Localization preserves prime ideals

1. Localizing $\mathbb Z$ at a prime. Let $R=\mathbb Z$ and $S=\mathbb Z\setminus(p)$, so $S^{-1}R=\mathbb Z_{(p)}$. The prime ideals of $\mathbb Z$ are $(0)$ and $(q)$ for primes $q$. One checks:

• $(p)\cap S=\varnothing$, so $(p)$ extends to the (unique) maximal ideal $p\mathbb Z_{(p)}$, which is prime.
• If $q\neq p$, then $(q)\cap S\neq\varnothing$ (since $q\in S$), so $(q)$ becomes the unit ideal after localization.
• Also $(0)\cap S=\varnothing$, so $(0)$ extends to $(0)$ in $\mathbb Z_{(p)}$.

1. Inverting a variable in a polynomial ring. Let $R=k[x,y]$ and $S=\{1,y,y^2,\dots\}$, so $S^{-1}R=k[x,y,y^{-1}]$. The ideal $\mathfrak p=(x)$ is prime and disjoint from $S$ (since no power of $y$ lies in $(x)$), hence $S^{-1}(x)$ is prime in $k[x,y,y^{-1}]$. By contrast, the prime ideal $(y)$ meets $S$ (it contains $y$), so $S^{-1}(y)=S^{-1}R$.

1. Localization at a prime and the resulting local ring. If $\mathfrak p$ is a prime ideal of $R$ and $S=R\setminus\mathfrak p$, then $S^{-1}R$ is the localization $R_{\mathfrak p}$, which is a local ring. The extension $S^{-1}\mathfrak p$ becomes the unique maximal ideal of $R_{\mathfrak p}$, and it is prime by the theorem.

For the full bijective correspondence between primes of $S^{-1}R$ and primes of $R$ disjoint from $S$, see the localization prime correspondence.