# The homomorphism defined by the system of genus characters

Let $F = ax^2 + bxy + cy^2$ be a binary quadratic form over $\mathbb{Z}$.
We say $D = b^2 – 4ac$ is the discriminant of $F$.
It is easy to see that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$).
If $D$ is not a square integer and gcd($a, b, c) = 1$, we say $ax^2 + bxy + cy^2$ is primitive.
Let $m$ be an integer.
If $m = ax^2 + bxy + cy^2$ has a solution in $\mathbb{Z}^2$, we say $m$ is represented by $F$.

Let $D$ be a non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$).
Let $\Phi_1,\dots,\Phi_{\mu}$ be the system of genus characters of discriminant $D$[see this question].
Let $n$ be an integer.
We denote by $[n]$ the image of $n$ by the canonical homomorphism $\mathbb{Z} \rightarrow \mathbb{Z}/D\mathbb{Z}$.
Let $(\mathbb{Z}/D\mathbb{Z})^\times$ be the group of invertible elements of the ring $\mathbb{Z}/D\mathbb{Z}$.

we define a homomorphism $\Phi\colon(\mathbb{Z}/D\mathbb{Z})^\times \rightarrow (\mathbb{Z}^\times)^\mu$ by
$\Phi([n]) = (\Phi_1(n),\dots,\Phi_{\mu}(n))$.

We would like to determine the kernel and the image of $\Phi$.
To do this, we need the following definition.

Definition
Let $D$ be a non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$).
If $D \equiv 0$ (mod $4$), $x^2 – \frac{D}{4}y^2$ is a primitive form of discriminant $D$.
If $D \equiv 1$ (mod $4$), $x^2 + xy + \frac{1 – D}{4}y^2$ is a primitive form of discriminant $D$.
We call $x^2 – \frac{D}{4}y^2$ or $x^2 + xy + \frac{1 – D}{4}y^2$ the principal form of discriminant $D$.

My 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 $4$).
Let $H = \{ [m] \in (\mathbb{Z}/D\mathbb{Z})^\times$; $m$ is represented by the principal form of discriminant $D\}$.
Then $\Phi$ is surjective and $H$ = Ker($\Phi$).

#### Solutions Collecting From Web of "The homomorphism defined by the system of genus characters"

It is true, the proof is easier for case that $D\equiv 1 \mod D$. There it is a easy congurence of the Chinese reminder theorem. In the case $-4n$ the proof is similar but requires case distinction. (see Cox, Primes of the form $x^2+ny^2$, Lemma 3.17)