# Climbing Mount Bourbaki

Thoughts on mathematics

November 14, 2010

## The Koszul complex II

Posted by Akhil Mathew under algebra, commutative algebra, MaBloWriMo | Tags: differential graded algebras, Koszul complex, regular sequences |
1 Comment

I’ve not been a very good MaBloWriMo participant this time around. Nonetheless, coursework does tend to sap the time and energy I have for blogging. I have been independently looking as of late at the formal function theorem in algebraic geometry, which can be phrased loosely by saying that the higher direct images under a proper morphism of schemes commute with formal completions. This is proved in Hartshorne for projective morphisms by first verifying it for the standard line bundles and then using a (subtle) exactness argument, but EGA III.4 presents an argument for general proper morphisms. The result is quite powerful, with applications for instance to Zariski’s main theorem (or at least a weak version thereof), and I would like to say a few words about it at some point, at least after I have a fuller understanding of it than I do now. So I confess to having been distracted by algebraic geometry.

For today, I shall continue with the story on the Koszul complex, and barely begin the connection between Koszul homology and regular sequences. Last time, we were trying to prove:

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, ${K_*(\lambda \oplus \lambda') \simeq K_*(\lambda) \oplus K_*(\lambda')}$. This is a condition on the differentials. 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.

Before starting the proof, let us talk about differential graded algebras. This is not really necessary, but the Koszul complex is a special case of a differential graded algebra.

Definition 25 A differential graded algebra is a graded unital associative algebra ${A}$ together with a derivation ${d: A \rightarrow A}$ of degree one (i.e. increasing the degree by one). This derivation is required to satisfy a graded version of the usual Leibnitz rule: ${d(ab) = (da)b + (-1)^{\mathrm{deg} a} a (db) }$. Moreover, ${A}$ is required to be a complex: ${d^2=0}$. So the derivation is a differential.

So the basic example to keep in mind here is the case of the Koszul complex. This is an algebra (it’s the exterior algebra). The derivation ${d}$ was immediately checked to be a differential. There is apparently a category-theoretic interpretation of DGAs, but I have not studied this.

Proof: As already stated, the graded algebra structures on ${K_*(\lambda), K_*(\lambda')}$ are the same. This is, I suppose, a piece of linear algebra, about exterior products, and I won’t prove it here. The point is that the differentials coincide. The differential on ${K_*(\lambda \oplus \lambda')}$ is given by extending the homomorphism ${L \oplus L' \stackrel{\lambda \oplus \lambda'}{\rightarrow} R}$ to a derivation. This extension is unique.

Now I claim that tensor product of two differential graded algebras with the product differential is a DGA itself. This says that the tensor product of the differentials is itself not only a differential, but a derivation on the tensor product. This is what we want, because then the product differential on ${K_*(\lambda) \otimes K_(\lambda')}$ is a derivation, and since the differential induced by ${\lambda + \lambda'}$ is one too, the two must coincide as they coincide in degree one. This is a routine computation, which is not suitable to blogging. So one should check that if ${(A, d_A), (B, d_B)}$ are DGAs, then ${(A \otimes B, d_{A \otimes B})}$, where ${A \otimes B}$ has the graded algebra structure and ${d_{A \otimes B}}$ is the product differential, is indeed a DGA.

0.7. Koszul homology and regular sequences

In general, the Koszul complex is not exact. The degree to which its homology vanishes does, however, say something. It tells you the length of regular sequences, or alternatively the depth.

First, we can compute the Koszul homology at the end. Let ${f_1, \dots, f_r \in R}$ for ${R}$ a commutative ring, and let ${M}$ be an ${R}$-module. Then ${K_1(\mathbf{f}) = R^r}$ and ${K_0(\mathbf{f}) = R}$ from the definitions. The differential ${K_1(\mathbf{f}) \rightarrow K_0(\mathbf{f})}$ is simply ${(a_i) \rightarrow \sum a_i r_i}$. In particular, we see that the homology of the Koszul complex at dimension zero is ${R/(f_1, \dots, f_r) R}$. More generally, this argument and the right-exactness of the tensor product shows that:

Proposition 26 We have$\displaystyle H_0(\mathbf{f}, M) = M/(f_1, \dots, f_r) M.$

So in general, the zeroth Koszul homology ${H_0}$ will be nonzero. But the higher ones vanish for regular sequences. We are aiming for:

Proposition 27 Let ${f_1, \dots, f_r}$ be an ${M}$-regular sequence. Then ${H_s(\mathbf{f}, M) = 0}$ for ${s \neq 0}$.

