In Joseph J. Rotman's An Introduction to Homological Algebra. It is stated (exercise 9.22, ii) that for a commutative ring $k$ and a group $G$, one has, for every integer $n$ and left $kG$ module $A$: $$ \mathrm{Tor}_n^{kG}(k,A) \simeq \mathrm{Tor}_n^{\mathbb{Z}G}(\mathbb{Z}, A) $$
This implies that group homology does not depend on the coefficient ring we chose. However, I am stuck with proving this isomorphism. By tensoring (over $\mathbb{Z}$) with the ring $k$ a $\mathbb{Z}G$-module, one can obtain a $kG$-module. So I think the idea is to tensor with $k$ a projective resolution of $\mathbb{Z}$ over $\mathbb{Z}G$ to obtain something over $kG$. However, there are several points I don't understand:
- Why will this construction still yield a projective resolution?
- Why taking the homology of the coinvariants of this resolution will yield isomorphic homology groups despite the operations that were made, i.e the tensoring by $k$, and taking the coinvariants of a $kG$ module instead of the coinvariants of a $\mathbb{Z}G$ module.
Take a free resolution $P_\bullet$ of $\mathbb Z$ (as $\mathbb ZG$-module). Note that as $\mathbb Z$-modules, this resolution splits (because $\mathbb Z$ is free over itself); hence applying $kG\otimes_{\mathbb ZG} - \cong k\otimes_\mathbb Z -$ leaves it exact. It is also clearly still free (as $kG$-modules), hence $P_\bullet\otimes_{\mathbb ZG} kG$ is a free resolution of $k$.
Moreover, we have an isomorphism of chain complexes $P_\bullet\otimes_{\mathbb ZG}kG \otimes_{kG} A\cong P_\bullet\otimes_{\mathbb ZG} A$. Taking homology yields the claim.