A number ring $R$ is singular over $p$ if and only if $p$ divides the index $[\tilde{R}:R]$. I'm learning about algebraic number theory and I stumbled upon the following statement:
Let $R$ be a number ring (i.e. a subring of a finite extension of $\mathbb{Q}$) with normalization $\tilde{R}$. Then $R$ is singular over a prime number $p$ if and only if $p|[\tilde{R}:R]$.
Using the fact that a prime ideal $\mathfrak{p}$ of $R$ is regular if and only if its mutliplier ring $r(\mathfrak{p})=\{x\in \mathrm{Quot}(R):x\mathfrak{p}\subset\mathfrak{p}\}$ equals $R$, I did one implication:
If $R$ has a singular prime $\mathfrak{p}$ over $p$, then there is some $x\in r(\mathfrak{p})\setminus R$. Now $\mathfrak{p}$ is finitely generated as $R$ is Noetherian, so $x\mathfrak{p}\subset\mathfrak{p}$ implies that $x\in\tilde{R}$. But then $px\in\mathfrak{p}\subset R$ shows that $\overline{x}\in\tilde{R}/R$ has order $p$, so that $p|[\tilde{R}:R]$.
Conversely, if $p$ divides the index then the finite abelian group $\tilde{R}/R$ has an element of order $p$ by Cauchy, i.e. there is some $x\in\tilde{R}\setminus R$ such that $px\in R$. As $px$ is not a unit in $R$ it is contained in a prime $\mathfrak{p}$ of $R$. If now i can show that $p\in\mathfrak{p}$, this implies that $\mathfrak{p}$ is singular since then $x\in r(\mathfrak{p})\setminus R$.
Is it true that $p\in\mathfrak{p}$? I have the feeling I'm missing something silly here. I know that $\mathfrak{p}$ must contain a unique prime number, but I don't see why it has to be $p$.
If someone is aware of some other proof for this then I would also love to hear it.
Thanks in advance!
Edit: after some thought I came up with a correct proof, which I'll post as an answer for future reference, I hope this is ok.
 A: We use localisations at $S=\mathbb{Z}\setminus p\mathbb{Z}$. Specifically, we prove that
\begin{equation}
R\text{ is regular over }p \,\Leftrightarrow \,R_{(p)}=\tilde{R}_{(p)}\,\Leftrightarrow \,p\!\!\not|\,[\tilde{R}:R].
\end{equation}
The prime ideals of $R_{(p)}$ correspond to the prime ideals of $R$ containing $p$, and this correspondence respects regularity: if $\mathfrak{p}|p$ in $R$ is a prime and $S^{-1}\mathfrak{p}$ the corresponding prime in $R_{(p)}$, then $R_{\mathfrak{p}}=(R_{(p)})_{S^{-1}\mathfrak{p}}$ since $R\setminus\mathfrak{p}\subset S$, so clearly $R_{\mathfrak{p}}$ is a DVR if and only if $(R_{(p)})_{S^{-1}\mathfrak{p}}$ is a DVR. Thus $R$ is regular over $p$ if and only if $R_{(p)}$ is Dedekind. As localization commutes with normalization, it follows that $R$ is regular over $p$ if and only if $R_{(p)}=\tilde{R}_{(p)}$. This establishes the first equivalence.
For the second equivalence, let $m=[\tilde{R}:R]$. If $p\!\!\not|\,m$, we have $m\in S$ so that if $x\in\tilde{R}_{(p)}$ we have $x=\frac{mx}{m}\in R_{(p)}$ and hence $R_{(p)}=\tilde{R}_{(p)}$. If $R_{(p)}=\tilde{R}_{(p)}$ then for any $\overline{x}\in\tilde{R}/R$ we have $x=\frac{y}{s}$ for some $y\in R$ and $s\in\mathbb{Z}$ not dividing $p$. But then $sx=y\in R$ implies that the order of $\overline{x}$ in $\tilde{R}/R$ divides $s$, hence the order cannot be $p$. Thus $\tilde{R}/R$ has no elements of order $p$ and hence $p\!\!\not|\,[\tilde{R}:R]$ because of Cauchy.
