Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

1 Answer

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.