Localization preserves Noetherianity

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

In particular, for any prime ideal $\mathfrak p$, the localization at $\mathfrak p$ is Noetherian: $R_{\mathfrak p}$ is a Noetherian local ring. Likewise, for any maximal ideal $\mathfrak m$, $R_{\mathfrak m}$ is a Noetherian local ring in the sense of local ring with maximal ideal.

1. Localizing the integers. $\mathbb Z$ is Noetherian, so for any prime $p$ the localization $\mathbb Z_{(p)}$ is Noetherian. Concretely, $\mathbb Z_{(p)}$ consists of fractions $a/b$ with $p\nmid b$, and its unique maximal ideal is generated by $p$.

1. Inverting a polynomial. If $R=k[x,y]$ with $k$ a field, then $R$ is Noetherian by Hilbert basis. Localizing at the multiplicative set $\{1,x,x^2,\dots\}$ gives $k[x,y]_x$, which is again Noetherian.

1. Localization can simplify quotients. In $A = k[x,y]/(xy)$, localize at powers of $x$. Since $x$ becomes invertible in $A_x$, the relation $xy=0$ forces $y=0$ in the localization, so $A_x \cong k[x,x^{-1}]$, a Noetherian ring.