For the group of integers $\mathbb Z$, we know that we have a finite free $\mathbb Z[t^{\pm 1}]-$resolution of $\mathbb Z$:

$$0\longrightarrow \mathbb Z[t^{\pm 1}] \stackrel{t-1}{\longrightarrow} \mathbb Z[t^{\pm 1}]\stackrel{\epsilon}{\longrightarrow} \mathbb Z\longrightarrow 0 $$ Where $\epsilon$ is the augmentation map.

Now what about $\mathbb Z\times \mathbb Z$, can we find a finite free $\mathbb Z[s^{\pm 1},t^{\pm 1}]-$resolution of $\mathbb Z\times \mathbb Z$. Thank you for your help!

  • 1
    $\begingroup$ It suffices to find a $\mathbb{Z}[s^{\pm},t^{\pm}]$-resolution of $\mathbb{Z}$. For this, see for example math.stackexchange.com/a/2618230. $\endgroup$ – Minseon Shin Sep 27 at 13:55
  • $\begingroup$ I use this definition: A resolution $F_∗$ for a group $G$ is an acyclic chain complex of $\mathbb ZG$-modules $$...F_2\stackrel{d_2}{\longrightarrow} F_1\stackrel{d_1}{\longrightarrow} F_0\stackrel{\epsilon}{\longrightarrow}F_{-1}=\mathbb Z\longrightarrow 0 $$ where $F_{−1} = \mathbb Z$ is considered a $\mathbb ZG$-module with the trivial action. So what do you mean here by it suffices to find a $\mathbb Z[s^{\pm 1},t^{\pm 1}]-$resolution of $\mathbb Z$ ? $\endgroup$ – palio Sep 27 at 18:30
  • $\begingroup$ Here $\mathbb{Z} \times \mathbb{Z}$ is also considered a $\mathbb{Z}G$-module with the trivial action, right? Then it would suffice to find a $\mathbb{Z}[s^{\pm},t^{\pm}]$-resolution of $\mathbb{Z}$, then take the direct sum of two copies of this resolution to get a resolution of $\mathbb{Z} \times \mathbb{Z}$, namely take $\dotsb \to F_{2} \oplus F_{2} \to F_{1} \oplus F_{1} \to F_{0} \oplus F_{0} \to \mathbb{Z} \oplus \mathbb{Z} \to 0$. $\endgroup$ – Minseon Shin Sep 27 at 19:16

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