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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.

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    I think you want to say that the prime divisors of the conductor divide the discriminant. Actually, isn't the conductor the radical of the discriminant?2012-07-23
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    A sketch of the proof: let $Q=P\cap A$ be the trace of $P$ in $A$. If $P$ contains (or divides) the ideal $f$, then $A_Q$ is not integrally closed. This implies that $Q$ divides the discriminant of $f(X)$ because it is known that $A/\mathbb Z$ is unramified outside of this discriminant. Hence $P$ divides the discriminant.2012-08-24

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