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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

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)

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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