Lemma 15.29.5. Let $R$ be a ring. Let $f_1, \ldots , f_ r \in R$. Let $M$ be an $R$-module. Let $H^ q$ be the $q$th cohomology module of the extended alternation Čech complex of $M$. Then

1. $H^ q = 0$ if $q \not\in [0, r]$,

2. for $x \in H^ i$ there exists an $n \geq 1$ such that $f_ i^ n x = 0$ for $i = 1, \ldots , r$,

3. the support of $H^ q$ is contained in $V(f_1, \ldots , f_ r)$,

4. if there is an $f \in (f_1, \ldots , f_ r)$ which acts invertibly on $M$, then $H^ q = 0$.

Proof. Part (1) follows from the fact that the extended alternating Čech complex is zero in degrees $< 0$ and $> r$. To prove (2) it suffices to show that for each $i$ there exists an $n \geq 1$ such that $f_ i^ n x = 0$. To see this it suffices to show that $(H^ q)_{f_ i} = 0$. Since localization is exact, $(H^ q)_{f_ i}$ is the $q$th cohomology module of the localization of the extended alternating complex of $M$ at $f_ i$. By Lemma 15.29.3 this localization is the extended alternating Čech complex of $M_{f_ i}$ over $R_{f_ i}$ with respect to the images of $f_1, \ldots , f_ r$ in $R_{f_ i}$. Thus we reduce to showing that $H^ q$ is zero if $f_ i$ is invertible, which follows from Lemma 15.29.4. Part (3) follows from the observation that $(H^ q)_{f_ i} = 0$ for all $i$ that we just proved. To see part (4) note that in this case $f$ acts invertibly on $H^ q$ and $H^ q$ is supported on $V(f)$ by (3). This forces $H^ q$ to be zero (small detail omitted). $\square$

There are also:

• 1 comment(s) on Section 15.29: The extended alternating Čech complex

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).