I wanted to understand the definition of the conductor for Dirichlet characters and I read https://en.wikipedia.org/wiki/Dirichlet_character: "We can formalize this differently by defining characters $\chi_1$ mod $N_1$ and $\chi_2$ mod $N_2$ to be co-trained if for some modulus $N$ such that $N_1$ and $N_2$ both divide $N$ we have $\chi_1(n) = \chi_2(n)$ for all $n$ coprime to $N$: that is, there is some character $\chi^*$ induced by each of $\chi_1$ and $\chi_2$. This is an equivalence relation on characters. A character with the smallest modulus in an equivalence class is primitive and this smallest modulus is the conductor of the characters in the class."

I am just really confused what they mean by it. Could someone possibly explain me what it means in a way that I can understand?


Every Dirichlet character $\chi$ modulo $k$ can be written as a product $$ \chi(n)=\psi(n)\chi_1(n), $$ where $\psi$ is a primitive character modulo $d$, where $d$ is the smallest positive divisor of $k$, which is an induced residue class order for $\chi$. This $d$ is called the conductor.

An example may be helpful. Let $k=9$ and $\chi$ given by $$ \chi(1)=1,\chi(2)=-1,\chi(3)=0,\chi(4)=1,\chi(5)=-1,\chi(6)=0,\chi(7)=1,\chi(8)=-1,\chi(9)=0. $$ This is obviously an extension of the character $\psi$ modulo $d=3$ with $\psi(1)=1,\psi(2)=-1,\psi(3)=0$. Furthermore, $d=3$ is an induced residue class order. It is the smallest, hence the conductor.

Recall the definition here:

Definition: A positive divisor $d\mid k$ is called induced residue class order for $\chi$, if $\chi(a)=1$ for all $a$ with $gcd(a,k)=1$ and $a\equiv 1 \bmod d$.

  • $\begingroup$ You say $d$ is the smallest positive divisor of $k$. But then doesn't that mean $d$ is always $1$? $\endgroup$ – Johnny T. Jan 22 '18 at 19:23
  • $\begingroup$ Is conductor the modulus of the primitive character inducing it? Is that all it is? $\endgroup$ – Johnny T. Jan 22 '18 at 19:24
  • $\begingroup$ No, not always $d=1$; it should also be an induced residue class. $\endgroup$ – Dietrich Burde Jan 22 '18 at 19:40

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