[proofplan] For a nonprincipal character, the point $t=0$ is handled by the real and nonreal Dirichlet nonvanishing theorems already proved in this chapter. For $t\neq0$, we use the de la Vallée Poussin positivity argument directly: the Euler products for $\zeta$, $L(s,\chi)$, and $L(s,\chi^2)$ imply that a certain product has modulus at least $1$ for real $\sigma>1$. If $L(1+it,\chi)$ vanished, the fourth power of that zero would dominate the cubic pole of $\zeta(\sigma)$ as $\sigma\downarrow1$, contradicting the lower bound. The principal case is then a finite Euler-factor comparison with the Riemann zeta function. [/proofplan]

[step:Prove the nonprincipal boundary nonvanishing assertion]Assume that $\chi$ is nonprincipal modulo $q$. Define \begin{align*} L_\chi:\{s\in\mathbb C:\operatorname{Re}(s)>1\}&\to\mathbb C\\ s&\mapsto \sum_{n=1}^{\infty}\frac{\chi(n)}{n^s}, \end{align*} and write $L(s,\chi)=L_\chi(s)$. Since $\chi$ is nonprincipal, its partial sums over complete periods vanish, so summation by parts gives a holomorphic continuation of $L_\chi$ to the half-plane $\operatorname{Re}(s)>0$. Let \begin{align*} \zeta:\{s\in\mathbb C:\operatorname{Re}(s)>1\}&\to\mathbb C\\ s&\mapsto \sum_{n=1}^{\infty}n^{-s} \end{align*} denote the Riemann zeta function, together with its meromorphic continuation. First take $t=0$. If $\chi$ is real-valued, then $L(1,\chi)\neq0$ by the [Dirichlet Nonvanishing Theorem for Real Characters](/theorems/TEMP-43). If $\chi$ is not real-valued, then $L(1,\chi)\neq0$ by [Dirichlet Nonvanishing for Nonreal Dirichlet Characters](/theorems/TEMP-44). Now fix $t\in\mathbb R\setminus\{0\}$ and suppose, for contradiction, that \begin{align*} L(1+it,\chi)=0. \end{align*} Let $m\geq1$ be the order of this zero. Let $\chi^2:\mathbb Z\to\mathbb C$ be the Dirichlet character $n\mapsto\chi(n)^2$. The function $L(s,\chi^2)$ is bounded in a neighbourhood of $s=1+2it$: if $\chi^2$ is nonprincipal this follows from the same summation-by-parts continuation, and if $\chi^2$ is principal then the finite Euler-factor comparison with $\zeta$ and the fact $2t\neq0$ show holomorphy there. For real $\sigma>1$, define \begin{align*} P_t(\sigma) := \left|\zeta(\sigma)^3L(\sigma+it,\chi)^4L(\sigma+2it,\chi^2)\right|. \end{align*} The absolutely convergent Euler products in the half-plane $\operatorname{Re}(s)>1$ give \begin{align*} \log P_t(\sigma) &= \sum_p\sum_{k=1}^{\infty} \frac{ 3+4\operatorname{Re}\!\left(\chi(p)^kp^{-ikt}\right) +\operatorname{Re}\!\left(\chi(p)^{2k}p^{-2ikt}\right) }{k p^{k\sigma}}, \end{align*} where the outer sum is over primes. For each prime $p$ and integer $k\geq1$, put \begin{align*} \alpha_{p,k}:=\chi(p)^kp^{-ikt}. \end{align*} Then $|\alpha_{p,k}|\leq1$, and the numerator is \begin{align*} 3+4\operatorname{Re}(\alpha_{p,k})+\operatorname{Re}(\alpha_{p,k}^2). \end{align*} Writing $\alpha_{p,k}=re^{i\theta}$ with $0\leq r\leq1$, this quantity equals \begin{align*} 3+4r\cos\theta+r^2\cos(2\theta) = 2r^2\cos^2\theta+4r\cos\theta+3-r^2. \end{align*} As a quadratic in $\cos\theta$ on $[-1,1]$, its minimum occurs at $\cos\theta=-1$ for $0\leq r\leq1$, and the minimum value is \begin{align*} 3-4r+r^2=(1-r)(3-r)\geq0. \end{align*} Thus $\log P_t(\sigma)\geq0$, so \begin{align*} P_t(\sigma)\geq1 \end{align*} for every $\sigma>1$. On the other hand, the [Hadamard-de la Vallée Poussin nonvanishing theorem for $\zeta$](/theorems/TEMP-30) gives a simple pole of $\zeta$ at $s=1$, so $\zeta(\sigma)^3=O((\sigma-1)^{-3})$ as $\sigma\downarrow1$. The assumed zero gives $L(\sigma+it,\chi)^4=O((\sigma-1)^{4m})$, and $L(\sigma+2it,\chi^2)=O(1)$. Therefore \begin{align*} P_t(\sigma)=O((\sigma-1)^{4m-3})\to0, \end{align*} because $m\geq1$. This contradicts $P_t(\sigma)\geq1$. Hence $L(1+it,\chi)\neq0$ for every $t\neq0$, and the nonprincipal case is proved.[/step]

