Is the ring of Laurent polynomials in $n$ noncommuting variables Noetherian?

Question

Suppose we have a Noetherian ring $R$. Is it true that the ring of Laurent polynomials $R\langle x_1,\,x_1^{-1},\ldots,\,x_n,\,x_n^{-1}\rangle$ in $n$ noncommuting variables is also Noetherian? If so, why?

Answer 1

It's not (left or right) Noetherian if $n>1$ (and $R\neq0$). It's the same as the group ring $RF_n$ of a free group on $n$ generators, and for a group ring to be Noetherian, the group must at least be Noetherian (which $F_n$ is not, since it contains an infinitely generated free subgroup).

To see that $RG$ is not Noetherian if $G$ is not, for any subgroup $H\leq G$, consider the right ideal $I_H$ of $RG$ generated by the augmentation ideal of $RH$ (or equivalently, $I_H$ is the set of $\sum_{g\in G}\lambda_gg\in RG$ such that $\sum_{g\in Hx}\lambda_g=0$ for every coset $Hx$ of $H$ in $G$). Then if $G$ has an infinite ascending chain $$H_0<H_1<H_2<\dots$$ of subgroups, $RG$ has an infinite ascending chain $$I_{H_0}<I_{H_1}<I_{H_2}<\dots$$ of right ideals.