Let $G$ be a torsion-free group and assume that the integral group ring $\mathbb{Z}G$ is torsion-free as well. Let $M$ be a torsion-free, finitely generated module over $\mathbb{Z}G$.


If we assume that $M \otimes\_{\mathbb{Z}G} \mathbb{Q}G$ is a projective $\mathbb{Q}G$-module, can we conclude that $M$ itself is projective?


(For finite groups $G$ a somewhat similar question was already asked here: [Looking for criterion for $\mathbb{Z}G$-modules to be projective](https://mathoverflow.net/q/91151/13356).)



Projectivity of torsion-free modules over integral group rings
That is already false when $G$ equals $\mathbb{Z}$. The group ring $\mathbb{Z}G$ is $\mathbb{Z}[t,t^{-1}]$. Let $p$ be a prime integer, and let $I\subset \mathbb{Z}G$ be the ideal $\langle p, t-1 \rangle$. Then $I\otimes\_{\mathbb{Z}G}\mathbb{Q}G$ is isomorphic to the principal ideal $\langle t-1 \rangle$, which is free of rank $1$. Yet associated to the short exact sequence, $$0 \to I \to \mathbb{Z}G \to \mathbb{Z}G/I \to 0,$$ consider the long exact sequence of $\text{Tor}\_\bullet^{\mathbb{Z}G}(\mathbb{Z}G/I,-)$. The second connecting map quickly gives that $$\text{Tor}\_1^{\mathbb{Z}G}(\mathbb{Z}G/I,I) = \text{Tor}\_2^{\mathbb{Z}G}(\mathbb{Z}G/I,\mathbb{Z}G/I) \cong \mathbb{Z}G/I$$ is nonzero. Therefore $I$ is not projective.


**Edit.** I corrected this from Ext (incorrect) to Tor (correct).


