As the tittle says, I am having a bad time thinking on the construction of a free resolution of $\mathbb{Z}$ as an $(\mathbb{Z}[x]/(x^n-1))$-module.
I know that I should give an exact sequence of the form (if $R= \mathbb{Z}[x]/(x^n-1)$):
$... \longrightarrow R^{I_n} \longrightarrow ...\longrightarrow R^{I_1} \longrightarrow R^{I_0} \longrightarrow \mathbb{Z} \longrightarrow 0$
being
$d_i : R^{I_i} \longrightarrow R^{I_{i-1}}$ for $i \ge 1 $
$d_0: R^{I_0} \longrightarrow \mathbb{Z}$
$e: \mathbb{Z} \longrightarrow 0$.
and $R^{I_i}$ being free $R$-modules.
Moreover they should satisfy the exact sequence condition: $Im(d_i)=Ker(d_{i-1})$.
I began thinking on the first application $d_0: R^{I_0} \longrightarrow \mathbb{Z}$. But I had some troubles finding it. I thought sending
$a_0 + a_1 \bar{x} + a_2 \bar{x}^2+...+a_{n-1} \bar{x}^{n-1} \mapsto a_0$
because I need $Im(d_0)=Ker(e) = \mathbb{Z}$. But it gets harder to find $d_1$.
Then I thought:
$a_0 + a_1 \bar{x} + a_2 \bar{x}^2+...+a_{n-1} \bar{x}^{n-1} \mapsto \sum a_i$
but again it gets hard.
Any hint/help?
EDIT: I chose $R^{I_0}$ to be $R$.
Consider the following sequence of $R$-modules:
$$ \ldots \longrightarrow R \longrightarrow R \longrightarrow R \longrightarrow \mathbb{Z} \rightarrow 0 . $$ We denote by $d_i$ the usual map. We define $d_0: R \rightarrow \mathbb{Z}: f(\overline{x}) \rightarrow f(1).$ Then, the kernel of $d_0$ corresponds to the polynomials in $R$ such that $f(1) = 0$. Consider now $d_1: R \rightarrow R: g(\overline{x})\rightarrow (\overline{x-1})g(\overline{x}). $
Notice that, per construction, the image of $d_1$ is the kernel of $d_0$. The kernel of $d_1$ is the set of functions such that $g(\overline{x})(\overline{x-1})=0$. Try to complete the sequence, using the fact that $(1+x+\ldots + x^{n-1})(x-1) = x^n-1.$