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Suppose I have $\chi$, a Dirichlet character modulo $q$. Let $q = p_1^{e_1} \cdots p_r^{e_r}$ be the prime factorization of $q$. Then do there exist characters $\chi_j$ modulo $p_j^{e_j}$ for each $1 \leq j \leq r$ such that $\chi = \chi_1 \cdots \chi_r$? And furthermore is this decomposition unique?

I thought the answer was yes for both questions but my neighbor insists that there is a problem when $2|q$ and I wanted to verify this with someone... I would greatly appreciate clarification on this. Thank you very much.

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    $\begingroup$ Yup, this exists and is unique. This should be covered in basically any textbook that discusses Dirichlet characters; it's basically equivalent to the Chinese remainder theorem. $\endgroup$ – Peter Humphries Aug 8 '17 at 9:09
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    $\begingroup$ A Dirichlet character modulo $q$ is the extension to $\mathbb{Z}$ of a character $(\mathbb{Z}/q\mathbb{Z})^\times \to \mathbb{C}^\times$. And the Chinese remainder theorem is basically the fact that $(\mathbb{Z}/q\mathbb{Z})^\times \simeq (\mathbb{Z}/p_1^{e_1}\mathbb{Z})^\times \times \ldots \times (\mathbb{Z}/p_r^{e_r}\mathbb{Z})^\times$ $\endgroup$ – reuns Aug 8 '17 at 9:19
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    $\begingroup$ What your neighbor was thinking to is probably that if $p$ is an odd prime then $(\mathbb{Z}/p^{e}\mathbb{Z})^\times$ is cyclic with $\varphi(p^e)$ elements, whereas if $p=2$ then $(\mathbb{Z}/2^{e}\mathbb{Z})^\times \simeq \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2^{e-2}\mathbb{Z}$. So in general the decomposition is $\chi(mq+(-1)^a g_0^{n_0} \prod_{i=1}^r g_i^{n_i}) = (-1)^{a k_0} \prod_{i=1}^r \zeta_{k_i}^{n_i}$ where $g_i$ are some generators of $(\mathbb{Z}/p_i^{e_i}\mathbb{Z})^\times$, $k_i | \varphi(p_i^{e_i})$ and $g_0$ is an element of order $2^{e_1-2}$ modulo $2^e_1$. $\endgroup$ – reuns Aug 8 '17 at 9:32
  • $\begingroup$ @PeterHumphries Thank you very much for this! $\endgroup$ – Johnny T. Aug 8 '17 at 14:29
  • $\begingroup$ @reuns Thank you very much for this! I was wondering should the product of $\zeta_{k_i}^{n_i}$ start from $i=0$ by any chance? $\endgroup$ – Johnny T. Aug 8 '17 at 14:30

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