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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 $L(1,\chi_{K})\neq 0$.

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You aren't giving us much to go on. You don't tell us what $\left({K/{\bf Q}\over p}\right)$ means. I guess "extends" means you can identify any prime $p$ with the element $r$ of ${\bf Z}/d{\bf Z}$ given by $p\equiv r\pmod d$, is that right? What's $\zeta_K$? Are you asking for a proof of the equation in (ii), or for a proof that the equation implies the inequality, or both? This would all be a lot easier if we knew what source you were following. And it would be a lot more pleasant if your accept rate weren't so low. –  Gerry Myerson Dec 4 '11 at 6:02
I apologize, by $\left(\frac{K/\mathbb{Q}}{p}\right)$ I mean the Frobenius authorphism. I will repair my accept rate in a minute since it seems I forgot to accept some really good answers. –  Anna Dec 5 '11 at 1:21
Also, $\zeta_{K}(s)= \prod_{p \in K} {\left(1-N(p)^{-s}\right)^{-1}}$. –  Anna Dec 5 '11 at 7:48

1 Answer 1

up vote 1 down vote accepted

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!)

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