Let $k$ be a field and $G$ be a torsion-free abelian group. Then $k[G]$ is an integral domain. If we denote its field of fractions by $F = k(G)$, is it true that $k(G \oplus \mathbb Z )\cong F(X)$?
If not true in general , is it true if $k=\mathbb Q$ ?
$k(G\oplus\mathbb{Z})$ is always isomorphic to $k(G)(X)$.
I'll write the abelian groups multiplicatively and let $X$ be a generator of $\mathbb{Z}$, so the elements of $G\oplus\mathbb{Z}$ have the form $gX^n$ for $g\in G$ and $n\in\mathbb{Z}$.
So $k[G\oplus\mathbb{Z}]$ is isomorphic to $k[G][X,X^{-1}]$.
Finally, for every integral domain $R$ with field of fractions $K$, the field of fractions of $R[X,X^{-1}]$ is isomorphic to $K(X)$, since certainly $R[X,X^{-1}]$ is a subring of $K(X)$, and every element of $K(X)$ is of the form $p(X)/q(X)$ where $p$ and $q$ are polynomials with coefficients in $K$, and multiplying numerator and denominator by a suitable element of $R$ we can assume that $p$ and $q$ have coefficients in $R$, so every element of $K(X)$ is in the field of fractions of $R[X,X^{-1}]$ (or even of $R[X]$).