[proofplan] We first reduce to the case where $\chi$ is primitive, because passing from a primitive character to an imprimitive induced character only multiplies the $L$-function by finitely many Euler factors that are nonzero at $s=1$. For primitive real $\chi$, we study the Dirichlet series $F(s)=\zeta(s)L(s,\chi)$ and show that its coefficients are nonnegative and are at least $1$ on square indices. If $L(1,\chi)=0$, then this zero cancels the simple pole of $\zeta$ at $s=1$, so $F$ is holomorphic on $\operatorname{Re}(s)>0$; Landau's theorem for Dirichlet series with nonnegative coefficients then contradicts the fact that the square-index lower bound forces the abscissa of convergence to be at least $1/2$. [/proofplan] [step:Reduce to the primitive character inducing $\chi$] Let $q_0 \in \mathbb{N}$ be the conductor of $\chi$, and let $\chi_0: \mathbb{Z} \to \{-1,0,1\}$ denote the primitive real Dirichlet character modulo $q_0$ inducing $\chi$. Since $\chi$ is nonprincipal, $\chi_0$ is nonprincipal. For $\operatorname{Re}(s)>1$, the Euler product comparison between an imprimitive character and its inducing primitive character gives \begin{align*} L(s,\chi) = L(s,\chi_0) \prod_{\substack{p \mid q \\ p \nmid q_0}} \left(1-\frac{\chi_0(p)}{p^s}\right). \end{align*} Here the product is finite and runs over primes $p$ dividing $q$ but not $q_0$. Since $\chi_0(p) \in \{-1,1\}$ for every prime $p \nmid q_0$, each factor at $s=1$ is either $1-p^{-1}$ or $1+p^{-1}$, hence is nonzero. Therefore $L(1,\chi)=0$ if and only if $L(1,\chi_0)=0$. It is enough to prove the theorem when $\chi$ itself is primitive. [guided] The only reason to separate primitive and imprimitive characters is that Euler products have their cleanest form for primitive characters. Let $q_0$ be the conductor of $\chi$, and let $\chi_0: \mathbb{Z} \to \{-1,0,1\}$ be the primitive character modulo $q_0$ inducing $\chi$. The original character $\chi$ is obtained from $\chi_0$ by deleting Euler factors at the primes that divide $q$ but not $q_0$. For $\operatorname{Re}(s)>1$, the standard [Euler product identity](/theorems/1694) for induced characters gives \begin{align*} L(s,\chi) = L(s,\chi_0) \prod_{\substack{p \mid q \\ p \nmid q_0}} \left(1-\frac{\chi_0(p)}{p^s}\right). \end{align*} This is a finite product, so it can be evaluated directly at $s=1$. If $p \nmid q_0$, then $\chi_0(p)$ is not $0$, and because $\chi_0$ is real we have $\chi_0(p) \in \{-1,1\}$. Thus \begin{align*} 1-\frac{\chi_0(p)}{p} \in \left\{1-\frac{1}{p},\,1+\frac{1}{p}\right\}, \end{align*} and neither number is zero. Hence the finite product is nonzero at $s=1$. Multiplication by a nonzero finite factor cannot create or remove a zero at $s=1$, so $L(1,\chi)=0$ exactly when $L(1,\chi_0)=0$. Therefore proving the theorem for primitive nonprincipal real characters proves it for all nonprincipal real characters. [/guided] [/step] [step:Build a Dirichlet series with nonnegative coefficients] Assume from now on that $\chi$ is primitive. Define \begin{align*} F: \{s \in \mathbb{C} : \operatorname{Re}(s)>1\} &\to \mathbb{C} \\ s &\mapsto \zeta(s)L(s,\chi). \end{align*} For $\operatorname{Re}(s)>1$, absolute convergence of the two Dirichlet series and multiplication by Dirichlet convolution give \begin{align*} F(s) = \sum_{n=1}^{\infty} \frac{a_n}{n^s}, \qquad a_n := \sum_{d \mid n} \chi(d). \end{align*} We show that $a_n \geq 0$ for every $n \in \mathbb{N}$. Since $\chi$ is completely multiplicative on integers coprime to its modulus and vanishes on integers not coprime to its modulus, the coefficients $a_n$ are multiplicative. It is therefore enough to compute $a_{p^k}$ for a prime $p$ and an integer $k \geq 0$. Put $x=p^{-s}$. The local Euler factor of $F$ at $p$ is \begin{align*} \left(1-x\right)^{-1}\left(1-\chi(p)x\right)^{-1}. \end{align*} If $\chi(p)=1$, this is $(1-x)^{-2}$ and its coefficients are $k+1$. If $\chi(p)=-1$, this is $(1-x^2)^{-1}$ and its coefficients are $1$ for even $k$ and $0$ for odd $k$. If $\chi(p)=0$, this is $(1-x)^{-1}$ and its coefficients are all $1$. Hence every local coefficient is nonnegative, and multiplicativity gives $a_n \geq 0$ for all $n$. Moreover, for every $m \in \mathbb{N}$, each prime-power factor of $m^2$ has even exponent, so the preceding local computation gives \begin{align*} a_{m^2} \geq 1. \end{align*} [/step] [step:Assume a zero at $s=1$ and continue $F$ past its abscissa of convergence] Suppose, for contradiction, that \begin{align*} L(1,\chi)=0. \end{align*} By the standard [analytic continuation](/page/Analytic%20Continuation) theorem for Dirichlet $L$-functions, $L(s,\chi)$ is entire because $\chi$ is nonprincipal. Also, by the standard meromorphic continuation theorem for the Riemann zeta function, $\zeta(s)$ is holomorphic on $\operatorname{Re}(s)>0$ except for a simple pole at $s=1$. Since $L(1,\chi)=0$, the product $\zeta(s)L(s,\chi)$ has a removable singularity at $s=1$. Therefore $F$ extends holomorphically to the whole half-plane \begin{align*} \{s \in \mathbb{C} : \operatorname{Re}(s)>0\}. \end{align*} Here we are citing results not yet in the wiki: the analytic continuation of nonprincipal Dirichlet $L$-functions and the meromorphic continuation of $\zeta(s)$ with a simple pole at $s=1$. [guided] Assume, toward a contradiction, that $L(1,\chi)=0$. The purpose of multiplying by $\zeta(s)$ was to create a Dirichlet series with nonnegative coefficients, but $\zeta(s)$ has a simple pole at $s=1$. The assumed zero of $L(s,\chi)$ exactly cancels that pole. The standard analytic continuation theorem for Dirichlet $L$-functions says that, because $\chi$ is nonprincipal, $L(s,\chi)$ extends to an entire function of $s$. The standard continuation theorem for the Riemann zeta function says that $\zeta(s)$ is holomorphic on the half-plane $\operatorname{Re}(s)>0$ except for a simple pole at $s=1$. Since $L(1,\chi)=0$, the product \begin{align*} F(s)=\zeta(s)L(s,\chi) \end{align*} has a removable singularity at $s=1$. Removing that singularity gives a holomorphic extension of $F$ to \begin{align*} \{s \in \mathbb{C} : \operatorname{Re}(s)>0\}. \end{align*} The cited analytic continuation facts are standard prerequisites not yet present in the wiki: analytic continuation of nonprincipal Dirichlet $L$-functions and meromorphic continuation of $\zeta(s)$ with a simple pole at $s=1$. [/guided] [/step] [step:Use square coefficients to force the abscissa of convergence to be at least $1/2$] Let $\sigma_c \in \mathbb{R}$ denote the abscissa of convergence of the Dirichlet series \begin{align*} \sum_{n=1}^{\infty} \frac{a_n}{n^s}. \end{align*} The series converges for every real $s>1$, so $\sigma_c \leq 1$. On the other hand, if $s \leq 1/2$ is real, then the nonnegativity of the coefficients and the square-index lower bound give \begin{align*} \sum_{n=1}^{\infty} \frac{a_n}{n^s} \geq \sum_{m=1}^{\infty} \frac{a_{m^2}}{(m^2)^s} \geq \sum_{m=1}^{\infty} \frac{1}{m^{2s}}. \end{align*} The final $p$-series diverges when $2s \leq 1$. Hence the Dirichlet series for $F$ cannot converge at any real $s \leq 1/2$, and therefore \begin{align*} \sigma_c \geq \frac{1}{2}. \end{align*} [/step] [step:Apply Landau's theorem to obtain the contradiction] Since $a_n \geq 0$ for every $n \in \mathbb{N}$ and the Dirichlet series has finite abscissa of convergence $\sigma_c \in [1/2,1]$, Landau's theorem for Dirichlet series with nonnegative coefficients applies: if such a Dirichlet series defines a [holomorphic function](/page/Holomorphic%20Function) on its half-plane of convergence, then the point $s=\sigma_c$ on the real axis is a singular point of the function and cannot be crossed by holomorphic continuation. This is a cited result not yet in the wiki: Landau's theorem for Dirichlet series with nonnegative coefficients. But under the assumption $L(1,\chi)=0$, the preceding continuation step showed that $F$ is holomorphic throughout $\operatorname{Re}(s)>0$. Since $\sigma_c \geq 1/2$, the point $s=\sigma_c$ lies inside this half-plane, so $F$ is holomorphic in a neighbourhood of $\sigma_c$. This contradicts Landau's theorem. Therefore the assumption $L(1,\chi)=0$ is false, and \begin{align*} L(1,\chi) \neq 0. \end{align*} [guided] We now use the special feature of the coefficients: they are nonnegative. Landau's theorem for Dirichlet series with nonnegative coefficients says the following. If \begin{align*} \sum_{n=1}^{\infty} \frac{a_n}{n^s} \end{align*} has $a_n \geq 0$ for every $n$, has finite abscissa of convergence $\sigma_c$, and represents a holomorphic function in its half-plane of convergence, then the real point $s=\sigma_c$ is a singular point of that function. In particular, the function cannot be holomorphically continued to any neighbourhood of $s=\sigma_c$. This theorem is a standard prerequisite not yet in the wiki. All hypotheses of Landau's theorem are satisfied here. We proved $a_n \geq 0$ for every $n \in \mathbb{N}$. The series converges for real $s>1$, so $\sigma_c \leq 1$, and the square-index estimate proved $\sigma_c \geq 1/2$; hence $\sigma_c$ is finite. Therefore Landau's theorem says that $s=\sigma_c$ must be a singular point of $F$. This contradicts the holomorphic continuation obtained from the assumed zero $L(1,\chi)=0$. Under that assumption, $F(s)=\zeta(s)L(s,\chi)$ is holomorphic on the entire half-plane \begin{align*} \operatorname{Re}(s)>0. \end{align*} Since $\sigma_c \geq 1/2$, the point $s=\sigma_c$ lies inside this half-plane, so $F$ is holomorphic in a neighbourhood of $\sigma_c$. That is exactly what Landau's theorem forbids. Hence the assumption $L(1,\chi)=0$ must be false, and we conclude \begin{align*} L(1,\chi) \neq 0. \end{align*} [/guided] [/step]