TITLE: $\mathfrak p$ be a prime ideal of a commutative ring $R$ , then $R_\mathfrak p/\mathfrak pR_\mathfrak p$ is the field of fractions of $R/\mathfrak p$?
QUESTION [9 upvotes]: Let $\mathfrak p$ be a prime ideal of a commutative ring $R$ with unity , then how to show that the field  $R_\mathfrak p/\mathfrak pR_\mathfrak p$ is isomorphic with the field of fractions of the integral domain $R/\mathfrak p$ ?
I can show that $R_\mathfrak p/\mathfrak pR_\mathfrak p \cong (R/\mathfrak p)_\mathfrak p$ as $R_\mathfrak p$ modules ... but I am not sure whether this implies the result I am asking or not

REPLY [2 votes]: Since localization preserves exact sequences we have that the exact sequence
$$ 0 \to \mathfrak p \to R \to R/\mathfrak p \to 0$$
becomes
$$0 \to \mathfrak p R_{\mathfrak p} \to R_{\mathfrak p} \to (R/\mathfrak p)_{\mathfrak p} \to 0$$
 Hence $(R/\mathfrak p)_{\mathfrak p}=R_{\mathfrak p}/\mathfrak pR_\mathfrak p$. From this you can conclude that it is the field of fractions since $R/\mathfrak p$ is a domain and localizing by $\mathfrak p$ means inverting all nonzero elements (there are no nonzero divisor in an integral domain).