Proofs of quadratic reciprocity

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In the mathematical field of 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.

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[edit] Proofs that are accessible

Of relatively elementary, combinatorial proofs, there are two which apply types of double counting. One by Ferdinand 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.

[edit] 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,

\left(\frac qp\right) = (-1)^{\sum_u \left \lfloor qu/p \right \rfloor},

where \left \lfloor x \right \rfloor 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,

\left(\frac 7{11}\right) = (-1)^{ \left \lfloor 14/11 \right \rfloor + \left \lfloor 28/11 \right \rfloor + \left \lfloor 42/11 \right \rfloor + \left \lfloor 56/11 \right \rfloor + \left \lfloor 70/11 \right \rfloor } = (-1)^{1 + 2 + 3 + 5 + 6} = (-1)^{17} = -1.

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 \Sigma_u \left \lfloor qu/p \right \rfloor counts the number of lattice points with even x-coordinate in the interior of the triangle ABC in the following diagram:

Lattice point diagram
Lattice point diagram
Example showing lattice points inside ABC with even x-coordinates, for p = 11 and q = 7
Example showing lattice points inside ABC with even x-coordinates, for p = 11 and q = 7

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:

The number of points with even x-coordinate inside BCYX (marked by O's) is equal modulo 2 to the number of such points in CZY (marked by X's)
The number of points with even x-coordinate inside BCYX (marked by O's) is equal modulo 2 to the number of such points in CZY (marked by X's)

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 number of points with even x-coordinate inside CZY is equal to the number of points with odd x-coordinate inside AXY
The number of points with even x-coordinate inside CZY is equal to the number of points with odd x-coordinate inside AXY

The conclusion is that

\left(\frac qp\right) = (-1)^\mu,

where μ is the total number of lattice points in the interior of AYX. Switching p and q, the same argument shows that

\left(\frac pq\right) = (-1)^\nu,

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

\left(\frac{p-1}2\right) \left(\frac{q-1}2\right),

we obtain finally

\left(\frac qp\right) \left(\frac pq\right) = (-1)^{\mu + \nu} = (-1)^{(p-1)(q-1)/4}.

[edit] 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

(-1)^{r(2)}2q \cdot (-1)^{r(4)}4q \cdot \cdots \cdot (-1)^{r(p-1)}(p-1)q \equiv 2 \cdot 4 \cdot \cdots \cdot (p-1) \pmod p.

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

q^{(p-1)/2} \equiv (-1)^{r(2) + r(4) + \cdots + r(p-1)} \pmod p.

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

\frac{qu}p = \left \lfloor \frac{qu}p\right \rfloor + \frac{r(u)}p,

and so since p is odd and u is even, we see that \left \lfloor qu/p \right \rfloor and r(u) are congruent modulo 2. Finally this shows that

q^{(p-1)/2} \equiv (-1)^{\sum_u \left \lfloor qu/p \right \rfloor} \pmod p.

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

[edit] 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.

[edit] Cyclotomic field setup

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

L = \mathbf Q(\zeta_p),

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

G = \operatorname{Gal}(L/\mathbf Q) \cong (\Z/p\Z)^\times,

which sends the automorphism σa satisfying

\sigma_a(\zeta_p) = \zeta_p^a

to the element

a \in (\Z/p\Z)^\times.

(This depends essentially on the irreducibility of the cyclotomic polynomial Φp(X), together with a little Galois theory.)

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

\mathbf Q(\sqrt{p^*}),

where

p^* = \begin{cases} p & \mbox{if } p = 1 \pmod 4, \\ -p & \mbox{if } p = 3 \pmod 4. \end{cases}

At this point we start to see a hint of quadratic reciprocity emerging from our framework. On one hand, the image of H in

(\mathbf Z/p\mathbf Z)^\times

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

\left(\frac qp\right) =1 \quad \iff \quad \sigma_q \in H \quad \iff \quad \sigma_q \mbox{ fixes } \mathbf Q(\sqrt{p^*}).

[edit] Enter the Frobenius

Choose any prime ideal β of the ring of integers OL lying over q, and let

\phi \in \operatorname{Gal}(L/\mathbf Q)

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

\phi(x) \equiv x^q \pmod \beta \,\!

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

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

\phi \mbox{ fixes } K \quad \iff \quad q \mbox{ splits completely in } K.

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

\phi(x) \equiv x^q \pmod \delta \,\!

for any x in OK, we see that

\phi\vert_K \in \operatorname{Gal}(K/\mathbf Q)

is a Frobenius for δ. A standard result concerning the Frobenius is that its order is equal to the corresponding inertial degree; that is,

\operatorname{ord}(\phi\vert_K) = [O_K/\delta O_K : \mathbf Z/q\mathbf Z].

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 (i.e. the polynomial Xp − 1 is separable in characteristic q), we must have

\phi(\zeta_p) = \zeta_p^q;

that is, the Frobenius coincides with the automorphism σq defined earlier. Taking K to be the quadratic field in which we are interested, we obtain the equivalence

\left(\frac qp\right) =1 \quad \iff \quad q \mbox{ splits completely in } \mathbf Q(\sqrt{p^*}).

[edit] Completing the proof

Finally we must show that

q \mbox{ splits completely in } \mathbf Q(\sqrt{p^*}) \quad \iff \quad \left(\frac{p^*}q\right) = 1.

Once we have done this, the law of quadratic reciprocity falls out immediately since

\left(\frac{p^*}q\right) = \left(\frac pq\right)

if p = 1 mod 4, and

\left(\frac{p^*}q\right) = \left(\frac{-p}q\right) = \left(\frac{-1}q\right)\left(\frac pq\right) = \begin{cases} +\left(\frac pq \right) & \mbox{if } q = 1 \pmod 4, \\ -\left(\frac pq\right) & \mbox{if } q = 3 \pmod 4\end{cases}

if p = 3 mod 4.

To show the last equivalence, suppose first that

\left(\frac{p^*}q\right) = 1.

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

x^2 \equiv p^* \pmod q, \,\!

say

x^2 - p^* = cq \,\!

for some integer c. Let

K = \mathbf Q(\sqrt{p^*}),

and consider the ideal

(x-\sqrt{p^*},q)

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

x-\sqrt{p^*}

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

(x+\sqrt{p^*}) = (x+\sqrt{p^*})(x-\sqrt{p^*},q) = (cq, q(x+\sqrt{p^*}))

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

(q) \subsetneq \beta,

so we may choose some

a+b\sqrt{p^*} \in \beta\setminus(q),

where a and b are in Q. Actually, since

p * = 1(mod 4),

elementary theory of quadratic fields implies that the ring of integers of K is precisely

\mathbf Z\left[\frac{1+\sqrt{p^*}}2\right],

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

a+b\sqrt{p^*} \in \beta\setminus(q),

where now a and b are in Z. In this case we have

(a+b\sqrt{p^*})(a-b\sqrt{p^*}) = a^2 - b^2p^* \in I \cap \mathbf Z = (q),

so

q \mid a^2 - b^2p^*.\,\!

However, q cannot divide b, since then also q divides a, which contradicts our choice of

a+b\sqrt{p^*}.

Therefore, we may divide by b modulo q, to obtain

p^* = (ab^{-1})^2 \pmod q\,\!

as desired.

[edit] References