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Let $R$ be a commutative ring. I am interested in the following two questions:

1) If $R$ is Noetherian, then is $R[u,u^{-1}]$ Noetherian? In fact, I know that this is true (the Hilbert basis theorem tells us $R[u]$ is Noetherian and then $R[u,u^{-1}]$ is a localisation of $R[u]$, which preserves the Noetherian property).

Is there a more insightful proof? I feel like just the above two facts, whilst telling us it is true, don't really offer much insight into why this is true.

2) If $R$ is local, then is $R[u,u^{-1}]$ still local? We are localising $R[u]$ at the multiplicative subset $S=\{ u^i \}_{i \ge 0}.$ The result would follow if I knew $S$ was a maximal multiplicative subset of $R[u]$, but is this true?

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"Insightful proof" is too subjective for me but I have answered question 2). – Georges Elencwajg Jan 29 '12 at 22:53
Dear Juan, The Hilbert Basis Theorem is enough in this case (see the answer by anonymous below); you don't need to think about localizations if you don't want to. (Although you can if you want to; your argument is perfectly correct.) The Hilbert Basis Theorem is a basic source for Noetherianness of rings; it's not clear what additional insight you want. Regards, – Matt E Jan 30 '12 at 1:13
@Matt - thanks for the comments. I see now that HBT is indeed required for this! – Juan S Jan 30 '12 at 1:28
up vote 6 down vote accepted

1) I think the best way to think of this ring is as $R[u,v] / (uv-1)$. With this presentation it is clear that it is noetherian.

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Ah ok - basically because we are adjoining a unit? – Juan S Jan 29 '12 at 23:14
@Juan: Dear Juan, No, because of the Hilbert Basis Theorem (it is a quotient of $R[u,v]$, which is Noetherian by HBT). Regards, – Matt E Jan 30 '12 at 1:04

2) No: $\mathbb C$ is local but $\mathbb C[u,u^{-1}]$ isn't.
Indeed, its maximal ideals are the principal ideals $(u-a) $'s with $a\in \mathbb C^*$.

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