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15.26. The Koszul complex

We define the Koszul complex as follows.

Definition 15.26.1. Let $R$ be a ring. Let $\varphi : E \to R$ be an $R$-module map. The Koszul complex $K_\bullet(\varphi)$ associated to $\varphi$ is the commutative differential graded algebra defined as follows:

  1. the underlying graded algebra is the exterior algebra $K_\bullet(\varphi) = \wedge(E)$,
  2. the differential $d : K_\bullet(\varphi) \to K_\bullet(\varphi)$ is the unique derivation such that $d(e) = \varphi(e)$ for all $e \in E = K_1(\varphi)$.

Explicitly, if $e_1 \wedge \ldots \wedge e_n$ is one of the generators of degree $n$ in $K_\bullet(\varphi)$, then $$ d(e_1 \wedge \ldots \wedge e_n) = \sum\nolimits_{i = 1, \ldots, n} (-1)^{i + 1} \varphi(e_i)e_1 \wedge \ldots \wedge \widehat{e_i} \wedge \ldots \wedge e_n. $$ It is straightforward to see that this gives a well defined derivation on the tensor algebra, which annihilates $e \otimes e$ and hence factors through the exterior algebra.

We often assume that $E$ is a finite free module, say $E = R^{\oplus n}$. In this case the map $\varphi$ is given by a sequence of elements $f_1, \ldots, f_n \in R$.

Definition 15.26.2. Let $R$ be a ring and let $f_1, \ldots, f_r \in R$. The Koszul complex on $f_1, \ldots, f_r$ is the Koszul complex associated to the map $(f_1, \ldots, f_r) : R^{\oplus r} \to R$. Notation $K_\bullet(f_\bullet)$, $K_\bullet(f_1, \ldots, f_r)$, $K_\bullet(R, f_1, \ldots, f_r)$, or $K_\bullet(R, f_\bullet)$.

Of course, if $E$ is finite locally free, then $K_\bullet(\varphi)$ is locally on $\mathop{\rm Spec}(R)$ isomorphic to a Koszul complex $K_\bullet(f_1, \ldots, f_r)$. This complex has many interesting formal properties.

Lemma 15.26.3. Let $\varphi : E \to R$ and $\varphi' : E' \to R$ be $R$-module maps. Let $\psi : E \to E'$ be an $R$-module map such that $\varphi' \circ \psi = \varphi$. Then $\psi$ induces a homomorphism of differential graded algebras $K_\bullet(\varphi) \to K_\bullet(\varphi')$.

Proof. This is immediate from the definitions. $\square$

Lemma 15.26.4. Let $f_1, \ldots, f_r \in R$ be a sequence. Let $(x_{ij})$ be an invertible $r \times r$-matrix with coefficients in $R$. Then the complexes $K_\bullet(f_\bullet)$ and $$ K_\bullet(\sum x_{1j}f_j, \sum x_{2j}f_j, \ldots, \sum x_{rj}f_j) $$ are isomorphic.

Proof. Set $g_i = \sum x_{ij}f_j$. The matrix $(x_{ji})$ gives an isomorphism $x : R^{\oplus r} \to R^{\oplus r}$ such that $(g_1, \ldots, g_r) = (f_1, \ldots, f_r) \circ x$. Hence this follows from the functoriality of the Koszul complex described in Lemma 15.26.3. $\square$

Lemma 15.26.5. Let $R$ be a ring. Let $\varphi : E \to R$ be an $R$-module map. Let $e \in E$ with image $f = \varphi(e)$ in $R$. Then $$ f = de + ed $$ as endomorphisms of $K_\bullet(\varphi)$.

Proof. This is true because $d(ea) = d(e)a - ed(a) = fa - ed(a)$. $\square$

Lemma 15.26.6. Let $R$ be a ring. Let $f_1, \ldots, f_r \in R$ be a sequence. Multiplication by $f_i$ on $K_\bullet(f_\bullet)$ is homotopic to zero, and in particular the cohomology modules $H_i(K_\bullet(f_\bullet))$ are annihilated by the ideal $(f_1, \ldots, f_r)$.

