Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We use the definitions of this question.

Let $D$ be a non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$). There exists a bijection $\psi\colon Cl^+(R) \rightarrow C(D)$ by the proposition of this question. We identify $C(D)$ with $Cl^+(R)$ by $\psi$. Hence $C(D)$ is an abelian group with this identification. Then the map $\Psi\colon C(D) \rightarrow$ Ker$(\chi)/H$ is a homomorphism. Let $G = (\mathbb{Z}/D\mathbb{Z})^\times$. Then Ker$(\chi)/H$ is a subgroup of $G/H$. By the proposition of this question, $G/H$ is isomorphic to $(\mathbb{Z}^\times)^\mu$. Hence $x^2 = 1$ for every element $x$ of Ker$(\chi)/H$. Hence $\Psi(C^2) = 1$ for every class $C \in C(D)$. Therefore $C(D)^2 \subset$ Ker$(\Psi)$. Gauss proved that $C(D)^2 =$ Ker$(\Psi)$. This is the main theorem of the genus theory of binary quadratic forms created by Gauss.

Since $\chi$ is surjective, $|G/$Ker($\chi)| = 2$. On the other hand, by the proposition of this question, $|G/H| = 2^\mu$. Hence |Ker$(\chi)/H| = 2^{\mu - 1}$. By the proposition of this question, the number of genera of discriminant $D$ is $2^{\mu - 1}$.Hence $\Psi$ is surjective. Hence, to prove $C(D)^2 =$ Ker$(\Psi)$, it suffices to prove that $[C(D) \colon C(D)^2] = 2^{\mu - 1}$. Let $A(D) = \{C \in C(D); C^2 = 1\}$.

There exists an exact sequence:

$$1 \rightarrow A(D) \rightarrow C(D) \rightarrow C(D)^2 \rightarrow 1$$

Hence $[C(D) \colon C(D)^2] = |A(D)|$. Hence it suffice to prove that $|A(D)| = 2^{\mu -1}$. To compute $|A(D)|$, we need a characterization of elements of $A(D)$.

Let $F = ax^2 + bxy + cy^2$ be a form of discriminant $D$. If $b \equiv 0$ (mod $a$), we say $F$ is an ambiguous form(Gauss D.A. art.163).

Let $C \in C(D)$. If $C$ contains an ambiguous form, $C$ is called an ambiguous class.

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

Proposition Let $C \in C(D)$. $C$ is an ambiguous class if and only if $C^2 = 1$.

share|cite|improve this question
@JohnSenior I think the most part of the theory of integral binary quadratic forms belongs to elementary number theory. – Makoto Kato Sep 9 '12 at 0:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.