proof of quadratic reciprocity rule
The quadratic reciprocity law is:
Theorem: (Gauss) Let p and q be distinct odd primes, and write p=2a+1 and q=2b+1. Then (pq)(qp)=(-1)ab.
((vw) is the Legendre symbol.)
Proof: Let R be the subset [-a,a]×[-b,b] of ℤ×ℤ. Let S be the interval
[-(pq-1)/2,(pq-1)/2] |
of ℤ.
By the Chinese remainder theorem, there exists a unique
bijection f:S→R such that, for any s∈S, if we
write f(s)=(x,y), then
and
.
Let be the subset of consisting of the values of on
.
contains, say, elements of the form
such that , and elements of the form
with . Intending to apply Gauss’s lemma,
we seek some kind of comparison between and .
has elements in the region , namely for all of the form with and . Thus
i.e.
(1) |
Swapping and , we have likewise
(2) |
Furthermore, for any , if then . It follows that for any other than , either or is in , but not both. Therefore
(3) |
Adding (1), (2), and (3) gives us
so
which, in view of Gauss’s lemma, is the desired conclusion.
For a bibliography of the more than 200 known proofs of the QRL, see http://www.rzuser.uni-heidelberg.de/ hb3/fchrono.htmlLemmermeyer .
Title | proof of quadratic reciprocity rule |
---|---|
Canonical name | ProofOfQuadraticReciprocityRule |
Date of creation | 2013-03-22 13:16:15 |
Last modified on | 2013-03-22 13:16:15 |
Owner | mathcam (2727) |
Last modified by | mathcam (2727) |
Numerical id | 12 |
Author | mathcam (2727) |
Entry type | Proof |
Classification | msc 11A15 |