Difference between revisions of "Quadratic reciprocity"
m (Fixed denominators.) |
(cleaned the LaTeX a bit; I'll add a proof later) |
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− | Let <math>p</math> be a [[prime number|prime]], and let <math>a</math> be any integer | + | Let <math>p</math> be a [[prime number|prime]], and let <math>a</math> be any integer. Then we can define the [[Legendre symbol]] |
+ | <cmath> \genfrac{(}{)}{}{}{a}{p} =\begin{cases} 1 & \text{if } a \text{ is a quadratic residue modulo } p, \ | ||
+ | 0 & \text{if } p \text{ divides } a, \ -1 & \text{otherwise}.\end{cases} </cmath> | ||
− | We say that <math>a</math> is a '''quadratic residue''' modulo <math>p</math> if there exists an integer <math>n</math> so that <math>n^2\equiv a\pmod p</math>. | + | We say that <math>a</math> is a '''quadratic residue''' modulo <math>p</math> if there exists an integer <math>n</math> so that <math>n^2\equiv a\pmod p</math>. |
+ | |||
+ | Equivalently, we can define the function <math>a \mapsto \genfrac{(}{)}{}{}{a}{p}</math> as the unique nonzero multiplicative [[homomorphism]] of <math>\mathbb{F}_p</math> into <math>\mathbb{R}</math>. | ||
== Quadratic Reciprocity Theorem == | == Quadratic Reciprocity Theorem == | ||
There are three parts. Let <math>p</math> and <math>q</math> be distinct [[odd integer | odd]] primes. Then the following hold: | There are three parts. Let <math>p</math> and <math>q</math> be distinct [[odd integer | odd]] primes. Then the following hold: | ||
− | + | <cmath> \begin{align*} | |
− | + | \genfrac{(}{)}{}{}{-1}{p} &= (-1)^{(p-1)/2} , \ | |
− | + | \genfrac{(}{)}{}{}{2}{p} &= (-1)^{(p^2-1)/8} , \ | |
− | + | \genfrac{(}{)}{}{}{p}{q} \genfrac{(}{)}{}{}{q}{p} &= (-1)^{(p-1)(q-1)/4} . | |
− | + | \end{align*} </cmath> | |
This theorem can help us evaluate Legendre symbols, since the following laws also apply: | This theorem can help us evaluate Legendre symbols, since the following laws also apply: | ||
− | + | * If <math>a\equiv b\pmod{p}</math>, then <math>\genfrac{(}{)}{}{}{a}{p} = \genfrac{(}{)}{}{}{b}{p}</math>. | |
− | * If <math>a\equiv b\pmod{p}</math>, then <math>\ | + | * <math>\genfrac{(}{)}{}{}{ab}{p}\right) = \genfrac{(}{)}{}{}{a}{p} \genfrac{(}{)}{}{}{b}{p}</math>. |
− | * <math>\ | ||
There also exist quadratic reciprocity laws in other [[ring of integers|rings of integers]]. (I'll put that here later if I remember.) | There also exist quadratic reciprocity laws in other [[ring of integers|rings of integers]]. (I'll put that here later if I remember.) |
Revision as of 21:09, 25 September 2008
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 nonzero multiplicative homomorphism of into .
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. Unknown error_msg).
There also exist quadratic reciprocity laws in other rings of integers. (I'll put that here later if I remember.)