Stupid question about conductors/Dirichlet Characters

Let $N$ be a natural number and $\mathbb{Z}_N := \mathbb{Z}/N\mathbb{Z}$. Let $\pi_{N'} : \mathbb{Z}_N \to \mathbb{Z}_{N'}$ be the projection, i.e. $\pi_{N'}(a + N\mathbb{Z}) := a + N'\mathbb{Z}$. A Dirichlet character (DC) mod $N$ is a group homomorphism $\chi : \mathbb{Z}_N^* \to \mathbb{C}^*$. For a divisor $N' | N$ we say that $\chi$ is defined mod $N'$ if for all $x, y \in \mathbb{Z}_N^*$ we have $\pi_{N'}(x) = \pi_{N'}(y) \Rightarrow \chi(x) = \chi(y)$. The conductor of $\chi$ is defined to be the smallest divisor $N'|N$ s.t. $\chi$ is defined modulo $N'$. Let me write $c(\chi)$ for its conductor. What i want to see now is the following:

$----$

If $N'|N$ is such that $\chi$ is defined modulo $N'$, then $c(\chi)|N'$.

$----$

I suspected the proof to go like this: Let us write $N' = vc(\chi) + r$ with $0 \leq r < c(\chi)$. We have to show $r=0$, so assume $r > 0$, then we claim that $\chi$ is defined modulo $r < c(\chi)$, a contradiction. For showing this, we simply compute for $a \in \mathbb{Z}_N^*$ and $k \in \mathbb{Z}$ such that $a + kr \in \mathbb{Z}_N^*$, that

$$\chi(a + kr) = \chi(a + k[N' - vc(\chi)]) = \chi( (a + kN') + (-kv)c(\chi)) = \chi(a + kN') = \chi(a)$$ where we have used that $\chi$ is defined modulo $N'$ and modulo $c(\chi)$. The problem with this proof is the following: It can happen that although $a$ and $a + kr$ are units in $\mathbb{Z}_N$, $a + kN'$ is not, so we cannot argue like this. I am still convinced that this assertion is true. Can somebody help?

Fabian Werner

• I think it is immediate and requires no proof. You said the conductor of $\chi$ defined mod $M$ is the smallest divisor of $M$ such that blah blah. Since $\chi$ is defined mod $N'$ just set $M=N'$ and you are done, no? – user58512 Feb 5 '13 at 22:01
• I am sorry but i do not understand what you mean. I cannot set $N$ (or $M$ as you call it), it is a fixed reference that i start with. – Fabian Werner Feb 5 '13 at 22:05

Is is indeed easy: for $N=p^e$, a power of a prime $p$, the assertion is clear. then one has to use the chinese remainder theorem, i.e. for $N = p_1^{e_1} \cdot ... \cdot p_r^{e_r}$

$$\mathbb{Z}_N^* \cong \mathbb{Z}_{p_1^{e_1}}^* \times ... \times \mathbb{Z}_{p_1^{e_1}}^*$$

The dual groups (i.e. the set of characters) go along nicely with this decomposition, i.e. every $\chi$ can be written as $\chi_1 \cdot ... \cdot \chi_r$ where the $\chi_j$ are DC's mod $p_j^{e_j}$. Then one has to use the commutativity of the diagram

$$\mathbb{Z}_N^* \cong \mathbb{Z}_{p_1^{e_1}}^* \times ... \times \mathbb{Z}_{p_1^{e_1}}^*$$

$$\downarrow ~~~~~~~~~~~~~~~~~~~~~~ \downarrow$$

$$\mathbb{Z}_A^* \cong \mathbb{Z}_{p_1^{a_1}}^* \times ... \times \mathbb{Z}_{p_1^{a_1}}^*$$

for a divisor $A = \prod p_i^{a_i}$ in order to see that $\chi$ is defined mod $A$ iff. every $\chi_i$ is defined mod $p_i^{a_i}$

best,

FW

• If you are happy with your answer, Fabian, you can accept it (click in the check mark next to it). That helps clear up the Unanswered Question list. – Gerry Myerson Dec 11 '13 at 11:52
• I did so. Thanks. – Fabian Werner Jan 6 '14 at 14:24