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

30.2.1.1
\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 (30.2.1.1) 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. $\square$

Comment #4273 by thonguyen on

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

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