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Let $G$ be a torsion free abelian group (infinitely generated to get anything interesting). The group algebra $\mathbb{Q}[G]$ is an integral domain. Let $\mathbb{Q}(G)$ be its field of fractions.

Are there non-isomorphic torsion free abelian groups $G$ and $H$ such that $\mathbb{Q}(G)\cong\mathbb{Q}(H)$? If so, are there easy examples?

This is motivated by this question. There are examples of torsion free abelian groups $G$ where $G\cong G\oplus\mathbb{Z}\oplus\mathbb{Z}$, but $G\not\cong G\oplus\mathbb{Z}$. Since $\mathbb{Q}(G\oplus\mathbb{Z}\oplus\mathbb{Z})\cong\mathbb{Q}(G)(X,Y)$ and $\mathbb{Q}(G\oplus\mathbb{Z})\cong\mathbb{Q}(G)(X)$, a negative answer to my question would give a positive answer to the motivating question.

I've found a little about this construction in the literature, but not much. For example, in

Gervasio G. Bastos and T. M. Viswanathan, MR 942063 Torsion-free abelian groups, valuations and twisted group rings, Canad. Math. Bull. 31 (1988), no. 2, 139--146.

it is proved that $\mathbb{Q}(G)$ is a purely transcendental extension of $\mathbb{Q}$ if and only if $G$ is free abelian.

Any other relevant references would be welcome. Also, there are obvious variants: e.g., using a different field, such as $\mathbb{F}_2$ instead of $\mathbb{Q}$.

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    $\begingroup$ Just a remark: if $G$ has no primitive element (in the sense that for every $g\in G$ there exists $n\ge 2$ such that $n^{-1}g\in G$) then I guess that $\pm G$ is characterized within $\mathbb{Q}(G)^*$ as those elements with roots of arbitrarily high order. (This holds if $G$ is a $\mathbb{Z}[1/p]$-module, or if $G$ is noncyclic of rank 1.) Since $G$ is isomorphic to $(\pm G,\cdot)$ modulo its torsion subgroup, it follows that $\mathbb{Q}(G)$ recognizes $G$ in this case. $\endgroup$ – YCor Jan 9 '17 at 11:55
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    $\begingroup$ Lovely question Jeremy. I only wish I could say something useful about it! $\endgroup$ – Nick Gill Jan 12 '17 at 10:57
  • $\begingroup$ This question could be related. $\endgroup$ – Watson Jun 27 '17 at 22:02

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