Квадраттық өзаралық заңның дәлелдемелерінің алуан түрлілігі

In number theory, the law of quadratic reciprocity, like the Pythagorean theorem, has lent itself to an unusually large number of proofs. Several hundred proofs of the law of quadratic reciprocity have been published.

Of the elementary combinatorial proofs, there are two which apply types of double counting. One by Gotthold Eisenstein counts lattice points. Another applies Zolotarev's lemma to , expressed by the Chinese remainder theorem as and calculates the signature of a permutation. The shortest known proof also uses a simplified version of double counting, namely double counting modulo a fixed prime.

The proof of Quadratic Reciprocity using Gauss sums is one of the more common and classic proofs. These proofs work by comparing computations of single values in two different ways, one using Euler's Criterion and the other using the Binomial theorem. As an example of how Euler's criterion is used, we can use it to give a quick proof of the first supplemental case of determining for an odd prime p: By Euler's criterion , but since both sides of the equivalence are ±1 and p is odd, we can deduce that .

Let , a primitive 8th root of unity and set Since and we see that Because is an algebraic integer, if p is an odd prime it makes sense to talk about it modulo p. (Formally we are considering the commutative ring formed by factoring the algebraic integers with the ideal generated by p. Because is not an algebraic integer, 1, 2, , p are distinct elements of .) Using Euler's criterion, it follows that We can then say that But we can also compute using the binomial theorem. Because the cross terms in the binomial expansion all contain factors of p, we find that We can evaluate this more exactly by breaking this up into two cases

These are the only options for a prime modulo 8 and both of these cases can be computed using the exponential form We can write this succinctly for all odd primes p as Combining these two expressions for and multiplying through by we find that Since both and are ±1 and 2 is invertible modulo p, we can conclude that

The idea for the general proof follows the above supplemental case: Find an algebraic integer that somehow encodes the Legendre symbols for p, then find a relationship between Legendre symbols by computing the qth power of this algebraic integer modulo q in two different ways, one using Euler's criterion the other using the binomial theorem. Let where is a primitive pth root of unity. This is a quadratic Gauss sum. A fundamental property of these Gauss sums is that where To put this in context of the next proof, the individual elements of the Gauss sum are in the cyclotomic field but the above formula shows that the sum itself is a generator of the unique quadratic field contained in L. Again, since the quadratic Gauss sum is an algebraic integer, we can use modular arithmetic with it. Using this fundamental formula and Euler's criterion we find thatThereforeUsing the binomial theorem, we also find that , If we let a be a multiplicative inverse of , then we can rewrite this sum as using the substitution , which doesn't affect the range of the sum. Since , we can then writeUsing these two expressions for , and multiplying through by givesSince is invertible modulo q, and the Legendre symbols are either ±1, we can then conclude that

The proof presented here is by no means the simplest known; however, it is quite a deep one, in the sense that it motivates some of the ideas of Artin reciprocity.

Suppose that p is an odd prime. The action takes place inside the cyclotomic field

where ζp is a primitive pth root of unity. The basic theory of cyclotomic fields informs us that there is a canonical isomorphism

which sends the automorphism σa satisfying to the element In particular, this isomorphism is injective because the multiplicative group of a field is a cyclic group:
Now consider the subgroup H of squares of elements of G. Since G is cyclic, H has index 2 in G, so the subfield corresponding to H under the Galois correspondence must be a quadratic extension of Q. (In fact it is the unique quadratic extension of Q contained in L.) The Gaussian period theory determines which one; it turns out to be , where

At this point we start to see a hint of quadratic reciprocity emerging from our framework. On one hand, the image of H in consists precisely of the (nonzero) quadratic residues modulo p. On the other hand, H is related to an attempt to take the square root of p (or possibly of −p). In other words, if now q is a prime (different from p), we have shown that