A real nonprincipal Dirichlet L-function is nonzero at one

Suppose $L(1,\chi )=0$, and set $G(s):=L(s,{\chi }_0)L(s,\chi )$. For $\mathrm{Re}s>1$, facts [L2] and [L3] give $G(s)={\prod }_{\chi (p)=1}(1-p^{-s}{)}^{-2}{\prod }_{\chi (p)=-1}(1-p^{-2s}{)}^{-1},$ because primes dividing $q$ contribute the trivial local factor $1$. Expanding the geometric series shows that $G(s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$ with nonnegative integer coefficients $a_n$. Moreover, if $(m,q)=1$, then the square coefficient $a_{m^2}\ge 1$: each local factor above has a positive integer coefficient at every even exponent occurring in $m^2$.

Step 1.1 implies ${\sum }_{n\ge 1}{a}_n{n}^{-1\mathrm{/}2}\ge {\sum }_{\begin{array}{c}m\ge 1\\ (m,q)=1\end{array}}\frac{a_{m^2}}{m}\ge {\sum }_{\begin{array}{c}m\ge 1\\ (m,q)=1\end{array}}\frac{1}{m}=\mathrm{\infty },$ so the abscissa of convergence ${\sigma }_c$ of $\sum a_n{n}^{-s}$ satisfies ${\sigma }_c\ge 1\mathrm{/}2$. On the other hand, the product identity in [L1] continues $L(s,{\chi }_0)$ to $\mathrm{Re}s>0$; [L1] and [L4] show that $G$ is holomorphic there, because the only possible singularity is the simple pole of $L(s,{\chi }_0)$ at $1$, canceled by the assumption of step 1.1. Since $G$ is represented by a Dirichlet series with nonnegative coefficients, [L5] forbids any positive abscissa of convergence. Thus ${\sigma }_c\le 0$, contradicting ${\sigma }_c\ge 1\mathrm{/}2$. Therefore $L(1,\chi )\mathrm{\ne }0$.