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If $\chi_1 , \chi _2 : \mathbb Z \to \mathbb C$ are two primitive Dirichlet characters modulo $n_1 , n_2$ respectively such that gcd$(n_1,n_2)=1$ , then is it true that the Dirichlet character $\chi_1 \chi_2$ , modulo $n_1n_2$ , is also primitive ? ( I can show the converse i.e. if product of two Dirichlet characters , modulo co-prime natural numbers , is primitive then so are each of them )

Please help . Thanks in advance

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Consider the unique non-principal $\chi$ character modulo $4$. Is it primitive? What about its square $\chi^2$?

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  • $\begingroup$ Sorry , I forgot to add that $\chi _i$ 's are Dirichlet characters modulo co-prime integers ... $\endgroup$ – user495643 Oct 25 '17 at 19:19
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A $\chi$ modulo $N$ is non-primitive iff $\chi(n) = \psi(n) 1_{(n,k)=1}$ for some $\psi$ modulo $N/k$ where $k | N$. Do you think it can happen if $\chi = \chi_1\chi_2$ two primitive characters modulo coprime $N_1,N_2$ ?

More generally any Dirichlet character $\chi_N$ modulo $N$ decomposes uniquely as $\chi_N = \prod_{p^e \| N} \chi_{p^e}$, which follows from the Chinese remainder theorem $(\mathbb{Z}/N\mathbb{Z})^\times \simeq\prod_{p^e \| N} (\mathbb{Z}/p^e\mathbb{Z})^\times$ ie. for every $a$ there is a $n \bmod N$ such that $n \equiv a \bmod p_1^{e_1}$ and $n\equiv 1 \bmod p_j^{e_j}$ for $j \ne 1$ so that $\chi_{p_1^{e_1}}(a)=\chi_N(n)$

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  • $\begingroup$ I can't prove that $\chi$ must be primitive if $\chi_1$ and $\chi_2$ are. $\endgroup$ – limitIntegral314 Nov 16 '17 at 0:52
  • $\begingroup$ @Mr.209 It is obvious once you know the definition of primitive $\endgroup$ – reuns Nov 16 '17 at 0:54
  • $\begingroup$ Well, it's not obvious for me... $\endgroup$ – limitIntegral314 Nov 16 '17 at 1:09
  • $\begingroup$ @Mr.209 Because you didn't write precisely what you want to show. Define $\chi$ in term of $\chi_1,\chi_2$, define "non-primitive", use what I said to define $\chi_1,\chi_2$ in term of $\chi$. $\endgroup$ – reuns Nov 16 '17 at 1:11
  • $\begingroup$ Clearly, we could write $\chi(n) = \chi_1(n) \cdot \chi_2(n) = \psi(n) \cdot 1_{(n, ab) = 1}(n)$ where $\psi$ is a character mod $d$, a divisor of $ab$. $\endgroup$ – limitIntegral314 Nov 16 '17 at 1:26

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