[proofplan] We use the defining construction of a Dirichlet character modulo $q$: it is obtained from a group homomorphism on the unit group $(\mathbb{Z}/q\mathbb{Z})^\times$ and extended by zero on residue classes not coprime to $q$. Periodicity follows because $\chi(n)$ depends only on the residue class of $n$ modulo $q$. Multiplicativity is the homomorphism law when both integers are coprime to $q$, and in the remaining case both sides vanish because a non-coprime factor makes the product non-coprime to $q$. [/proofplan] [step:Unpack the defining data of the Dirichlet character] Let \begin{align*} \pi_q: \mathbb{Z} &\to \mathbb{Z}/q\mathbb{Z} \\ n &\mapsto n+q\mathbb{Z} \end{align*} be the quotient map. Since $\chi$ is a Dirichlet character modulo $q$, there is a group homomorphism \begin{align*} \widetilde{\chi}: (\mathbb{Z}/q\mathbb{Z})^\times &\to \mathbb{C}^\times \end{align*} such that, for every $n \in \mathbb{Z}$, \begin{align*} \chi(n) = \begin{cases} \widetilde{\chi}(\pi_q(n)), & \gcd(n,q)=1,\\ 0, & \gcd(n,q)>1. \end{cases} \end{align*} Here $(\mathbb{Z}/q\mathbb{Z})^\times$ denotes the multiplicative group of invertible residue classes modulo $q$, and $\pi_q(n) \in (\mathbb{Z}/q\mathbb{Z})^\times$ exactly when $\gcd(n,q)=1$. [/step] [step:Read off the vanishing criterion from the zero extension] For every $n \in \mathbb{Z}$, the displayed definition gives $\chi(n)=0$ when $\gcd(n,q)>1$. If $\gcd(n,q)=1$, then $\pi_q(n) \in (\mathbb{Z}/q\mathbb{Z})^\times$, so $\widetilde{\chi}(\pi_q(n)) \in \mathbb{C}^\times$ and hence is nonzero. Therefore \begin{align*} \chi(n)=0 \iff \gcd(n,q)>1. \end{align*} [/step] [step:Prove periodicity from equality of residue classes] Let $n \in \mathbb{Z}$. Since \begin{align*} \pi_q(n+q)=\pi_q(n), \end{align*} the integers $n+q$ and $n$ have the same residue class modulo $q$. Therefore they are either both coprime to $q$ or both not coprime to $q$. If $\gcd(n,q)=1$, then \begin{align*} \chi(n+q)=\widetilde{\chi}(\pi_q(n+q)) =\widetilde{\chi}(\pi_q(n)) =\chi(n). \end{align*} If $\gcd(n,q)>1$, then also $\gcd(n+q,q)>1$, so \begin{align*} \chi(n+q)=0=\chi(n). \end{align*} Thus $\chi(n+q)=\chi(n)$ for every $n \in \mathbb{Z}$. [/step] [step:Compute the value at the identity] Since $\gcd(1,q)=1$, the defining formula gives \begin{align*} \chi(1)=\widetilde{\chi}(\pi_q(1)). \end{align*} The residue class $\pi_q(1)$ is the identity element of $(\mathbb{Z}/q\mathbb{Z})^\times$. Because $\widetilde{\chi}$ is a group homomorphism into $\mathbb{C}^\times$, it sends the identity element to the identity element $1 \in \mathbb{C}^\times$. Hence \begin{align*} \chi(1)=1. \end{align*} [/step] [step:Prove multiplicativity when both factors are coprime to the modulus] Let $m,n \in \mathbb{Z}$ and suppose $\gcd(m,q)=1$ and $\gcd(n,q)=1$. Then $\gcd(mn,q)=1$, so $\pi_q(m)$, $\pi_q(n)$, and $\pi_q(mn)$ all lie in $(\mathbb{Z}/q\mathbb{Z})^\times$. Since $\pi_q$ respects multiplication of residue classes, \begin{align*} \pi_q(mn)=\pi_q(m)\pi_q(n). \end{align*} Using the homomorphism law for $\widetilde{\chi}$ gives \begin{align*} \chi(mn) &=\widetilde{\chi}(\pi_q(mn))\\ &=\widetilde{\chi}(\pi_q(m)\pi_q(n))\\ &=\widetilde{\chi}(\pi_q(m))\widetilde{\chi}(\pi_q(n))\\ &=\chi(m)\chi(n). \end{align*} [guided] We first handle the case in which the zero extension is not involved. Assume $\gcd(m,q)=1$ and $\gcd(n,q)=1$. Then both residue classes $\pi_q(m)$ and $\pi_q(n)$ are units modulo $q$. Their product is also a unit, and it is precisely the residue class of the integer product: \begin{align*} \pi_q(mn)=\pi_q(m)\pi_q(n). \end{align*} Since $\chi$ agrees with $\widetilde{\chi}$ on coprime residue classes, we may compute entirely inside the group $(\mathbb{Z}/q\mathbb{Z})^\times$. The homomorphism property of \begin{align*} \widetilde{\chi}: (\mathbb{Z}/q\mathbb{Z})^\times \to \mathbb{C}^\times \end{align*} gives \begin{align*} \chi(mn) &=\widetilde{\chi}(\pi_q(mn))\\ &=\widetilde{\chi}(\pi_q(m)\pi_q(n))\\ &=\widetilde{\chi}(\pi_q(m))\widetilde{\chi}(\pi_q(n))\\ &=\chi(m)\chi(n). \end{align*} This proves multiplicativity in the case where both factors are invertible modulo $q$. [/guided] [/step] [step:Prove multiplicativity when at least one factor is not coprime to the modulus] Now suppose at least one of $\gcd(m,q)$ or $\gcd(n,q)$ is greater than $1$. Then $\chi(m)\chi(n)=0$ by the vanishing criterion, because at least one factor on the right-hand side is zero. It remains to check that $\chi(mn)=0$. If $\gcd(m,q)>1$, choose a prime $p$ dividing both $m$ and $q$. Then $p$ divides $mn$ and $q$, so $\gcd(mn,q)>1$. The same argument applies if $\gcd(n,q)>1$. Hence $\gcd(mn,q)>1$, and the vanishing criterion gives \begin{align*} \chi(mn)=0=\chi(m)\chi(n). \end{align*} Together with the coprime case, this proves \begin{align*} \chi(mn)=\chi(m)\chi(n) \end{align*} for all $m,n \in \mathbb{Z}$. [/step] [step:Collect the four identities] The preceding steps prove, for all $m,n \in \mathbb{Z}$, \begin{align*} \chi(mn) &= \chi(m)\chi(n),\\ \chi(n+q) &= \chi(n),\\ \chi(1) &= 1, \end{align*} and \begin{align*} \chi(n)=0 \iff \gcd(n,q)>1. \end{align*} These are exactly the stated basic properties of the Dirichlet character $\chi$ modulo $q$. [/step]