## 15.29 The extended alternating Čech complex

Let $R$ be a ring. Let $f_1, \ldots , f_ r \in R$. The extended alternating Čech complex of $R$ is the cochain complex

$R \to \bigoplus \nolimits _{i_0} R_{f_{i_0}} \to \bigoplus \nolimits _{i_0 < i_1} R_{f_{i_0}f_{i_1}} \to \ldots \to R_{f_1\ldots f_ r}$

where $R$ is in degree $0$, the term $\bigoplus _{i_0} R_{f_{i_0}}$ is in degre $1$, and so on. The maps are defined as follows

1. The map $R \to \bigoplus \nolimits _{i_0} R_{f_{i_0}}$ is given by the canonical maps $R \to R_{f_{i_0}}$.

2. Given $1 \leq i_0 < \ldots < i_{p + 1} \leq r$ and $0 \leq j \leq p + 1$ we have the canonical localization map

$R_{f_{i_0} \ldots \hat f_{i_ j} \ldots f_{i_{p + 1}}} \to R_{f_{i_0} \ldots f_{i_ p}}$
3. The differentials use the canonical maps of (2) with sign $(-1)^ j$.

If $M$ is any $R$-module, the extended alternating Čech complex of $M$ is the similarly constructed cochain complex

$M \to \bigoplus \nolimits _{i_0} M_{f_{i_0}} \to \bigoplus \nolimits _{i_0 < i_1} M_{f_{i_0}f_{i_1}} \to \ldots \to M_{f_1\ldots f_ r}$

where $M$ is in degree $0$ as before.

Lemma 15.29.1. The extended alternating Čech complexes defined above are complexes of $R$-modules.

Proof. Omitted. $\square$

Lemma 15.29.2. Let $R$ be a ring. Let $f_1, \ldots , f_ r \in R$. Let $M$ be an $R$-module. The extended alternating Čech complex of $M$ is the tensor product over $R$ of $M$ with the extended alternating Čech complex of $R$.

Proof. Omitted. $\square$

Lemma 15.29.3. Let $R$ be a ring. Let $f_1, \ldots , f_ r \in R$. Let $M$ be an $R$-module. Let $R \to S$ be a ring map, denote $g_1, \ldots , g_ r \in S$ the images of $f_1, \ldots , f_ r$, and set $N = M \otimes _ R S$. The extended alternating Čech complex constructed using $S$, $g_1, \ldots , g_ r$, and $N$ is the tensor product of the extended alternating Čech complex of $M$ with $S$ over $R$.

Proof. Omitted. $\square$

Lemma 15.29.4. Let $R$ be a ring. Let $f_1, \ldots , f_ r \in R$. Let $M$ be an $R$-module. If there exists an $i \in \{ 1, \ldots , r\}$ such that $f_ i$ is a unit, then the extended alternating Čech complex of $M$ is homotopy equivalent to $0$.

Proof. We will use the following notation: a cochain $x$ of degree $p + 1$ in the extended alternating Čech complex of $M$ is $x = (x_{i_0 \ldots i_ p})$ where $x_{i_0 \ldots i_ p}$ is in $M_{f_{i_0} \ldots f_{i_ p}}$. With this notation we have

$d(x)_{i_0 \ldots i_{p + 1}} = \sum \nolimits _ j (-1)^ j x_{i_0 \ldots \hat i_ j \ldots i_ p}$

As homotopy we use the maps

$h : \text{cochains of degree }p + 2 \to \text{cochains of degree }p + 1$

given by the rule

$h(x)_{i_0 \ldots i_ p} = 0 \text{ if } i \in \{ i_0, \ldots , i_ p\} \text{ and } h(x)_{i_0 \ldots i_ p} = (-1)^ j x_{i_0 \ldots i_ j i i_{j + 1} \ldots i_ p} \text{ if not}$

Here $j$ is the unique index such that $i_ j < i < i_{j + 1}$ in the second case; also, since $f_ i$ is a unit we have the equality

$M_{f_{i_0} \ldots f_{i_ p}} = M_{f_{i_0} \ldots f_{i_ j} f_ i f_{i_{j + 1}} \ldots f_{i_ p}}$

which we can use to make sense of thinking of $(-1)^ j x_{i_0 \ldots i_ j i i_{j + 1} \ldots i_ p}$ as an element of $M_{f_{i_0} \ldots f_{i_ p}}$. We will show by a computation that $d h + h d$ equals the negative of the identity map which finishes the proof. To do this fix $x$ a cochain of degree $p + 1$ and let $1 \leq i_0 < \ldots < i_ p \leq r$.

