Localization of a ring

We will denote by $S^{-1}R$, where $S$ is a multiplicatively closed subset of $R$, the ring of fractions of $R$ with respect to $S$ (see definition \cite{atiyah} page 36).
(By the way, if $R$ is an integrity domain, $S=R-\{0\}$ is multiplicatively closed and $S^{-1}R$ is the field of fractions of $R$.)
If $\mathfrak{p}$ is a prime ideal of $R$ then $S=R-\mathfrak{p}$ is multiplicatively closed, and we denote

It is called the localization of $R$ at $\mathfrak{p}$. The elements of the form $a/s$ with $a\in \mathfrak{p}$ and $s\notin \mathfrak{p}$ form an ideal $\mathfrak{m}$ of $R_{\mathfrak{p}}$ which is maximal. Take $b/s\notin \mathfrak{m}$, i.e., $b\notin \mathfrak{p}$.
Then, clearly, $b/s$ is a unit so $\mathfrak{m}$ is maximal and is unique. In other words: $R_{\mathfrak{p}}$ is a local ring.