# The inverse class of the class represented by a primitive binary quadratic form of discriminant $D$

We use the definitions of this question.

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 $4$). 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. Let $[F]$ be a class of $C(D)$ represented by a primitive form $F = ax^2 + bxy + cy^2$. Let $G = ax^2 - bxy + cy^2$. Then $[F][G] = 1$.

Let $x_1,\cdots, x_n$ be a sequence of elements of $R$. We denote by $[x_1,\cdots,x_n]$ the $\mathbb{Z}$-submodule of $R$ generated by $x_1,\cdots, x_n$.
Let $I = [a, (-b + \sqrt D)/2]$. Let $J = [a, (-b + \sqrt D)/2]$. $I$ and $J$ are primitive ideals of $R$.
$IJ = [a, (-b + \sqrt D)/2][a, (b + \sqrt D)/2] = [a^2, a(b + \sqrt D)/2, a(-b + \sqrt D)/2, (D - b^2)/4] = [a^2, ab, a(b + \sqrt D)/2, a(-b + \sqrt D)/2, ac] = a[a, b, (b + \sqrt D)/2, (-b + \sqrt D)/2, c] = aR$