2

We aim to show that if $\chi$ is a complex Dirichlet character mod $m$, then $L(1, \chi) \neq 0$. Assuming otherwise, we easily prove that if $F(s) = \prod_{\chi}L(s, \chi)$, where the product is over all Dirichlet characters mod $m$, then $F(1) = 0$. However, in the previous proposition, we showed that $F(s) \geq 1$ for all $s > 1$, and the authors state this is a contradiction.

Why? Prior to this, we have only shown that $L(s, \chi)$ analytically continues to $\Re s > 0$ if $\chi$ is nontrivial, so how can we conclude? Do we know that $\lim_{s \to 1}F(s) = F(1)$?

Bill Dubuque
  • 282,220
Johnny Apple
  • 4,571

1 Answers1

2

When $\chi$ is trivial, $L(s, \chi)$ is essentially $\zeta(s)$ (except missing Euler factors at primes $p$ dividing the modulus of $\chi$). You should know that $\zeta(s)$ has meromorphic continuation and a simple pole at $s = 1$. Combined with the continuation of $L(s, \chi)$ to $\mathrm{Re} s > 0$ for nontrivial $\chi$, you know that $F(s)$ has meromorphic continuation for $\mathrm{Re} s > 0$.

The point is then that if $\chi$ is a nontrivial, non-real character with $L(1, \chi) = 0$, then $L(1, \overline{\chi}) = 0$ too and $\overline{\chi} \neq \chi$. Evaluating $F(s)$ at $s = 1$ has a simple pole from $\zeta(s)$ and a double zero, one from $L(s, \chi)$ and one from $L(s, \overline{\chi})$. Thus $F(s)$ is actually analytic for $\mathrm{Re} s > 0$. Hence $\lim_{s \to 1} F(s) = F(1)$.

davidlowryduda
  • 94,345