# 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 $$n\in \mathbb{Z}$$

• I am not sure if this is a research-level question, btw. – Bugs Bunny Feb 15 '20 at 18:27
• Why "faithfully flat" in the title? it doesn't reappear in the question. – YCor Feb 17 '20 at 2:13

## 2 Answers

No way. Let $$G={\mathbb Z}$$ so that $${\mathbb Z}[G]={\mathbb Z}[x,x^{-1}]$$. Then use the complex $$\ldots \rightarrow 0 \rightarrow 0 \rightarrow {\mathbb Z}[x,x^{-1}] \xrightarrow{1-x+x^2} {\mathbb Z}[x,x^{-1}] \rightarrow 0 \rightarrow 0 \rightarrow \ldots$$

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

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.

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