Prime correspondence under localization

Let $R$ be a commutative ring and $S\subseteq R$ a multiplicative set . Write $S^{-1}R$ for the localization .

Theorem (prime correspondence)

There is an inclusion-preserving bijection

given by:

• contraction: $\mathfrak P \mapsto \mathfrak p = \{\,r\in R : r/1\in \mathfrak P\,\}$,
• extension: $\mathfrak p \mapsto S^{-1}\mathfrak p = \{\,r/s : r\in\mathfrak p,\, s\in S\,\}$.

In particular, if $\mathfrak p\subset R$ is prime and we localize at $\mathfrak p$ (so $R_{\mathfrak p}$ is the localization at \(\mathfrak p\) ), then prime ideals of $R_{\mathfrak p}$ correspond exactly to prime ideals $\mathfrak q\subseteq \mathfrak p$ in $R$.

Maximal ideals correspond similarly: maximal ideals of $S^{-1}R$ correspond to maximal ideals of $R$ disjoint from $S$.

Examples

1. Invert 2 in $\mathbb{Z}$.
   Let $S=\{1,2,4,8,\dots\}$. Then $S^{-1}\mathbb{Z}=\mathbb{Z}[1/2]$.
   Prime ideals of $\mathbb{Z}[1/2]$ correspond to primes $\mathfrak p\subset \mathbb{Z}$ with $\mathfrak p\cap S=\varnothing$, i.e. $(0)$ and $(p)$ for odd primes $p$. The prime $(2)$ disappears because it meets $S$.

2. Localize $\mathbb{Z}$ at $(p)$.
   In $R=\mathbb{Z}$, take $S=\mathbb{Z}\setminus (p)$, so $S^{-1}R=\mathbb{Z}_{(p)}$.
   Prime ideals of $\mathbb{Z}_{(p)}$ correspond to primes of $\mathbb{Z}$ contained in $(p)$, namely $(0)$ and $(p)$. Thus $\mathrm{Spec}(\mathbb{Z}_{(p)})=\{(0),(p)\}$.

3. Localize $k[x,y]$ at powers of $x$.
   Let $R=k[x,y]$ and $S=\{1,x,x^2,\dots\}$. Then $S^{-1}R=R_x$.
   Prime ideals in $R_x$ correspond to prime ideals of $R$ not containing $x$. For instance, $(y)$ survives (it does not contain $x$), while $(x,y)$ does not (it contains $x$, hence meets $S$).