Proofs of quadratic reciprocity

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.

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.

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",: $left(frac\; qp\; ight)\; =\; (-1)^\{sum\_u\; left\; lfloor\; qu/p\; ight\; floor\},$where $left\; lfloor\; x\; ight\; floor$ 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,: $left(frac\; 7\{11\}\; ight)\; =\; (-1)^\{\; left\; lfloor\; 14/11\; ight\; floor\; +\; left\; lfloor\; 28/11\; ight\; floor\; +\; left\; lfloor\; 42/11\; ight\; floor\; +\; left\; lfloor\; 56/11\; ight\; floor\; +\; left\; lfloor\; 70/11\; ight\; floor\; \}\; =\; (-1)^\{1\; +\; 2\; +\; 3\; +\; 5\; +\; 6\}\; =\; (-1)^\{17\}\; =\; -1.$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 $Sigma\_u\; left\; lfloor\; qu/p\; ight\; floor$ 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: $left(frac\; qp\; ight)\; =\; (-1)^mu,$where μ is the "total" number of lattice points in the interior of AYX. Switching "p" and "q", the same argument shows that: $left(frac\; pq\; ight)\; =\; (-1)^\; u,$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: $left(frac\{p-1\}2\; ight)\; left(frac\{q-1\}2\; ight),$we obtain finally: $left(frac\; qp\; ight)\; left(frac\; pq\; ight)\; =\; (-1)^\{mu\; +\; u\}\; =\; (-1)^\{(p-1)(q-1)/4\}.$

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,: $frac\{qu\}p\; =\; left\; lfloor\; frac\{qu\}p\; ight\; floor\; +\; frac\{r(u)\}p,$and so since "p" is odd and "u" is even, we see that $left\; lfloor\; qu/p\; ight\; floor$ and "r"("u") are congruent modulo 2. Finally this shows that: $q^\{(p-1)/2\}\; equiv\; (-1)^\{sum\_u\; left\; lfloor\; qu/p\; ight\; floor\}\; pmod\; p.$We are finished because the left hand side is just an alternative expression for ("q"/"p").

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:$L\; =\; mathbf\; Q(zeta\_p),$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:$sigma\_a(zeta\_p)\; =\; zeta\_p^a$to the element:

(This depends essentially on the irreducibility of the cyclotomic polynomial Φ_p("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:

At this point we start to see a hint of quadratic reciprocity emerging from our framework. On one hand, the image of "H" in:$(mathbf\; Z/pmathbf\; Z)^\; imes$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:$left(frac\; qp\; ight)\; =1\; quad\; iff\; quad\; sigma\_q\; in\; H\; quad\; iff\; quad\; sigma\_q\; mbox\{\; fixes\; \}\; mathbf\; Q(sqrt\{p^*\}).$

Enter the Frobenius

Choose any prime ideal β of the ring of integers "O"_"L" 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: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 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 "O"_"K" below β (and hence above "q"). Then, since:$phi(x)\; equiv\; x^q\; pmod\; delta\; ,!$for any "x" in "O"_"K", we see that:$phivert\_K\; in\; operatorname\{Gal\}(K/mathbf\; Q)$is a Frobenius for δ. A standard result concerning the Frobenius is that its order is equal to the corresponding inertial degree; that is,:$operatorname\{ord\}(phivert\_K)\; =\; [O\_K/delta\; O\_K\; :\; mathbf\; Z/qmathbf\; Z]\; .$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 "p"^{th roots of unity are distinct modulo β (i.e. the polynomial "X"p − 1 is separable in characteristic "q"), we must have:$phi(zeta\_p)\; =\; zeta\_p^q;$that is, the Frobenius coincides with the automorphism σ_"q" defined earlier. Taking "K" to be the quadratic field in which we are interested, we obtain the equivalence:$left(frac\; qp\; ight)\; =1\; quad\; iff\; quad\; q\; mbox\{\; splits\; completely\; in\; \}\; mathbf\; Q(sqrt\{p^*\}).$}

Completing the proof

Finally we must show that:$q\; mbox\{\; splits\; completely\; in\; \}\; mathbf\; Q(sqrt\{p^*\})\; quad\; iff\; quad\; left(frac\{p^*\}q\; ight)\; =\; 1.$Once we have done this, the law of quadratic reciprocity falls out immediately since:$left(frac\{p^*\}q\; ight)\; =\; left(frac\; pq\; ight)$if "p" = 1 mod 4, and:if "p" = 3 mod 4.

To show the last equivalence, suppose first that:$left(frac\{p^*\}q\; ight)\; =\; 1.$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:so we may choose some:where "a" and "b" are in Q. Actually, since:$p^*\; =\; 1\; pmod\; 4,$elementary theory of quadratic fields implies that the ring of integers of "K" is precisely:$mathbf\; Zleft\; [frac\{1+sqrt\{p^*2\; ight]\; ,$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:$(a+bsqrt\{p^*\})(a-bsqrt\{p^*\})\; =\; a^2\; -\; b^2p^*\; in\; I\; cap\; mathbf\; Z\; =\; (q),$so:$q\; mid\; a^2\; -\; b^2p^*.,!$However, "q" cannot divide "b", since then also "q" divides "a", which contradicts our choice of:$a+bsqrt\{p^*\}.$Therefore, we may divide by "b" modulo "q", to obtain:$p^*\; =\; (ab^\{-1\})^2\; pmod\; q,!$as desired.

References

* [http://www.rzuser.uni-heidelberg.de/~hb3/fchrono.html Chronology of Proofs of the Quadratic Reciprocity Law] (221 proofs!)
* G. Rousseau. "On the Quadratic Reciprocity Law", "J. Austral. Math. Soc. (Series A)", v51, 1991, 423–425. ( [http://anziamj.austms.org.au/JAMSA/V51/Part3/Rousseau.html online] )
* L. Washington. "Introduction to Cyclotomic Fields", 2nd ed.