In the previous post we showed that the splitting behavior of a rational prime in the ring of cyclotomic integers depends only on the residue class of . This is suggestive enough of quadratic reciprocity that now would be a good time to give a full proof.
The key result is the following fundamental observation.
Proposition: Let be an odd prime. Then contains .
Quadratic reciprocity has a function field version over finite fields which David Speyer explains the geometric meaning of in an old post. While this is very much in line with what we’ve been talking about, it’s a little over my head, so I’ll leave it for the interested reader to peruse.
Proof of QR assuming the proposition
The point of the proposition is that it implies that the action of the Frobenius map on in depends only on its action on roots of unity, hence only on the residue class of as in the cyclotomic case. Before we discuss the proposition, let’s show how it implies QR without much extra effort.
First we dispense of the supplements. is just Euler’s criterion. To determine , we compute explicitly that , hence
If is an odd prime, the above equation still makes sense in the splitting field of over , so the action of the Frobenius map on is determined by its action on . In particular, it fixes if and only if either or , i.e. if and only if . This is usually written
Now let be an odd prime. Knowing only that , it follows that any element of either fixes or sends it to its negative, hence is precisely the fixed field of the subgroup of quadratic residues.
If is an odd prime relatively prime to , then the Frobenius map acts on the splitting field of , in which also splits. But according to what we just said above, the Frobenius map will fix if and only if is a square in . Thus
This is equivalent to the usual statement of quadratic reciprocity.
Proof 1 of the proposition
From the right perspective the proposition is “almost” obvious. If is odd, has a subgroup of index given by the quadratic residues, and the fixed field of this subgroup must therefore be some quadratic extension of called the quadratic subfield of . Since we know that is the only prime that ramifies in , it must be the only prime that ramifies in the quadratic subfield, so the quadratic subfield must be either or . The only question is, which?
Edit, 1/26/10: As it turns out, what I said above about ramification is actually enough to determine the sign under the square root, but I haven’t yet given the correct definition of ramification with respect to a number field. With the correct definition, which I will give eventually, is unramified at if and only if – this is precisely when exists in – and this is enough to determine the sign without any computations!
One elementary argument, which I learned from Ireland and Rosen, is simply to write
Since is odd, we can write the above as and then pair up factors as follows: , which gives
and the result follows. (The elements are all units in , so the factorization alone is already enough to conclude that is a unit times a square.)
Proof 2 of the proposition
Another classical approach is to write down the fixed field of explicitly. It’s not hard to see that every element of or is a linear combination of
the quadratic Gauss sums, where is some quadratic non-residue in . Now, is equal to its complex conjugate (hence real) if and only if is a quadratic residue , hence if and only if ; otherwise, is equal to the negative of its complex conjugate (hence complex). Since , it’s therefore reasonable to conjecture that . Since we know that the signs of all these expressions match by the above argument, it’s enough to show that their absolute values match, too.
Here’s a Fourier-theoretic approach. The function
is a function of period in which behaves as follows: if is a quadratic residue, and if is a quadratic non-residue. We can therefore compute the Fourier coefficient of in as
Specializing to and rearranging terms, this gives
which gives the desired result. Incidentally, although this argument tells us that , it doesn’t tell us which square root is which. This is harder, as David Speyer explains. The identity necessary to determine the sign is essentially what connects this proof to the first proof.
Remark 1: The discrete Fourier transform matrix of order satisfies , so its eigenvalues are of the form and its trace is precisely . One way to interpret the computations we did above is that Gauss sums are essentially eigenfunctions of the discrete Fourier transform, and one of the eigenfunctions of the usual Fourier transform on is none other than the normal distribution , whose integral bears an uncanny resemblance to a Gauss sum. What this all means I have no idea, but I asked about it on Math Overflow.
Remark 2: If we suppose that quadratic residues behave more or less randomly, should behave like a random walk after unit steps, which heuristically should be somewhere around a distance of units from the origin. Isn’t it funny that the heuristic is exact in this case?
The geometric picture
There is a geometric way to interpret the statement of quadratic reciprocity going back at least to Hilbert which goes something like this: has a “compactification” given by adding a point whose residue field is (in some formal sense). There is a “meromorphic function” we can define on the resulting “compact Riemann surface” whose only “poles” are at , and quadratic reciprocity is equivalent to the statement that the sum of the “residues” at the “poles” of this “function” are equal to zero (which is true of any meromorphic function on a compact Riemann surface)!
Apparently people know how to make this rigorous, but I’m certainly not one of them. The function in question is related to the Hilbert symbol, which I might talk about in the future along with the mysterious point (or, as Conway calls it, ).