Normality of a certain localization of an order of an algebraic number field Let $f(X) \in \mathbb{Z}[X]$ be a monic irreducible polynomial.
Let $\theta$ be a root of $f(X)$.
Let $A = \mathbb{Z}[\theta]$.
Let $K$ be the field of fractions of $A$.
Let $p$ be a prime number.
Suppose $pA = \alpha^n A$, where $\alpha \in A$ and $n = [K : \mathbb{Q}]$.
Let $A_p$ be the localization of $A$ with respect to $S = \mathbb{Z} - p\mathbb{Z}$.
My question: Is $A_p$ integrally closed?
Motivation
Let $p$ be an odd prime number.
Let $\theta$ be a $p$-th primitive root of unity.
Let $\alpha = 1 - \theta$. Then it is well known that $pA = \alpha^n A$.
It is also well known that $A_p$ is integrally closed.
This is a related question.
 A: Proposition
Let $\theta$ be a root of $f(X)$.
Let $A = \mathbb{Z}[\theta]$.
Let $p$ be a prime number.
Suppose $pA = \alpha^n A$, where $\alpha \in A$.
Let $A_p$ be the localization of $A$ with respect to $S = \mathbb{Z} - p\mathbb{Z}$.
Then $A_p$ is a discrete valuation ring.
Hence $A_p$ is integrally closed.
Moreover $P = \alpha A$ is a prime ideal and $A_p = A_P$.
Proof:
By Lemma 5 of my answer to this question, $\alpha A$ is a maximal ideal of $A$.
Let $P = \alpha A$.
Let $Q$ be a prime ideal of $A$ lying over $p\mathbb{Z}$.
Since $\alpha \in Q$, $P \subset Q$.
Since $P$ is maximal, $P = Q$.
Hence $P$ is the only prime ideal lying over $p\mathbb{Z}$.
Hence $A_p$ is a local ring with the maximal ideal $PA_p$.
Since $PA_p$ is generated by $\alpha A$, $A_p$ is a discrete valuation ring by Lemma 2 of my answer to this question.
Since $A_P$ is a local ring and it dominates a valuation ring $A_p$, $A_p = A_P$.
QED
A: Here's a proof sketch:
Let $\mathcal{O}$ be the ring of algebraic integers in $K$. Given the hypotheses,
$$ p \mathcal{O} = (\alpha \mathcal{O})^n $$
is a prime factorization, and so $\alpha\mathcal{O}$ is a degree 1 prime ideal and $\mathcal{O}_p = \mathcal{O}_{\alpha}$. There isn't any ways for $A$ to be non-regular at $\alpha A$, so $A_p = A_\alpha = \mathcal{O}_\alpha$ is integrally closed.

One way to see that $\alpha \mathcal{O}$ is a prime ideal of degree $1$ is as follows. The sum of the degrees of the factors of $p \mathcal{O}$, counted with multiplicity, must add up to $n$. The only way this can happen is if the prime factorization of $\alpha \mathcal{O}$ consists of a single prime ideal of degree 1. Therefore that prime ideal is $\alpha \mathcal{O}$ itself.

One way to check $A_\alpha = \mathcal{O}_\alpha$ is that it is implied by the fact the composite
$$A \to \mathcal{O} \to \mathcal{O} / \alpha \mathcal{O} \cong \mathbb{F}_p $$
is surjective. 
Another fact that implies $A_\alpha = \mathcal{O}_\alpha$ is the fact that $\alpha A$ is an invertible prime ideal of $A$.
Alas, for both of these I forget how to argue why it's implied things from first principles. :(
