3
$\begingroup$

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

$\endgroup$
6
+50
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.