Localization of a Noetherian Ring is Noetherian

Statement

Let $R$ be a Noetherian ring and let $S\subseteq R$ be a multiplicative set . Then the localized ring S^{-1}R is again Noetherian.

Equivalently: every ideal of $S^{-1}R$ is finitely generated.

A common strengthening: if $M$ is a finitely generated $R$-module, then $S^{-1}M$ is a finitely generated $S^{-1}R$-module (compare localization of modules ).

See also: localization Noetherian corollary .

Examples

1. Integers localized at a prime.
   $\mathbb Z$ is Noetherian. For a prime $p$, the localization $\mathbb Z_{(p)}$ (invert all integers not divisible by $p$) is therefore Noetherian. In fact it is a PID, so every ideal is principal.

2. Inverting one variable in a polynomial ring.
   $R=k[x,y]$ is Noetherian. Localizing at $S=\{1,x,x^2,\dots\}$ gives

   $$R_x \cong k[x,x^{-1},y],$$

   which is still Noetherian.

3. Principal open subsets stay Noetherian (geometric intuition).
   If $R$ is Noetherian and $f\in R$, then $R_f$ is Noetherian. Algebraically this is the special case $S=\{1,f,f^2,\dots\}$; geometrically it corresponds to restricting to a principal open set in Spec(R) with the Zariski topology .