[guided]Let $\chi$ be nonprincipal. Its $L$-function is initially \begin{align*} L_\chi:\{s\in\mathbb C:\operatorname{Re}(s)>1\}&\to\mathbb C\\ s&\mapsto \sum_{n=1}^{\infty}\frac{\chi(n)}{n^s}. \end{align*} Because the character sum over one full period is zero, summation by parts extends this function holomorphically to $\operatorname{Re}(s)>0$. Let \begin{align*} \zeta:\{s\in\mathbb C:\operatorname{Re}(s)>1\}&\to\mathbb C\\ s&\mapsto \sum_{n=1}^{\infty}n^{-s} \end{align*} denote the Riemann zeta function and its meromorphic continuation. At $t=0$, the already proved Dirichlet nonvanishing theorems apply: the real-valued case is [Dirichlet Nonvanishing Theorem for Real Characters](/theorems/TEMP-43), and the nonreal-valued case is [Dirichlet Nonvanishing for Nonreal Dirichlet Characters](/theorems/TEMP-44). Now take $t\neq0$ and assume toward a contradiction that $L(1+it,\chi)=0$. If the zero has order $m\geq1$, then near $\sigma=1$, \begin{align*} |L(\sigma+it,\chi)|=O((\sigma-1)^m). \end{align*} Define the auxiliary character \begin{align*} \chi^2:\mathbb Z&\to\mathbb C\\ n&\mapsto \chi(n)^2. \end{align*} Its $L$-function is bounded near $1+2it$: if $\chi^2$ is nonprincipal this is again summation by parts, while if $\chi^2$ is principal then the only possible pole is at \begin{align*} s=1, \end{align*} not at \begin{align*} s=1+2it, \end{align*} because \begin{align*} 2t\neq0. \end{align*} For real $\sigma>1$, define \begin{align*} P_t(\sigma) := \left|\zeta(\sigma)^3L(\sigma+it,\chi)^4L(\sigma+2it,\chi^2)\right|. \end{align*} Euler products give the logarithmic expansion \begin{align*} \log P_t(\sigma) &= \sum_p\sum_{k=1}^{\infty} \frac{ 3+4\operatorname{Re}\!\left(\chi(p)^kp^{-ikt}\right) +\operatorname{Re}\!\left(\chi(p)^{2k}p^{-2ikt}\right) }{k p^{k\sigma}}. \end{align*} For each prime $p$ and integer $k\geq1$, define \begin{align*} \alpha_{p,k}:=\chi(p)^kp^{-ikt}. \end{align*} Then \begin{align*} |\alpha_{p,k}|\leq1 \end{align*} and the numerator in the logarithmic expansion is \begin{align*} 3+4\operatorname{Re}(\alpha_{p,k})+\operatorname{Re}(\alpha_{p,k}^2). \end{align*} Writing \begin{align*} \alpha_{p,k}=re^{i\theta}, \qquad 0\leq r\leq1, \end{align*} this numerator becomes \begin{align*} 3+4r\cos\theta+r^2\cos(2\theta) = 2r^2\cos^2\theta+4r\cos\theta+3-r^2. \end{align*} Its minimum on $-1\leq\cos\theta\leq1$ occurs at $\cos\theta=-1$, and the minimum value is \begin{align*} 3-4r+r^2=(1-r)(3-r)\geq0. \end{align*} Thus \begin{align*} \log P_t(\sigma)\geq0, \qquad P_t(\sigma)\geq1. \end{align*} But the simple pole of $\zeta$ contributes at most order $3$, while the fourth power of the assumed zero contributes order at least $4$. The remaining factor is bounded. Hence the product tends to $0$ as $\sigma\downarrow1$, contradicting the lower bound. Thus no such zero exists. More explicitly, the pole and zero estimates are \begin{align*} \zeta(\sigma)^3&=O((\sigma-1)^{-3}),\\ L(\sigma+it,\chi)^4&=O((\sigma-1)^{4m}),\\ L(\sigma+2it,\chi^2)&=O(1), \end{align*} so \begin{align*} P_t(\sigma)=O((\sigma-1)^{4m-3})\to0, \end{align*} because \begin{align*} m\geq1. \end{align*} This contradicts \begin{align*} P_t(\sigma)\geq1. \end{align*} Therefore \begin{align*} L(1+it,\chi)\neq0. \end{align*}[/guided]

