# How can one deduce quadratic reciprocity from Hilbert reciprocity?

Hilbert reciprocity says the following:

Define $(a,b)_p$ to be $1$ if there is a non-trivial solution in $\mathbb{Q}_p$ to $z^2=ax^2+by^2$, and $-1$ if there isn't. Then $\prod_p (a,b)_p =1$, where the product runs also over the infinite prime (and where $\mathbb{Q}_{\infty}$ is $\mathbb{R}$).

Quadratic reciprocity says: Define $\left( \frac{p}{q}\right)$ to be $1$ if $p$ is a square in $\mathbb{F}_q$, and $-1$ otherwise. Then for every two different odd primes $p$ and $q$, $\left( \frac{p}{q}\right)\left( \frac{q}{p}\right)=(-1)^{\frac{(p-1)(q-1)}{4}}$.

I have heard that people think of Hilbert reciprocity as a generalization of quadratic reciprocity. How does one deduce it as a consequence?

-

This is from page 46 in Rational Quadratic Forms by J. W. S. Cassels. Let $p,q$ denote fixed (positive) odd primes with $p \neq q.$ Let $r$ run over all primes including $2,\infty,$ and simply take the product of all the $(p,q)_r$ and see what happens. The values I will be showing below are tabulated on page 43 for $r = 2$ and $r$ (positive) odd, while $r = \infty$ is on page 44. There are five cases:

For (positive) $r$ odd, $r \neq p,q$ we have $(p,q)_r = 1.$

For $r = \infty,$ $(p,q)_\infty = 1.$

For $r = q,$ $(p,q)_q = (p | q).$

For $r = p,$ $(p,q)_p = (q | p).$

For $r = 2,$ $(p,q)_2 = (-1)^{(p-1)(q-1)/4}.$

That's it, you just need to confirm all five in your mind. The fact that the product of all five turns out to be 1 is, indeed, the statement that $$(p | q) (q | p) = (-1)^{(p-1)(q-1)/4}.$$

-
Thanks! I thought it might be something with $a=p$ and $b=q$. I can't quite convince myself of all of the facts above... Is there a proof of these identities in Cassels? –  Nicole Oct 30 '11 at 23:53
Proofs, yes, and Cassels wanted mostly to demonstrate what happens in the ordinary integers $\mathbf Z,$ so the description is about as technically simple as is possible for this material. –  Will Jagy Oct 31 '11 at 0:14

Given the level of some of your other questions, I think you might enjoy going through the exercises in the back of Cassels and Fröhlich (which are excellent, having been written by Serre and Tate). Exercise 2.11 there is to prove quadratic reciprocity using Hilbert symbols and the product formula. However, they are building off of class field theory, so perhaps this isn't what you had in mind.

If you go this route, you will have to do some work in $\mathbf{Q}_2$ in order to establish $$(p, q)_2 = (-1)^{(p - 1)(q - 1)/4},$$ but this is not so bad (because of Hensel's lemma), and is also done in Chapter III of Serre's A Course in Arithmetic.

-

I happened to post on this (one hour before the question!) in another answer:

The proof is quite easy.

For places $r$ different from $p, q, 2$, and $\infty$, the local Hilbert symbol is $+1$ because the quadratic form $x^2 -py^2 - qz^2$ has rank higher than 2. More combinatorially, let $z=1$ and observe that the sets of values of $x^2$ and of $py^2$ have cardinality more than $|r|/2$ so that their sum has to cover all possibilities mod $r$. Because the degree $2$ is less than $r$, Hensel lifting shows that an $r$-adic solution can be found from a mod $r$ solution.

The values of the local symbol at $p$ and $q$ are, by definition, the Legendre symbols. (The existence of a solution mod $p$ or $q$ is the information recorded in a Legendre symbol, and Hensel lifting of the modular square roots to p-adic square roots can always be done for odd primes.)

The value of the the local symbol at $\infty$ is $+1$ because the form is not positive definite or negative definite. Or write down a particular real solution such as $y=z=1, x = \sqrt{p+q}$.

This leaves the value of the local symbol at $2$. The condition for $py^2 + qz^2$ to be a square 2-adically is that it be a square mod $8$, i.e., congruent to $0,1$, or $4$, and a primitive nonzero solution has $y$ or $z$ odd. Case analysis is not difficult:

1. The symbol is +1 if $p$ or $q$ is $1 \mod 8$. If $p=1$ there is a solution with $y=1$ and $z=0$.
2. The symbol is +1 if $p$ or $q$ is $5 \mod 8$. If $p=5$ there is a solution with $y=1$ and $z=2$.
3. The symbol is -1 if $p$ and $q$ are both $-1 \mod 4$. Both $y$ and $z$ must be even in this case.

This pattern of signs is the same as the $(-1)^{(p-1)(q-1)/4}$ in the quadratic reciprocity formula.

Finally, the product of all the local symbols is 1:

$$(1) (\frac{p}{q}) (\frac{q}{p}) (-1)^{(p-1)(q-1)/4} = 1$$

-