The n-Category Café

John, I didn’t attempt to work through the details of the proof you sketched for your course, but have you looked at the postscript in John H. Conway’s The Sensual (Quadratic) Form? There seems to some sort of family resemblance between the elegant proof of quadratic reciprocity given by Conway and the Eisenstein proof, but you may find the Conway proof much more illuminating.

He starts off very sensibly by giving Zolotarev’s definition of the Jacobi symbol: If $a$ is prime to $n$, the symbol $(a/n)$ is defined to be the sign of the permutation of multiplying by $a$ on the set $\mathbb{Z}$ mod $n$. He comments, “This definition leads to an extremely simple proof of the quadratic reciprocity theorem. It is remarkable that this proof does not use either the notion of prime number, or even that of square number. We shall however use he fact that the sign of a permutation is multiplicative.”

Of course, one has to relate the Zolotarev definition of $(a/p)$ to the usual definition, but that’s not hard (just uses the fact that the multiplicative group of $\mathbb{Z}$ mod $p$ is cyclic).

Hi, here some articles putting it into an interesting geometrical context:

Hill, Richard
A geometric proof of a reciprocity law.
Nagoya Math. J. 137 (1995), 77–144.
‘… the author uses singular homology groups and homotopy theory’

Grant, David
Geometric proofs of reciprocity laws.
J. Reine Angew. Math. 586 (2005), 91–124.
‘This beautiful and self-contained paper contributes to our understanding of the relationship between power reciprocity laws and the geometry of abelian varieties.’

Ash, Avner; Gross, Robert
Generalized non-abelian reciprocity laws: a context for Wiles’ proof.
Bull. London Math. Soc. 32 (2000), no. 4, 385–397.

Not in the library here, but its title sounds very interesting:
Kubota, Tomio
A foundation of class field theory applying properties of spatial figures
Sugaku Expositions 8 (1995), no. 1, 1–16.

This is basically just a fleshing out of what Noah said above. Let q be an integer which is 1 modulo 4. The essence of quadratic reciprocity is that (q on p) is periodic with period q as a function of p. The fact that this periodic function has the specific form given by quadratic reciprocity is not so hard given that it is periodic; one first shows that (q on p) is a character of (Z/q)^* (Conway does this very nicely, IIRC) and then notes that there aren’t a lot of options.

So, how to prove periodicity? Both the elementary proof you are considering and the cyclotomic proofs Noah mentions start in the same place with Euler’s criterion – (q on p)=q^{(p-1)/2} modulo p. I like to understand this through Galois theory of finite fields. Working in K, an algebraic extension of $F_p$ where $x^2-q$ splits, we know that $\sqrt(q)^p$ is a Galois conjugate of $\sqrt(q)$ and is hence $\pm \sqrt(q)$. Moreover, q is a square in $F_p$ if and only if $\sqrt(q)$ is fixed by $Gal(K/F_p)$, which is generated by the Frobenius, we see that (q on p) is $\sqrt(q)^p/\sqrt(q)$, which can be rewritten as $q^{(p-1)/2}$.

In the elementary proof, $q^{(p-1)/2}$ is rewritten as $(q*1)(q*2) \cdots (q*(p-1)/2)/(1*2*\cdots (p-1)/2)$. One looks at each of the terms in the numerator to see it lies in {1,2,(p-1)/2}+p*Z or whether it lies in {-1,-2,…,-(p-1)/2}+p*Z. I’ll do the example q=-3. For k=1,2,…,[(p-1)/6], we have $q*k \in \{-1,-2,...,-(p-1)/2\}$ and for k=[(p+1)/3], …, (p-1)/2 we have $q*k \in \{-1,-2,...,-(p-1)/2\}-p$. So (up to boundary errors on my part) (q on p) depends only on the parity of $[(p-1)/2]-[(p+1)/3]+[(p-1)/6]$. It is fairly clear that this is periodic modulo $12$ and, if I didn’t screw up, the actual period is $3$. In general, you should be able to convince yourself that the (q on p) has period 4q and, if you are careful, get rid of the 4.

