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 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)
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