The Stacks Project


Tag 062A

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

Lemma 15.26.9. Let $R$ be a ring. Let $A_\bullet$ be a complex of $R$-modules. Let $f, g \in R$. Let $C(f)_\bullet$ be the cone of $f : A_\bullet \to A_\bullet$. Define similarly $C(g)_\bullet$ and $C(fg)_\bullet$. Then $C(fg)_\bullet$ is homotopy equivalent to the cone of a map $$ C(f)_\bullet[1] \longrightarrow C(g)_\bullet $$

Proof. We first prove this if $A_\bullet$ is the complex consisting of $R$ placed in degree $0$. In this case the map we use is $$ \xymatrix{ 0 \ar[r] \ar[d] & 0 \ar[r] \ar[d] & R \ar[r]^f \ar[d]^1 & R \ar[r] \ar[d] & 0 \ar[d] \\ 0 \ar[r] & R \ar[r]^g & R \ar[r] & 0 \ar[r] & 0 } $$ The cone of this is the chain complex consisting of $R \oplus R$ placed in degrees $1$ and $0$ and differential (15.26.6.1) $$ \left( \begin{matrix} g & 1 \\ 0 & -f \end{matrix} \right) : R^{\oplus 2} \longrightarrow R^{\oplus 2} $$ We leave it to the reader to show this this chain complex is homotopic to the complex $fg : R \to R$. In general we write $C(f)_\bullet$ and $C(g)_\bullet$ as the total complex of the double complexes $$ (R \xrightarrow{f} R) \otimes_R A_\bullet \quad\text{and}\quad (R \xrightarrow{g} R) \otimes_R A_\bullet $$ and in this way we deduce the result from the special case discussed above. Some details omitted. $\square$

    The code snippet corresponding to this tag is a part of the file more-algebra.tex and is located in lines 5671–5681 (see updates for more information).

    \begin{lemma}
    \label{lemma-cone-squared}
    Let $R$ be a ring. Let $A_\bullet$ be a complex of $R$-modules.
    Let $f, g \in R$. Let $C(f)_\bullet$ be the cone of
    $f : A_\bullet \to A_\bullet$. Define similarly $C(g)_\bullet$ and
    $C(fg)_\bullet$. Then $C(fg)_\bullet$ is homotopy equivalent to the
    cone of a map
    $$
    C(f)_\bullet[1] \longrightarrow C(g)_\bullet
    $$
    \end{lemma}
    
    \begin{proof}
    We first prove this if $A_\bullet$ is the complex consisting of $R$ placed
    in degree $0$. In this case the map we use is
    $$
    \xymatrix{
    0 \ar[r] \ar[d] &
    0 \ar[r] \ar[d] &
    R \ar[r]^f \ar[d]^1 &
    R \ar[r] \ar[d] & 0 \ar[d] \\
    0 \ar[r] & R \ar[r]^g & R \ar[r] & 0 \ar[r] & 0
    }
    $$
    The cone of this is the chain complex consisting of $R \oplus R$ placed in
    degrees $1$ and $0$ and differential (\ref{equation-differential-cone})
    $$
    \left(
    \begin{matrix}
    g & 1 \\
    0 & -f
    \end{matrix}
    \right) :
    R^{\oplus 2} \longrightarrow R^{\oplus 2}
    $$
    We leave it to the reader to show this this chain complex is
    homotopic to the complex $fg : R \to R$. In general we
    write $C(f)_\bullet$ and $C(g)_\bullet$
    as the total complex of the double complexes
    $$
    (R \xrightarrow{f} R) \otimes_R A_\bullet
    \quad\text{and}\quad
    (R \xrightarrow{g} R) \otimes_R A_\bullet
    $$
    and in this way we deduce the result from the special case discussed above.
    Some details omitted.
    \end{proof}

    Comments (2)

    Comment #2773 by Darij Grinberg (site) on August 19, 2017 a 2:21 am UTC

    "this this".

    Comment #2776 by Darij Grinberg (site) on August 19, 2017 a 2:32 am UTC

    This said, I really wouldn't mind a more mundane proof by explicit description of the map and of the quasi-isomorphisms... I'm not sure if I can follow the proof above.

    Add a comment on tag 062A

    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 lower-right corner).

    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 box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

    This captcha seems more appropriate than the usual illegible gibberish, right?