Conductor of a quadratic order We need some definitions to state the problem.
Let $B$ be a commutative ring, $A$ its subring.
We denote by $(A : B)$ the set $\{x \in B | xB \subset A\}$.
$(A : B)$ is an ideal of $B$.
It is contained $A$, hence it is also an ideal of $A$.
It is called the conductor of the ring extention $B/A$.
Let $K$ be an algebraic number field of degree $n$.
An order of $K$ is a subring $R$ of $K$ such that $R$ is a free $\mathbb{Z}$-module of rank $n$.
Now let $K$ be a quadratic number field.
Let $\mathcal{O}_K$ be the ring of algebraic integers in $K$.
Let $R$ be an order of $K$.
Let $\mathfrak{f} = (R : \mathcal{O}_K)$.
The ideal $\mathfrak{f}$ is important for the ideal theory of $R$ as shown in this question.
Since $R$ is a finitely generated $\mathbb{Z}$-module, $R \subset \mathcal{O}_K$.
Since both $\mathcal{O}_K$ and $R$ are free $\mathbb{Z}$-modules of rank $2$,
the $\mathbb{Z}$-module $\mathcal{O}_K/R$ is finite.
Let $f$ be the order of $\mathcal{O}_K/R$.
I came up with the following proposition.
Proposition
$\mathfrak{f} = f\mathcal{O}_K$.
Outline of my proof
Let $d$ be the discriminant of $\mathcal{O}_K$.Then by this question, 1, $\omega = \frac{d + \sqrt d}{2}$ is a basis of $\mathcal{O}_K$ as a $\mathbb{Z}$-module. It is easy to see that $R = \mathbb{Z} + \mathbb{Z}f\omega$.Let $\alpha = a + bf\omega \in (R : \mathcal{O}_K)$. I deduce that $a$ is divisible by $f$ from $\alpha\omega \in R$ using $\omega^2 = d\omega - \frac{d(d-1)}{4}$.
A full proof was given below as an answer.
My question
How do you prove the proposition?
I would like to know other proofs based on different ideas from mine.
I welcome you to provide as many different proofs as possible.
I wish the proofs would be detailed enough for people who have basic knowledge of introductory algebraic number theory to be able to understand.
 A: Let $d$ be the discriminant of $\mathcal{O}_K$.Then by this question, 1, $\omega = \frac{d + \sqrt d}{2}$ is a basis of $\mathcal{O}_K$ as a $\mathbb{Z}$-module.
We first prove that $R = \mathbb{Z} + \mathbb{Z}f\omega$.
Clearly $f\omega \in R$.
Let $c$ be the smallest positive rational integer such that $c\omega \in R$.
Let $a + b\omega \in R$, where $a, b$ are rational integers.
Since $a \in R, b\omega \in R$.
Hence $b$ is divisible by $c$.
Hence $R = \mathbb{Z} + \mathbb{Z}c\omega$.
Since the order of $\mathcal{O}_K/R$ is $c$, $c = f$ as desired.
Since $f\mathcal{O}_K \subset (R : \mathcal{O}_K)$,
it suffices to prove that $(R : \mathcal{O}_K) \subset f\mathcal{O}_K$.
Let $\alpha = a + bf\omega \in (R : \mathcal{O}_K)$, where $a, b$ are rational integers.
We denote $\sigma(\omega)$ by $\omega'$, where $\sigma$ is the unique non-identity automorphism of $K/\mathbb{Q}$.
Clealy $\omega' = \frac{d - \sqrt d}{2}$.
Then $\omega$ is a root of the polynomial
$(x - \omega)(x - \omega') = x^2 - dx + \frac{d(d-1)}{4} \in \mathbb{Z}[x]$.
Hence $\omega^2 = d\omega - \frac{d(d-1)}{4}$,
Hence $\alpha\omega = a\omega + bf\omega^2 = a\omega + bfd\omega - bf\frac{d(d-1)}{4} \in R$.
Hence $a + bfd \equiv 0$ (mod $f$).
Hence $a \equiv 0$ (mod $f$).
Hence $\alpha \in f\mathcal{O}_K$ as deisred.
QED
