Editing Quadratic reciprocity (section)

===Imaginary quadratic fields===
The above laws are special cases of more general laws that hold for the [[ring of integers]] in any [[Quadratic field|imaginary quadratic number field]]. Let ''k'' be an imaginary quadratic number field with ring of integers <math>\mathcal{O}_k.</math> For a [[Number field#Prime ideals|prime ideal]] <math>\mathfrak{p} \subset \mathcal{O}_k </math> with odd norm <math>\mathrm{N} \mathfrak{p}</math> and <math>\alpha\in \mathcal{O}_k,</math> define the quadratic character for <math>\mathcal{O}_k </math> as

:<math>\left[\frac{\alpha}{\mathfrak{p} }\right]_2 \equiv \alpha^{\frac{\mathrm{N} \mathfrak{p} - 1}{2}} \bmod{\mathfrak{p}} = 
\begin{cases}
1 &\alpha\not\in \mathfrak{p} \text{ and } \exists \eta \in \mathcal{O}_k \text{ such that } \alpha - \eta^2 \in \mathfrak{p} \\
-1 & \alpha\not\in \mathfrak{p} \text{ and there is no such } \eta \\
0 & \alpha\in \mathfrak{p}
\end{cases}</math>

for an arbitrary ideal <math>\mathfrak{a} \subset \mathcal{O}_k</math> factored into prime ideals <math>\mathfrak{a} = \mathfrak{p}_1 \cdots \mathfrak{p}_n</math> define

:<math>\left [\frac{\alpha}{\mathfrak{a}}\right ]_2 = \left[\frac{\alpha}{\mathfrak{p}_1 }\right]_2\cdots \left[\frac{\alpha}{\mathfrak{p}_n }\right]_2,</math>

and for <math>\beta \in \mathcal{O}_k</math> define

:<math>\left [\frac{\alpha}{\beta}\right ]_2 = \left [\frac{\alpha}{\beta \mathcal{O}_k}\right ]_2. </math>

Let <math>\mathcal{O}_k = \Z \omega_1\oplus \Z \omega_2,</math> i.e. <math>\left\{\omega_1, \omega_2\right\}</math> is an [[Number field#Integral basis|integral basis]] for <math>\mathcal{O}_k.</math> For <math>\nu \in \mathcal{O}_k</math> with odd norm <math>\mathrm{N}\nu,</math> define (ordinary) integers ''a'', ''b'', ''c'', ''d'' by the equations,

:<math>\begin{align}
\nu\omega_1&=a\omega_1+b\omega_2\\
\nu\omega_2&=c\omega_1+d\omega_2
\end{align}</math>

and a function

:<math>\chi(\nu) := \imath^{(b^2-a+2)c+(a^2-b+2)d+ad}.</math>

If ''m'' = ''Nμ'' and ''n'' = ''Nν'' are both odd, Herglotz proved<ref>Lemmermeyer, Thm 8.15, p.256 ff</ref>

:<math> \left [\frac{\mu}{\nu}\right ]_2 \left[\frac{\nu}{\mu}\right]_2 = (-1)^{\frac{m-1}{2}\frac{n-1}{2}} \chi(\mu)^{m\frac{n-1}{2}} \chi(\nu)^{-n\frac{m-1}{2}}. </math>

Also, if

:<math> \mu \equiv\mu' \bmod{4} \quad \text{and} \quad \nu \equiv\nu' \bmod{4}</math>

Then<ref>Lemmermeyer Thm. 8.18, p. 260</ref>

:<math> \left [\frac{\mu}{\nu}\right ]_2 \left[\frac{\nu}{\mu}\right]_2 = \left [\frac{\mu'}{\nu'}\right ]_2 \left[\frac{\nu'}{\mu'}\right]_2.</math>