On L functions and primitive Dirichlet characters I'm walking through the proof (a proof, better said) of Dirichlet's theorem and I'm having trouble explaining this. I'll state it as an exercise.
First, say $K$ is a quadratic extension of $\mathbb{Q}$ with discriminant $d$. Prove that:
1) the map on primes $p$ that do not divide the discriminant: $p \rightarrow \left(\frac{K/\mathbb{Q}}{p}\right)$ extends to a non-trivial quadratic character $\chi \ :\ (\mathbb{Z}/d)^{\times} \rightarrow \left\{\pm 1\right\} ( = Gal(K/\mathbb{Q}) )$. Is $\chi_{K}$ primitive?
(ii) $\zeta_{K}=\zeta(s)L(s,\chi_{K})$; from this, infer that $L(1,\chi_{K})\neq 0$.
 A: I'll get you started for (1):
For each prime $p$, we have three possibilities: $p$ divides $d$ and hence ramifies in $\mathcal{O}_K$, or $p \nmid d$ in which case $p$ splits completely or remains inert. In either case, picking a prime $Q$ in $\mathcal{O}_K$ above $p\mathbb{Z}$ supplies us with a residual field $F_Q=\mathcal{O}_K/Q$ over $\mathbb{F}_p$. The 'local' Galois group $\text{Gal}(F_Q/\mathbb{F}_p)$ embeds into the 'global' Galois group $\text{Gal}(K/\mathbb{Q})$ at the unramified primes. Since $|\text{Gal}(F_Q/\mathbb{F}_p)| = [F_Q : \mathbb{F}_p]=\log_p|F_Q|$, this embedding $\text{Gal}(F_Q/\mathbb{F}_p) \hookrightarrow \text{Gal}(K/\mathbb{Q})$ is the trivial map if and only if $F_Q/\mathbb{F}_p$ is a trivial extension, which is the case if and only if $p$ splits. In any case, the image of the canonical generator of $\text{Gal}(F_Q/\mathbb{F}_p)$ (which is cyclic) is defined to be the Frobenius at $p$, $\text{Frob } p$. It depends only up to conjugation on the choice of $Q$ lying over $p$.
In the case of a quadratic extension, $\text{Gal}(K/\mathbb{Q})=\{1, \sigma\}$, and we get the following characterization of the Frobenius:
$$\text{Frob }p= \begin{cases}\sigma & \text{if }p \text{ is inert} \\ 1 & \text{if }p\text{ splits}\end{cases}$$
Now if we are given a representation $\rho: \text{Gal}(K/\mathbb{Q}) \to \text{GL}(V)$, we get a map on the primes of $\mathbb{Z}$ to $\text{GL}(V)$ by composing with the Frobenius: $p \mapsto \rho(\text{Frob p})$. This map is well-defined up to conjugation in $\text{GL}(V)$. It extends by multiplicativity to all positive integers - this is the Artin map.
In the case where the global Galois group is abelian, or the representation one-dimensional, everything is well-defined. This is the case you are dealing with. 
There is only one nontrivial irreducible representation $\rho$ of the cyclic group on two elements, and it's the obvious map to $\{\pm 1\}$. This is the origin of the Dirichlet character associated to a quadratic extension: it is the Artin map associated to the representation $\rho$.
Now how do we bridge the gap with the usual definition of a Dirichlet character, as an element of the dual of $\mathbb{Z}/d\mathbb{Z}$? This is the content of the theorem of Quadratic Reciprocity. 
For (2), recall that the Dedekind zeta-function of a number field always has a simple pole at $s=1$. Also, $\zeta(s)$ has a pole at $s=1$. Comparing orders at $s=1$ on both sides yields the result (which has immediate implications!)
