This is a detailed problem, let me write down the problem and the process I have done:

Assume that $k=k_1k_2\cdots k_r$ where $k_i$ and $k_j$ are relatively prime for $i\neq j$.

Let $\chi$ be a Dirichlet character mod $k$, it is known that (as what I have done):

1) Given any integer $a$, there is an unique integer $a_i$ modulo $k$ such that $$a_i\equiv a\pmod {k_i}\text{ and } a_i\equiv 1\pmod {k_j} \text{ for } j\neq i.$$ 2) Let $\chi$ be a character mod $k$. Define $\chi_i$ by the equation $$\chi_i(a)=\chi(a_i)$$ where $a_i$ is the integer in statement (1), then $\chi_i$ is a character modulo $k_i$.

3)Every character $\chi$ mod $k$ can be factorized uniquely into $\chi=\chi_1\chi_2\cdots \chi_r$, which is $$\chi(a)=\chi_1(a)\cdots \chi_r(a) \quad\text{ for all integer $a$. }$$

The definition of Induced modulus: Let $d$ be a divisor of $k$, we say $d$ is an induced modulus of $\chi$ mod $k$ provided that for each $a$ having the property $$\gcd(a,k)=1, \quad a\equiv 1\pmod d$$ then it implies that $\chi(a)=1$.

The definition of the conductor: The smallest induced modulus of character $\chi$.

Above are all of the information we need, I hope, then the problem is:

Let $f(\chi)$ denote the conductor of $\chi$, if $\chi$ has the factorization $$\chi=\chi_1\cdots \chi_r$$ prove that the conductor of $\chi$ is the product of the conductors of $\chi_i$, which is to prove that $$f(\chi)=f(\chi_1)\cdots f(\chi_r).$$

My proof: For each index $i$ with $1\leq i\leq r$, let $d_i$ denote the conductor of $\chi_i$, then for all $a$ having the property that $$(a,k_i)=1, \quad a\equiv 1\pmod {d_i}$$ we have $\chi_i(a)=1$.

By solving the congruent system and denote $d=d_1\cdots d_r$, we can find a unique $a$ modulo $d$ such that $$a\equiv 1\pmod d.$$ and of course it should satisfy $(a,k)=1$ since $(a,k_i)=1$ for each index $i$.

By using statement (3), we can factorize $\chi(a)$ into $$\chi(a)=\chi_1(a)\cdots \chi_r(a).$$ where $\chi_i$ is defined as in statement (2).

As I have mentioned just now, $\chi_i(a)=1$ for all index $i$. So we have $\chi(a)=1$. So I have shown that $d$ is an induced modulus of $\chi$, and that $d$ is the conductor of $\chi$ because each $d_i$ is the conductor of $\chi_i$.

Is there any problem in my proof? This proof is a bit longer, thanks for spending your precious time on checking this.


For $\chi$ completely multiplicative and $k$-periodic, factorize $k = \prod_{i=1}^r k_i$ where $k_i = p_i^{e_i}$ and the $p_i$ are distinct primes.

Let $b_i \equiv 1 \bmod k_i, b_i\equiv 0 \bmod k_j$ and let $\chi_i(n) = \chi(nb_i + 1-b_i)$. It is completely multiplicative $k_i$-periodic.

Then $\prod_{i=1}^r(nb_i + 1-b_i) \equiv n \bmod k$ so $\prod_{i=1}^r \chi_i(n) = \chi(\prod_{i=1}^r(nb_i + 1-b_i)) = \chi(n)$.

Finally for the conductor : it is not hard to see that $\chi$ is not $d$ periodic for any $d < k$ iff each $\chi_i$ is not $d_i$ periodic for any $d_i < k_i$

Note the Chinese remainder theorem is the isomorphism of rings $\Bbb{Z}/k\Bbb{Z} \cong \Bbb{Z}/k_1\Bbb{Z} \times \ldots \times \Bbb{Z}/k_r\Bbb{Z}$ the isomorphism is given by $n \mapsto \sum_{i=1}^r n b_i = \prod_{i=1}^r (n b_i + 1-b_i)$

  • $\begingroup$ I don't know why we should write $n$ as $nb_i+1-b_i$, I know it is true, but why is it necessary? $\endgroup$ – kelvin hong 方 Mar 2 at 3:55
  • 1
    $\begingroup$ @kelvinhong方 You said so in your post because it is $n \mapsto (1 \bmod k_1, \ldots, n\bmod k_i, 1 \bmod k_{i+1},\ldots)$. Also note $b_i^2 = b_i \bmod k$ implies $(nb_i+1-b_i)(mb_i+1-b_i) = nm b_i^2 + (n+m)b_i(1-b_i)+(1-b_i)^2)=nm b_i+1-b_i \bmod k$ and $\chi_i(n) \chi_i(m)= \chi_i(nm)$ $\endgroup$ – reuns Mar 2 at 17:18

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