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Tag 0628

Chapter 15: More on Algebra > Section 15.26: The Koszul complex

Lemma 15.26.7. 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 first 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')}(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. $\square$

    The code snippet corresponding to this tag is a part of the file more-algebra.tex and is located in lines 5597–5610 (see updates for more information).

    \begin{lemma}
    \label{lemma-cone-koszul-abstract}
    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
    $$
    f :
    K_\bullet(\varphi)
    \longrightarrow
    K_\bullet(\varphi).
    $$
    \end{lemma}
    
    \begin{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 first 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')}(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.
    \end{proof}

    Comments (1)

    Comment #2772 by Darij Grinberg (site) on August 19, 2017 a 2:16 am UTC

    "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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