Equivalent condition for being a regular prime ideal $\newcommand{\p}{\mathfrak{p}}$ 
$\newcommand{\tp}{\tilde{\mathfrak{p}}}$ 
$\newcommand{\tA}{\tilde{A}}$
I have a question about Neukirch, Algebraic Number Theory, page 92.
The problem is to show the following.

Let $A$ be a one-dimensional Noetherian domain, and $\tA$ be its normalization. Let $\p$ be a prime ideal of $A$, and $\p\tA = \tp_1^{e_1} \dots \tp_r^{e_r}$ be a prime decomposition.
  Then, $\p$ is regular (i.e. $A_\p$ is a DVR) $\Leftrightarrow$ $r=1$, $e_1=1$ and $f_1 = (\tA/\tp_1 : A/\p) = 1$.

For ($\Rightarrow$), I proved $r = 1$ by $A_\p =\tilde{A_\p} = \tA_{\tp_i}$,
but I can't prove $e_1 = f_1 = 1$. 
I have no idea about ($\Leftarrow$).
 A: "$\Rightarrow$" Let $S=A\setminus\mathfrak p$. Then $S^{-1}\tilde A=\tilde A_{\mathfrak p}=A_{\mathfrak p}$. If prime ideal $\tilde{\mathfrak p}\subset\tilde A$ lies over $\mathfrak p$ then it survives in $A_{\mathfrak p}$, so there is only one with this property. Then $\mathfrak p\tilde A=\tilde{\mathfrak p}^e$, so $S^{-1}(\mathfrak p\tilde A)=S^{-1}\tilde{\mathfrak p}^e$. This entails $\mathfrak p(S^{-1}\tilde A)=(S^{-1}\tilde{\mathfrak p})^e$, that is, $\mathfrak pA_{\mathfrak p}=(\mathfrak pA_{\mathfrak p})^e$ and thus we get $e=1$. It follows  $\mathfrak p\tilde A=\tilde{\mathfrak p}$. Now it is obvious that $f_1=1$.
"$\Leftarrow$" From $f_1=1$ we get $\tilde A=A+\mathfrak p\tilde A$. Now localize at $\mathfrak p$ and get $\tilde A_{\mathfrak p}=A_{\mathfrak p}+\mathfrak p\tilde A_{\mathfrak p}$. At this point I have to invoke a condition which appears in Neukirch on page 78:

(*) $\tilde A$ is a finitely generated $A$-module.

Then $\tilde A_{\mathfrak p}$ is a finitely generated $A_{\mathfrak p}$-module and by Nakayama lemma we get $\tilde A_{\mathfrak p}=A_{\mathfrak p}$, so $A_{\mathfrak p}$ is integrally closed.
