# Localization of Dedekind domain at a prime ideal is a P.I.D

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.

• 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? – Mindlack Jan 4 at 13:28
• @Mindlack I know the proof that two definitions of yours are equivalent, so either would be OK to me. – scd Jan 4 at 13:31
• 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. – leoli1 Jan 4 at 13:32

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 $$A$$.