# Group ring $R(G)$ isomorphic to polynomial ring $R[x]$

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?

• Try $R[x,\frac1x]\cong R[x,y]/(xy-1)$ instead. – Arthur Nov 18 '16 at 17:51
• What is wrong with $f : R(G) \to R[x]$ given by $$f(r_0 +r_1g + r_2g^2 + \cdots +r_ng^n) = r_0 + r_1x +r_2x^2 + \cdots + r_nx^n$$ – Fly by Night Nov 18 '16 at 17:57
• @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. – rschwieb Nov 18 '16 at 18:01
• @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.) – Alex Wertheim Nov 18 '16 at 18:05
• @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? – 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 $\mathbb F[x]$.