# Quadratic reciprocity

Let be a prime, and let be any integer. Then we can define the Legendre symbol

We say that is a **quadratic residue** modulo if there exists an integer so that .

Equivalently, we can define the function as the unique nontrivial multiplicative homomorphism of into $\mathbb{R}^\\times$ (Error compiling LaTeX. ! Missing { inserted.), extended by .

## Quadratic Reciprocity Theorem

There are three parts. Let and be distinct odd primes. Then the following hold:

This theorem can help us evaluate Legendre symbols, since the following laws also apply:

- If , then .
- $\genfrac{(}{)}{}{}{ab}{p}\right) = \genfrac{(}{)}{}{}{a}{p} \genfrac{(}{)}{}{}{b}{p}$ (Error compiling LaTeX. ! Extra \right.).

There also exist quadratic reciprocity laws in other rings of integers. (I'll put that here later if I remember.)

## Proof

**Theorem 1.** Let be an odd prime. Then .

*Proof.* It suffices to show that if and only if is a quadratic residue mod .

Suppose that is a quadratic residue mod . Then , for some residue mod , so by Fermat's Little Theorem.

On the other hand, suppose that . Then is even, so is an integer. Since every nonzero residue mod is a root of the polynomial and the nonzero residues cannot all be roots of the polynomial , it follows that for some residue , Therefore is a quadratic residue mod , as desired.

Now, let and be distinct odd primes, and let be the splitting field of the polynomial over the finite field . Let be a primitive th root of unity in . We define the Gaussian sum

**Lemma.**

*Proof.* By definition, we have
Letting , we have
Now, is a root of the polynomial
it follows that for ,
while for , we have
Therefore
But since there are nonsquares and nonzero square mod , it follows that
Therefore
by Theorem 1.

**Theorem 2.** .

*Proof.* We compute the quantity in two different ways.

We first note that since in , Since , Thus

On the other hand, from the lemma,

\[\tau_q^p = (\tau_q^2)^{(p-1)/2} \cdot \tau_q = \bigl[ q (-1)^{(q-1)/2} \bigr]^{(p-1)/2} \tau_q = q^{(p-1)/2} (-1)^{(p-1)(q-1)/4 \tau_q .\] (Error compiling LaTeX. ! Missing } inserted.)

Since , we then have Since is evidently nonzero and we therefore have as desired.

**Theorem 3.** .

*Proof.* Let be the splitting field of the polynomial
over ; let be a root of the polynomial
in .

We note that So

On the other hand, since is a field of characteristic , Thus Now, if , then and , so , and On the other hand, if , then and , so Thus the theorem holds in all cases.

## References

- Helmut Koch,
*Number Theory: Algebraic Numbers and Functions,*American Mathematical Society 2000. ISBN 0-8218-2054-0 begin_of_the_skype_highlighting 0-8218-2054-0 end_of_the_skype_highlighting begin_of_the_skype_highlighting 0-8218-2054-0 end_of_the_skype_highlighting begin_of_the_skype_highlighting 0-8218-2054-0 end_of_the_skype_highlighting.