How would I be able to describe all units of the group ring $\mathbb{Q}(G)$ where $G$ is specifically an infinite cyclic group?


You could realize that this group algebra is isomorphic to $\Bbb Q[x,x^{-1}]$, the Laurent polynomials over $\Bbb Q$.

Since it is just the localization of $\Bbb Q[x]$ at the powers of x, the units are easy to describe.

  • $\begingroup$ Thanks, but I'm not familiar with Laurent polynomials, so I'll be sure to check this out further. I know that the rationals form an integral domain (and $\mathbb{Q}$ is a field), so in this case (from my brief reading thus far), I can describe the units having the form $ux^{k}$ where $u$ is a nonzero element of $\mathbb{Q}$? $\endgroup$ – user0990 Dec 14 '15 at 2:18
  • $\begingroup$ @user0990 Yes, and k is any integer. $\endgroup$ – rschwieb Dec 14 '15 at 3:37
  • $\begingroup$ Great, thanks so much; marking as answered $\endgroup$ – user0990 Dec 14 '15 at 3:51
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    $\begingroup$ I don't understand the other comment very much. It is certainly not isomorphic to a field because it is not a field. $\endgroup$ – rschwieb Dec 14 '15 at 5:01
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    $\begingroup$ @ALannister 1. Yes. 2. You can solve the whole problem by working with Laurent polynomials and no mention of localization. 3. The integral powers of $x$ are obviously a group isomorphic to $\mathbb Z$. The definition of $\mathbb Q[x,x^{-1}]$ and $\mathbb Q[\{x^n\mid n\in \mathbb Z\}$ amount to the same thing! The reason you need two generators is that ring multiplication only generates a monoid, not a group. You need to throw in $x^{-1}$ to finish generating the group. $\endgroup$ – rschwieb Apr 26 '17 at 3:00

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