Proofs of quadratic reciprocity


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

Proofs that are accessible

Of relatively 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.

Eisenstein's proof

Eisenstein's proof of quadratic reciprocity is a simplification of Gauss's third proof. It is more geometrically intuitive and requires less technical manipulation.
The point of departure is "Eisenstein's lemma", which states that for distinct odd primes p, q,
where denotes the floor function, and where the sum is taken over the even integers u = 2, 4, 6,..., p−1. For example,
This result is very similar to Gauss's lemma, and can be proved in a similar fashion.
Using this representation of, the main argument is quite elegant. The sum counts the number of lattice points with even x-coordinate in the interior of the triangle ABC in the following diagram:
Because each column has an even number of points, the number of such lattice points in the region BCYX is the same modulo 2 as the number of such points in the region CZY:
Then by flipping the diagram in both axes, we see that the number of points with even x-coordinate inside CZY is the same as the number of points inside AXY having odd x-coordinates:
The conclusion is that
where μ is the total number of lattice points in the interior of AYX. Switching p and q, the same argument shows that
where ν is the number of lattice points in the interior of WYA. Since there are no lattice points on the line AY itself, and since the total number of points in the rectangle WYXA is
we obtain finally

Proof of Eisenstein's lemma

For an even integer u in the range 1 ≤ u ≤ p−1, denote by r the least positive residue of qu modulo p. The numbers rr, again treated as least positive residues modulo p, are all even Furthermore, they are all distinct, because if rr ≡ rr, then we may divide out by q to obtain u ≡ ±t. This forces u ≡ t, because both u and t are even, whereas p is odd. Since there exactly /2 of them and they are distinct, they must be simply a rearrangement of the even integers 2, 4,..., p−1. Multiplying them together, we obtain
Dividing out successively by 2, 4,..., p−1 on both sides and rearranging, we have
On the other hand, by the definition of r and the floor function,
and so since p is odd and u is even, we see that and r are congruent modulo 2. Finally this shows that
We are finished because the left hand side is just an alternative expression for.

Proof using Quadratic Gauss Sums

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.

The Second Supplemental Case

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. 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 general case

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

Proof using algebraic number theory

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.

Cyclotomic field setup

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. 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 quadratic residues modulo p. On the other hand, H is related to an attempt to take the square root of p. In other words, if now q is a prime, we have shown that

The Frobenius automorphism

In the ring of integers, choose any unramified prime ideal β of lying over q, and let be the Frobenius automorphism associated to β; the characteristic property of is that
The key fact about that we need is that for any subfield K of L,
Indeed, let δ be any ideal of OK below β. Then, since for any, we see that is a Frobenius for δ. A standard result concerning is that its order is equal to the corresponding inertial degree; that is,
The left hand side is equal to 1 if and only if φ fixes K, and the right hand side is equal to one if and only q splits completely in K, so we are done.
Now, since the pth roots of unity are distinct modulo β, we must have
that is, coincides with the automorphism σq defined earlier. Taking K to be the quadratic field in which we are interested, we obtain the equivalence

Completing the proof

Finally we must show that
Once we have done this, the law of quadratic reciprocity falls out immediately since
and
for.
To show the last equivalence, suppose first that In this case, there is some integer x such that say for some integer c. Let
and consider the ideal of K. It certainly divides the principal ideal. It cannot be equal to, since is not divisible by q. It cannot be the unit ideal, because then
is divisible by q, which is again impossible. Therefore must split in K.
Conversely, suppose that splits, and let β be a prime of K above q. Then so we may choose some
Actually, since elementary theory of quadratic fields implies that the ring of integers of K is precisely so the denominators of a and b are at worst equal to 2. Since q ≠ 2, we may safely multiply a and b by 2, and assume that where now a and b are in Z. In this case we have
so However, q cannot divide b, since then also q divides a, which contradicts our choice of Therefore, we may divide by b modulo q, to obtain as desired.