Case I: $i \in \{ i_0, \ldots , i_ p\}$. Say $i = i_ t$. Then we have $h(d(x))_{i_0 \ldots i_ p} = 0$. On the other hand we have

$d(h(x))_{i_0 \ldots i_ p} = \sum (-1)^ j h(x)_{i_0 \ldots \hat i_ j \ldots i_ p} = (-1)^ t h(x)_{i_0 \ldots \hat i \ldots i_ p} = (-1)^ t (-1)^{t - 1} x_{i_0 \ldots i_ p}$

Thus $(dh + hd)(x)_{i_0 \ldots i_ p} = -x_{i_0 \ldots i_ p}$ as desired.

Case II: $i \not\in \{ i_0, \ldots , i_ p\}$. Let $j$ be such that $i_ j < i < i_{j + 1}$. Then we see that

\begin{align*} h(d(x))_{i_0 \ldots i_ p} & = (-1)^ j d(x)_{i_0 \ldots i_ j i i_{j + 1} \ldots i_ p} \\ & = \sum \nolimits _{j' \leq j} (-1)^{j + j'} x_{i_0 \ldots \hat i_{j'} \ldots i_ j i i_{j + 1} \ldots i_ p} - x_{i_0 \ldots i_ p} \\ & + \sum \nolimits _{j' > j} (-1)^{j + j' + 1} x_{i_0 \ldots i_ j i i_{j + 1} \ldots \hat i_{j'} \ldots i_ p} \end{align*}

On the other hand we have

\begin{align*} d(h(x))_{i_0 \ldots i_ p} & = \sum \nolimits _{j'} (-1)^{j'} h(x)_{i_0 \ldots \hat i_{j'} \ldots i_ p} \\ & = \sum \nolimits _{j' \leq j} (-1)^{j' + j - 1} x_{i_0 \ldots \hat i_{j'} \ldots i_ j i i_{j + 1} \ldots i_ p} \\ & + \sum \nolimits _{j' > j} (-1)^{j' + j} x_{i_0 \ldots i_ j i i_{j + 1} \ldots \hat i_{j'} \ldots i_ p} \end{align*}

Adding these up we obtain $(dh + hd)(x)_{i_0 \ldots i_ p} = - x_{i_0 \ldots i_ p}$ as desired. $\square$

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). $\square$

Lemma 15.29.6. Let $R$ be a ring. Let $f_1, \ldots , f_ r \in R$. The extended alternating Čech complex

$R \to \bigoplus \nolimits _{i_0} R_{f_{i_0}} \to \bigoplus \nolimits _{i_0 < i_1} R_{f_{i_0}f_{i_1}} \to \ldots \to R_{f_1\ldots f_ r}$

is a colimit of the Koszul complexes $K(R, f_1^ n, \ldots , f_ r^ n)$; see proof for a precise statement.

Proof. We urge the reader to prove this for themselves. Denote $K(R, f_1^ n, \ldots , f_ r^ n)$ the Koszul complex of Definition 15.28.2 viewed as a cochain complex sitting in degrees $0, \ldots , r$. Thus we have

$K(R, f_1^ n, \ldots , f_ r^ n) : 0 \to \wedge ^ r(R^{\oplus r}) \to \wedge ^{r - 1}(R^{\oplus r}) \to \ldots \to R^{\oplus r} \to R \to 0$

with the term $\wedge ^ r(R^{\oplus r})$ sitting in degree $0$. Let $e^ n_1, \ldots , e^ n_ r$ be the standard basis of $R^{\oplus r}$. Then the elements $e^ n_{j_1} \wedge \ldots \wedge e^ n_{j_{r - p}}$ for $1 \leq j_1 < \ldots < j_{r - p} \leq r$ form a basis for the term in degree $p$ of the Koszul complex. Further, observe that

