation we have studied in Algebra, Lemma 10.24.1) is exact. It suffices to show that ( is exact after localizing at a prime $\mathfrak p$, see Algebra, Lemma 10.23.1. In fact we will show that the extended complex localized at $\mathfrak p$ is homotopic to zero.

There exists an index $i$ such that $f_ i \not\in \mathfrak p$. Choose and fix such an element $i_{\text{fix}}$. Note that $M_{f_{i_{\text{fix}}}, \mathfrak p} = M_{\mathfrak p}$. Similarly for a localization at a product $f_{i_0} \ldots f_{i_ p}$ and $\mathfrak p$ we can drop any $f_{i_ j}$ for which $i_ j = i_{\text{fix}}$. Let us define a homotopy

\[ h : \prod \nolimits _{i_0 \ldots i_{p + 1}} M_{f_{i_0} \ldots f_{i_{p + 1}}, \mathfrak p} \longrightarrow \prod \nolimits _{i_0 \ldots i_ p} M_{f_{i_0} \ldots f_{i_ p}, \mathfrak p} \]

by the rule

\[ h(s)_{i_0 \ldots i_ p} = s_{i_{\text{fix}} i_0 \ldots i_ p} \]

(This is “dual” to the homotopy in the proof of Cohomology, Lemma 20.10.4.) In other words, $h : \prod _{i_0} M_{f_{i_0}, \mathfrak p} \to M_\mathfrak p$ is projection onto the factor $M_{f_{i_{\text{fix}}}, \mathfrak p} = M_{\mathfrak p}$ and in general the map $h$ equal projection onto the factors $M_{f_{i_{\text{fix}}} f_{i_1} \ldots f_{i_{p + 1}}, \mathfrak p} = M_{f_{i_1} \ldots f_{i_{p + 1}}, \mathfrak p}$. We compute

\begin{align*} (dh + hd)(s)_{i_0 \ldots i_ p} & = \sum \nolimits _{j = 0}^ p (-1)^ j h(s)_{i_0 \ldots \hat i_ j \ldots i_ p} + d(s)_{i_{\text{fix}} i_0 \ldots i_ p}\\ & = \sum \nolimits _{j = 0}^ p (-1)^ j s_{i_{\text{fix}} i_0 \ldots \hat i_ j \ldots i_ p} + s_{i_0 \ldots i_ p} + \sum \nolimits _{j = 0}^ p (-1)^{j + 1} s_{i_{\text{fix}} i_0 \ldots \hat i_ j \ldots i_ p} \\ & = s_{i_0 \ldots i_ p} \end{align*}

This proves the identity map is homotopic to zero as desired. $\square$

Comments (2)

Comment #4273 by thonguyen on

On the line 17 of the proof, I think there is a typo on it should be Thanks.

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 01X9. Beware of the difference between the letter 'O' and the digit '0'.