[step:Compare the principal character with the zeta function]Assume that $\chi=\chi_0$ is the [principal Dirichlet character](/page/Principal%20Dirichlet%20Character) modulo $q$. By the [Euler-factor formula for the principal Dirichlet character](/theorems/TEMP-37), for $\operatorname{Re}(s)>1$ and by meromorphic continuation, \begin{align*} L(s,\chi_0)=\zeta(s)\prod_{p\mid q}\left(1-p^{-s}\right). \end{align*} Define the finite Euler factor \begin{align*} E_q: \mathbb C &\to \mathbb C \\ s &\mapsto \prod_{p\mid q}\left(1-p^{-s}\right). \end{align*} At $s=1$ each factor $1-p^{-1}$ is nonzero, so $E_q(1)\neq 0$. Since $\zeta$ has a simple pole at $s=1$ and $E_q$ is holomorphic and nonzero at $s=1$, the product $L(s,\chi_0)=\zeta(s)E_q(s)$ has a simple pole at $s=1$. Now let $t\in\mathbb R$. If $t=0$, then $s=1$ is the pole just proved and hence is not a zero. If $t\neq 0$, the zeta boundary theorem gives $\zeta(1+it)\neq 0$, and for every prime $p\mid q$, \begin{align*} |p^{-(1+it)}|=p^{-1}<1, \end{align*} so $1-p^{-(1+it)}\neq 0$. Therefore $L(1+it,\chi_0)\neq 0$ for every $t\neq 0$, and the principal-character case follows.[/step]

[guided]For the principal character, the inducing primitive character is the principal character modulo $1$, so the associated primitive $L$-function is the Riemann zeta function. The finite difference between $L(s,\chi_0)$ and $\zeta(s)$ is exactly the deletion of Euler factors at primes dividing $q$. The [Euler-factor formula for the principal Dirichlet character](/theorems/TEMP-37) gives \begin{align*} L(s,\chi_0)=\zeta(s)\prod_{p\mid q}\left(1-p^{-s}\right) \end{align*} for $\operatorname{Re}(s)>1$, and meromorphic continuation extends the identity to the boundary line. Define the finite Euler factor as the map \begin{align*} E_q: \mathbb C &\to \mathbb C \\ s &\mapsto \prod_{p\mid q}\left(1-p^{-s}\right). \end{align*} This is an entire function because it is a finite product of entire functions. At $s=1$, \begin{align*} E_q(1)=\prod_{p\mid q}\left(1-p^{-1}\right)\neq 0. \end{align*} The zeta boundary theorem says that $\zeta$ has a simple pole at $s=1$. Multiplying a [meromorphic function](/page/Meromorphic%20Function) with a simple pole by a [holomorphic function](/page/Holomorphic%20Function) that is nonzero at the pole preserves the order of the pole. Hence $L(s,\chi_0)=\zeta(s)E_q(s)$ has a simple pole at $s=1$. It remains to rule out zeros on the rest of the boundary line. Let $t\in\mathbb R$ and set $s=1+it$. If $t=0$, then $s=1$ is a pole, so it is not a zero. If $t\neq 0$, the zeta boundary theorem gives $\zeta(1+it)\neq 0$. For each prime $p\mid q$, \begin{align*} |p^{-(1+it)}|=p^{-1}<1, \end{align*} so $p^{-(1+it)}\neq 1$ and therefore $1-p^{-(1+it)}\neq 0$. Thus $E_q(1+it)\neq 0$, and the product formula gives $L(1+it,\chi_0)\neq 0$ for $t\neq 0$. This proves both the simple pole and the boundary nonvanishing assertion in the principal case.[/guided]