Localization of a Noetherian ring is Noetherian

Remarks

Let $\lambda \colon R \to S^{-1}R$ be the canonical map. Each ideal $J$ of $S^{-1}R$ is the extension of its contraction: $J = S^{-1}\lambda^{-1}(J)$. Indeed, if $a/s \in J$, then $a/1 = s \cdot (a/s) \in J$, so $a \in \lambda^{-1}(J)$. Hence a strictly ascending chain of ideals in $S^{-1}R$ gives a strictly ascending chain of their contractions in $R$, which cannot be infinite. The same argument shows that $S^{-1}M$ is a Noetherian $S^{-1}R$-module when $M$ is a Noetherian $R$-module.

Example: $\Z_{(p)}$, $\Z[1/n]$, $\Q$ and the local rings $k[x_1, \ldots, x_n]_{\mathfrak p}$ of a polynomial ring over a field $k$ are Noetherian.

The converse fails: Noetherianity is not a local property. An infinite product $R = \prod_{i \in \N} k$ of copies of a field $k$ is not Noetherian, but each localization $R_{\mathfrak p}$ at a prime ideal is a field.

Going deeper

For a student

This fact is used all the time with $S = R \setminus \mathfrak p$: the local rings $R_{\mathfrak p}$ of a Noetherian ring are Noetherian local rings. Thus the theory of Noetherian local rings, such as Krull dimension, the embedding dimension and regular local rings, applies to every prime of a Noetherian ring. Together with Hilbert's basis theorem and passage to quotients, it shows that the coordinate rings of affine varieties and all their localizations are Noetherian.