$k$ is a number field, $R$ its ring of integers and $\mathfrak p$ a nonzero prime ideal of $R$. Let $R_\mathfrak p$ be the localization of $R$ at $\mathfrak p$. Is it true that $R_\mathfrak p/{\mathfrak pR_\mathfrak p}\cong (R/\mathfrak p)_{\mathfrak pR/\mathfrak p}$?
Many thanks in advance.
Note that $\mathfrak p R/\mathfrak p = 0$, so $(R/\mathfrak p)_{\mathfrak pR/\mathfrak p}$ is the field of fractions of $R/\mathfrak p$, which I'll call $R'$. We have a map $\phi: R_\mathfrak p \rightarrow R'$ given by $\frac{a}{b} \mapsto \frac{a+\mathfrak p}{b+\mathfrak p}$. Note $b \notin \mathfrak p$, so this fraction makes sense. Moreover, if $\frac{a}{b} = \frac{a'}{b'}$, then $a'b=b'a$. This implies $(a'+ \mathfrak p)(b + \mathfrak p)=(b'+\mathfrak p)(a+\mathfrak p)$, so $\phi$ is well-defined.
It's clear that $\phi$ is surjective, and $\phi(\frac{a}{b})=0$ if and only if $\phi(\frac{a}{1})=0$ if and only if $a\in \mathfrak p$. Hence $\ker \phi = \mathfrak p R_{\mathfrak p}$, and $R_{\mathfrak p} / \ker\phi = R_{\mathfrak p}/ \mathfrak p R_{\mathfrak p} \simeq R'$.