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}}
$
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.
 A: $
\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.
A: 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 $|\mathcal{O}_{\mathfrak{p}}/\alpha\mathcal{O}_{\mathfrak{p}}|$ $=$ $|\mathcal{O}_{K,\mathfrak{p}}/\alpha\mathcal{O}_{K,\mathfrak{p}}|$.
