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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.

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  • $\begingroup$ Which definition are you using for a Dedekind domain? Unique factorization of ideals? Noetherian, integral, normal and dimension 1? Any other custom variant of these? $\endgroup$ – Mindlack Jan 4 at 13:28
  • $\begingroup$ @Mindlack I know the proof that two definitions of yours are equivalent, so either would be OK to me. $\endgroup$ – scd Jan 4 at 13:31
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    $\begingroup$ What is the prime ideal factorization of $xA_{\mathfrak{p}}$? Notice that there is only one non-zero prime ideal in the localization $A_{\mathfrak{p}}$. More generally one can similarly show (using the Chinese remainder theorem) that any Dedekind domain with only finitely many prime ideals is a PID. $\endgroup$ – leoli1 Jan 4 at 13:32
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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 $A$.

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