In the cyclotomic proofs, one tries to lift the Frobenius map to a Galois autmorphism in some characteristic zero setting. Specifically, one shows (using that q=1 mod 4) that $\sqrt(q)$ is contained in $L_q := Q[e^{2 \pi i/q}]$. Writing $\zeta=e^{2 \pi i/q}$, $L_q$ has a Galois automorphism $\sigma_p=\zeta \to \zeta^p$ which clearly only depends on what $p$ is modulo $q$. One then considers the ring of integers O inside $L_q$. There is a quotient of O which is a field $K$ as in our second paragraph and such that $\sigma_p$ descends to the Frobenius on this quotient. Since $\sigma_p$ only depended on p modulo q, we have the periodicity again.

Just an advertisement – I have a post on the secret blogging seminar where I explain what the finite field version of QR says, and how to think about it geometrically.

John,

Here is a sketch of the proof of quadratic reciprocity given in algebraic number theory books nowadays (I think essentially that given by David Speyer). In his lectures on algebraic number theory Hermann Weyl says “There certainly exist more elementary proofs of the reciprocity law, but hardly one that is less artificial and goes straight to the root of the phenomenon.”

Let p and q be primes and let
q* be defined by
q* = q (for q = 1 (mod 4)
q* = -q (for q = 3 (mod 4)

Let w be a primitive q’th root of unity.

Then the following statements are true:
1) 1 = (q*/p) p splits in Q(sqrt(q*))
(This is easy)

2) p splits in Q(sqrt(q*)) p splits into an even number of ideals in Q(w)

(This step requires Galois theory. The basic point is that Q(sqrt(q*)) is a subfield of Q(w) and that if p splits in Q(sqrt(q*)) then each of the two prime factors of p splits into an equal number of factors when we move to Q(w) and that conversely if p splits into an even number of factors in Q(w) then they can be split into two groups that separately multiply to two factors in Q(sqrt(q*)).

3)The number of ideals in Q(w) that p splits into is given by
r = (q - 1)/f

where f is the minimum positive integer such that

p^f = 1 (mod q)

(Deriving this formula requires algebraic number theory)

Once we have these three facts we see that (q*/p) = 1 iff r is even but by our formula this is so iff f | (q - 1)/2)

this is so iff

p^[(q - 1)/2] = 1 (mod q)

which is (by Euler criterion) the same as (p/q) = 1.

——————-

I hope the above was sufficiently concise and not too technical. I don’t know whether or not there’s a known relationship between this proof and Eisenstein’s lattice point counting proof.

hi,
i know there is like about 227 proofs of the law of quadratic reciprocity,
but i was wondering can anyone tell me the trigonometric proof for the law of quadratic reciprocity?
i cant seems to find any website or online resources for that.
many many thanks.

Dear John, One can perhaps explain the elementary proofs of quadratic reciprocity conceptually by using the following result.

Proposition. If a cyclic group C of odd order acts freely on a finite set X, and if f is a permutation of X commuting with the action of C, then the sign of f is the same as the sign of the induced permutation on the quotient set X/C.

In particular, let p and q be distinct odd primes, let X be the set of integers between -pq/2 and pq/2, let C be a cyclic group of order q acting on X so that its generator sends k to k+p mod pq, and let f be the permutation of X such that

f(i+pj)=qi+pj mod pq

for i between -p/2 and p/2 and for j between -q/2 and q/2. The induced permutation on X/C is essentially multiplication by q on the ring of integers modulo p, so its sign is the Legendre symbol (q/p) as interpreted by Zolotarev. By the Proposition, (q/p) is also the sign of f.

Similarly (p/q) is the sign of the permutation g of X given by

g(qi+j)=qi+pj mod pq

for i between -p/2 and p/2 and for j between -q/2 and q/2.

The Legendre symbols (q/p) and (p/q) have now been interpreted in a conceptual way as the signs of two permutations of the same set X, so their ratio can be naturally computed as the sign of the composite c=hf, where h is the inverse of g; thus

c(i+pj)=qi+j

for i between -p/2 and p/2 and for j between -q/2 and q/2. One has c(-k) = -c(k) for all k in X, so the sign of c is -1 raised to the power m, where m is the number of positive integers k in X such that c(k) is negative. It turns out that

m = (p-1)(q-1)/4,

so this gives one quadratic reciprocity.

Excuse a pure amateur who is trying to understand QR for butting in, but isn’t the following statement, which I have highlighted from the opening page of this discussion, incorrect?

I cannot paste it. It says if p and q are both 4n+3 type primes, then they have opposite answers, in other words are not mutually quadratic residues in each other’s set.

But 19 and 23 when squared are both equal to 16 mod the other one.

Please explain where I have gone wrong, or repair your site.

Respectfully,

Jesse Baird