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 $B$ be the ring of algebraic integers in K.
Let $\mathfrak{f}$ = $\{\alpha \in B\mid \alpha B \subset A\}$. $\mathfrak{f}$ is an ideal of both $A$ and $B$. $\mathfrak{f}$ is called the conductor of $A$.
My question: Is the following theorem true? If yes, how would you prove this?
Theorem If a prime ideal $P$ of $B$ divides $\mathfrak{f}$, then $P$ divides the discriminant of $f(X)$.
Motivation Let $p$ be a prime number. Suppose $p$ does not divide the discriminant of $f(X)$. By the lying-over theorem, there exists a prime ideal $P$ of $A$ lying over $p$. It is well known that $A_P$ is integrally closed if and only if $P$ does not divide $\mathfrak{f}$. If the theorem is correct, $A_P$ is integrally closed. Hence $A_P$ is a discrete valuation ring. Hence we can prove this.