$R_{\mathfrak{p}}$ is local with unique maximal ideal $\mathfrak{p}{R}_{\mathfrak{p}}$

The ring is nonzero: if $1\mathrm{/}1=0$, [F3] gives $u\mathrm{\notin }\mathfrak{p}$ with $u=0$, impossible because $0\in \mathfrak{p}$.

By [F2], if $r\mathrm{\notin }\mathfrak{p}$, then $r\mathrm{/}s$ is a unit by taking $a=1$. If $r\in \mathfrak{p}$ and $ar$ lay outside $\mathfrak{p}$, the ideal property would be contradicted; hence $r\mathrm{/}s$ is not a unit. Thus the displayed set is exactly the set of nonunits.

Membership in the displayed set is independent of the chosen fraction: if $r\mathrm{/}s=r^{\mathrm{\prime }}\mathrm{/}{s}^{\mathrm{\prime }}$ with $r\in \mathfrak{p}$, then [F3] gives $u\mathrm{\notin }\mathfrak{p}$ with $u(r{s}^{\mathrm{\prime }}-r^{\mathrm{\prime }}s)=0$. Hence $u{r}^{\mathrm{\prime }}s=ur{s}^{\mathrm{\prime }}\in \mathfrak{p}$; primality and $u,s\mathrm{\notin }\mathfrak{p}$ force $r^{\mathrm{\prime }}\in \mathfrak{p}$.

The displayed set is an ideal: common-denominator addition and multiplication by arbitrary fractions preserve numerator membership in $\mathfrak{p}$. It is proper because $1\mathrm{\notin }\mathfrak{p}$, so $1\mathrm{/}1$ is not in it. Every proper ideal contains only nonunits, so step 1.2 makes every proper ideal lie inside it. It is therefore the unique maximal ideal, and [F4] makes $R_{\mathfrak{p}}$ local.