Difference between revisions of "Quadratic reciprocity"
(added proof for 2) |
(Event his is better than the one single align... And actually, it's unnatural to call the supplementary laws part if QR) |
||
Line 5: | Line 5: | ||
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 | + | Equivalently, we can define the function <math>a \mapsto \genfrac{(}{)}{}{}{a}{p}</math> as the unique nontrivial multiplicative [[homomorphism]] of <math>\mathbb{F}_p^\times</math> into <math>\mathbb{R}^\times</math>, extended by <math>0 \mapsto 0</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: | ||
− | < | + | * <math>\genfrac{(}{)}{}{}{-1}{p} = (-1)^{(p-1)/2},</math> |
− | \genfrac{(}{)}{}{}{-1}{p} | + | * <math> \genfrac{(}{)}{}{}{2}{p} = (-1)^{(p^2-1)/8},</math> |
− | \genfrac{(}{)}{}{}{2}{p} | + | * <math>\genfrac{(}{)}{}{}{p}{q} \genfrac{(}{)}{}{}{q}{p} = (-1)^{(p-1)(q-1)/4} .</math> |
− | \genfrac{(}{)}{}{}{p}{q} \genfrac{(}{)}{}{}{q}{p} | ||
− | |||
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>\genfrac{(}{)}{}{}{a}{p} = \genfrac{(}{)}{}{}{b}{p}</math>. | ||
Line 112: | Line 110: | ||
== 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 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. |
[[Category:Number theory]] | [[Category:Number theory]] |
Revision as of 06:59, 15 August 2010
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. Unknown error_msg), 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. Unknown error_msg).
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. Unknown error_msg)
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.