Proof. Special case of Lemma 15.26.5. $\square$

In Derived Categories, Section 13.9 we defined the cone of a morphism of cochain complexes. The cone $C(f)_\bullet$ of a morphism of chain complexes $f : A_\bullet \to B_\bullet$ is the complex $C(f)_\bullet$ given by $C(f)_n = B_n \oplus A_{n - 1}$ and differential \begin{equation} \tag{15.26.6.1} d_{C(f), n} = \left( \begin{matrix} d_{B, n} & f_{n - 1} \\ 0 & -d_{A, n - 1} \end{matrix} \right) \end{equation} It comes equipped with canonical morphisms of complexes $i : B_\bullet \to C(f)_\bullet$ and $p : C(f)_\bullet \to A_\bullet[-1]$ induced by the obvious maps $B_n \to C(f)_n \to A_{n - 1}$.

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. $\square$

Lemma 15.26.8. Let $R$ be a ring. Let $f_1, \ldots, f_r$ be a sequence of elements of $R$. The complex $K_\bullet(f_1, \ldots, f_r)$ is isomorphic to the cone of the map of complexes $$ f_r : K_\bullet(f_1, \ldots, f_{r - 1}) \longrightarrow K_\bullet(f_1, \ldots, f_{r - 1}). $$

Proof. Special case of Lemma 15.26.7. $\square$

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. $\square$

Lemma 15.26.10. Let $R$ be a ring. Let $\varphi : E \to R$ be an $R$-module map. Let $f, g \in R$. Set $E' = E \oplus R$ and define $\varphi'_f, \varphi'_g, \varphi'_{fg} : E' \to R$ by $\varphi$ on $E$ and multiplication by $f, g, fg$ on $R$. The complex $K_\bullet(\varphi'_{fg})$ is isomorphic to the cone of a map of complexes $$ K_\bullet(\varphi'_f)[1] \longrightarrow K_\bullet(\varphi'_g). $$

Proof. By Lemma 15.26.7 the complex $K_\bullet(\varphi'_f)$ is isomorphic to the cone of multiplication by $f$ on $K_\bullet(\varphi)$ and similarly for the other two cases. Hence the lemma follows from Lemma 15.26.9. $\square$

Lemma 15.26.11. Let $R$ be a ring. Let $f_1, \ldots, f_{r - 1}$ be a sequence of elements of $R$. Let $f, g \in R$. The complex $K_\bullet(f_1, \ldots, f_{r - 1}, fg)$ is homotopy equivalent to the cone of a map of complexes $$ K_\bullet(f_1, \ldots, f_{r - 1}, f)[1] \longrightarrow K_\bullet(f_1, \ldots, f_{r - 1}, g) $$

Proof. Special case of Lemma 15.26.10. $\square$

Lemma 15.26.12. Let $A$ be a ring. Let $f_1, \ldots, f_r$, $g_1, \ldots, g_s$ be elements of $A$. Then there is an isomorphism of Koszul complexes $$ K_\bullet(A, f_1, \ldots, f_r, g_1, \ldots, g_s) = \text{Tot}(K_\bullet(A, f_1, \ldots, f_r) \otimes_A K_\bullet(A, g_1, \ldots, g_s)). $$

Proof. Omitted. Hint: If $K_\bullet(A, f_1, \ldots, f_r)$ is generated as a differential graded algebra by $x_1, \ldots, x_r$ with $\text{d}(x_i) = f_i$ and $K_\bullet(A, g_1, \ldots, g_s)$ is generated as a differential graded algebra by $y_1, \ldots, y_s$ with $\text{d}(y_j) = g_j$, then we can think of $K_\bullet(A, f_1, \ldots, f_r, g_1, \ldots, g_s)$ as the differential graded algebra generated by the sequence of elements $x_1, \ldots, x_r, y_1, \ldots, y_r$ with $\text{d}(x_i) = f_i$ and $\text{d}(y_j) = g_j$. $\square$

Lemma 15.26.13. Let $R$ be a ring. Let $f_1, \ldots, f_r \in R$. The extended alternating Čech complex $$ R \to \prod\nolimits_{i_0} R_{f_{i_0}} \to \prod\nolimits_{i_0 < i_1} R_{f_{i_0}f_{i_1}} \to \ldots \to R_{f_1\ldots f_r} $$ is a colimit of the Koszul complexes $K(R, f_1^n, \ldots, f_r^n)$.

