Dirichlet characters and quadratic fields

Solution 1:

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:

1. $\chi_D$ is completely multiplicative.
2. $\chi_D(0) = 0$ and $\chi_D(1) = 1$.
3. For every odd prime $p$, $\chi_D(p)$ is equal to the Legendre symbol mod $p$.
4. $\chi_D(2) = \begin{cases}0 & \text{ if } D \equiv 2 \pmod{2}\\ 1 &\text{ if } D \equiv 1 \pmod{8}\\ -1 &\text{ if } D \equiv 5 \pmod{8} \end{cases}$.
5. $\chi_D(-1) = \begin{cases}1 & \text{ if } D > 0 \\ -1 & \text{ if } D < 0 \end{cases}$.

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.