I'm trying to understand the process of lifting a Dirichlet character $ \chi$, where a Dirichlet character modulo $N$ being defined as a homomorphism from $(\mathbb{Z} / N \mathbb{Z})^*$ to $\mathbb{C}^*$.
Now my book states that I lift a Dirichlet character as follows and gives also an example on how this works:
However, the distribution of these elements seems purely abitrary to me and I simply cannot see the logic behind this (Maybe there is more to the process of lifting a Dirichlet character than I grasped from this explanation). It would be amazing if anyone could help me out! Thanks!
I really don't like how Dirichlet characters are introduced in number theory books. Tell me if this makes more sense to you :
Dirichlet character : completely multiplicative and $m$ periodic
If $\chi$ is a Dirichlet character modulo $d$ then $\psi(n) = \chi(n) 1_{\gcd(n,k)=1}$ is the product of two Dirichlet characters modulo $d$ and $k$ thus is a Dirichlet character modulo $m = \text{lcm}(d,k)$.
Write $m= \prod_{j=1}^l p_j^{e_j}$. By the Chinese remainder theorem, for every residue classes $a_j \bmod p_j^{e_j}$ there exists $b \bmod m$ such that $\forall j, b \equiv a_j \bmod p_j^{e_j}$. Therefore any $\chi$ modulo $m$ decomposes uniquely as $\chi = \prod_{j=1}^l \chi_{p_j^{e_j}}$ where $\chi_{p_i^{e_i}}$ is a Dirichlet character modulo $p_i^{e_j}$ defined by $\chi_{p_i^{e_i}}(n) = \chi(b)$ with $b \equiv n \bmod p_i^{e_i}, \forall j \ne i, b \equiv 1 \bmod p_j^{e_j}$.
$\chi$ is primitive means none of the $\chi_{p_i^{e_i}}$ is the trivial character $1_{\gcd(n,p_i)=1}$.
That is to say for $\gcd(m,k)=1$, the map $(\chi, \varphi) \mapsto \chi\varphi$ is an isomorphism from the pair of Dirichlet characters modulo $m,k$ to the Dirichlet characters modulo $mk$.