# Difference between revisions of "Quadratic reciprocity"

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On the other hand, suppose that <math>(-1)^{(p-1)/2} = 1</math>. Then <math>(p-1)/2</math> is even, so <math>(p-1)/4</math> is an integer. Since every nonzero residue mod <math>p</math> is a root of the polynomial | On the other hand, suppose that <math>(-1)^{(p-1)/2} = 1</math>. Then <math>(p-1)/2</math> is even, so <math>(p-1)/4</math> is an integer. Since every nonzero residue mod <math>p</math> is a root of the polynomial | ||

− | <cmath> (x^{p-1} - 1 = (x^{(p-1)/2} + 1)(x^{(p-1)/2} - 1) , </cmath> | + | <cmath> (x^{p-1} - 1) = (x^{(p-1)/2} + 1)(x^{(p-1)/2} - 1) , </cmath> |

and the <math>p-1</math> nonzero residues cannot all be roots of the polynomial <math>x^{(p-1)/2} - 1</math>, it follows that for some residue <math>k</math>, | and the <math>p-1</math> nonzero residues cannot all be roots of the polynomial <math>x^{(p-1)/2} - 1</math>, it follows that for some residue <math>k</math>, | ||

<cmath> \bigl(k^{(p-1)/2}\bigr)^2 = k^{(p-1)/2} = -1 . </cmath> | <cmath> \bigl(k^{(p-1)/2}\bigr)^2 = k^{(p-1)/2} = -1 . </cmath> | ||

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== References == | == References == | ||

− | * Helmut Koch, ''Number Theory: Algebraic Numbers and Functions,'' American Mathematical Society 2000. ISBN 0-8218-2054-0 | + | * Helmut Koch, ''Number Theory: Algebraic Numbers and Functions,'' American Mathematical Society 2000. ISBN 0-8218-2054-0 |

[[Category:Number theory]] | [[Category:Number theory]] |

## Revision as of 21:24, 29 March 2013

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 , 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