Explicit reciprocity law

Over the complex numbers (a, b) is always 1. Over the reals, the Hilbert symbol of odd degree is trivial, and the Hilbert symbol of even degree is given by (a, b)_∞ is +1 if at least one of a or b is positive, and −1 if both are negative.

In the unramified case, when the order of the Hilbert symbol is coprime to the residue characteristic of the local field, the tame Hilbert symbol is given by[1]

${\displaystyle (a,b)=\omega ((-1)^{\operatorname {ord} (a)\operatorname {ord} (b)}b^{\operatorname {ord} (a)}/a^{\operatorname {ord} (b)})^{(q-1)/n}}$

where ω(a) is the (q – 1)-th root of unity congruent to a and ord(a) is the value of the valuation of the local field, and n is the degree of the Hilbert symbol, and q is the order of the residue class field. The number n divides q – 1 because the local field contains the nth roots of unity by assumption.

As a special case, over the p-adics with p odd, writing ${\displaystyle a=p^{\alpha }u}$ and ${\displaystyle b=p^{\beta }v}$, where u and v are integers coprime to p, we have for the quadratic Hilbert symbol

${\displaystyle (a,b)_{p}=(-1)^{\alpha \beta \varepsilon (p)}\left({\frac {u}{p}}\right)^{\beta }\left({\frac {v}{p}}\right)^{\alpha }}$, where ${\displaystyle \varepsilon (p)=(p-1)/2}$

and the expression involves two Legendre symbols.

The simplest example of a Hilbert symbol in the ramified case is the quadratic Hilbert symbol over the 2-adic integers. Over the 2-adics, again writing ${\displaystyle a=2^{\alpha }u}$ and ${\displaystyle b=2^{\beta }v}$, where u and v are odd numbers, we have for the quadratic Hilbert symbol

${\displaystyle (a,b)_{2}=(-1)^{\epsilon (u)\epsilon (v)+\alpha \omega (v)+\beta \omega (u)}}$, where ${\displaystyle \omega (x)=(x^{2}-1)/8}$ and ${\displaystyle \epsilon (x)=(x-1)/2.}$