$ \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}} $
Going through Keith Conrads notes on non-maximal orders and the conductor ideal, I stumbled upon an equality that was supposed to be proven by the reader $$ [\O_\p : \alpha \O_\p] = [\O_{\K,\p} : \alpha \O_{\K,\p}], \tag{1} $$
where $\O \subseteq \O_\K$ is an order inside a maximal order of a number field $\K$, $\p\subseteq \O$ a prime ideal of $\O$, $\O_\p = (\O \setminus \p)^{-1}\O$ the localization of $\O$ at $\p$, $\O_{\K,\p}=(\O \setminus \p)^{-1}\O_{\K}$ the "localization of $\O_\K$ at $\p\subseteq\O$", and $\alpha$ an element of $\O$.
The theorem actually states that if $\a\subseteq \O_\K$ is an ideal such that $\a \cap \O$ is an invertible $\O$ ideal, then
$$ \O/(\a\cap\O) \cong \O_K/\a $$
In the above isomorphism the injectivity is natural and the surjectivity follows from $\O_\K = \O + \a$ which we prove by showing $\O_{\K,\p} = \O_\p + \a_\p$ over all primes $\p \subseteq \O$. Locally $\a_\p \cap \O_\p = \alpha\O_\p$ for some $\alpha\in \O$ and then we have a chain of inequalities $$ [\O_\p : \alpha \O_\p] \leq [\O_{\K,\p} : \a_\p] \leq [\O_{\K,\p} : \alpha \O_{\K,\p}] $$ the first one following from $\O_\p/\alpha\O_\p \hookrightarrow \O_{\K,\p}/\a_\p$, and the second one from $\alpha\O_{\K,\p} \subseteq \a_p \subseteq \O_{\K,\p}$. One then proves the equation (1), and obtains an isomorphism from the injective map above, proving the theorem.
Is the equation (1) true in general? It's obvious if $\alpha \not\in \p$, but how does one go about proving it otherwise? Is it built upon the dependence of $\alpha$ on $\a\subseteq\O_\K$, or is it true for all $\alpha \in \O$?
I believe that the following is not true $$ \O_\p/\alpha\O_\p \cong \O_{\K,\p}/\alpha \O_{\K,\p}. $$ How would one go about for proving (1)?
EDIT:
One idea is to use the isomorphisms \begin{equation} \O/\alpha\O \cong \bigoplus_{\p \supseteq \alpha\O}\O_\p/\alpha\O_\p \\ \O_\K/\alpha\O_\K \cong \bigoplus_{\q \supseteq \alpha\O_\K}\O_{\K,\q}/\alpha\O_{\K_\q} \end{equation} with $\q$ being ideals of $\O_\K$. Cardinalities of the above sets are equal to the norm of $\alpha$, and it's just left to relate $\O_{\K,\p}$ with $\O_{\K,\q}$. The problem is that $\p$ is an ideal of $\O$ and not of $\O_{\K}$ and I don't see an obvious way to link that to $\q\subseteq \O_{K}$. One may say that $\p\O_\K = \p_1^{e_1}\cdots\p_s^{e_s}$ where the $\p_i$'s are $\O_\K$ prime ideals, and they're all different over different prime ideals $\p$ of $\O$. Ramifications can happen.
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 $|\mathcal{O}_{\mathfrak{p}}/\alpha\mathcal{O}_{\mathfrak{p}}|$ $=$ $|\mathcal{O}_{K,\mathfrak{p}}/\alpha\mathcal{O}_{K,\mathfrak{p}}|$.