A localization of a Dedekind domain is Dedekind or a field

By [F1] and [L1], the localisation $S^{-1}R$ is Noetherian. Because $R$ is a domain and $0\mathrm{\notin }S$, so is $S^{-1}R$. By [F1] and [L2], it is integrally closed.

Suppose $S^{-1}R$ has a nonzero prime ideal $\mathfrak{n}$. By [L3], it is the extension of a nonzero prime ideal $\mathfrak{p}\subset R$ disjoint from $S$. In a Dedekind domain every nonzero prime is maximal, so every nonzero prime of $S^{-1}R$ is maximal as well by the same correspondence [L3]. Therefore $S^{-1}R$ has dimension $1$, and step 1.1 makes it a Dedekind domain.

Suppose instead that $S^{-1}R$ has no nonzero prime ideals. Then [L3] says that the only prime ideal of $S^{-1}R$ is $(0)$. In a domain, $(0)$ is maximal exactly when the ring is a field. Hence $S^{-1}R$ is a field.