L-Function does not Vanish at One

Theorem
Let $\psi$ be a non-trivial Dirichlet character modulo $q$.

Let $\map L {s, \chi}$ be the Dirichlet $L$-function associated to $\chi$.

Then $\map L {1, \chi} \ne 0$.

Proof
Let $G^*$ be the group of characters modulo $q$.

Let $\ds \map {\zeta_q} s = \prod_{\chi \mathop \in G^*} \map L {s, \chi} R$.

Among the factors of $\zeta_q$ is the the $L$-function associated to the trivial character, which by Analytic Continuation of Dirichlet L-Function we know to have a simple pole at $s = 1$.

$\map L {1, \psi} = 0$.

Then the zero of this factor kills the pole of the principal $L$-function.

So by Analytic Continuation of Dirichlet L-Function $\zeta_q$ is analytic on $\map \Re s > 0$.

For $\map \Re s > 1$ we have:

For any prime $p$, let $f_p$ be the order of $p$ mod $q$.

Then:


 * $\map \chi p^{f_p} = \map \chi {p^{f_p} } = \map \chi 1 = 1$

So $\map \chi p$ is an $f_p$th root of unity.

Moreover by the Orthogonality Relations for Characters each distinct such root occurs $\map \phi q / f_p$ times among the numbers $\map \chi p$ where $\chi \in G^*$.

Also, letting $\xi$ be a primitive $f_p$th root of unity we find that for any $u \in \C$:


 * $\ds \prod_{i \mathop = 0}^{f_p} \paren {1 - \xi^i u} = 1 - u^\xi$

Putting these facts together:


 * $\ds \prod_{\chi \mathop \in G^*} \frac 1 {1 - \map \chi p p^{-s} } = \paren {\frac 1 {1 - p^{-f_p s} } }^{\map \phi q / f_p}$

Therefore:

Also, if $\chi_0$ is the trivial character modulo $q$, by definition of Euler product we have:


 * $\ds \map L {\map \phi q s, \chi_0} = \prod_{p \mathop \nmid q} \paren {1 + p^{-\map \phi q s} + p^{-2 \map \phi q s} + \cdots}$

from which we see that for $s \in \R_{\ge 0}$:
 * $\map {\zeta_q} s \ge \map L {\map \phi q s, \chi_0}$

However, by Analytic Continuation of Dirichlet L-Function, $\map L {\map \phi q s, \chi_0}$ diverges for $s = \map \phi q^{-1}$, and therefore so does $\map {\zeta_q} s$.

But we showed above that $\map {\zeta_q} s$ converges for $\map \Re s > 0$, a contradiction.