## 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:

the underlying graded algebra is the exterior algebra $K_\bullet (\varphi ) = \wedge (E)$,

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

15.28.6.1
\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 (15.28.6.1)

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

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 isomorphic 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 (0)