Consider the free ring $R = \mathbb{Z}\left<x_1,\ldots,x_n\right>$ where $x_1,\ldots,x_n$ are non-commuting intermediates. Is the ring $R$ Noetherian?
More generally, let $S$ be a commutative Noetherian ring. Is the non-commutative ring $R_S = S\left<x_1,\ldots,x_n\right>$ Noetherian?
(In some sense, this question is asking for a non-commutative variant of Hilbert's basis theorem).
Best Answer
No, it is not Noetherian. Even for any field, $k\langle x,y\rangle$ is known to be neither left nor right Noetherian, and you have quotients of this ring that produce such ring (say, the quotient by the ideal $(2)$.) This ring cannot be right Noetherian because it would mean the quotient would be right Noetherian.
One way to see it is not Noetherian is that you can take the quotient by $xy-1$ and have a ring where $x$ has a one-sided inverse. This is not possible in a Noetherian ring.