Dirichlet characters and quadratic fields Let $\chi$ be a Dirichlet character. This is,
$$\chi: (\mathbb{Z}/m\mathbb{Z})^* \to S^1$$
I'm trying to understand the relation between those characters and fiel extensions.
For example, the principal (trivial) character $\chi_0$ mod $1$ is naturally associated to $\mathbb{Q}$, in the sense that $L(s,\chi_0)=\zeta(s)$.
In particular I'm interested in the following. Let $K$ be a quadratic extensions. We know that:
$$\zeta_K(s)=\zeta(s)L(s,\chi_K)$$
Let's call quadratic characters to those who appear that way. There should be a bijection between quadratic characters and quadratic extensions (right?).

Are all non-principal characters quadratic?

But perhaps the most natural question is,

Given a Dirichlet character $\chi$, how do we get a field extension
  $K$ from it? Is $K$ quadratic always? Is this process unique?

Thanks for any information. I'd appreciate references also.
 A: This is probably not a complete answer to your question but here it is anyways.
Usually, one calls a Dirichlet character $\chi$ quadratic, if it has order 2 in the dual of $(\mathbb{Z}/m\mathbb{Z})^\ast$, i.e. if $\chi^2 = 1$ and $\chi$ is not principal. Notice that every qudratic character takes only the values $\pm 1$ and $-1$ at least once. Equivalently, a quadratic characater is a real and non-principal character.
Next, call an integer $D$ a quadratic discriminant, if it arises as the discriminant of a quadratic number field. Equivalently, $D$ is a quadratic discrimant, if $D$ is $\neq 1$, square-free and $D \equiv 1 \mod{4}$ or if $D  = 4d$ for some square-free integer $d \equiv 2$ or $3 \mod 4$.
Attached to every quadratic discriminant $D$ is a special quadratic character, called the Kronecker symbol, sometimes denoted by $\chi_D(n)$ or $\left( \frac{D}{n} \right)$ where $n$ is the argument. One can define it by requiring:

*

*$\chi_D$ is completely multiplicative.

*$\chi_D(0) = 0$ and $\chi_D(1) = 1$.

*For every odd prime $p$, $\chi_D(p)$ is equal to the Legendre symbol mod $p$.

*$\chi_D(2) = \begin{cases}0 & \text{ if } D \equiv 2  \pmod{2}\\ 1 &\text{ if } D \equiv 1 \pmod{8}\\ -1 &\text{ if } D \equiv 5 \pmod{8} \end{cases}$.

*$\chi_D(-1) = \begin{cases}1 & \text{ if } D > 0 \\ -1 & \text{ if } D < 0 \end{cases}$.

One has then the following Theorem, whose proof uses a lot of the theory of Dirichlet characters and Gauss' quadratic reciprocity law.
Theorem. For every quadratic discrimant $D$, the Kronecker symbol $\chi_D$ defines a primitive quadratic character mod $|D|$. Conversely, every primitve quadratic character is given by a Kronecker symbol and hence every quadratic character is induced by some Kronecker symbol.
In my opinion, Montgomery's and Vaughan's ''Multiplicative number theory'' or Davenport's classical ''Multiplicative number theory'' is a good reference for this.
Furthermore, the following formula holds for quadratic fields $K$:
$$\zeta_K(s) = \zeta(s)L(s, \chi_D).$$
In your notation $\chi_D = \chi_K$.
The answer to your first question is no with both your and 'my' defintion of quadratic.
