Wednesday, 17 December 2008

ac.commutative algebra - How to prove that the subrings of the rational numbers are noetherian?

I will break this up into two steps, each of which is a standard exercise:



1) Let $R$ be a principal ideal domain with fraction field $K$. Every overring of $R$ -- i.e., every ring $S$ with $R subset S subset K$ -- is the localization of $R$ at a multiplicative subset.



2) If $R$ is a Noetherian ring and $S$ is a multiplicative subset, then the localization $S^{-1} R$ is a Noetherian ring.

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