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This question already has an answer here:

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.

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marked as duplicate by Stahl, user26857 abstract-algebra Nov 16 '16 at 8:20

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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.

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  • $\begingroup$ Thank you for the suggestion. The latter is the one I'm looking for. I still don't know how to prove that. $\endgroup$ – user358174 Nov 16 '16 at 11:13