We are now going to discuss another mechanism for determining the length of maximal ${M}$-sequences, namely the Koszul complex. This is going to be super-useful in a whole bunch of ways. For one thing, it is integral in the proof that regular local rings are of finite global dimension, because the Koszul complex becomes a free resolution of the residue field.

Another one, which has excited me as of late, is that if you have a suitable scheme (say, quasi-compact and quasi-separated) and a quasi-coherent sheaf on it, then its Cech cohomology is in fact a direct limit of Koszul cohomologies! So properties of Koszul cohomology can be used to compute the cohomology of projective space as in Hartshorne, and thus to prove the fundamental theorem that higher direct images by projective (and, eventually, proper) morphisms preserve coherence. But this is getting rather far afield of what I want to talk about today.

Let ${L}$ be a finitely generated ${R}$-module. Consider the graded commutative algebra ${K = \bigwedge L = \bigoplus \wedge^i L}$ with the product given by the wedge product; the graded commutativity is similar to the cup-product in cohomology, and implies that

$\displaystyle x \wedge y = (-1)^{\deg x \deg y} y \wedge x.$

Given ${\lambda: L \rightarrow R}$, we can define a differential on ${K}$ as follows. Namely, we define

$\displaystyle d( x_1 \wedge \dots \wedge x_n) = \sum_i (-1)^i\lambda(x_i) x_1 \wedge \dots \wedge \hat{x_i} \wedge \dots \wedge x_n.$

(More precisely, this clearly defines an alternating map ${L^n \rightarrow \wedge^{n-1} L}$, and this thus factors through the alternating product by the universal property.) It is very easy to see that ${d \circ d = 0}$. Moreover, ${d}$ is an anti-derivation. If ${x,y \in K}$ are homogeneous elements of the graded algebra, then

$\displaystyle d(x\wedge y) = d(x)\wedge y + (-1)^x x \wedge d(y)$

Definition 20 The complex, together with the multiplicative structure, just defined is called the Koszul complex and is denoted ${K_*(\lambda)}$.

The special case we shall care the most about is when ${L = R^n}$ and ${\lambda: R^n \rightarrow R}$ is given by the dot product with a vector ${\mathbf{f}=(f_1, \dots, f_n) \in R^n}$. Then we shall write

$\displaystyle K_*(\mathbf{f})$

for the Koszul complex. Now that we have a complex, we can define its homology (and cohomology).

Definition 21 Let ${M}$ be an ${R}$-module. We write ${K_*(\lambda, M)}$ for the complex ${K(\lambda) \otimes M}$ (and similarly ${K_*(\mathbf{f}, M)}$ when ${\mathbf{f}: R^n \rightarrow R}$ is ${\lambda}$). The homology of this complex is called the Koszul homology of ${M}$ and is denoted ${H_i(\lambda, M)}$ (or ${H_i(\mathbf{f}, M)}$).

One of the basic facts about this is that ${K_*(\lambda, \cdot)}$ is an exact functor if ${L}$ is flat. In particular, ${K_*(\mathbf{f}, \cdot)}$ is an exact functor as each of the terms of the complex are free. In particular,

Theorem 22 If ${L}$ is a flat module, then ${H_i(\lambda, M)}$ is a ${\delta}$-functor.

We can also dualize everything.

Definition 23 Let ${M}$ be an ${R}$-module. We write ${K^*(\lambda, M)}$ for the complex ${\hom(K(\lambda), M)}$. The cohomology of this cochain complex is called the Koszul cohomology with ${M}$ coefficients and is denoted ${H^i(\lambda, M)}$. (Similarly we define ${K^*(\mathbf{f}, M)}$ when ${L}$ is a free module, and write ${H^i(\mathbf{f}, M)}$ for the Koszul cohomology.)

It is similarly easy to see that when ${L}$ is projective, then ${K^*(\lambda, M)}$ will be an exact functor in ${M}$, and the Koszul cohomology will be a ${\delta}$-functor where the connecting homomorphisms raise the degree.

So what can we do with this? Well, as we will see the Koszul complex detects the regularity of sequences. This is a nontrivial fact, and it will basically rely on the fact that, first of all, the Koszul complex ${K(f)}$ for ${f \in R}$ (i.e. the complex of the free module ${R}$ together with the functional ${R \stackrel{f}{\rightarrow} R}$) is very simple; it’s

$\displaystyle 0 \rightarrow R \stackrel{f}{\rightarrow} R \rightarrow 0.$

It will rely on this simple observation and the fact, proved next, that Koszul complexes behave nicely with respect to tensoring. Given graded ${R}$-algebras ${A, B}$, we can form the graded tensor product ${A \otimes_R B}$. By definition, this is just ${A \otimes_R B}$ as a graded module, but the multiplicative structure is slightly different. Namely, we define the product of ${a \otimes b, a' \otimes b'}$ as

$\displaystyle (-1)^{\deg a' \deg b} aa' \otimes bb'$

if the elements in question are homogeneous. This is a fairly common construction. When one has finite-dimensional CW complexes ${X,Y}$, for instance, the cohomology ring of ${X \times Y}$ with coefficients in a field is the graded tensor product of the cohomology rings of ${X}$ and ${Y}$. Another example, more relevant here, is that if ${N,N'}$ are two modules, then the exterior algebra ${\bigwedge (N \oplus N')}$ is the graded tensor product of ${\bigwedge N }$ and ${\bigwedge N'}$.

Proposition 24 Let ${\lambda: L \rightarrow R, \lambda': L' \rightarrow R}$ be linear functionals. Then the Koszul complex ${K_*(\lambda \oplus \lambda')}$ is the tensor product ${K_*(\lambda) \otimes K_*(\lambda')}$ as differential graded algebras.

So in other words, not only is the algebra structure preserved by taking the tensor product, but when you think of them as chain complexes, ${K_*(\lambda \oplus \lambda')}$ Here ${\lambda \oplus \lambda'}$ is the functional ${L \oplus L' \stackrel{\lambda \oplus \lambda'}{\rightarrow} R \oplus R \rightarrow R}$ where the last map is addition. So for instance this implies that ${K_*(\mathbf{f}) \otimes K_*(\mathbf{f}') \simeq K_*(\mathbf{f}, \mathbf{f}')}$ for two tuples ${\mathbf{f} = (f_1, \dots, f_i), \mathbf{f}' = (f'_1, \dots, f'_j)}$. This implies that in the case we care about most, catenation of lists of elements corresponds to the tensor product.

I think I’m going to prove this result tomorrow. It’s all right; we don’t have to rush.