In trying to understand the connection between binary quadratic forms and ideals in the ring of algebraic integers, I came across a problem. I know that every equivalence class of positive forms contains precisely one reduced form. But e.g. Richard A. Mollin in his book "Quadratics" says that:
If $I_i=[a_i,(b_i+\sqrt{D})/2]$ (for $i=1,2$) are two distinct, equivalent, reduced ideals in $\mathcal{O}_D$, with $\vert b_i\vert \leq a_i$, then $a_1=a_2=a$, $\vert b_1 \vert =\vert b_2\vert $. (This tells us that there are at most two reduced ideals in any class of $\mathcal{O}_D$ and when two distinct such ideals are in a class, then they are conjugates of one another.)
So there are at most two reduced ideals in any class of $\mathcal{O}_D$.
But the group of classes of binary quadratic forms of discriminant $-D$ is isomorphic to the class group of the quadratic number field $\mathbb{Q}(\sqrt{D})$.
So since they are isomorphic, should not the number of reduced forms respectively reduced ideals be the same?