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I have read that there is an isomorphism between a total ring of fractions of the group ring $\Bbb Q(G)$ and the field of rational functions over $\Bbb Q$ with one variable, where $G$ is an infinite cyclic group. How is this so?

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    $\begingroup$ For what $G$? This is certainly not true for all $G$. $\endgroup$ – knsam Dec 18 '15 at 6:05
  • $\begingroup$ @knsam where $G$ is an infinite cyclic group (I added it to my question as well, thanks) $\endgroup$ – user0990 Dec 19 '15 at 23:54
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This is true if $G=\mathbb Z$; in that case the group ring is isomorphic to $\mathbb Q[x,x^{-1}]$, the ring of Laurent polynomials in one variable, and the field of fractions is the same as the field of fractions of the ordinary polynomial ring, which is the field of rational functions in one variable.

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  • $\begingroup$ How are they the same? $\endgroup$ – user0990 Dec 20 '15 at 21:53
  • $\begingroup$ Clear denominators to get the same canonical fraction. The ring of Laurent polynomials is obtained from the polynomial ring by adjoining an inverse of $x$. This still gives you a subring of $\mathbb Q(x)$. $\endgroup$ – Matt Samuel Dec 20 '15 at 21:55
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For $G = 1$, the group ring is just $\mathbf{Q}$ and its total ring of fractions is also $\mathbf{Q}$.

This is false in general.

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