Proof Verification: Let $f(\chi)$ be the conductor of $\chi$, proof that $f(\chi)=f(\chi_1)\cdots f(\chi_r)$.

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 $$n \mapsto \sum_{i=1}^r n b_i = \prod_{i=1}^r (n b_i + 1-b_i)$$
• 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? – kelvin hong 方 Mar 2 at 3:55
• @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 $\chi_i(n) \chi_i(m)= \chi_i(nm)$ – reuns Mar 2 at 17:18