Localization Preserves Primality

Statement

Let $R$ be a commutative ring and $S\subseteq R$ a multiplicative set . Let $\mathfrak p$ be a prime ideal of $R$.

• If $\mathfrak p\cap S=\varnothing$, then the extended ideal

  $$S^{-1}\mathfrak p \subseteq S^{-1}R$$

  is a prime ideal of $S^{-1}R$.

• If $\mathfrak p\cap S\neq\varnothing$, then $S^{-1}\mathfrak p = S^{-1}R$ (the whole ring), so it is not a proper prime ideal.

This is one direction of the full prime correspondence for localization (primes of $S^{-1}R$ correspond to primes of $R$ disjoint from $S$). Special case: localization at a prime .

Examples

1. Localizing $\mathbb Z$ at a prime $p$.
   Take $R=\mathbb Z$ and $S=\mathbb Z\setminus p\mathbb Z$, so $S^{-1}R=\mathbb Z_{(p)}$.
   The primes of $\mathbb Z$ disjoint from $S$ are $(0)$ and $(p)$. Their extensions are

   $$S^{-1}(0)=(0)\subset \mathbb Z_{(p)},\qquad S^{-1}(p)=p\mathbb Z_{(p)},$$

   both prime in $\mathbb Z_{(p)}$.

2. Inverting $x$ in $k[x,y]$.
   Let $R=k[x,y]$ and $S=\{1,x,x^2,\dots\}$. Then $S^{-1}R\cong k[x,x^{-1},y]$.

   • The prime $(y)$ satisfies $(y)\cap S=\varnothing$, so its extension $(y)S^{-1}R$ is prime.
   • The prime $(x)$ satisfies $(x)\cap S\neq\varnothing$ (since $x\in S$), so $(x)S^{-1}R = S^{-1}R$.

3. Prime becomes maximal in a local ring.
   If $S=R\setminus \mathfrak p$, then $S^{-1}R = R_{\mathfrak p}$ is local and the extension $\mathfrak pR_{\mathfrak p}$ is the unique maximal ideal, hence prime.