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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.)

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    $\begingroup$ Group rings are always torsion-free. $\endgroup$ Aug 7, 2015 at 17:14
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    $\begingroup$ @QiaochuYuan: Group rings over positive characteristic fields need not be torsion-free. For a cyclic group $G = \langle g \rangle$ of order $p$ and for a field $k$ of characteristic $p$, $kG$ contains the nonzero element $1\underline{g} - 1 \underline{e}$ whose $p^{\text{th}}$ power equals $0$. $\endgroup$ Aug 7, 2015 at 17:21
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    $\begingroup$ @Jason: is that what torsion-free means for a ring? I interpreted it to mean that the underlying abelian group is torsion-free. I don't think I would call nilpotents torsion. $\endgroup$ Aug 7, 2015 at 17:47
  • $\begingroup$ Oh, sorry. I actually meant "ZG is a domain", not "ZG is torsion-free" (i.e., I assume the Kaplansky conjecture to be true). But let me leave it as it is, since it doesn't change that much (see Jason's answer). $\endgroup$
    – AlexE
    Aug 7, 2015 at 17:54
  • $\begingroup$ @QiaochuYuan: "Is that what torsion-free means for a ring?" Of course this depends on your convention. I am interpreting "torsion elements" to mean "nonzero zero-divisors". $\endgroup$ Aug 7, 2015 at 19:03

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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).

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