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In the previous post we showed that the splitting behavior of a rational prime $p$ in the ring of cyclotomic integers $\mathbb{Z}[\zeta_n]$ depends only on the residue class of $p \bmod n$. 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 $q$ be an odd prime. Then $\mathbb{Z}[\zeta_q]$ contains $\sqrt{ q^{*} } = \sqrt{ (-1)^{ \frac{q-1}{2} } q}$.

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 $x \mapsto x^p$ on $\sqrt{ q^{*} }$ in $\mathbb{F}_p$ depends only on its action on $q^{th}$ roots of unity, hence only on the residue class of $p \bmod q$ 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. $\left( \frac{-1}{p} \right) = (-1)^{ \frac{p-1}{2} }$ is just Euler’s criterion. To determine $\left( \frac{2}{p} \right)$, we compute explicitly that $\zeta_8 = \frac{1}{ \sqrt{2} } + \frac{i}{ \sqrt{2} }$, hence

$\sqrt{2} = \zeta_8 + \zeta_8^{-1}$.

If $p$ is an odd prime, the above equation still makes sense in the splitting field of $\Phi_8(x) = x^4 + 1$ over $\mathbb{F}_p$, so the action of the Frobenius map on $\sqrt{2}$ is determined by its action on $\zeta_8$. In particular, it fixes $\sqrt{2}$ if and only if either $\zeta_8^p \equiv \zeta_8 \bmod p$ or $\zeta_8^p \equiv \zeta_8^{-1} \bmod p$, i.e. if and only if $p \equiv \pm 1 \bmod 8$. This is usually written

$\displaystyle \left( \frac{2}{p} \right) = (-1)^{ \frac{p^2 - 1}{8} }$.

Now let $q$ be an odd prime. Knowing only that $\sqrt{ q^{*} } \in \mathbb{Z}[\zeta_q]$, it follows that any element of $\text{Gal}(\mathbb{Q}[\zeta_q]/\mathbb{Q}) \simeq (\mathbb{Z}/q\mathbb{Z})^{*}$ either fixes $\sqrt{ q^{*} }$ or sends it to its negative, hence $\mathbb{Q}[\sqrt{ q^{*} } ]$ is precisely the fixed field of the subgroup of quadratic residues.

If $p$ is an odd prime relatively prime to $q$, then the Frobenius map acts on the splitting field of $\Phi_q(x)$, in which $x^2 - q^{*}$ also splits. But according to what we just said above, the Frobenius map will fix $\sqrt{ q^{*} }$ if and only if $p$ is a square in $(\mathbb{Z}/q\mathbb{Z})^{*}$. Thus

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

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 $q$ is odd, $(\mathbb{Z}/q\mathbb{Z})^{*}$ has a subgroup $G$ of index $2$ given by the quadratic residues, and the fixed field of this subgroup must therefore be some quadratic extension of $\mathbb{Q}$ called the quadratic subfield of $\mathbb{Q}[\zeta_q]$. Since we know that $q$ is the only prime that ramifies in $\mathbb{Z}[\zeta_q]$, it must be the only prime that ramifies in the quadratic subfield, so the quadratic subfield must be either $\mathbb{Q}[ \sqrt{q} ]$ or $\mathbb{Q}[ \sqrt{-q} ]$. 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, $\mathbb{Q}[\sqrt{d}]$ is unramified at $2$ if and only if $d \equiv 1 \bmod 4$ – this is precisely when $\sqrt{d}$ exists in $\mathbb{Q}_2$ – 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

$\displaystyle q = \Phi_q(1) = \prod_{i=1}^{q-1} (1 - \zeta_q^i)$.

Since $q$ is odd, we can write the above as $\prod_{i=1}^{q-1} (1 - \zeta_q^{2i})$ and then pair up factors as follows: $(1 - \zeta_q^{2i})(1 - \zeta_q^{-2i}) = 2 - \zeta_q^{2i} - \zeta_q^{-2i} = - (1 - \zeta_q^i)^2$, which gives

$\displaystyle q = (-1)^{ \frac{q-1}{2} } \prod_{i=1}^{ \frac{q-1}{2} } \left( 1 - \zeta_q^i \right)^2$

and the result follows. (The elements $\frac{1 - \zeta_q^i}{1 - \zeta_q}$ are all units in $\mathbb{Z}[\zeta_q]$, so the factorization alone is already enough to conclude that $q$ is a unit times a square.)

