Let $f(X) \in \mathbb{Z}[X]$ be a monic irreducible polynomial of degree $n$. Let $\theta$ be a root of $f(X)$. Let $A = \mathbb{Z}[\theta]$. Let $p$ be a prime number. Suppose the discriminant of $f(X)$ is a power of $p$. Suppose $pA = \alpha^n A$, where $\alpha \in A$.
My question: Is $A$ 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 the discriminant of the minimal polynomial of $\theta$ is a power of $p$ and $pA = \alpha^n A$. It is also well known that $A$ is integrally closed.
This is a related question.