Localizing a PID at a nonzero prime

Let $a\in R$ be nonzero. If $a\in \mathfrak{p}$, then $a=\pi a_1$ because $\mathfrak{p}=(\pi )$. If $a_1\in \mathfrak{p}$, divide by $\pi$ again, and continue. This process stops, for otherwise $(a)⊊(a_1)⊊(a_2)⊊\cdots$ would be a strict ascending chain of principal ideals, contradicting [L1]. Thus $a=u{\pi }^{n}b$ with $n\ge 0$, $u$ a unit of $R$, and $b\mathrm{\notin }\mathfrak{p}$.

Every nonzero element of $R_{\mathfrak{p}}$ is therefore $(a\mathrm{/}s)=((ub)\mathrm{/}s)(\pi \mathrm{/}1{)}^n$ with $s\mathrm{\notin }\mathfrak{p}$ and $b\mathrm{\notin }\mathfrak{p}$. Since both $b$ and $s$ are in the denominator set, $(ub)\mathrm{/}s$ is a unit of $R_{\mathfrak{p}}$. Hence $R_{\mathfrak{p}}$ is local with nonzero maximal ideal generated by $\pi \mathrm{/}1$, and every ideal is principal by the same minimum-exponent argument as in a DVR.

The ring $R_{\mathfrak{p}}$ is not a field because $\pi \mathrm{/}1$ lies in its maximal ideal and is not a unit. So [L2] applies and shows that $R_{\mathfrak{p}}$ is a discrete valuation ring with uniformiser $\pi \mathrm{/}1$.