How to prove the following version of the Hilbert basis theorem:
$R$ is Noetherian if and only if $R[|x|]$ is Noetherian.
Of course, in view of the isomorphism: $$\frac{R[|x|]}{(x)~R[|x|]} \simeq R$$ one direction follows. I'm struggling to come up with a proof for the converse. Any help is much appreicated.
You could use the usual Hilbert basis theorem to show that $R[x]$ is Noetherian, and then note that $R[[x]]$ is the $(x)$-adic completion of the Noetherian ring $R[x]$ and hence Noetherian.
I think you can also directly adapt the usual argument about killing leading terms of polynomials.