TITLE: Localization at a multiplicative set is a localization at a prime ideal if local
QUESTION [6 upvotes]: Let $R$ be an integral domain. Assume $S$ is a multiplicative set in $R$ such that the localization $S^{-1}R$ is a local ring. Does there exist a prime ideal $\mathfrak{p}\subset R$ such that $S^{-1}R\simeq R_{\mathfrak{p}}$ as $R$-algebras (if not, at least as unital rings)?

REPLY [4 votes]: $R$ is an integral domain and $A=S^{-1} R$. That it is a local domain  means it has a unique maximal ideal $m\subset A - A^\times$.
For every non-zero $a \in A - A^\times$ then $(a)$ is a proper ideal of $A$ so it is contained in a maximal ideal which must be $m$, thus $a \in m$ and $m = A-A^\times$.
Let $$P = m \cap R= (A-A^\times) \cap R = R - (A^\times \cap R)$$ which is a prime ideal of $R$. 
Whence $$A = (A^\times \cap R)^{-1}R=(R-P)^{-1} R=R_P$$