Editing Quadratic reciprocity (section)

===Hilbert symbol===
The quadratic reciprocity law can be formulated in terms of the [[Hilbert symbol]] <math>(a,b)_v</math> where ''a'' and ''b'' are any two nonzero rational numbers and ''v'' runs over all the non-trivial absolute values of the rationals (the Archimedean one and the ''p''-adic absolute values for primes ''p''). The Hilbert symbol <math>(a,b)_v</math> is 1 or −1. It is defined to be 1 if and only if the equation <math>ax^2+by^2=z^2</math> has a solution in the [[Completion (ring theory)|completion]] of the rationals at ''v'' other than <math>x=y=z=0</math>. The Hilbert reciprocity law states that <math>(a,b)_v</math>, for fixed ''a'' and ''b'' and varying ''v'', is 1 for all but finitely many ''v'' and the product of <math>(a,b)_v</math> over all ''v'' is 1. (This formally resembles the residue theorem from complex analysis.)

The proof of Hilbert reciprocity reduces to checking a few special cases, and the non-trivial cases turn out to be equivalent to the main law and the two supplementary laws of quadratic reciprocity for the Legendre symbol. There is no kind of reciprocity in the Hilbert reciprocity law; its name simply indicates the historical source of the result in quadratic reciprocity. Unlike quadratic reciprocity, which requires sign conditions (namely positivity of the primes involved) and a special treatment of the prime 2, the Hilbert reciprocity law treats all absolute values of the rationals on an equal footing. Therefore, it is a more natural way of expressing quadratic reciprocity with a view towards generalization: the Hilbert reciprocity law extends with very few changes to all [[global field]]s and this extension can rightly be considered a generalization of quadratic reciprocity to all global fields.
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The next example should be beefed up a bit and moved to [[Jacobi symbol]]. There it would be possible to see much more clearly how the algorithm looks just like the Euclidean algorithm. ~~~~

For example taking ''p'' to be 11 and ''q'' to be 19, we can relate <math>\left(\frac{11}{19}\right)</math> to <math>\left(\frac{19}{11}\right)</math>, which is <math>\left(\frac{8}{11}\right)</math> or <math>\left(\frac{-3}{11}\right)</math>. To proceed further we may need to know supplementary laws for computing <math>\left(\frac{3}{q}\right)</math> and <math>\left(\frac{-1}{q}\right)</math> explicitly. For example,

:<math>\left(\frac{-1}{q}\right) = (-1)^{(q-1)/2}.</math>

Using this we relate <math>\left(\frac{-3}{11}\right)</math> to <math>\left(\frac{3}{11}\right)</math> to <math>\left(\frac{11}{3}\right)</math> to <math>\left(\frac{2}{3}\right)</math> to <math>\left(\frac{-1}{3}\right)</math>, and can complete the initial calculation.

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