Proof 2 of the proposition

Another classical approach is to write down the fixed field of $G$ explicitly. It’s not hard to see that every element of $\mathbb{Q}[\zeta_n]^G$ or $\mathbb{Z}[\zeta_n]^G$ is a linear combination of

$\displaystyle g_1(q) = \sum_{i=0}^{q-1} \zeta_q^{i^2}$.

and

$\displaystyle g_2(q) = \sum_{i=0}^{q-1} \zeta_q^{ai^2}$,

the quadratic Gauss sums, where $a$ is some quadratic non-residue in $(\mathbb{Z}/q\mathbb{Z})^{*}$. Now, $g_1(q)$ is equal to its complex conjugate (hence real) if and only if $-1$ is a quadratic residue $\bmod q$, hence if and only if $q \equiv 1 \bmod 4$; otherwise, $g_1(q)$ is equal to the negative of its complex conjugate (hence complex). Since $g_1(q) + g_2(q) = 0$, it’s therefore reasonable to conjecture that $g_1(q)^2 = g_2(q)^2 = -g_1 g_2 = q^{*}$. 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

$\displaystyle g(q, a) = \sum_{i=0}^{q-1} \zeta_q^{ai^2}$

is a function of period $q$ in $a$ which behaves as follows: $g(q, 0) = q, g(q, a) = g_1(q)$ if $a$ is a quadratic residue, and $g(q, a) = g_2(q)$ if $a$ is a quadratic non-residue. We can therefore compute the Fourier coefficient of $\zeta_q^k$ in $g(q, a)$ as

$\displaystyle \left( \frac{k}{q} \right) + 1 = \frac{1}{q} \sum_{a=0}^{q-1} g(q, a) \zeta_q^{-ka}$.

Specializing to $k = 1$ and rearranging terms, this gives

$\displaystyle q = g_1(q) \left( \frac{\overline{g_1(q)} - 1}{2} \right) + g_2(q) \left( \frac{\overline{g_2(q)} - 1}{2} \right) = \frac{|g_1(q)|^2 + |g_2(q)|^2}{2}$

which gives the desired result. Incidentally, although this argument tells us that $(x - g_1(q))(x - g_2(q)) = x^2 - q^{*}$, 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 $\mathbf{D}_q$ of order $q$ satisfies $\mathbf{D}_q^4 = q^2 \mathbf{I}_q$, so its eigenvalues are of the form $\pm \sqrt{\pm q}$ and its trace is precisely $g_1(q)$. 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 $\mathbb{R}$ is none other than the normal distribution $e^{-x^2}$, 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, $g_1(q)$ should behave like a random walk after $q$ unit steps, which heuristically should be somewhere around a distance of $\sqrt{q}$ 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: $\text{Spec } \mathbb{Z}$ has a “compactification” given by adding a point $\infty$ whose residue field is $\mathbb{R}$ (in some formal sense). There is a “meromorphic function” we can define on the resulting “compact Riemann surface” whose only “poles” are at $2, p, q, \infty$, 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 $\infty$ (or, as Conway calls it, $-1$).

### 2 Responses

but here is what I really want to know.
Perhaps you can point me in the right direction.
The rules of quadratic reciprocity, and their extension to the jacobi symbol,
are understood for integers, eisenstein integers (adjoin cube roots of 1),
and gaussian integers (adjoin fourth roots of 1).
These can all be found on wikipedia under quadratic reciprocity.
I would like to know the rules, if they are known,
for the ring of Z adjoin the fifth roots of 1, or perhaps the eighth roots of 1.
These are cyclotomic ufds, so we can talk about complex primes p and q,
and whether p is a quadratic residue mod q, or q a residue mod p.
I don’t need any proofs here,
I wouldn’t understand them anyways; I just need the rules.
Like p\q = q\p when they are primary primes, or whatever it turns out to be.

Many thanks.

• My impression is that one should be able to recover such things from the Artin reciprocity law, but I haven’t studied class field theory so I wouldn’t be able to tell you how this works. You might try asking this question on math.stackexchange.com.