# Different version of the Hilbert basis theorem [duplicate]

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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.

## marked as duplicate by Stahl, user26857 abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 16 '16 at 8:20

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.