Proof. The transition maps $K(R, f_1^n, \ldots, f_r^n) \to K(R, f_1^{n + 1}, \ldots, f_r^{n + 1})$ are the maps sending $e_{i_1} \wedge \ldots \wedge e_{i_p}$ to $f_{i_{p + 1}} \ldots f_{i_r} e_{i_1} \wedge \ldots \wedge e_{i_p}$ where the indices are such that $\{1, \ldots, r\} = \{i_1, \ldots, i_r\}$. In particular the transition maps are always $1$ in degree $r$ and equal to $f_1\ldots f_r$ in degree $0$. The terms of the colimit are equal to the terms of the extended alternating Čech complex by Algebra, Lemma 10.9.9. $\square$

    The code snippet corresponding to this tag is a part of the file more-algebra.tex and is located in lines 5416–5765 (see updates for more information).

    \section{The Koszul complex}
    \label{section-koszul}
    
    \noindent
    We define the Koszul complex as follows.
    
    \begin{definition}
    \label{definition-koszul}
    Let $R$ be a ring. Let $\varphi : E \to R$ be an $R$-module map. The
    {\it Koszul complex} $K_\bullet(\varphi)$ associated to $\varphi$
    is the commutative differential graded algebra defined as follows:
    \begin{enumerate}
    \item the underlying graded algebra is the exterior algebra
    $K_\bullet(\varphi) = \wedge(E)$,
    \item the differential $d : K_\bullet(\varphi) \to K_\bullet(\varphi)$
    is the unique derivation such that $d(e) = \varphi(e)$ for all
    $e \in E = K_1(\varphi)$.
    \end{enumerate}
    \end{definition}
    
    \noindent
    Explicitly, if $e_1 \wedge \ldots \wedge e_n$ is one of the generators of
    degree $n$ in $K_\bullet(\varphi)$, then
    $$
    d(e_1 \wedge \ldots \wedge e_n) =
    \sum\nolimits_{i = 1, \ldots, n} (-1)^{i + 1}
    \varphi(e_i)e_1 \wedge \ldots \wedge \widehat{e_i} \wedge \ldots \wedge e_n.
    $$
    It is straightforward to see that this gives a well defined derivation
    on the tensor algebra, which annihilates $e \otimes e$ and hence factors
    through the exterior algebra.
    
    \medskip\noindent
    We often assume that $E$ is a finite free module, say $E = R^{\oplus n}$.
    In this case the map $\varphi$ is given by a sequence of elements
    $f_1, \ldots, f_n \in R$.
    
    \begin{definition}
    \label{definition-koszul-complex}
    Let $R$ be a ring and let $f_1, \ldots, f_r \in R$. The
    {\it Koszul complex on $f_1, \ldots, f_r$} is the Koszul complex
    associated to the map $(f_1, \ldots, f_r) : R^{\oplus r} \to R$.
    Notation $K_\bullet(f_\bullet)$, $K_\bullet(f_1, \ldots, f_r)$,
    $K_\bullet(R, f_1, \ldots, f_r)$, or $K_\bullet(R, f_\bullet)$.
    \end{definition}
    
    \noindent
    Of course, if $E$ is finite locally free, then $K_\bullet(\varphi)$ is
    locally on $\Spec(R)$ isomorphic to a Koszul complex
    $K_\bullet(f_1, \ldots, f_r)$.
    This complex has many interesting formal properties.
    
