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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 $C^2 = 1$.

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@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

Yes, it is true. Gauss states this result in Section 249 of Disquisitiones, although his argument is scattered through some earlier sections.

The main thing to know is that the class of an ambiguous form has the property that the opposite of any form in the class will also be in the class, and any class with this property contains an ambiguous form.

Accepting this, the "only if" part of your proposition follows from the more general observation that opposite forms lie in inverse classes. This is checked just computing the composition of a form with its opposite and seeing that the result is in the principal class -- Gauss does this in Section 243 of Disquisitiones.

For the "if" part, suppose that $C^2=1$, and suppose $D$ is the class containing the opposite of some form in $C$. Then $C = CCD = D$. The first equality is because opposite forms lie in inverse classes, and the second equality comes from our assumption. Thus, the class C contains a form and its opposite, and so it contains an ambiguous form by the "main thing to know" that I describe above.

Gauss proves the "main thing to know" in Sections 163-165. The easier part is showing that a class containing an ambiguous form is closed under taking opposites, essentially because this means that the class is invariant under improper equivalence. To show that a class which is closed under taking opposites contains an ambiguous form, he first observes that any two forms in the class must by assumption be both properly and improperly equivalent. The fact that this implies that they are properly equivalent to an ambiguous form is the topic of Section 164. He actually shows how to construct this ambiguous form with a lengthy and difficult computation, and then with further difficult computations proves that it is ambiguous.

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