Dirichlet characters mapping integers to given roots of unity Given $a_1, a_2, ... a_m > 1$ pairwise coprime, and $\mu_1, \mu_2, ... \mu_m$ arbitrary roots of unity, does there exist a Dirichlet character $\chi$ with $\chi(a_i) = \mu_i$ for all i?
EDIT: The proposition clearly holds if and only if it holds whenever $a_i = p_i$ prime. The forward direction is obvious. Conversely, writing $a_i = p_{i,1}^{\alpha_{i,1}}...p_{i,k_{i}}^{\alpha_{i,k_{i}}}$ and solving the corresponding problem with $\chi(p_{i,j}) = 1$ for $j > 1$ and $\chi(p_{i,1}) = \mu_i^{\alpha_{i,1}^{-1}}$ gives the desired result.
The problem then also has the following reformulation: given $p_1, ... p_m$ distinct prime numbers, $\mu$ an arbitrary root of unity, do their exist infinitely many distinct primitive Dirichlet characters with $\chi(p_1) = \mu$ and $\chi(p_i) = 1$ for $i > 1$.
Indeed, if the original proposition holds, let $\chi_1$ be any primitive Dirichlet character satisfying the given hypotheses. Now let $\chi_2$ be another primitive Dirichlet character satisfying the given hypotheses, together with the assumption that $\chi(q_2) = e^{2 \pi i /q_2}$ for some $q_2$ prime greater than the modulus of $\chi_1$ (the important thing is that we added a condition that invalidates our former solution. Proceed in like manner to form $\chi_3$, $\chi_4$, ... distinct characters all satisfying our hypotheses.
Conversely, suppose that given $p_1, ... p_m$ distinct prime numbers, $\mu$ an arbitrary root of unity, there is a character with $\chi(p_1) = \mu$, $\chi(p_i) = 1$ for $i > 1$. Now for each $1 \leq i \leq m$, let $\psi_i$ be a primitive Dirichlet character with $\psi_i(p_i) = \mu_i$ and $\psi_i(p_j) = 1$ for $j \neq i$. Then $\chi = \psi_1 ... \psi_m$ is a Dirichlet character with the desired properties.
However, these reformulations fall far short of actually verifying or disproving the existence of such Dirichlet characters.
 A: The answer is positive.
Let $K = \mathbf{Q}(\zeta_N)$, and let
$$L = \mathbf{Q}(\zeta_N,\sqrt[N]{p_i}).$$
Then $\mathrm{Gal}(L/K) = (\mathbf{Z}/N \mathbf{Z})^m$ (this uses that $N$ is odd).
By the Cebotarev Density Theorem, there exists a prime $q$ such that the Frobenius
element $\sigma = \sigma_{\mathfrak{q}}$ of a prime $\mathfrak{q}$ above $q$ in $L$ is equal to
$$(1,0,0,\ldots) \in \mathrm{Gal}(L/K) \subset \mathrm{Gal}(L/\mathbf{Q}).$$
Thinking about what the Frobenius element actually is, we make the following deductions:

*

*Since $\sigma$ is trivial in the quotient $\mathrm{Gal}(K/\mathbf{Q})$, we have $q \equiv 1 \mod N$.

*For all primes $p_i$ with $i > 1$, the order $p_i$ in $\mathbf{F}^{\times}_q$ is divisible by $N$, or equivalently that $p_i$ is a perfect $N$th power modulo $q$. This is because Frobenius is also trivial in $\mathrm{Gal}(K(\sqrt[N]{p_i})/\mathbf{Q})$.

*The order of $p_1$ in $\mathbf{F}^{\times}_q$ is is prime to $N$.

Now let $E$ denote the (unique) degree $N$ extension of $\mathbf{Q}$ contained in $\mathbf{Q}(\zeta_q)$.
The conditions above imply that Frobenius at $p_i$  in $E$ is trivial for $i > 1$ and has order exactly $N$ for $i=1$.
Thus $E$ gives rise to the corresponding Dirichlet Character (of conductor $q$).
If $N$ is even, one has to modify this argument slightly by changing $L$ to $L = \mathbf{Q}(\zeta_N,\sqrt[2N]{p_i})$, the extra factor of $2$ allowing one to account for the fact that $\sqrt{p_i}$ may lie inside $\mathbf{Q}(\zeta_N)$.