    \begin{lemma}
    \label{lemma-functorial}
    Let $\varphi : E \to R$ and $\varphi' : E' \to R$ be $R$-module maps.
    Let $\psi : E \to E'$ be an $R$-module map such that
    $\varphi' \circ \psi = \varphi$. Then $\psi$ induces a
    homomorphism of differential graded algebras
    $K_\bullet(\varphi) \to K_\bullet(\varphi')$.
    \end{lemma}
    
    \begin{proof}
    This is immediate from the definitions.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-change-basis}
    Let $f_1, \ldots, f_r \in R$ be a sequence.
    Let $(x_{ij})$ be an invertible $r \times r$-matrix with
    coefficients in $R$. Then the complexes
    $K_\bullet(f_\bullet)$ and
    $$
    K_\bullet(\sum x_{1j}f_j, \sum x_{2j}f_j, \ldots, \sum x_{rj}f_j)
    $$
    are isomorphic.
    \end{lemma}
    
    \begin{proof}
    Set $g_i = \sum x_{ij}f_j$.
    The matrix $(x_{ji})$ gives an isomorphism $x : R^{\oplus r} \to R^{\oplus r}$
    such that $(g_1, \ldots, g_r) = (f_1, \ldots, f_r) \circ x$.
    Hence this follows from the functoriality of the Koszul complex
    described in
    Lemma \ref{lemma-functorial}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-homotopy-koszul-abstract}
    Let $R$ be a ring. Let $\varphi : E \to R$ be an $R$-module map.
    Let $e \in E$ with image $f = \varphi(e)$ in $R$. Then
    $$
    f = de + ed
    $$
    as endomorphisms of $K_\bullet(\varphi)$.
    \end{lemma}
    
    \begin{proof}
    This is true because $d(ea) = d(e)a - ed(a) = fa - ed(a)$.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-homotopy-koszul}
    Let $R$ be a ring. Let $f_1, \ldots, f_r \in R$ be a sequence.
    Multiplication by $f_i$ on $K_\bullet(f_\bullet)$ is homotopic to
    zero, and in particular the cohomology modules $H_i(K_\bullet(f_\bullet))$
    are annihilated by the ideal $(f_1, \ldots, f_r)$.
    \end{lemma}
    
    \begin{proof}
    Special case of
    Lemma \ref{lemma-homotopy-koszul-abstract}.
    \end{proof}
    
    \noindent
    In
    Derived Categories, Section \ref{derived-section-cones}
    we defined the cone of a morphism of cochain complexes.
    The cone $C(f)_\bullet$ of a morphism of chain complexes
    $f : A_\bullet \to B_\bullet$ is the complex $C(f)_\bullet$ given by
    $C(f)_n = B_n \oplus A_{n - 1}$ and differential
    \begin{equation}
    \label{equation-differential-cone}
    d_{C(f), n} =
    \left(
    \begin{matrix}
    d_{B, n} & f_{n - 1} \\
    0 & -d_{A, n - 1}
    \end{matrix}
    \right)
    \end{equation}
    It comes equipped with canonical morphisms of complexes
    $i : B_\bullet \to C(f)_\bullet$ and
    $p : C(f)_\bullet \to A_\bullet[-1]$
    induced by the obvious maps $B_n \to C(f)_n \to A_{n - 1}$.
    
    \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}
    
    \begin{lemma}
    \label{lemma-cone-koszul}
    Let $R$ be a ring. Let $f_1, \ldots, f_r$ be a sequence of elements
    of $R$. The complex $K_\bullet(f_1, \ldots, f_r)$ is isomorphic to the
    cone of the map of complexes
    $$
    f_r :
    K_\bullet(f_1, \ldots, f_{r - 1})
    \longrightarrow
    K_\bullet(f_1, \ldots, f_{r - 1}).
    $$
    \end{lemma}
    
    \begin{proof}
    Special case of
    Lemma \ref{lemma-cone-koszul-abstract}.
    \end{proof}
    
