# The field of fractions of the rational group algebra of a torsion free abelian group

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.

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 $\mathbb{Q}$.
• 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. – YCor Jan 9 '17 at 11:55