Currently, I am learning about the Koszul complex as presented in the text "Binomial Ideals" by Herzog, Hibi and Ohsugi. To aid my understanding, I wanted to view an explicit computation of the homology of a Koszul complex, but I was unable to find anything helpful. If anybody could provide one I would be very grateful, otherwise I would just like to know where I'm going wrong in the example below.
Let $S = \mathbb{K}[x,y,z]$ be the standard graded polynomial ring in $3$ indeterminates over the field $\mathbb{K}$, $\mathbf{x} = x,y,z $ a regular sequence and $S/I$ a finitely generated graded $S$-module, where $I=(xy - z^2)$ is an ideal in $S$. The text states that the Koszul complex of $S/I$ with respect to $\mathbf{x}$ is given by $K(\mathbf{x};S/I) = K(\mathbf{x};S) \otimes_{S} S/I$, where $K(\mathbf{x};S)$ is the usual Koszul complex with respect to $\mathbf{x}$. So I would like to compute $H_{i}(\mathbf{x};S/I) $ for all $ i$.
If $F$ is a free $S$-module with basis $e_1,e_2,e_3$, then $K(\mathbf{x};S)$ can be written as follows. $$ 0 \longrightarrow \bigwedge\nolimits^3 F \xrightarrow{\begin{pmatrix} z \\ -y \\ x \end{pmatrix}} \bigwedge\nolimits^2 F \xrightarrow{\begin{pmatrix} -y & -z & 0\\ x & 0 & -z\\ 0 & x & y \end{pmatrix}} F \xrightarrow{\begin{pmatrix} x & y & z \end{pmatrix}} S \xrightarrow{\partial_0} 0 $$ So now if we consider the complex $K(\mathbf{x};S/I)$, then the following homologies are easy to compute:
$H_{0}(K(\mathbf{x};S/I)) = \mathop{Ker}(\partial_0 \otimes_S 1)/\mathop{Im}(\partial_1 \otimes_S 1) \cong S/(x,y,z)$ $H_{2}(K(\mathbf{x};S/I)) = \mathop{Ker}(\partial_2 \otimes_S 1)/\mathop{Im}(\partial_3 \otimes_S 1) \cong 0$ $H_{3}(K(\mathbf{x};S/I)) = \mathop{Ker}(\partial_3 \otimes_S 1)/\mathop{Im}(\partial_4 \otimes_S 1) = 0$
But for $H_{1}(K(\mathbf{x};S/I))$ my result is not correct. It is clear that $\mathop{Ker}(\partial_1) = \mathop{Im}(\partial_2)$, so wouldn't this mean that $\mathop{Ker}(\partial_1 \otimes_{S} 1) = \mathop{Im}(\partial_2 \otimes_S 1)$?
A theorem in the above text states that the $\mathbb{K}$ dimension of these homologies is equal to the graded betti numbers of $S/I$, which I've already determined using Macaulay2 and by computing the minimal free $S$-resolution of $S/I$. From this I know the homology for $i=0,1$ should have $\mathbb{K}$-dimension $1$. Yet my computation of $H_{1}(K(\mathbf{x};S/I))$ does not agree with this.
Any help at all would be appreciated.