Question: Let $m \in \mathbb{Z}_{\neq 0, 1}$ be squarefree integer. Define $N \in \mathbb{N}$ as $$N := \begin{cases} |m| &\ \text{if}\ m \equiv 1\pmod 4\\ 4 |m| &\ \text{if}\ m \equiv 2,3 \pmod 4 \end{cases}$$ and define the Dirichlet character as $\chi_m : (\mathbb{Z}/N \mathbb{Z})^\times \to \{\pm 1\} \subset \mathbb{C}^\times$ by $$\chi_m(a \pmod N) := \left( \prod_{\substack{l :\ \text{odd prime}\\ l \mid m\ \text{in}\ \mathbb{Z}}} \left( \frac{a}{l} \right) \right) \theta_m(a),$$ where $\left(\frac{\cdot}{l}\right)$ is the Legendre symbol and $\theta_m(a)$ is defined as
- If $m \equiv 1 \pmod 4$ then $\theta_m(a) = 1$.
- If $m \equiv 3 \pmod 4$ then $$\theta_m(a) = \begin{cases} 1 &\ \text{if}\ a \equiv 1 \pmod 4\\ -1 &\ \text{otherwise} \end{cases}$$
- If $m$ is even then $$\theta_m(a) = \begin{cases} 1 &\ \text{if}\ a \equiv 1\ \text{or}\ a\equiv 1-m \pmod 8\\ -1 &\ \text{otherwise} \end{cases}$$ How do you show that this Dirichlet character modulo $N$ is a primitive Dirichlet character modulo $N$?
A necessary condition for Dirichlet character $\chi$ modulo $N$ is primitive is that the Gauss sum $$G(\chi, \zeta_N) := \sum_{a \in (\mathbb{Z}N\mathbb{Z})^\times}\chi(a) \zeta_N^{a}$$ where, $\zeta_N$ is a primitive $N$-th root of unity, satisfy $|G(\chi, \zeta_N)| = \sqrt{N}$. However, the converse direction need not hold. Is there any way to do so via the definition of primitive Dirichlet character? I am to prove via contradiction for small cases of $m$, directly via definition, however I am not sure how to generalise this result for all such squarefree $m$.
Source: Proposition 5.17 of Kato et al's Introduction to Class Field Theory.