L-Function does not Vanish at One

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

Let $L(s,\chi)$ be the $L$-function associated to $\chi$.

Then $L(1,\chi) \neq 0$.

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

Let $\displaystyle \zeta_q(s) = \prod_{\chi \in G^*}L(s,\chi)$.

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

Suppose that $L(1,\psi) = 0$.

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

So by Analytic Continuation of Dirichlet L-Functions $\zeta_q$ is analytic on $\Re(s) > 0$.

For $\Re(s) > 1$ we have

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

Then


 * $\chi(p)^{f_p} = \chi(p^{f_p}) = \chi(1) = 1$

So $\chi(p)$ is an $f_p^\text{th}$ root of unity.

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

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


 * $\displaystyle \prod_{i = 0}^{f_p} \left(1 - \xi^i u\right) = 1 - u^\xi$

Putting these facts together we have


 * $\displaystyle \prod_{\chi \in G^*} \frac{1}{1 - \chi(p)p^{-s} } = \left( \frac1{1 - p^{-f_p s}} \right)^{\phi(q)/f_p}$

Therefore,

Also, if $\chi_0$ is the trivial character modulo $q$, by Euler Product we have


 * $L(\phi(q)s,\chi_0) = \prod_{p\nmid q}\left( 1 + p^{-\phi(q)s} + p^{-2\phi(q)s} + \cdots \right)$

From which we see that for $s \in \R$, $s \geq 0$, $\zeta_q(s) = L(\phi(q)s,\chi_0)$.

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

But we showed above that $\zeta_q(s)$ converges for $\Re(s) > 0$, a contradiction.