Lemma 36.6.12. With $X$, $f_1, \ldots , f_ c \in \Gamma (X, \mathcal{O}_ X)$, and $\mathcal{F}$ as in Remark 36.6.4. Let $a_{ji} \in \Gamma (X, \mathcal{O}_ X)$ for $1 \leq i, j \leq c$ and set $g_ j = \sum _{i = 1, \ldots , c} a_{ji}f_ i$. Assume $g_1, \ldots , g_ c$ scheme theoretically cut out $Z$. If $\mathcal{F}$ is quasi-coherent, then
\[ c_{f_1, \ldots , f_ c} = \det (a_{ji}) c_{g_1, \ldots , g_ c} \]
where $c_{f_1, \ldots , f_ c}$ and $c_{g_1, \ldots , g_ c}$ are as in Remark 36.6.10.
Proof.
We will prove that $c_{f_1, \ldots , f_ c}(s) = \det (a_{ij}) c_{g_1, \ldots , g_ c}(s)$ as global sections of $\mathcal{H}_ Z(\mathcal{F})$ for any $s \in \mathcal{F}(X)$. This is sufficient since we then obtain the same result for section over any open subscheme of $X$. To do this, for $1 \leq i_0 < \ldots < i_ p \leq c$ and $1 \leq j_0 < \ldots < j_ q \leq c$ we denote $U_{i_0 \ldots i_ p} \subset X$, $V_{j_0 \ldots j_ q} \subset X$, and $W_{i_0 \ldots i_ p, j_0 \ldots j_ q} \subset X$ the open subscheme where $f_{i_0} \ldots f_{i_ p}$ is invertible, $g_{j_0} \ldots g_{j_ q}$ is invertible, and where $f_{i_0} \ldots f_{i_ p}g_{j_0} \ldots g_{j_ q}$ is invertible. We denote $\mathcal{F}_{i_0 \ldots i_ p}$, resp. $\mathcal{F}'_{j_0 \ldots j_ q}$ $\mathcal{F}''_{i_0 \ldots i_ p, j_0 \ldots j_ q}$ the pushforward to $X$ of the restriction of $\mathcal{F}$ to $U_{i_0 \ldots i_ p}$, resp. $V_{j_0 \ldots j_ q}$, resp. $W_{i_0 \ldots i_ p, j_0 \ldots j_ q}$. Then we obtain three extended alternating Čech complexes
\[ \mathcal{F}^\bullet : \mathcal{F} \to \bigoplus \nolimits _{i_0} \mathcal{F}_{i_0} \to \bigoplus \nolimits _{i_0 < i_1} \mathcal{F}_{i_0i_1} \to \ldots \]
and
\[ (\mathcal{F}')^\bullet : \mathcal{F} \to \bigoplus \nolimits _{j_0} \mathcal{F}'_{j_0} \to \bigoplus \nolimits _{j_0 < j_1} \mathcal{F}'_{j_0j_1} \to \ldots \]
and
\[ (\mathcal{F}'')^\bullet : \mathcal{F} \to \bigoplus \nolimits _{i_0} \mathcal{F}_{i_0} \oplus \bigoplus \nolimits _{j_0} \mathcal{F}'_{j_0} \to \bigoplus \nolimits _{i_0 < i_1} \mathcal{F}_{i_0i_1} \oplus \bigoplus \nolimits _{i_0, j_0} \mathcal{F}''_{i_0, j_0} \oplus \bigoplus \nolimits _{j_0 < j_1} \mathcal{F}'_{j_0j_1} \to \ldots \]
whose differentials are those used in defining (36.6.4.1). There are maps of complexes
\[ (\mathcal{F}'')^\bullet \to \mathcal{F}^\bullet \quad \text{and}\quad (\mathcal{F}'')^\bullet \to (\mathcal{F}')^\bullet \]
given by the projection maps on the terms (and hence inducing the identity map in degree $0$). Observe that by Lemma 36.6.7 each of these complexes represents $i_*R\mathcal{H}_ Z(\mathcal{F})$ and these maps represent the identity on this object. Thus it suffices to find an element
\[ \sigma \in H^ c((\mathcal{F}'')^\bullet (X)) \]
mapping to $c_{f_1, \ldots , f_ c}(s)$ and $\det (a_{ji})c_{g_1, \ldots , g_ c}(s)$ by these two maps. It turns out we can explicitly give a cocycle for $\sigma $. Namely, we take
\[ \sigma _{1 \ldots c} = \frac{s}{f_1 \ldots f_ c} \in \mathcal{F}_{1 \ldots c}(X) \quad \text{and}\quad \sigma '_{1 \ldots c} = \frac{\det (a_{ji})s}{g_1 \ldots g_ c} \in \mathcal{F}'_{1 \ldots c}(X) \]
and we take
\[ \sigma _{i_0 \ldots i_ p, j_0 \ldots j_{c - p - 2}} = \frac{\lambda (i_0 \ldots i_ p, j_0 \ldots j_{c - p - 2})s}{f_{i_0} \ldots f_{i_ p}g_{j_0} \ldots g_{j_{c - p - 2}}} \in \mathcal{F}''_{i_0 \ldots i_ p, j_0 \ldots j_{c - p - 2}}(X) \]
where $\lambda (i_0 \ldots i_ p, j_0 \ldots j_{c - p - 2})$ is the coefficient of $e_1 \wedge \ldots \wedge e_ c$ in the formal expression
\[ e_{i_0} \wedge \ldots \wedge e_{i_ p} \wedge (a_{j_01} e_1 + \ldots + a_{j_0c}e_ c) \wedge \ldots \wedge (a_{j_{c - p - 2}1} e_1 + \ldots + a_{j_{c - p - 2}c}e_ c) \]
To verify that $\sigma $ is a cocycle, we have to show for $1 \leq i_0 < \ldots < i_ p \leq c$ and $1 \leq j_0 < \ldots < j_{c - p - 1} \leq c$ that we have
\begin{align*} 0 & = \sum \nolimits _{a = 0, \ldots , p} (-1)^ a f_{i_ a} \lambda (i_0 \ldots \hat i_ a \ldots i_ p, j_0 \ldots j_{c - p - 1}) \\ & + \sum \nolimits _{b = 0, \ldots , c - p - 1} (-1)^{p + b + 1}g_{j_ b} \lambda (i_0 \ldots i_ p, j_0 \ldots \hat j_ b \ldots j_{c - p - 1}) \end{align*}
The easiest way to see this is perhaps to argue that the formal expression
\[ \xi = e_{i_0} \wedge \ldots \wedge e_{i_ p} \wedge (a_{j_01} e_1 + \ldots + a_{j_0c}e_ c) \wedge \ldots \wedge (a_{j_{c - p - 1}1} e_1 + \ldots + a_{j_{c - p - 1}c}e_ c) \]
is $0$ as it is an element of the $(c + 1)$st wedge power of the free module on $e_1, \ldots , e_ c$ and that the expression above is the image of $\xi $ under the Koszul differential sending $e_ i \to f_ i$. Some details omitted.
$\square$
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