1
$\begingroup$

Let $ax^2 + bxy + cy^2$ be a binary quadratic form over $\mathbb{Z}$. Let $D = b^2 - 4ac$ be its discriminant. It is easy to see that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$).

Conversely suppose $D$ is a non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$). Then there exists the principal form of discriminant $D$(see this question).

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 $n$. Let $\alpha_1, \cdots, \alpha_n$ be a basis of $R$ as a $\mathbb{Z}$-module. Let $d = det(Tr_{K/\mathbb{Q}}(\alpha_i\alpha_j))$. It is easy to see that $d$ is independent of a choice of a basis of $R$. We call $d$ the discriminant of $R$.

Is the following proposition true? If yes, how do we prove it?

Proposition Let $D$ be a non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$). Then there exists an order $R$ of a quadratic number field $K$ such that the discriminant of $R$ is $D$. $K$ and $R$ are uniquely determined by $D$.

Conversely let $R$ be an order of a quadratic number field $K$. Let $D$ be the discriminant of $R$. Then $D$ is a non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$).

$\endgroup$
1
$\begingroup$

By my answer to this question, $D$ can be written uniquely as $D = f^2 d$, where $f \gt 0$ is an integer and $d$ is the discriminant of a unique quadratic number field $K$. By my answer to this question, there exists a unique order $R$ of $K$ such that $f$ is the order of the group $\mathcal{O}_K/R$ and its discriminant is $D = f^2d$.

It remains to prove the last assertion. Let $1, \omega$ be the canonical integral basis of $K$(see the definition in my answer to this question). Let $f$ be the order of the group $\mathcal{O}_K/R$. By my answer to this question, $D = f^2 d$. By Lemma 8 in my answer to this question, $d$ is a non-square integer. Hence $D$ is a non-square integer. By Lemma 7 in my answer to this question, $d \equiv 0$ (mod $4$) or $d \equiv 1$ (mod $4$). Since $f^2 \equiv 0$ (mod $4$) or $f^2 \equiv 1$ (mod $4$), $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$) as desired.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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