# Prime divisors of the conductor 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 $B$ 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? – Harry Jul 23 '12 at 20:27
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. – user18119 Aug 24 '12 at 7:04

Let $d$ be the discriminant of $f(X)$. Let $\mathfrak{D}_{K/\mathbb{Q}}$ be the different. We will prove in the following proposition that $f'(\theta)B = \mathfrak{f}\mathfrak{D}_{K/\mathbb{Q}}$

Therefore, if a prime ideal $P$ of $B$ divides $\mathfrak{f}$, it divides $f'(\theta)$. Since $|N(f'(\theta))| = |d|$, $P$ divides $d$.

We first prove Euler's formula(Lemma 4) on a polynomial.

Lemma 1 Let $K$ be a field. Let $n \geq 1$ be an integer. Let $\alpha_1,\dots,\alpha_n$ be distinct elements of $K$. Let $f(X), g(X) \in K[X]$ be polynomials of degree $\leq n - 1$. Suppose $f(\alpha_i) = g(\alpha_i)$ for $i = 1,\dots,n$. Then $f(X) = g(X)$.

Proof: Let $h(X) = f(X) - g(x)$. $h(\alpha_i) = 0$ for $i = 1,\dots,n$. Since deg $h(X) \leq n - 1$, $h(X) = 0$. QED

Lemma 2(Lagrange's interpolation formula) Let $K$ be a field. Let $n \geq 1$ be an integer. Let $\alpha_1,\dots,\alpha_n$ be distinct elements of $K$. Let $f(X) = (X - \alpha_1)\cdots(X - \alpha_n)$. Let $g(X) \in K[X]$ be a polynomial of degree $\leq n - 1$. Then $\sum_i g(\alpha_i)f(X)/f'(\alpha_i)(X - \alpha_i) = g(X)$.

Proof: Let $H(X) = \sum_i g(\alpha_i)f(X)/f'(\alpha_i)(X - \alpha_i)$. Let $h_i(X) = f(X)/(X - \alpha_i)$. Then $h_i(\alpha_i) = f'(\alpha_i)$ and $h_i(\alpha_j) = 0$ for $i \neq j$. Hence $H(\alpha_i) = g(\alpha_i)$ for all $i$. Hence $H(X) = g(X)$ by Lemma 1. QED

Lemma 3 Let $K$ be a field. Let $n \geq 1$ be an integer. Let $\alpha_1,\dots,\alpha_n$ be distinct elements of $K$. Let $f(X) = (X - \alpha_1)\cdots(X - \alpha_n)$. Then

$\sum_i \alpha_i^r f(X)/f'(\alpha_i)(X - \alpha_i) = X^r$ for $0 \leq r \leq n - 1$. $\sum_i \alpha_i^n f(X)/f'(\alpha_i)(X - \alpha_i) = X^n - f(X)$

Proof: This follows immediately from Lemma 2.

Lemma 4(Euler's formula) Let $K$ be a field. Let $n \geq 1$ be an integer. Let $\alpha_1,\dots,\alpha_n$ be distinct elements of $K$. Let $f(X) = (X - \alpha_1)\cdots(X - \alpha_n)$. Then

$\sum_i \alpha_i^{k}/f'(\alpha_i) = 0$ for $0 \leq k \leq n - 2$.

$\sum_i \alpha_i^{n-1}/f'(\alpha_i) = 1$

Proof: Suppose $0 \leq k \leq n - 2$. By Lemma 3, $\frac{X^{k+1}}{f(X)} = \sum_i \frac{\alpha_i^{k+1}}{f'(\alpha_i)(X - \alpha_i)}$. Substituting $X = 0$, we get

$\sum_i \frac{\alpha_i^k}{f'(\alpha_i)} = 0$.

Note that this is valid when $f(0) = 0$.

By Lemma 3, $\frac{X^n}{f(X)} = 1 + \sum_i \frac{\alpha_i^n}{f'(\alpha_i)(X - \alpha_i)}$. Substituting $X = 0$, we get $0 = 1 - \sum_i \frac{\alpha_i^{n-1}}{f'(\alpha_i)}$.

Hence $\sum_i \frac{\alpha_i^{n-1}}{f'(\alpha_i)} = 1$.

Note that this is valid when $f(0) = 0$. QED

Fixed symbols

We fix the following symbols.

Let $A$ be a Dedekind domain.

Let $K$ be the field of fractions of $A$.

Let $L$ be a finite separable extension of $K$.

Let $B$ be the integral closure of $A$ in $L$.

We denote the trace of $x \in L$ over $K$ by $Tr(x)$.

