[proofplan] We first pass from Dirichlet characters to characters of the finite unit group $(\mathbb{Z}/q\mathbb{Z})^\times$. Using the Chinese remainder decomposition of this unit group into prime-power factors, we find the smallest prime-power modulus through which each local character factors. Multiplying these local conductors gives a divisor $f \mid q$, and the kernel-triviality criterion for descent produces a character modulo $f$ inducing $\chi$. Minimality of the local exponents gives primitivity, and the same local minimality proves uniqueness. [/proofplan] [step:Translate induction of Dirichlet characters into descent of unit-group characters] Let \begin{align*} G_m := (\mathbb{Z}/m\mathbb{Z})^\times \end{align*} for each $m \in \mathbb{N}$, with the convention that $G_1$ is the trivial group. Let \begin{align*} \widetilde{\chi}: G_q &\to \mathbb{C}^\times \end{align*} be the group character obtained by restricting $\chi$ to reduced residue classes modulo $q$. For divisors $d \mid q$, define the reduction homomorphism \begin{align*} \rho_{q,d}: G_q &\to G_d,\\ a \bmod q &\mapsto a \bmod d. \end{align*} This map is well-defined because $\gcd(a,q)=1$ implies $\gcd(a,d)=1$. [claim:Descent through a divisor is equivalent to triviality on the reduction kernel] Let $d \mid q$. A group character $\widetilde{\chi}:G_q \to \mathbb{C}^\times$ factors through $\rho_{q,d}$ if and only if \begin{align*} \widetilde{\chi}(u)=1 \end{align*} for every $u \in \ker \rho_{q,d}$. [/claim] [proof] If $\widetilde{\chi}=\psi \circ \rho_{q,d}$ for a character $\psi:G_d\to\mathbb{C}^\times$, then for $u \in \ker\rho_{q,d}$, \begin{align*} \widetilde{\chi}(u)=\psi(\rho_{q,d}(u))=\psi(1)=1. \end{align*} Conversely, suppose $\widetilde{\chi}$ is trivial on $\ker\rho_{q,d}$. Since $d \mid q$, the reduction map $\rho_{q,d}$ is surjective. Define \begin{align*} \psi:G_d &\to \mathbb{C}^\times,\\ v &\mapsto \widetilde{\chi}(u), \end{align*} where $u \in G_q$ is any element satisfying $\rho_{q,d}(u)=v$. This is well-defined: if $u_1,u_2 \in G_q$ both map to $v$, then $u_1u_2^{-1}\in \ker\rho_{q,d}$, hence \begin{align*} \widetilde{\chi}(u_1)\widetilde{\chi}(u_2)^{-1} = \widetilde{\chi}(u_1u_2^{-1}) = 1. \end{align*} Thus $\widetilde{\chi}(u_1)=\widetilde{\chi}(u_2)$. The map $\psi$ is a group homomorphism because $\widetilde{\chi}$ and $\rho_{q,d}$ are group homomorphisms. Therefore $\widetilde{\chi}=\psi\circ\rho_{q,d}$. [/proof] [/step] [step:Decompose the character into prime-power components] Write the prime factorization of $q$ as \begin{align*} q=\prod_{p \in P} p^{\alpha_p}, \end{align*} where $P$ is the finite set of primes dividing $q$ and $\alpha_p \in \mathbb{N}$ for each $p \in P$. By the [Chinese Remainder Theorem](/theorems/734) for unit groups (citing a result not yet in the wiki: Chinese [Remainder Theorem](/theorems/1707)), the map \begin{align*} \Theta_q:G_q &\to \prod_{p \in P} G_{p^{\alpha_p}},\\ a \bmod q &\mapsto (a \bmod p^{\alpha_p})_{p \in P} \end{align*} is an isomorphism. For each $p \in P$, define the local character \begin{align*} \widetilde{\chi}_p:G_{p^{\alpha_p}} &\to \mathbb{C}^\times \end{align*} as follows. Given $u_p \in G_{p^{\alpha_p}}$, let $\iota_p(u_p)\in \prod_{\ell \in P}G_{\ell^{\alpha_\ell}}$ be the element whose $p$-component is $u_p$ and whose $\ell$-component is $1$ for every $\ell \neq p$. Set \begin{align*} \widetilde{\chi}_p(u_p) := \widetilde{\chi}\bigl(\Theta_q^{-1}(\iota_p(u_p))\bigr). \end{align*} Then $\widetilde{\chi}_p$ is a group character on $G_{p^{\alpha_p}}$. Since every element of the product decomposes uniquely as a product of its prime-power components, for every $u \in G_q$ with \begin{align*} \Theta_q(u)=(u_p)_{p \in P}, \end{align*} we have \begin{align*} \widetilde{\chi}(u) = \prod_{p \in P}\widetilde{\chi}_p(u_p). \end{align*} [/step] [step:Choose the minimal local exponent through which each component descends] For each $p \in P$ and each integer $\beta$ with $0 \leq \beta \leq \alpha_p$, define the reduction homomorphism \begin{align*} \rho_{p,\alpha_p,\beta}:G_{p^{\alpha_p}} &\to G_{p^\beta},\\ a \bmod p^{\alpha_p} &\mapsto a \bmod p^\beta. \end{align*} Here $p^0=1$, so $G_{p^0}=G_1$ is the trivial group. The set of exponents $\beta$ such that $\widetilde{\chi}_p$ is trivial on $\ker\rho_{p,\alpha_p,\beta}$ is nonempty, since $\beta=\alpha_p$ belongs to it. Let $\beta_p$ be the smallest such exponent: \begin{align*} \beta_p := \min\left\{ \beta \in \{0,\dots,\alpha_p\}: \widetilde{\chi}_p(u)=1 \text{ for every } u \in \ker\rho_{p,\alpha_p,\beta} \right\}. \end{align*} Define \begin{align*} f:=\prod_{p \in P}p^{\beta_p}. \end{align*} Then $f \mid q$. [/step] [step:Construct the inducing character modulo the product of the local conductors] We prove that $\widetilde{\chi}$ factors through $\rho_{q,f}:G_q\to G_f$. Let $u\in\ker\rho_{q,f}$. Write \begin{align*} \Theta_q(u)=(u_p)_{p \in P}. \end{align*} Since $u\equiv 1 \pmod f$, each component satisfies \begin{align*} u_p \in \ker \rho_{p,\alpha_p,\beta_p}. \end{align*} By the definition of $\beta_p$, \begin{align*} \widetilde{\chi}_p(u_p)=1 \end{align*} for every $p \in P$. Therefore \begin{align*} \widetilde{\chi}(u) = \prod_{p \in P}\widetilde{\chi}_p(u_p) = 1. \end{align*} By the descent criterion, there exists a unique group character \begin{align*} \widetilde{\chi}^*:G_f &\to \mathbb{C}^\times \end{align*} such that \begin{align*} \widetilde{\chi}=\widetilde{\chi}^*\circ\rho_{q,f}. \end{align*} Define the Dirichlet character \begin{align*} \chi^*:\mathbb{Z}&\to\mathbb{C} \end{align*} modulo $f$ by \begin{align*} \chi^*(n) = \begin{cases} \widetilde{\chi}^*(n \bmod f), & \gcd(n,f)=1,\\ 0, & \gcd(n,f)>1. \end{cases} \end{align*} If $\gcd(n,q)=1$, then $n\bmod q\in G_q$, and \begin{align*} \chi(n) = \widetilde{\chi}(n\bmod q) = \widetilde{\chi}^*(n\bmod f) = \chi^*(n). \end{align*} If $\gcd(n,q)>1$, then $\chi(n)=0$ by the definition of a Dirichlet character modulo $q$, and the character induced from $\chi^*$ to modulus $q$ is also $0$ at $n$. Hence $\chi^*$ induces $\chi$ modulo $q$. [/step] [step:Use minimality of the local exponents to prove primitivity] Suppose, for contradiction, that $\chi^*$ is not primitive modulo $f$. Then there exists a proper divisor $d \mid f$ and a Dirichlet character modulo $d$ inducing $\chi^*$. On unit groups, this means that $\widetilde{\chi}^*:G_f\to\mathbb{C}^\times$ factors through the reduction map \begin{align*} \rho_{f,d}:G_f&\to G_d. \end{align*} Write \begin{align*} d=\prod_{p \in P}p^{\delta_p}, \end{align*} where $0\leq \delta_p\leq \beta_p$ for each $p \in P$. Since $d$ is a proper divisor of $f$, there is a prime $p_0\in P$ such that \begin{align*} \delta_{p_0}<\beta_{p_0}. \end{align*} Because $\widetilde{\chi}=\widetilde{\chi}^*\circ\rho_{q,f}$ and $\widetilde{\chi}^*$ factors through $\rho_{f,d}$, the character $\widetilde{\chi}$ factors through $\rho_{q,d}$. Restricting this factorization to the $p_0$-component in the Chinese remainder decomposition shows that $\widetilde{\chi}_{p_0}$ is trivial on \begin{align*} \ker\rho_{p_0,\alpha_{p_0},\delta_{p_0}}. \end{align*} This contradicts the minimal definition of $\beta_{p_0}$, because $\delta_{p_0}<\beta_{p_0}$. Therefore $\chi^*$ is primitive modulo $f$. [/step] [step:Identify any other primitive inducing character with the same local minimal exponents] Let $g \mid q$, and let $\psi$ be a primitive Dirichlet character modulo $g$ inducing $\chi$ modulo $q$. Write \begin{align*} g=\prod_{p \in P}p^{\gamma_p}, \end{align*} where $0\leq \gamma_p\leq \alpha_p$ for each $p \in P$. Let \begin{align*} \widetilde{\psi}:G_g&\to\mathbb{C}^\times \end{align*} be the unit-group character associated to $\psi$. Since $\psi$ induces $\chi$, the character $\widetilde{\chi}:G_q\to\mathbb{C}^\times$ factors through $\rho_{q,g}$. Restricting to the $p$-component shows that $\widetilde{\chi}_p$ is trivial on \begin{align*} \ker\rho_{p,\alpha_p,\gamma_p}. \end{align*} By the minimality of $\beta_p$, we obtain \begin{align*} \beta_p\leq \gamma_p \end{align*} for every $p\in P$. We now prove the reverse inequality. Suppose that $\gamma_{p_0}>\beta_{p_0}$ for some prime $p_0\in P$. Let \begin{align*} g' := g/p_0. \end{align*} Because $\gamma_{p_0}\geq 1$, this is an integer divisor of $g$. The inequality $\beta_{p_0}\leq \gamma_{p_0}-1$ and the inequalities $\beta_p\leq\gamma_p$ for $p\neq p_0$ imply that $f\mid g'$. Since $\widetilde{\chi}$ factors through $G_f$, it also factors through $G_{g'}$. The character $\widetilde{\chi}$ also factors through $G_g$ via $\widetilde{\psi}$. Hence $\widetilde{\psi}$ is trivial on $\ker\rho_{g,g'}$: if $v\in\ker\rho_{g,g'}$, choose $u\in G_q$ with $\rho_{q,g}(u)=v$; then $\rho_{q,g'}(u)=1$, so the factorization through $G_{g'}$ gives \begin{align*} \widetilde{\psi}(v) = \widetilde{\chi}(u) = 1. \end{align*} By the descent criterion, $\widetilde{\psi}$ factors through $G_{g'}$. Therefore $\psi$ is induced from a character modulo the proper divisor $g'$ of $g$, contradicting the primitivity of $\psi$. Thus $\gamma_p\leq\beta_p$ for every $p\in P$. Combining both inequalities gives $\gamma_p=\beta_p$ for every $p\in P$, hence $g=f$. [/step] [step:Conclude equality of the inducing primitive characters] It remains to prove that the primitive character modulo $f$ is unique. Let $\psi$ be a primitive character modulo $f$ inducing $\chi$ modulo $q$, and let $\widetilde{\psi}:G_f\to\mathbb{C}^\times$ be its unit-group character. We already proved that any primitive inducing modulus must equal $f$. Let $a\in G_f$. Choose an integer $n\in\mathbb{Z}$ such that $n\equiv a\pmod f$ and $\gcd(n,q)=1$. Such an $n$ exists by the Chinese Remainder Theorem, by additionally requiring $n\equiv 1\pmod p$ for every prime $p\mid q$ with $p\nmid f$. Since both $\chi^*$ and $\psi$ induce $\chi$ modulo $q$, we have \begin{align*} \widetilde{\chi}^*(a) = \chi^*(n) = \chi(n) = \psi(n) = \widetilde{\psi}(a). \end{align*} Thus $\widetilde{\chi}^*=\widetilde{\psi}$ on $G_f$, and therefore $\chi^*=\psi$ as Dirichlet characters modulo $f$. This proves existence and uniqueness of the primitive character inducing $\chi$, and the corresponding modulus $f$ is the conductor of $\chi$. [/step]