TITLE: Shortest/Most elegant proof for $L(1,\chi)\neq 0$
QUESTION [64 upvotes]: Let $\chi$ be a Dirichlet character and $L(1,\chi)$ the associated L-function evaluated at $s=1$. What would be the 'shortest' proof of the non-vanishing of $L(1,\chi)$?
Background: The non-vanishing of $L(1,\chi)$ plays an essential role in the proof of Dirichlet´s theorem on primes in arithmetic progressions. In his "Introduction to analytic number theory", T. M. Apostol gives an elementary proof of the above fact estimating various sums in a few lemmas in the context of a proof of the aforementioned Dirichlet theorem. While his approach has the advantage of being self-contained and not requiring much of a background, it is quite lenghty. In their "Analytic number theory", H. Iwaniec and E. Kowalski remark that in Dirichlet´s original proof the non-vanishing of $L(1,\chi)$ for real Dirichlet characters is a simple consequence of Dirichlet´s class number formula. However, in both approaches it is necessary to distinguish between real and complex Dirichlet characters. Hence my two "sub"-questions:
1) Is there a proof that avoids the distinction between the complex and real case?
2) Are there in general other proof strategies for $L(1,\chi)\neq 0$ that can be considered shorter and/or more elegant than the two mentioned above?

REPLY [3 votes]: An algebraic number theory argument? At least when $\chi(n)$ is the quadratic character $\bmod p$ an odd prime.
Up to a constant it is its own discrete Fourier transform:
$$\chi(n) = \frac{G(\chi)}{p} \sum_{k=1}^p \chi(k) e^{2i\pi kn/p}$$ so that
$$L(1,\chi) =\sum_{n\ge 1}\frac{\chi(n)}{n}= -\frac{G(\chi)}{p}\sum_{k=1}^p \chi(k) \log(1-e^{2i\pi k/p}) $$
$L(1,\chi)=0$ is the same as $$\prod_{k=1,\chi(k)=1}^p (1-e^{2i\pi k/p})= \prod_{k=1,\chi(k)=1}^p (1-e^{2i\pi a k/p})\tag{1}$$
for a quadratic non-residue $a$.
$\prod_{k=1,\chi(k)=1}^p (1-e^{2i\pi a k/p})$ is the only Galois conjugate of $\prod_{k=1,\chi(k)=1}^p (1-e^{2i\pi k/p})$.
Whence $(1)$ is the same as $$\prod_{k=1,\chi(k)=1}^p (1-e^{2i\pi k/p})\in \Bbb{Q}$$
This is impossible because the $p$-adic valuation of $1-e^{2i\pi k/p}$ is $\frac1{p-1}$ so $$v_p\left(\prod_{k=1,\chi(k)=1}^p (1-e^{2i\pi k/p})\right)=\frac12$$