Show that localizing a ring at a prime ideal produces a local ring.
Solution
Let \(S=R\setminus\mathfrak p\), so \(R_{\mathfrak p}=S^{-1}R\).
1. The extended ideal \(\mathfrak pR_{\mathfrak p}\) is proper.
Suppose \(1\in\mathfrak pR_{\mathfrak p}\). Then \(1=a/s\) for some \(a\in\mathfrak p\) and \(s\notin\mathfrak p\). Equality in the localization implies that for some \(t\notin\mathfrak p\), \[ t(s-a)=0. \] Hence \(ts=ta\in\mathfrak p\). Since \(\mathfrak p\) is prime while \(t,s\notin\mathfrak p\), this is impossible. Thus \(\mathfrak pR_{\mathfrak p}\) is proper.
2. A fraction \(a/s\in R_{\mathfrak p}\) lies in \(\mathfrak pR_{\mathfrak p}\) if and only if \(a\in\mathfrak p\).
If \(a\in\mathfrak p\), then \(a/s\in\mathfrak pR_{\mathfrak p}\). Conversely, suppose \(a/s\in\mathfrak pR_{\mathfrak p}\). Then \(a/s=b/u\) for some \(b\in\mathfrak p\) and \(u\notin\mathfrak p\). Thus for some \(t\notin\mathfrak p\), \[ t(au-bs)=0, \] so \(tau=tbs\in\mathfrak p\). Since \(t,u\notin\mathfrak p\) and \(\mathfrak p\) is prime, it follows that \(a\in\mathfrak p\).
3. Every element outside \(\mathfrak pR_{\mathfrak p}\) is a unit.
Let \(a/s\notin\mathfrak pR_{\mathfrak p}\). By 2, \(a\notin\mathfrak p\), so \(a\in S\). Therefore \(s/a\) is a valid element of \(R_{\mathfrak p}\) and \[ \frac{a}{s}\frac{s}{a}=1. \] Thus \(a/s\) is a unit.
4. Therefore \(R_{\mathfrak p}\) is local with unique maximal ideal \(\mathfrak pR_{\mathfrak p}\).
By 1, \(\mathfrak pR_{\mathfrak p}\) is proper. By 3, every element outside it is a unit. Hence every proper ideal is contained in \(\mathfrak pR_{\mathfrak p}\), so this ideal is maximal and is the unique maximal ideal.