Proof: The argument, as expected, will be inductive. The first step is the core of the idea, though. The Koszul complex for ${\mathbf{f}}$ consisting of one element is

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

It is clear that the homology of this complex detects the nonzerodivisorness of ${f}$ on ${M}$. In general, the result is an inductivization of the above observation. When ${r=1}$, the above proposition is true. Let us assume it true for ${r-1}$, and we prove it for ${r}$. So let ${f_1, \dots, f_r}$ be an ${M}$-regular sequence, and let ${\mathbf{f}' = (f_1, \dots, f_{r-1})}$. We know that the homology of the complex

$\displaystyle K(\mathbf{f}', M)$

vanishes in dimension ${\neq 0}$, and is ${M/(\mathbf{f}'M)}$ for dimension zero. This is “close” to what we want as ${K(\mathbf{f}', M)}$ and ${K(\mathbf{f}, M)}$ are “similar,” but we need a way of going between them. So far, we know that

$\displaystyle K(\mathbf{f}, M) = K(\mathbf{f}',M) \otimes K(f_r, M),$

and that ${f_r}$ is a nonzerodivisor on ${H_0(K(\mathbf{f}', M))}$. That way is provided by:

Lemma 28 Let ${\left\{C_n\right\}_{n \geq 0}}$ be a chain complex of ${R}$-modules such that ${C_*}$ is exact in positive dimension. Suppose ${y \in R}$ is a nonzerodivisor on ${H_0(C)}$. Then ${C_* \otimes K(y, R)}$ is acyclic in positive dimension.

Proof: Because I’m in the mood to use a sledgehammer, let’s deduce this from a spectral sequence. We know that there is a double complex ${\left\{C_p \otimes K_q(y, R)\right\}_{p,q \geq 0}}$.

There are two spectral sequences that converge to the same thing. For the first homology, we take the horizontal homology, and then the vertical homology of the horizontal homology. But since ${K(y,R)}$ is just ${R}$ and zero, the horizontal homology is zero except in dimension zero, where it’s ${H_0(C)}$ located at ${(0,0)}$ and ${(0,1)}$. The vertical differential is multiplication by ${y}$. When we take the next page in this spectral sequence, the fact that ${y}$ is ${H_0(C)}$-regular implies that it is ${H_0(C)/y H_0(C)}$ at the origin and nothing elsewhere. In particular, the second ${E_2}$ page of this spectral sequence is centered at the origin. This spectral sequence converges to the total homology of the double complex. That total homology is ${H_*(C \otimes K(y,R))}$.

But the spectral sequence obviously collapses at ${E_2}$, and the convergent limit ${E_\infty = E_2}$ as a result. But from the thus calculated ${E_\infty}$ page of the spectral sequence, we find that there is nothing about the nonzero diagonals, and consequently since the sequence converges ${H_*(C \otimes K(y,R))}$, we see that ${C \otimes K(y,R)}$ is acyclic in positive dimensions.

Now, with the lemma established, the result is clear. I should note that the lemma can be proved slightly less conceptually but more elementarily (without spectral sequences) if one writes some exact sequences.

This result is very far from the best we can do. The Koszul homology may very well be zero without the initial sequence being a regular sequence. The more natural result, which can be proved using more sophisticated refinements of the above reasoning, is that Koszul homology ${H_*(\mathbf{f}, M)}$ detects the length of a maximal ${M}$-sequence in the ideal ${(\mathbf{f}) \subset R}$. I want to get to this result, but first there are some interesting things one can do with what’s been proved alone in algebraic geometry.
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