I was reading Apostol's Number Theory Book and I came across this question(I have rephrased it here for the sake of clarity):
Let $f(\chi)$ denote the conductor of $\chi \mod k$ where $k = k_1 \cdots k_r$ and $(k_i , k_j) = 1$ for $i \neq j$. If $\chi = \chi_1 \cdots \chi_r$ where $\chi_i$ is a character modulo $k_i$, prove that $f(\chi) = f(\chi_1) \cdots f(\chi_r).$
Here goes my attempt:
Let $r = 2$ and $a = mf(\chi_1)f(\chi_2) + 1$ where $m \in \mathbb{Z}$ $$\chi(a) = \chi_1(a)\chi_2(a) = 1 \implies f(\chi) | f(\chi_1)f(\chi_2)$$ Now, take $b_n = nf(\chi) + 1$ where $n$ is an integer. $\chi(b_n) = 1$ for all $n$ because $f(\chi)$ is the conductor of $\chi$. We can see that $$\chi(b_{kf(\chi_2)}) = \chi_1(b_{kf(\chi_2)})\chi_2(b_{kf(\chi_2)}) = \chi_1(kf(\chi)f(\chi_2) + 1) = 1$$ for all integers $k$. This means $f(\chi_1)|f(\chi)f(\chi_2)$. Since $(k_1, k_2) = (f(\chi_1), f(\chi_2)) = 1$, $f(\chi_1)|f(\chi)$. Repeating these steps for $b_n = b_{kf(\chi_1)}$ yields $f(\chi_2)|f(\chi)$. By linearity, $f(\chi_1)f(\chi_2) | f(\chi)$. But we have shown that $f(\chi)|f(\chi_1)f(\chi_2).$ Thus, $f(\chi) = f(\chi_1)f(\chi_2).$
Now take $r \geq 3$. Let $$\chi^\prime = \prod_{i = 1}^{r-1} \chi_i. $$Since $(\prod_{i = 1}^{r-1}k_i , k_r) = 1$, $$f(\chi) = f(\chi^\prime \chi_r)= f(\chi^\prime)f(\chi_r)$$ Inducting on $\chi^\prime$, we get $$f(\chi) = \prod_{i = 1}^{r} f(\chi_i).$$
This looks right to me. But I haven't really "grokked" the whole concept of Dirichlet characters and their properties. Plus, in my opinion, this solution seems more effecient than the one on Hurst's Solution Guide, a book which impressed me with very elegant solutions to some of the hardest questions on Apostol's number theory book, which I deemed impossible at first. All these reasons add up to make uncertain about the validity of my solution. Can anyone check it for me please? Thanks!