Proofs of quadratic reciprocity

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 Z/pqZ expressed by the Chinese remainder theorem as Z/pZ×Z/qZ, 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 ⌊ x ⌋ denotes the floor function (the largest integer less than or equal to x), 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 (proof given below).

Using this representation of (q/p), the main argument is quite elegant. The sum Σ u ⌊ q u / p ⌋ 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 (namely q−1 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 (because p and q are relatively prime), 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(u) the least positive residue of qu modulo p. (For example, for p = 11, q = 7, we allow u = 2, 4, 6, 8, 10, and the corresponding values of r(u) are 3, 6, 9, 1, 4.) The numbers (−1)^r(u)r(u), again treated as least positive residues modulo p, are all even (in our running example, they are 8, 6, 2, 10, 4.) Furthermore, they are all distinct, because if (−1)^r(u)r(u) ≡ (−1)^r(t)r(t) (mod p), then we may divide out by q to obtain u ≡ ±t (mod p). This forces u ≡ t (mod p), because both u and t are even, whereas p is odd. Since there exactly (p−1)/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 (which is permissible since none of them are divisible by p) and rearranging, we have

On the other hand, by the definition of r(u) and the floor function,

and so since p is odd and u is even, we see that ⌊ q u / p ⌋ and r(u) are congruent modulo 2. Finally this shows that

We are finished because the left hand side is just an alternative expression for (q/p).

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 ( − 1 p ) for an odd prime p: By Euler's criterion ( − 1 p ) ≡ ( − 1 ) p − 1 2 ( mod p ) , but since both sides of the equivalence are ±1 and p is odd, we can deduce that ( − 1 p ) = ( − 1 ) p − 1 2 .

The Second Supplemental Case

Let ζ 8 = e 2 π i / 8 , a primitive 8th root of unity and set τ = ζ 8 + ζ 8 − 1 . Since ζ 8 2 = i and ζ 8 − 2 = − i we see that τ 2 = 2 . Because τ 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 A with the ideal generated by p. Because p − 1 is not an algebraic integer, 1, 2, ..., p are distinct elements of A / p A .) Using Euler's criterion, it follows that

These are the only options for a prime modulo 8 and both of these cases can be computed using the exponential form ζ 8 = e 2 π i 8 . We can write this succinctly for all odd primes p as

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

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 ζ_p is a primitive p^th root of unity. The basic theory of cyclotomic fields informs us that there is a canonical isomorphism

which sends the automorphism σ_a satisfying

to the element

(This is because the morphism of reduction from Z to Z/qZ is injective on the set of p-th roots of unity)

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 an prime (different from p), we have so far shown that

The Frobenius automorphism

Choose any prime ideal β of the ring of integers O_L lying over q, which is unramified, and let

be the Frobenius automorphism associated to β; the characteristic property of ϕ is that

for any x in O_L. (The existence of such a Frobenius element depends on quite a bit of algebraic number theory machinery.)

The key fact about ϕ that we need is that for any subfield K of L,

Indeed, let δ be any ideal of O_K below β (and hence above q). Then, since

for any x in O_K, we see that

is a Frobenius for δ. A standard result concerning ϕ 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 p^th roots of unity are distinct modulo β (i.e. the polynomial X^p − 1 is separable in characteristic q), we must have

that is, ϕ coincides with the automorphism σ_q 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

if p ≡ 1 (mod 4), and

if p ≡ -1 (mod 4).

To show the last equivalence, suppose first that

In this case, there is some integer x (not divisible by q) such that

say

for some integer c. Let

and consider the ideal

of K. It certainly divides the principal ideal (q). It cannot be equal to (q), since

is not divisible by q. It cannot be the unit ideal, because then

is divisible by q, which is again impossible. Therefore (q) must split in K.

Conversely, suppose that (q) splits, and let β be a prime of K above q. Then

so we may choose some

where a and b are in Q. 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.