Is the ring $R[[x]][\frac{1}{x}]$ Noetherian if $R$ is Noetherian? Let $R$ be a Noetherian ring.
Then its extended polynomial ring $R[x]$ and power series ring $R[[x]]$ both are Noetherian as well.

Is the ring $R[[x]][\frac{1}{x}]$ Noetherian ?


I know if $R$ is Noethetian ring with zero nilradical and if $S$ is the set of regular elements of $R$ (i.e., $S=\{r \in R: rs=0 \Rightarrow s=0\}$), then the ring $S^{-1}R$ is also Noetherian.
So if $R$ is in addition an integral (e.g., $R=\mathbb Z_p=$ ring $p$-adic integers), then $R[[x]][x^{-1}]$ is also Noetherian.
Am I correct ?

You can think of $R[[x]][x^{-1}]$ as the polynomial ring $R[[x]][y]$ modulo the ideal generated by $xy - 1$ (so that $y = x^{-1}$). Since quotients of Noetherian rings are Noetherian, $R[[x]][x^{-1}]$ is Noetherian.
