# 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.

• It would be more interesting to prove this result for any number field (finite over $\mathbb Q$). – Cantlog Nov 6 '13 at 21:41
• @Cantlog I guess the proposition does not hold in an algebraic number field in general. – Makoto Kato Nov 6 '13 at 21:58
• The question is local: let $A$ be a local noetherian domain of dimension $1$ and let $B$ be its normalization. What is the conductor $(A:B)$ ? – Cantlog Nov 6 '13 at 22:02
• @Cantlog I have no idea except its definition. Could you tell me what it is and how it relates to the problem? – Makoto Kato Nov 6 '13 at 22:51
• I had no opportunities to think about it. The conductor ideal is the product of the local conductor ideals, so it is enough to compute locally. Have you tried some examples ? – Cantlog Nov 7 '13 at 16:46

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