proof of quadratic reciprocity rule


The quadratic reciprocity law is:

Theorem: (Gauss) Let p and q be distinct odd primes, and write p=2⁢a+1 and q=2⁢b+1. Then (pq)⁢(qp)=(-1)a⁢b.

((vw) is the Legendre symbolMathworldPlanetmath.)

Proof: Let R be the subset [-a,a]×[-b,b] of ℤ×ℤ. Let S be the interval

[-(p⁢q-1)/2,(p⁢q-1)/2]

of ℤ. By the Chinese remainder theoremMathworldPlanetmathPlanetmathPlanetmath, there exists a unique bijection f:S→R such that, for any s∈S, if we write f⁢(s)=(x,y), then x≡s(modp) and y≡s(modq). Let P be the subset of R consisting of the values of f on [1,(p⁢q-1)/2]. P contains, say, u elements of the form (x,0) such that x<0, and v elements of the form (0,y) with y<0. Intending to apply Gauss’s lemma, we seek some kind of comparison between u and v.

We define three subsets of P by

R0 = {(x,y)∈P|x>0,y>0}
R1 = {(x,y)∈P|x<0,y≥0}
R2 = {(x,y)∈P|x≥0,y<0}

and we let Ni be the cardinal of Ri for each i.

P has a⁢b+b elements in the region y>0, namely f⁢(m) for all m of the form k+l⁢q with 1≤k≤b and 0≤l≤a. Thus

N0+N1=a⁢b+b-(b-v)+u

i.e.

N0+N1 = a⁢b+u+v. (1)

Swapping p and q, we have likewise

N0+N2 = a⁢b+u+v. (2)

Furthermore, for any s∈S, if f⁢(s)=(x,y) then f⁢(-s)=(-x,-y). It follows that for any (x,y)∈R other than (0,0), either (x,y) or (-x,-y) is in P, but not both. Therefore

N1+N2 = a⁢b+u+v. (3)

Adding (1), (2), and (3) gives us

0≡a⁢b+u+v(mod2)

so

(-1)a⁢b=(-1)u⁢(-1)v

which, in view of Gauss’s lemma, is the desired conclusionMathworldPlanetmath.

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