    \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}
    
    \begin{lemma}
    \label{lemma-koszul-mult-abstract}
    Let $R$ be a ring. Let $\varphi : E \to R$ be an $R$-module map.
    Let $f, g \in R$. Set $E' = E \oplus R$ and define
    $\varphi'_f, \varphi'_g, \varphi'_{fg} : E' \to R$
    by $\varphi$ on $E$ and multiplication by $f, g, fg$ on $R$.
    The complex $K_\bullet(\varphi'_{fg})$ is isomorphic to the
    cone of a map of complexes
    $$
    K_\bullet(\varphi'_f)[1]
    \longrightarrow
    K_\bullet(\varphi'_g).
    $$
    \end{lemma}
    
    \begin{proof}
    By
    Lemma \ref{lemma-cone-koszul-abstract}
    the complex $K_\bullet(\varphi'_f)$ is isomorphic to the cone of
    multiplication by $f$ on $K_\bullet(\varphi)$ and similarly
    for the other two cases. Hence the lemma follows from
    Lemma \ref{lemma-cone-squared}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-koszul-mult}
    Let $R$ be a ring. Let $f_1, \ldots, f_{r - 1}$ be a sequence of elements
    of $R$. Let $f, g \in R$. The complex
    $K_\bullet(f_1, \ldots, f_{r - 1}, fg)$
    is homotopy equivalent to the cone of a map of complexes
    $$
    K_\bullet(f_1, \ldots, f_{r - 1}, f)[1]
    \longrightarrow
    K_\bullet(f_1, \ldots, f_{r - 1}, g)
    $$
    \end{lemma}
    
    \begin{proof}
    Special case of
    Lemma \ref{lemma-koszul-mult-abstract}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-join-sequences-koszul-complex}
    Let $A$ be a ring.
    Let $f_1, \ldots, f_r$, $g_1, \ldots, g_s$ be elements of $A$.
    Then there is an isomorphism of Koszul complexes
    $$
    K_\bullet(A, f_1, \ldots, f_r, g_1, \ldots, g_s) =
    \text{Tot}(K_\bullet(A, f_1, \ldots, f_r) \otimes_A
    K_\bullet(A, g_1, \ldots, g_s)).
    $$
    \end{lemma}
    
    \begin{proof}
    Omitted. Hint: If $K_\bullet(A, f_1, \ldots, f_r)$ is generated as a
    differential graded algebra by $x_1, \ldots, x_r$ with $\text{d}(x_i) = f_i$
    and $K_\bullet(A, g_1, \ldots, g_s)$ is generated as a
    differential graded algebra by $y_1, \ldots, y_s$ with $\text{d}(y_j) = g_j$,
    then we can think of $K_\bullet(A, f_1, \ldots, f_r, g_1, \ldots, g_s)$
    as the differential graded algebra generated by the sequence of elements
    $x_1, \ldots, x_r, y_1, \ldots, y_r$ with $\text{d}(x_i) = f_i$
    and $\text{d}(y_j) = g_j$.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-extended-alternating-Cech-is-colimit-koszul}
    Let $R$ be a ring. Let $f_1, \ldots, f_r \in R$.
    The extended alternating {\v C}ech complex
    $$
    R \to \prod\nolimits_{i_0} R_{f_{i_0}} \to
    \prod\nolimits_{i_0 < i_1} R_{f_{i_0}f_{i_1}} \to
    \ldots \to R_{f_1\ldots f_r}
    $$
    is a colimit of the Koszul complexes $K(R, f_1^n, \ldots, f_r^n)$.
    \end{lemma}
    
    \begin{proof}
    The transition maps $K(R, f_1^n, \ldots, f_r^n) \to
    K(R, f_1^{n + 1}, \ldots, f_r^{n + 1})$ are the maps
    sending $e_{i_1} \wedge \ldots \wedge e_{i_p}$ to
    $f_{i_{p + 1}} \ldots f_{i_r} e_{i_1} \wedge \ldots \wedge e_{i_p}$
    where the indices are such that $\{1, \ldots, r\} = \{i_1, \ldots, i_r\}$.
    In particular the transition maps are always $1$ in degree $r$
    and equal to $f_1\ldots f_r$ in degree $0$.
    The terms of the colimit are equal to the terms of the
    extended alternating {\v C}ech complex by
    Algebra, Lemma \ref{algebra-lemma-localization-colimit}.
    \end{proof}

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