# Discriminant of a binary quadratic form and an order of a quadratic number field

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 $4$).

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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.