Lemma 15.28.7. Let $R$ be a ring. Let $\varphi : E \to R$ be an $R$-module map. Let $f \in R$. Set $E' = E \oplus R$ and define $\varphi ' : E' \to R$ by $\varphi$ on $E$ and multiplication by $f$ on $R$. The complex $K_\bullet (\varphi ')$ is isomorphic to the cone of the map of complexes

$f : K_\bullet (\varphi ) \longrightarrow K_\bullet (\varphi ).$

Proof. Denote $e_0 \in E'$ the element $1 \in R \subset R \oplus E$. By our definition of the cone above we see that

$C(f)_ n = K_ n(\varphi ) \oplus K_{n - 1}(\varphi ) = \wedge ^ n(E) \oplus \wedge ^{n - 1}(E) = \wedge ^ n(E')$

where in the last $=$ we map $(0, e_1 \wedge \ldots \wedge e_{n - 1})$ to $e_0 \wedge e_1 \wedge \ldots \wedge e_{n - 1}$ in $\wedge ^ n(E')$. A computation shows that this isomorphism is compatible with differentials. Namely, this is clear for elements of the first summand as $\varphi '|_ E = \varphi$ and $d_{C(f)}$ restricted to the first summand is just $d_{K_\bullet (\varphi )}$. On the other hand, if $e_1 \wedge \ldots \wedge e_{n - 1}$ is in the second summand, then

$d_{C(f)}(0, e_1 \wedge \ldots \wedge e_{n - 1}) = fe_1 \wedge \ldots \wedge e_{n - 1} - d_{K_\bullet (\varphi )}(e_1 \wedge \ldots \wedge e_{n - 1})$

and on the other hand

\begin{align*} & d_{K_\bullet (\varphi ')}(0, e_0 \wedge e_1 \wedge \ldots \wedge e_{n - 1}) \\ & = \sum \nolimits _{i = 0, \ldots , n - 1} (-1)^ i \varphi '(e_ i)e_0 \wedge \ldots \wedge \widehat{e_ i} \wedge \ldots \wedge e_{n - 1} \\ & = fe_1 \wedge \ldots \wedge e_{n - 1} + \sum \nolimits _{i = 1, \ldots , n - 1} (-1)^ i \varphi (e_ i)e_0 \wedge \ldots \wedge \widehat{e_ i} \wedge \ldots \wedge e_{n - 1} \\ & = fe_1 \wedge \ldots \wedge e_{n - 1} - e_0 \left(\sum \nolimits _{i = 1, \ldots , n - 1} (-1)^{i + 1} \varphi (e_ i)e_1 \wedge \ldots \wedge \widehat{e_ i} \wedge \ldots \wedge e_{n - 1}\right) \end{align*}

which is the image of the result of the previous computation. $\square$

Comment #2772 by on

"in the first summand" should be "in the second summand", right?

Also, I'd replace "d_{K_\bullet (\varphi)} (e_1 \wedge ... \wedge e_{n-1}" by "(0, d_{K_\bullet (\varphi)} (e_1 \wedge ... \wedge e_{n-1})".

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