Definition 1 Let $M$ be an $A$-submodule of $L$. We denote by $M^*$ the set {$\alpha \in L; Tr(\alpha M) \subset A$}. Clearly $M^*$ is an $A$-submodule of $L$. If $M$ is a $B$-submodule of $L$, clearly $M^*$ is a $B$-submodule of $L$.

Lemma 5 $B^*$ is a fractional ideal of $L$.

Proof: Let $S:L\times L \rightarrow K$ be a symmetric bilinear form defined by $S(x,y) = Tr(xy)$. It is well known that $S$ is non-degenerate. Let $e_1,\dots,e_n$ be a basis of $L$ over $K$ such that every $e_i \in B$. There exists the dual basis $e_1^*,\dots,e_n^*$ of $e_1,\dots,e_n$ in $L$. That is, $e_1^*,\dots,e_n^*$ is a basis of $L$ over K such that $Tr(e_i e_j^*) = \delta_{ij}$. Let $M$ be a free $A$-submodule of $L$ generated by $e_1,\dots,e_n$. It is easy to see that $M^*$ is a free $A$-submodule of $L$ generated by $e_1^*,\dots,e_n^*$. Clearly $M \subset B \subset B^* \subset M^*$. Let $a$ be a non-zero element of $A$ such that $ae_i^* \in B$. Since $B^*$ is a $B$-submodule of $L$ and $aB^* \subset aM^* \subset B$, $B^*$ is a fractional ideal of $L$. QED

Definition 2 By Lemma 1, $B^*$ is a fractional ideal of $L$. Since $B \subset B^*$, $(B^*)^{-1} \subset B$. Hence $(B^*)^{-1}$ is an ideal of $B$. $(B^*)^{-1}$ is called the different of $L/K$ and we denote it by $\mathfrak{D}_{L/K}$.

Definition 3 Let $\theta \in B$ be such that $L = K(\theta)$. Let $\mathfrak{f}$ = {$\alpha \in B$; $\alpha B \subset A[\theta]$}. $\mathfrak{f}$ is an ideal of $B$ contained in $A[\theta]$. We call it the conductor of $A[\theta]$.

Proposition Let $\theta \in B$ be such that $L = K(\theta)$. Let $f(X)$ be the minimal polynomial of $\theta$ over $K$. Let $\mathfrak{f}$ be the conductor of $A[\theta]$. Then $f'(\theta)B = \mathfrak{f}\mathfrak{D}_{L/K}$.

Proof: Let $n = [L : K]$. Let $\sigma_1,\dots,\sigma_n$ be $n$ distinct $K$-embeddings of $L$ into an extension field of $L$. We assume $\sigma_1$ is the identity map.

Let $f(X)/(X - \theta) = X^{n-1} + \beta_{n-2} X^{n-2} +\cdots+ \beta_1 X + \beta_0$, where $\beta_i \in B$. Let $\mu \in B^*$.

Let $g(X)$ = $Tr(\mu f(X)/(X - \theta)) = Tr(\mu) X^{n-1} + Tr(\mu\beta_{n-2}) X^{n-2} +\cdots+ Tr(\mu\beta_1) X+Tr(\mu\beta_0)$.

Since $\mu \in B^*$, $g(X) \in A[X]$.

Since $Tr(\mu f(X)/(X - \theta)) = \sum_i \sigma_i(\mu) f(X)/(X - \sigma_i(\theta))$, substituting $X$ by $\theta$, we get $g(\theta) = \mu f'(\theta)$. Hence $B^* f'(\theta) \subset A[\theta]$. Let $I = B^* f'(\theta)$. It suffices to prove that $I = \mathfrak{f}$.

Let $\mu \in B^*$. Let $\beta \in B$. Since $\beta\mu \in B^*$, $\beta\mu f'(\theta) \in A[\theta]$. Hence $\mu f'(\theta) \in \mathfrak{f}$. Hence $I \subset \mathfrak{f}$.

Let $\lambda \in \mathfrak{f}$. Let $\beta \in B$. $\lambda\beta = \alpha_0 + \alpha_1 \theta +\cdots+\alpha^{n-1} \theta^{n-1}$, where $\alpha_i \in A$. $\lambda\beta/f'(\theta) = \alpha_0/f'(\theta) + \alpha_1 \theta/f'(\theta) +\cdots+\alpha^{n-1} \theta^{n-1}/f'(\theta)$.

By Lemma 4(Euler's formula), $Tr(\lambda\beta/f'(\theta)) = \alpha_0Tr(1/f'(\theta)) + \alpha_1 Tr(\theta/f'(\theta)) + \cdots+ \alpha_{n-1} Tr(\theta^{n-1}/f'(\theta)) = \alpha_{n-1}$.

Hence $\lambda/f'(\theta) \in B^*$. Hence $\lambda \in B^*f'(\theta) = I$. QED

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