$d(e^ n_{j_1} \wedge \ldots \wedge e^ n_{j_{r - p}}) = \sum (-1)^{a + 1} f_{j_ a}^ n e^ n_{j_1} \wedge \ldots \wedge \hat e^ n_{j_ a} \wedge \ldots \wedge e^ n_{j_{r - p}}$

by our construction of the Koszul complex in Section 15.28. The transition maps of our system

$K(R, f_1^ n, \ldots , f_ r^ n) \to K(R, f_1^{n + 1}, \ldots , f_ r^{n + 1})$

are given by the rule

$e^ n_{j_1} \wedge \ldots \wedge e^ n_{j_{r - p}} \longmapsto f_{i_0} \ldots f_{i_{p - 1}} e^{n + 1}_{j_1} \wedge \ldots \wedge e^{n + 1}_{j_{r - p}}$

where the indices $1 \leq i_0 < \ldots < i_{p - 1} \leq r$ are such that $\{ 1, \ldots r\} = \{ i_0, \ldots , i_{p - 1}\} \amalg \{ j_1, \ldots , j_{r - p}\}$. We omit the short computation that shows this is compatible with differentials. Observe that the transition maps are always $1$ in degree $0$ and equal to $f_1 \ldots f_ r$ in degree $r$.

Denote $K^ p(R, f_1^ n, \ldots , f_ r^ n)$ the term of degree $p$ in the Koszul complex. Observe that for any $f \in R$ we have

$R_ f = \mathop{\mathrm{colim}}\nolimits (R \xrightarrow {f} R \xrightarrow {f} R \to \ldots )$

Hence we see that in degree $p$ we obtain

$\mathop{\mathrm{colim}}\nolimits K^ p(R, f_1^ n, \ldots f_ r^ n) = \bigoplus \nolimits _{1 \leq i_0 < \ldots < i_{p - 1} \leq r} R_{f_{i_0} \ldots f_{i_{p - 1}}}$

Here the element $e^ n_{j_1} \wedge \ldots \wedge e^ n_{j_{r - p}}$ of the Koszul complex above maps in the colimit to the element $(f_{i_0} \ldots f_{i_{p - 1}})^{-n}$ in the summand $R_{f_{i_0} \ldots f_{i_{p - 1}}}$ where the indices are chosen such that $\{ 1, \ldots r\} = \{ i_0, \ldots , i_{p - 1}\} \amalg \{ j_1, \ldots , j_{r - p - 2}\}$. Thus the differential on this complex is given by

$d(1\text{ in }R_{f_{i_0} \ldots f_{i_{p - 1}}}) = \sum \nolimits _{i \not\in \{ i_0, \ldots , i_{p - 1}\} } (-1)^{i - t}\text{ in } R_{f_{i_0} \ldots f_{i_ t} f_ i f_{i_{t + 1}} \ldots f_{i_{p - 1}}}$

Thus if we consider the map of complexes given in degree $p$ by the map

$\bigoplus \nolimits _{1 \leq i_0 < \ldots < i_{p - 1} \leq r} R_{f_{i_0} \ldots f_{i_{p - 1}}} \longrightarrow \bigoplus \nolimits _{1 \leq i_0 < \ldots < i_{p - 1} \leq r} R_{f_{i_0} \ldots f_{i_{p - 1}}}$

determined by the rule

$1\text{ in }R_{f_{i_0} \ldots f_{i_{p - 1}}} \longmapsto (-1)^{i_0 + \ldots + i_{p - 1} + p}\text{ in }R_{f_{i_0} \ldots f_{i_{p - 1}}}$

then we get an isomorphism of complexes from $\mathop{\mathrm{colim}}\nolimits K(R, f_1^ n, \ldots , f_ r^ n)$ to the extended alternating Čech complex defined in this section. We omit the verification that the signs work out. $\square$

## Comments (1)

Comment #6205 by Owen on

In the definition of the differentials of the extended alternating Čech complex, the 'canonical localization map' of (2) lands in $R_{f_{i_0}\ldots f_{i_{p+1}}}$ rather than in $R_{f_{i_0}\ldots f_{i_p}}$.

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