# Clarification regarding a claim in Heilbronn’s 1934 paper

I was reading Heilbronn’s 1934 paper where he proves that $$H(d) \to \infty$$ as $$d \to -\infty$$, where $$H(d)$$ is the ideal class number of the imaginary quadratic field with discriminant $$d$$. I couldn't reason out one of his claims in the proof to a lemma. I have attached an image of this part below.

I am struggling with the claim that ".. $$a^H$$ is at least four times representable by the principal form". I suppose that he is appealing to a lemma which is mentioned earlier in the paper without proof, I state it below in a slightly modified way preceded briefly by some background.

Let $$I$$ be an ideal in the ring of integers of $$K = \mathbb{Q}[\sqrt{d}]$$ with $$\mathbb{Z}$$-basis $$(\alpha, \beta)$$. Then we may associate $$I$$ with the form $$f_{I, (\alpha, \beta)}(x, y) = \frac{Nm(\alpha x + \beta y)}{Nm(I)}$$. It can be proven that this $$f_{I, (\alpha, \beta)}$$ is an integral primitive binary quadratic form with discriminant $$d$$, and further that this induces a group isomorphism between the classgroup of $$K$$ and the proper equivalence classes of forms with discriminant $$d$$ under certain restricrtions to include only positive definite forms.

The following is the statement of the lemma: If $$J$$ is any ideal in the same class as $$I$$ then $$Nm(J)$$ can be represented by $$f_{I, (\alpha, \beta)}$$ in two distinct ways.

The following is my proof of this fact: W.l.o.g we may assume that $$J = I$$. Now since $$Nm(I) \in I$$, $$Nm(I) = \alpha x + \beta y$$ for some $$x, y \in \mathbb{Z}$$. Then $$f_{I, (\alpha, \beta)}(x, y) = \frac{Nm(\alpha x + \beta y)}{Nm(I)} = \frac{Nm(Nm(I))}{Nm(I)} = Nm(I)$$. And $$f_{I, (\alpha, \beta)}(-x, -y)$$ is another representation.

Going back to Heilbronn’s claim, I am assuming that since $$\mathfrak{a}^H \neq \mathfrak{a'}^H$$ and the norm of each of them is representable in at least two different ways by the principal form (based on the lemma above) $$a^H = Nm(\mathfrak{a}^H) = Nm(\mathfrak{a'}^H)$$ is representable in at least 4 ways.

But my question is that, couldn't it be the case that these representations overlap. For example, suppose $$(\alpha, \beta)$$ is a $$\mathbb{Z}$$-basis for $$\mathfrak{a}^H$$ then $$(\alpha', \beta')$$ is a $$\mathbb{Z}$$-basis for $$\mathfrak{a'}^H$$. Here I use $$'$$ to denote conjugation. Isn't it possible that $$a^H = Nm(\mathfrak{a}^H) = \alpha x + \beta y = \alpha' x + \beta' y = Nm(\mathfrak{a'}^H) = a^H$$. In that case, going through the proof of the lemma above, it is easy to see that the representations overlap.

Probably I am missing something here. Would be really helpful if someone could provide a clarification.

One can see more directly that $$a^H$$ has a representation by the principal form with $$y\neq 0$$ (hence also with $$y>0$$). Indeed, let $$\omega:=\begin{cases}\sqrt{d},&d\equiv 0\pmod{4}; \\ (1+\sqrt{d})/2,&d\equiv 1\pmod{4}.\end{cases}$$ Then the ring of integers is $$\mathfrak{o}=\mathbb{Z}+\omega\mathbb{Z}$$, and the principal form is $$(x+\omega y)(x+\omega'y)$$. By assumption, $$\mathfrak{a}^H$$ is a principal ideal, say $$\mathfrak{a}^H=\lambda\mathfrak{o}$$ with $$\lambda\in\mathfrak{o}$$. In addition, $$\mathfrak{a}^H\neq \mathfrak{a}'^H$$, hence $$\lambda\neq\lambda'$$. Let us write $$\lambda$$ as $$x+y\omega$$ with $$x,y\in\mathbb{Z}$$, then $$y\neq 0$$, and $$a^H=N(\mathfrak{a}^H)=\lambda\lambda'=(x+\omega y)(x+\omega'y).$$ Done.
P.S. When constructing $$f_{I,(\alpha,\beta)}(x,y)$$, one needs to use an oriented basis $$(\alpha,\beta)$$ of $$I$$. Oriented means that $$(\alpha'\beta-\alpha\beta')/\sqrt{d}>0$$, or equivalently $$(\alpha'\beta-\alpha\beta')/\sqrt{d}=N(I)$$.