The Stacks project

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 #6206 by Owen on

Toward the end of the proof it is written the indices are chosen so that . I think is meant.

There are also:

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

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 0913. Beware of the difference between the letter 'O' and the digit '0'.