Let $A$ be a Dedekind domain and $\mathfrak{p}\subset A$ be a prime ideal. Then the localization $A_\mathfrak{p}$ is also a Dedekind domain. I can show it has a unique maximal ideal $\mathfrak{p}':=\mathfrak{p}A_\mathfrak{p}$. Thus any ideal in $A_\mathfrak{p}$ can be expressed as $\mathfrak{p}'^n$. I want to show that for any $x\in\mathfrak{p}'\setminus \mathfrak{p}'^2$, we have $\mathfrak{p}'^n=x^nA_\mathfrak{p}$. This in particular implies $A_\mathfrak{p}$ is a P.I.D.
Any help or hint is much appreciated.
Best Answer
Given the comments, the question is actually as follows: why is a local Dedekind domain a PID?
Let $A$ be a local Dedekind domain, and let $m$ be its maximal ideal. For each $x \in A \backslash \{0\}$, define $v(x) \in \mathbb{N}$ as the nonnegative integer such that $m^{v(x)}=xA$.
Let $x \in m$ be such that $v(x)$ is minimal: then, for each $y$, $yA=m^{v(y)} \leq m^{v(x)}=xA$ so that $y \in xA$. Therefore $m \subset xA$ and $m$ is a principal ideal, thus so are its powers, and therefore so is every ideal of $A$.