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In mathematics, given a local field K, such as the fields of reals or p-adic numbers, whose multiplicative group of non-zero elements is K×, the Hilbert symbol is an algebraic construction, extracted from reciprocity laws, and important in the formulation of local class field theory. As the name suggests, it was in some sense introduced by David Hilbert, although it would be anachronistic to say that of the local field formulation.

Explicitly, it is the function (–, –) from K× × K× to {−1,1} defined by

$$(a,b)=\begin{cases}1,&\mbox{ if }z^2=ax^2+by^2\mbox{ has a non-zero solution }(x,y,z)\in K^3;\\-1,&\mbox{ if not.}\end{cases}$$

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Properties

The following three properties follow directly from the definition, by choosing suitable solutions of the diophantine equation above:

• If a is a square, then (a, b) = 1 for all b.
• For all a,b in K×, (a, b) = (b, a).
• For any a in K× such that a−1 is also in K×, we have (a, 1−a) = 1.

The (bi)multiplicativity, i.e.,

(a, b1b2) = (a, b1)·(a, b2)

for any a, b1 and b2 in K× is, however, more difficult to prove, and requires the development of local class field theory.

The third property ensures that the Hilbert symbol factors over the second Milnor K-group $$K^M_2 (K)$$, which is by definition

K×K× / (a ⊗ 1−a, aK× \ {1})

By the first property it even factors over $$K^M_2 (K) / 2$$. This is the first step towards the Milnor conjecture.
Interpretation as an algebra

The Hilbert symbol can also be used to denote the central simple algebra over K with basis 1,i,j,k and multiplication rules $$i^2=a, j^2=b, ij=-ji=k$$. In this case the algebra represents an element of order 2 in the Brauer group of K, which is identified with -1 if it is a division algebra and +1 if it is isomorphic to the algebra of 2 by 2 matrices.
Hilbert symbols over the rationals

For a place v of the rational number field and rational numbers a, b we let (a, b)v denote the value of the Hilbert symbol in the corresponding completion Qv. As usual, if v is the valuation attached to a prime number p then the corresponding completion is the p-adic field and if v is the infinite place then the completion is the real number field.

Over the reals, (a, b) is +1 if at least one of a or b is positive, and −1 if both are negative.

Over the p-adics with p odd, writing $$a = p^{\alpha} u$$ and $$b = p^{\beta} v$$, where u and v are integers coprime to p, we have

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

and the expression involves two Legendre symbols.

Over the 2-adics, again writing $$a = 2^\alpha u and b = 2^\beta v$$, where u and v are odd numbers, we have

$$(a,b)_2 = (-1)^{\epsilon(u)\epsilon(v) + \alpha\omega(v) + \beta\omega(u)}$$ , where $$\omega(x) = (x^2-1)/8.$$

It is known that if v ranges over all places, (a, b)v is 1 for almost all places. Therefore the following product formula

$$\prod_v (a,b)_v = 1$$

makes sense. It is equivalent to the law of quadratic reciprocity.