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Is every stably free module of commutative group ring free? if not can you give me an example of commutative group ring with nonfree stably free module.

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    $\begingroup$ Do you mean the integral group ring $\mathbb{Z}G$? And are you interested in arbitrary (infinitely generated) abelian groups? For a finitely generated abelian group $G$, every stably free module for $\mathbb{Z}G$ is free. (F.E.A. Johnson, "Stably free cancellation for abelian group rings", Arch. Math. 102, p. 7-10 (2014). link.springer.com/article/10.1007/s00013-013-0599-8 ) $\endgroup$
    – Jeremy Rickard
    Commented Apr 15, 2017 at 10:20
  • $\begingroup$ Surely this is true for all torsion-free abelian groups, as they (and the corresponding group rings) are direct limits of finitely generated free ones (and their group rings), for which Quillen-Suslin applies? SF implies F is preserved by direct limits, so in general it would be enough to do it for finitely generated abelian groups anyway. $\endgroup$
    – David Handelman
    Commented Apr 15, 2017 at 18:10

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