The Stacks project

15.28 The Koszul complex

We define the Koszul complex as follows.

Definition 15.28.1. 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 15.28.2. 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 $\mathop{\mathrm{Spec}}(R)$ isomorphic to a Koszul complex $K_\bullet (f_1, \ldots , f_ r)$. This complex has many interesting formal properties.

Lemma 15.28.3. 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 15.28.4. 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 15.28.3. $\square$

Lemma 15.28.5. 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 15.28.6. 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 15.28.5. $\square$

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} \label{more-algebra-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 $B_ n \to C(f)_ n \to A_{n - 1}$.

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. $\square$

Lemma 15.28.8. 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 15.28.7. $\square$

Lemma 15.28.9. 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 complex $C(f)_\bullet $ is the complex

\[ \ldots \to 0 \to R \xrightarrow {f} R \to 0 \to \ldots \]

with $R$ placed in (homological) degrees $1$ and $0$. The map of complexes 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 $R^{\oplus 2}$ placed in degrees $1$ and $0$ and differential (

\[ \left( \begin{matrix} g & 1 \\ 0 & -f \end{matrix} \right) : R^{\oplus 2} \longrightarrow R^{\oplus 2} \]

To see this chain complex is homotopic to $C(fg)_\bullet $, i.e., to $R \xrightarrow {fg} R$, consider the maps of complexes

\[ \xymatrix{ R \ar[d]_{(1, -g)} \ar[r]_{fg} & R \ar[d]^{(0, 1)} \\ R^{\oplus 2} \ar[r] & R^{\oplus 2} } \quad \quad \xymatrix{ R^{\oplus 2} \ar[d]_{(1, 0)} \ar[r] & R^{\oplus 2} \ar[d]^{(f, 1)} \\ R \ar[r]^{fg} & R } \]

with obvious notation. The composition of these two maps in one direction is the identity on $C(fg)_\bullet $, but in the other direction it isn't the identity. We omit writing out the required homotopy.

To see the result holds in general, we use that we have a functor $K_\bullet \mapsto \text{Tot}(A_\bullet \otimes _ R K_\bullet )$ on the category of complexes which is compatible with homotopies and cones. Then 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 15.28.10. 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 homotopy equivalent to the cone of a map of complexes

\[ K_\bullet (\varphi '_ f)[1] \longrightarrow K_\bullet (\varphi '_ g). \]

Proof. By Lemma 15.28.7 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 15.28.9. $\square$

Lemma 15.28.11. 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 15.28.10. $\square$

Lemma 15.28.12. Let $R$ be a ring. Let $f_1, \ldots , f_ r$, $g_1, \ldots , g_ s$ be elements of $R$. Then there is an isomorphism of Koszul complexes

\[ K_\bullet (R, f_1, \ldots , f_ r, g_1, \ldots , g_ s) = \text{Tot}(K_\bullet (R, f_1, \ldots , f_ r) \otimes _ R K_\bullet (R, g_1, \ldots , g_ s)). \]

Proof. Omitted. Hint: If $K_\bullet (R, f_1, \ldots , f_ r)$ is generated as a differential graded algebra by $x_1, \ldots , x_ r$ with $\text{d}(x_ i) = f_ i$ and $K_\bullet (R, g_1, \ldots , g_ s)$ 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 $K_\bullet (R, f_1, \ldots , f_ r, g_1, \ldots , g_ s)$ as the differential graded algebra generated by the sequence of elements $x_1, \ldots , x_ r, y_1, \ldots , y_ s$ with $\text{d}(x_ i) = f_ i$ and $\text{d}(y_ j) = g_ j$. $\square$

Comments (2)

Comment #8389 by Peng Du on

In line above tag/0628, better to denote the map by p: C(f)⟶A[1] (rather than [-1]). This might depend on your convention, but using [1] is consistent with all texts, both homotopically (e.g. suspension in homotopy theory become [1] in triangulated categories) and homologically/cohomologically. The moral is, [1] always shifts a chain/cochain 1 step to the left (provided you draw all arrows rightward), i.e. to the opposite of the direction of a chain/cochain.

Comment #8999 by on

OK, I am not going to change the convention of shifting for chain complexes. However, for cochain complexes what you say is the convention we use. Almost all of the complexes in the Stacks project are cochain complexes.

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