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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 $n\in \mathbb{Z}$

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  • $\begingroup$ I am not sure if this is a research-level question, btw. $\endgroup$ – Bugs Bunny Feb 15 '20 at 18:27
  • $\begingroup$ Why "faithfully flat" in the title? it doesn't reappear in the question. $\endgroup$ – YCor Feb 17 '20 at 2:13
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No way. Let $G={\mathbb Z}$ so that ${\mathbb Z}[G]={\mathbb Z}[x,x^{-1}]$. Then use the complex $$\ldots \rightarrow 0 \rightarrow 0 \rightarrow {\mathbb Z}[x,x^{-1}] \xrightarrow{1-x+x^2} {\mathbb Z}[x,x^{-1}] \rightarrow 0 \rightarrow 0 \rightarrow \ldots$$

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    $\begingroup$ The complex $M_{\bullet}\otimes _{\mathbb{Z}[G]}\mathbb{Z}$ doesn't seem to be exact. $\endgroup$ – abx Feb 15 '20 at 18:38
  • $\begingroup$ Point taken. It is just wrong polynomial. We need $f(1)=1$, not $f(1)=0$. $\endgroup$ – Bugs Bunny Feb 16 '20 at 10:08
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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.

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  • $\begingroup$ As Bugs Bunny's corrected answer shows, this is unnecessarily complicated. $\endgroup$ – Jeremy Rickard Feb 16 '20 at 10:52

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