## free resolution of $\mathbb{Z}$ as an $\mathbb{Z}[x]/(x^n-1)$-module

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?

Posted on Categories proxies