Let $R$ be a ring and $G$ an infinite multiplicative cyclic group with generator $g$. Is the group ring $R(G) \cong R[x]$?

My guess is that for $R = \mathbb{Q}$ this is not true. Am I on the right track?

  • 1
    $\begingroup$ Try $R[x,\frac1x]\cong R[x,y]/(xy-1)$ instead. $\endgroup$ – Arthur Nov 18 '16 at 17:51
  • 1
    $\begingroup$ What is wrong with $f : R(G) \to R[x]$ given by $$f(r_0 +r_1g + r_2g^2 + \cdots +r_ng^n) = r_0 + r_1x +r_2x^2 + \cdots + r_nx^n$$ $\endgroup$ – Fly by Night Nov 18 '16 at 17:57
  • $\begingroup$ @Arthur $R[x,\frac{1}{x}]\cong R[\mathbb Z]$ by mapping integers to integer powers of $x$, right? Seems the same as what the OP is describing. $\endgroup$ – rschwieb Nov 18 '16 at 18:01
  • 2
    $\begingroup$ @FlybyNight: $g \in R(G)$ is a unit, but your map then takes $g$ to $x \in R[x]$, which is not a unit. (More pertinently, the infinite cyclic group generated by $g$ also contains $g^{-1}, g^{-2}, \ldots$, so your definition is incomplete.) $\endgroup$ – Alex Wertheim Nov 18 '16 at 18:05
  • $\begingroup$ @AlexWertheim Yes, of course, an infinite cyclic group doesn't "wrap around", so the negative powers need considering. I guess a Laurent series is the place to start, would that make $x$ a unit? $\endgroup$ – Fly by Night Nov 18 '16 at 18:12

You're on the right track, yes. $\mathbb F[\mathbb Z]$ is isomorphic to the Laurent polynomials $\mathbb F[x;x^{-1}]$, and this is never isomorphic to $\mathbb F[x]$.

  • $\begingroup$ Got it! Thank you for the help $\endgroup$ – user2640211 Nov 18 '16 at 18:48

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.