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Let $R = \mathbb{Z}[x_1,...,x_6]$ be a polynomial ring. Then we may form the Koszul complex $K(x_1,...,x_6)$ which looks something like:

$$ R \xrightarrow{d_6} R^6 \xrightarrow{d_5} R^{15} \xrightarrow{} ... \xrightarrow{d_2} R^6 \xrightarrow{d_1} R $$

where, for example, $$d_1 = (x_1,x_2,...,x_6)$$

$$d_2 = \begin{pmatrix} -x_2 &-x_3 &-x_4 &-x_5 &-x_6 & 0 &0& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ x_1 & 0 & 0 & 0 & 0 &-x_3 &-x_4& -x_5& -x_6& 0& 0& 0 & 0 & 0& 0\\ 0 & x_1 & 0 & 0 & 0 & x_2 & 0 & 0 & 0& -x_4& -x_5& -x_6 & 0 & 0 & 0\\ 0 &0& x_1& 0 & 0 & 0 & x_2 & 0 & 0 & x_3 & 0 & 0 &-x_5 &-x_6 & 0\\ 0 & 0 & 0 & x_1 & 0 & 0 & 0 & x_2 & 0 & 0 & x_3 & 0 & x_4 &0 &-x_6\\ 0 & 0 & 0 & 0 & x_1 & 0 & 0 & 0 & x_2 & 0 & 0 & x_3 & 0 & x_4 & x_5\\ \end{pmatrix}$$

etc. Now, each of the $x_i$ can be expressed as a polynomial in $\mathbb{Z}[t]$ (what is happening here is that the $x_i$ are characters of a group, and in writing them as polynomials in $\mathbb{Z}[t]$ we are restricting the characters, but this is not so important).

So let's say for example we have:

$$x_1 = (t-2)(t^2 + 1)$$ $$x_2 = (t-2)(t+2)(t^3 + 4t + 8)$$ $$\dots$$

then by tensoring above complex with $\mathbb{Z}[t]$ (i.e. $- \otimes_R \mathbb{Z}[t]$) we form what I am calling the Koszul complex in one variable:

$$ \mathbb{Z}[t] \xrightarrow{d_6} \mathbb{Z}[t]^6 \xrightarrow{d_5} \mathbb{Z}[t]^{15} \xrightarrow{} ... \mathbb{Z}[t]^6 \xrightarrow{d_1} \mathbb{Z}[t] $$

where the $d_i$ are just the $d_i$ above, but each of the $x_i$ are replaced with their expression/restriction in $\mathbb{Z}[t]$.

I have found that sagemath has a KoszulComplex function which gives the Koszul complex in terms of the $x_i$, but I don't believe there any any implement for using this to compute homology groups.

I cannot seem to find a prebuilt way to compute the homology groups here, although it seems like this should exist. Finally, I will add that if Euclidean-domainness is an issue, I am happy to replace $\mathbb{Z}[t]$ with $F[t]$ where $F = \mathbb{F}_p$ or $\mathbb{Q}$.

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I do not know any implementations of "homology of a chain complex" over any ring except $R = k$ a field or $R = \Bbb Z$, and the latter is much less developed.

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  • $\begingroup$ Welcome to MSE. This seems to be a comment rather than an answer. $\endgroup$ Jul 17, 2022 at 13:06
  • $\begingroup$ @user1079171, Thanks - unfortunately this is also the conclusion I was getting from my online searches. Perhaps I can do it by hand in the cases I'm interested in! $\endgroup$
    – Dylan
    Jul 18, 2022 at 10:58

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