Norm of an element at a localization

$ \newcommand{\K}{\mathbb{K}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\F}{\mathbb{F}} \newcommand{\O}{\mathcal{O}} \newcommand{\p}{\mathfrak{p}} \newcommand{\q}{\mathfrak{q}} \newcommand{\a}{\mathfrak{a}} \require{AMScd} \DeclareMathOperator{\coker}{coker} $ I believe I've come up with an answer. A confirmation of the correctness or a simpler proof would be appreciated. I'd also like to see a counterexample to the isomorphism (or a proof if it's true).

We have the following exact diagram:

\begin{CD} @. @. @. 0 @. \\ @. @. @. @VVV @.\\ @. 0 @. 0 @>>> \ker{\phi} @>>> ...\\ @. @VVV @VVV @VVV @.\\\ 0 @>>> \alpha\O_\p @>>> \O_\p @>>> \O_\p/\alpha\O_\p @>>> 0\\ @. @VVV @VVV @VV\phi V \\ 0 @>>> \alpha \O_{\K,\p} @>>> \O_{\K,\p} @>>> \O_{\K,\p}/\alpha \O_{\K,\p} @>>> 0\\ @. @VVV @VVV @VVV @.\\ ... @>>> \alpha \O_{\K,\p}/\alpha\O_\p @>>> \O_{\K,\p}/\O_\p @>>> \coker{\phi} @>>> 0\\ @. @VVV @VVV @VVV @. \\ @. 0 @. 0 @. 0 \end{CD}

One can check that the multiplication by $\alpha$ map leads to $\O_{\K,\p}/\O_\p \cong \alpha \O_{\K,\p}/\alpha\O_\p$. If these are finite (I believe they are), then due to the snake lemma \begin{CD} 0 @>>> \ker{\phi} @>>> \alpha \O_{\K,\p}/\alpha\O_\p @>>> \O_{\K,\p}/\O_\p @>>> \coker{\phi} @>>> 0 \end{CD} we have $\#\ker{\phi} = \#\coker\phi$, and so we have $\#\O_\p/\alpha\O_\p = \# \O_{\K,\p}/\alpha \O_{\K,\p}$ from the exactness of the third column.

There are two doubts I have left. Is $\O_{\K,\p}/\O_\p$ finite? It should be since $[\O_\K : \O_\p] < \infty$, but I'm not 100% sure that $\O\setminus\p$ in the denominator can't create problems. And is $\phi$ in general an isomorphism ? I have a feeling like it's not, but I couldn't find a counterexample.

Very nice!

$\mathcal{O}_K/\mathcal{O}$ is a finite set, hence an $\mathcal{O}$-module of finite length, so $\mathcal{O}_K/\mathcal{O}$ $\cong$ $\prod_{\mathfrak{p}}\mathcal{O}_{K,\mathfrak{p}}/\mathcal{O}_{\mathfrak{p}}$ (see e.g. Eisenbud, Th. 2.13). The quotients $\mathcal{O}_{K,\mathfrak{p}}/\mathcal{O}_{\mathfrak{p}}$ are therefore surely finite.

If $\phi$ is surjective then $\mathcal{O}_{K,\mathfrak{p}}=\mathcal{O}_{\mathfrak{p}}+\alpha\mathcal{O}_{K,\mathfrak{p}}$. This fails for $K=\mathbb{Q}(\sqrt{5})$, $\mathcal{O}=\mathbb{Z}[\sqrt{5}]$, $\alpha=2$, with $\mathfrak{p}$ the $\mathcal{O}$-prime above $2\mathbb{Z}$. So generally $\phi$ is not an isomorphism.

Edit: less dramatically, $[\mathcal{O}_{K,\mathfrak{p}}:\mathcal{O}_{\mathfrak{p}}][\mathcal{O}_{\mathfrak{p}}:\alpha\mathcal{O}_{\mathfrak{p}}]$ $=$ $[\mathcal{O}_{K,\mathfrak{p}}:\alpha\mathcal{O}_{K,\mathfrak{p}}][\alpha\mathcal{O}_{K,\mathfrak{p}}:\alpha\mathcal{O}_{\mathfrak{p}}]$ $<\infty$ directly yields $|\mathcal{O}_{\mathfrak{p}}/\alpha\mathcal{O}_{\mathfrak{p}}|$ $=$ $|\mathcal{O}_{K,\mathfrak{p}}/\alpha\mathcal{O}_{K,\mathfrak{p}}|$.