So last time, when (partially) computing the cohomology of affine space, we used a fact about the Koszul complex. Namely, I claimed that the Koszul complex is acyclic when the elements in question generate the unit ideal. This was swept under the rug, and logically I should have covered that before getting to yesterday’s bit of algebraic geometry. So today, I will backtrack into the elementary properties of the Koszul complex, and prove a more general claim.

0.9. A chain-homotopy on the Koszul complex

Before proceeding, we need to invoke a basic fact about the Koszul complex. If ${K_*(\mathbf{f})}$ is a Koszul complex, then multiplication by anything in ${(\mathbf{f})}$ is chain-homotopic to zero. In particular, if ${\mathbf{f}}$ generates the unit ideal, then ${K_*(\mathbf{f})}$ is homotopically trivial, thus exact. This is one reason we should restrict our definition of “regular sequence” (as we do) to sequences that do not generate the unit ideal, or the connection with the exactness of the Koszul complex wouldn’t work as well.

Proposition 33 Let ${g \in (\mathbf{f})}$. Then the multiplication by ${g}$ map ${K_*(\mathbf{f}) \rightarrow K_*(\mathbf{f})}$ is chain-homotopic to zero.

Proof: Let ${\mathbf{f} = (f_1, \dots, f_r)}$ and let ${g = \sum g_i f_i}$. Then there is a vector ${q_g = (g_1, \dots, g_r) \in R^r}$. We can define a map of degree one

$\displaystyle H: K_*(\mathbf{f}) \rightarrow K_*(\mathbf{f})$

which is the interior product with ${q_g}$, i.e. sending ${v \in K_*(\mathbf{f}) = \bigwedge R^r}$ to ${q_g \wedge v}$. Now, however, we know that the differential, which we’ll call ${d}$, is a derivation on the Koszul algebra ${K_*(\mathbf{f})}$. In particular,

$\displaystyle d Hx = d (q_g \wedge x) =( d q_g) \wedge x + (-1) q_g \wedge dx .$

as ${q_g}$ has degree one. But ${dq_g}$, from its definition, is just ${g}$ times the unit of the Koszul algebra and ${-q_g \wedge dx = Hdx}$. In particular, we find

$\displaystyle dHx + Hdx = gx.$

This implies that multiplication by ${g}$ is homotopically trivial.

As an example, let ${(R, \mathfrak{m})}$ be a local ring. Let ${x_1, \dots, x_r}$ generate ${\mathfrak{m}}$. Then any element acts in a way that is homotopically trivial on ${K_*(\mathbf{x})}$. In particular, the homology groups are vector spaces (finite dimensional!) over the residue field ${R/\mathfrak{m}}$. From this, we can easily prove:

Corollary 34 If ${M}$ is any ${R}$-module, and ${g \in (\mathbf{f})}$, then ${g}$ acts by zero on ${H_*(\mathbf{f}, M)}$ and ${H^*(\mathbf{f}, M)}$. In particular, if ${(\mathbf{f})=1}$, then the Koszul homology and cohomology vanish identically for any module.

This is now clear, because the chain-homotopy on the Koszul complex passes to a chain homotopy on the Koszul complex with coefficients or the dual Koszul complex.