# Koszul Homology vs Koszul Cohomology

Let $R$ be a ring and $x \in R$. The Koszul complex $K_\bullet(x)$ is then $0 \rightarrow R \stackrel{x}{\rightarrow} R \rightarrow 0$. Given $x_1,\dots,x_n \in R$ the Koszul complex $K_\bullet(x_1,\dots,x_n)$ is defined to be $K_\bullet(x_1) \otimes \cdots \otimes K_\bullet(x_n)$, where $\otimes$ is the tensor product of complexes. Then the Koszul homology of an $R$-module $M$ is the homology of the complex $K_\bullet(x_1,\dots,x_n) \otimes M$.

In his notes, Craig Huneke is defining a different Koszul complex $K^\bullet(x)$ by $0 \rightarrow R \rightarrow R_x \rightarrow 0$, where the middle arrow is the canonical map of localization. As above, he defines $K^\bullet(x_1,\dots,x_n) = K^\bullet(x_1) \otimes \cdots \otimes K^\bullet(x_n)$ and the Koszul cohomology of an $R$-module $M$ is the cohomology of the complex $K^\bullet(x_1,\dots,x_n) \otimes M$. In these notes it is then shown that the local cohomology $H^i_I(M)$ of $M$ with respect to an ideal $I=(x_1,\dots,x_n)$ is in fact the Koszul cohomology of $M$.

Question: I have seen before the Koszul homology in Matsumura and Bruns and Herzog, but Koszul cohomology is new to me. The dual denominations Koszul homology-cohomology suggest that the two Koszul complexes $K_\bullet(x_1,\dots,x_n)$ and
$K^\bullet(x_1,\dots,x_n)$ described above are in some sense dual. Is this the case and in what sense?

• First of all, it is Craig Huneke. I believe his Koszul complex is also called as Cech complex. Please google Cech complex. Oct 9, 2014 at 8:01

As Youngsu says, what you have written as Koszul cohomology is sometimes called Cech cohomology instead (geometrically, the terms of degree $> 0$ are exactly those of the Cech complex for the open sets $D(x_i)$ - I will not comment on Huneke's use of the term though).
Here is a way to relate Cech cohomology with Koszul homology: given the Koszul complex $K^\bullet(x) : 0 \to R \xrightarrow{x} R \to 0$, one can form a direct limit of complexes
$$\require{AMScd} \begin{CD} 0 @. 0 @. 0\\ @VVV @VVV @VVV\\ R @>=>> R @>=>> R @>=>> ...\\ @VVxV @VVx^2V @VVx^3V\\ R @>x>> R @>x>> R @>x>> ...\\ @VVV @VVV @VVV\\ 0 @. 0 @. 0 \end{CD} \qquad \implies \qquad \begin{CD} 0 \\ @VVV \\ R \\ @VVV \\ R_x \\ @VVV \\ 0 \end{CD}$$
where the vertical columns are $K^\bullet(x^j)$, and the limit is the Cech complex, which I will call $C^\bullet(x) = \varinjlim K^\bullet(x^j)$. Tensoring such sequences together gives (since $\varinjlim$ and $\otimes$ commute) $C^\bullet(x_1, \ldots, x_n; M) \cong \varinjlim K^\bullet(x_1^j, \ldots, x_n^j; M)$. One can show that the cohomology of this complex computes local cohomology, i.e. if $I := (x_1, \ldots, x_n)$, then $H^i_I(M) \cong H^i(C^\bullet(x_1, \ldots, x_n; M))$, and since $\varinjlim$ is exact, this is $\varinjlim H^i(K^\bullet(x_1^j, \ldots, x_n^j; M))$ (in the case that $x_1, \ldots, x_n$ is a regular sequence, one can see this directly by noting that the Koszul complex on $(x_1^j, \ldots, x_n^j)$ is a free resolution of $R/(x_1^j, \ldots, x_n^j)$, thus can be used to compute $\text{Ext}$, and local cohomology is a limit of $\text{Ext}$'s.