# Imprimitive Dirichlet Characters

I've started to read the fifth chapter of "Multiplicative Number Theory" by Harold Davenport and I got stuck at some point. Let me elaborate the part that i didn't quite understand.

Let $\chi$ be any Dirichlet character to the modulus $q$ other than the principal character. It has been stated that if the function $\chi(n)$ may have a period less than $q$ for the values of $n$ restricted by the condition that $(n,q) = 1$, let us say the least period is $m$, clearly $m|q$.

But I didn't understand clearly what is being said here, because for the values of $n$ , $(n,m) =1$ but $(n,q)>1$, we can find integer $t$ such that $(q,n+tm) = 1$, In other words $\chi(n) = 0$ but $\chi(n+tm) \neq 0$ because $(n+tm,q) =1$, however $\chi(n+tm) = \chi(n)$ .

I didn't understand that construction.

## 1 Answer

The issue is that the Statement $\chi(n+tm) = \chi(n)$ is not true in that case. This holds only if $(n,q)=(n+tm,q) = 1$. The least period $m$ only occurs with respect to the units $n$ such that $(n,q)=1$.

The idea is to start with a Dirichlet character $\chi :\mathbb Z \to \mathbb C$, with period $q$, so you can view is as a map $$\chi :\mathbb Z/q\mathbb Z \to \mathbb C.$$

We have $\chi (n)=0$ whenever $(n,q) >1$, and more over $\chi(nm) = \chi(n)\chi(m)$, so we can actually understand $\chi$ as coming from a homomorphism $$\chi' :(\mathbb Z/q\mathbb Z)^{\times} \to \mathbb C.$$

Now, here it can be the case that there exists $m<q$ such that for $n,n'\in (\mathbb Z/q\mathbb Z)^{\times}$ such that $n\equiv n'\mod m$, then $\chi'(n)=\chi'(n')$.

The smallest such $m$ is the least period. If this is true, then we get $m|q$, and there exists a homomphism $\overline{\chi}': (\mathbb Z/m\mathbb Z)^{\times} \to \mathbb C$, from which we can get a Dirichlet character $$\overline{\chi}: \mathbb Z \to \mathbb C,$$ with modulus $m$. This is the primitive Dirichlet character associated to $\chi$.

So from this you see that the $m$-periodicity of the original $\chi$ is only visible for $(n,q)=(n+tm,q) = 1$.