Localization at a prime ideal

Remarks

The ring $R_{\mathfrak p}$ is a local ring. Its unique maximal ideal is $\mathfrak p R_{\mathfrak p} = \set{a/s : a \in \mathfrak p, \ s \notin \mathfrak p}$. Indeed, an element $a/s$ with $a \notin \mathfrak p$ has the inverse $s/a$, so every element outside $\mathfrak p R_{\mathfrak p}$ is a unit. The residue field $R_{\mathfrak p} / \mathfrak p R_{\mathfrak p}$ is called the residue field of $R$ at $\mathfrak p$.

Example: for $R = \Z$ and $\mathfrak p = p\Z$ with $p$ prime, $\Z_{(p)}$ is the ring of rationals $a/b$ in lowest terms with $p \nmid b$. Its maximal ideal is $p \Z_{(p)}$ and its residue field is $\Z/p\Z$. For $\mathfrak p = (0)$ in an integral domain $R$, the localization $R_{(0)}$ is the field of fractions of $R$.

Non-example: the construction needs $\mathfrak p$ to be prime. For the ideal $6\Z$ of $\Z$, the complement $\Z \setminus 6\Z$ contains $2$ and $3$ but not $6$, so it is not a multiplicative set.

Notation: for an element $f \in R$, the ring $R_f$ is the localization at the powers of $f$, which differs from $R_{\mathfrak p}$.