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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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    @Matt - thanks for the comments. I see now that HBT is indeed required for this!2012-01-30

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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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    @Juan: Dear Juan, No, because of the Hilbert Basis Theorem (it is a quotient of $R[u,v]$, which is Noetherian by HBT). Regards,2012-01-30
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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 $a\in \mathbb C^*$.