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
