Faithfully flat modules over a group algebra Suppose we have the following data:
1) A group ring $\mathbb{Z}[G]$, where $G$ is a torsion free group.
2) $M_{\bullet}$ a bounded (above and below) chain complex of $\mathbb{Z}[G]$-modules such that each $M_{i}$ is a finitely generated free $\mathbb{Z}[G]$-module. 
My question is the following:
If the homology of $M_{\bullet}\otimes_{\mathbb{Z}[G]}\mathbb{Z}$ is trivial ie 
$H_{n}(M_{\bullet}\otimes_{\mathbb{Z}[G]}\mathbb{Z})=0$ for all $n\in \mathbb{Z}$ does it imply that $$H_{n}(M_{\bullet})=0$$ for all $n\in \mathbb{Z}$
 A: No way. Let $G={\mathbb Z}$ so that ${\mathbb Z}[G]={\mathbb Z}[x,x^{-1}]$. Then use the complex
$$\ldots \rightarrow 0 \rightarrow 0 \rightarrow {\mathbb Z}[x,x^{-1}] \xrightarrow{1-x+x^2} {\mathbb Z}[x,x^{-1}] \rightarrow 0 \rightarrow 0 \rightarrow \ldots$$
A: Let $G$ be infinite cyclic, generated by $x$.
Let $M_\bullet$ be a free resolution of the $\mathbb{Z}[G]$-module $U=\mathbb{Z}/3\mathbb{Z}$ with $x$ acting by multiplication by $-1$. For example, take $M_\bullet$ to be
$$\dots\to0\to\mathbb{Z}[G]\stackrel{\pmatrix{-3\\x+1}}{\longrightarrow}\mathbb{Z}[G]\oplus\mathbb{Z}[G]\stackrel{\pmatrix{x+1&3}}{\longrightarrow}\mathbb{Z}[G]\to0\to\dots$$
Then $M_\bullet$ is not acyclic, but the homology of $M_\bullet\otimes_{\mathbb{Z}[G]}\mathbb{Z}$ is $\text{Tor}^{\mathbb{Z}[G]}_\bullet(U,\mathbb{Z})$, which is zero, since tensoring $U$ with the projective resolution
$$\dots\to0\to\mathbb{Z}[G]\stackrel{x-1}{\longrightarrow}\mathbb{Z}[G]\to0\to\dots$$
of $\mathbb{Z}$ gives the complex
$$\dots\to0\to\mathbb{Z}/3\mathbb{Z}\stackrel{-2}{\longrightarrow}\mathbb{Z}/3\mathbb{Z}\to0\to\dots,$$
which is acyclic.
