Prime correspondence under localization

The map

induces an inclusion-preserving bijection between:

• prime ideals $\mathfrak p\subseteq R$ with $\mathfrak p\cap S=\varnothing$, and
• prime ideals $\mathfrak q\subseteq S^{-1}R$.

The inverse bijection is contraction:

In other words, “primes survive localization exactly when they do not meet the set of denominators.” This refines the fact that localization preserves primality.

Geometric form

On the prime spectrum, this correspondence identifies $\operatorname{Spec}(S^{-1}R)$ with the open subset

for the Zariski topology. For $S=\{1,f,f^2,\dots\}$, this is the basic open set $D(f)$.

Special case: localization at a prime

If $S=R\setminus \mathfrak p$ for a prime ideal $\mathfrak p$, then $S^{-1}R$ is the ring $R_\mathfrak p$. The primes of $R_\mathfrak p$ correspond exactly to primes $\mathfrak q\subseteq R$ with $\mathfrak q\subseteq \mathfrak p$.

1. Inverting a prime in $\mathbb{Z}$. Take $R=\mathbb{Z}$ and $S=\{1,p,p^2,\dots\}$. Then $S^{-1}R=\mathbb{Z}[1/p]$. A prime ideal of $\mathbb{Z}$ meets $S$ iff it contains $p$, so the primes of $\mathbb{Z}[1/p]$ correspond to $(0)$ and $(\ell)$ for primes $\ell\neq p$.

1. Localizing away from a hypersurface. Let $R=k[x,y]$ and $S=\{1,x,x^2,\dots\}$, so $S^{-1}R=R_x$. Primes of $R_x$ correspond to primes of $k[x,y]$ that do not contain $x$. For instance, $(y)$ survives (since $x\notin(y)$), while $(x,y)$ does not survive (since it contains $x$).

1. Localization at a maximal ideal. In $R=k[x,y]$, localize at $\mathfrak m=(x,y)$ to get $R_\mathfrak m$. The primes of $R_\mathfrak m$ correspond to primes contained in $(x,y)$, namely $(0)$, $(x)$, $(y)$, and $(x,y)$.