Difference between revisions of "Quadratic reciprocity"
(added proof of the main result) |
(added proof for 2) |
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<cmath> \genfrac{(}{)}{}{}{p}{q} \genfrac{(}{)}{}{}{q}{p} = (-1)^{(p-1)(q-1)/4}, </cmath> | <cmath> \genfrac{(}{)}{}{}{p}{q} \genfrac{(}{)}{}{}{q}{p} = (-1)^{(p-1)(q-1)/4}, </cmath> | ||
as desired. <math>\blacksquare</math> | as desired. <math>\blacksquare</math> | ||
+ | |||
+ | '''Theorem 3.''' <math>\genfrac{(}{)}{}{}{2}{p} = (-1)^{(p^2 - 1)/8}</math>. | ||
+ | |||
+ | ''Proof.'' Let <math>K</math> be the splitting field of the polynomial | ||
+ | <math>x^8-1</math> over <math>\mathbb{F}_p</math>; let <math>\zeta</math> be a root of the polynomial | ||
+ | <math>x^4+1</math> in <math>K</math>. | ||
+ | |||
+ | We note that | ||
+ | <cmath> (\zeta + \zeta^{-1})^2 = \zeta^2 + 2 + \zeta^{-2} = 2 + \zeta^{-2} (\zeta^{4} + 1) = 2 . </cmath> | ||
+ | So | ||
+ | <cmath> (\zeta + \zeta^{-1})^p = (\zeta + \zeta^{-1}) 2^{(p-1)/2} = (\zeta + \zeta^{-1}) \genfrac{(}{)}{}{}{2}{p}. </cmath> | ||
+ | |||
+ | On the other hand, since <math>K</math> is a field of characteristic <math>p</math>, | ||
+ | <cmath> (\zeta + \zeta^{-1})^p = \zeta^p + \zeta^{-p} . </cmath> | ||
+ | Thus | ||
+ | <cmath> \zeta^p + \zeta^{-p} = (\zeta + \zeta^{-1})^p = (\zeta + \zeta^{-1} \genfrac{(}{)}{}{}{2}{p} . </cmath> | ||
+ | Now, if <math>p \equiv 4 \pm 1 \pmod{8}</math>, then | ||
+ | <cmath> \zeta^{p} + \zeta^{-p} = - ( \zeta + \zeta^{-1} ) </cmath> | ||
+ | and <math>p^2 - 1 \equiv 8 \pmod{16}</math>, so <math>(-1)^{(p^2-1)/8} = -1</math>, | ||
+ | and | ||
+ | <cmath> \genfrac{(}{)}{}{}{2}{p} = -1 = (-1)^{(p^2 - 1)/8} . </cmath> | ||
+ | On the other hand, if <math>p \equiv \pm 1 \pmod{8}</math>, then | ||
+ | <cmath> \zeta^p + \zeta^{-p} = \zeta + \zeta^{-1}, </cmath> | ||
+ | and <math>p^2 -1 \equiv 0 \pmod{16}</math>, so | ||
+ | <cmath> \genfrac{(}{)}{}{}{2}{p} = 1 = (-1)^{p^2-1} . </cmath> | ||
+ | Thus the theorem holds in all cases. <math>\blacksquare</math> | ||
+ | |||
== References == | == References == |
Revision as of 21:53, 24 March 2009
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.)
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.