Difference between revisions of "Quadratic reciprocity"
(Event his is better than the one single align... And actually, it's unnatural to call the supplementary laws part if QR) |
Astroroman (talk | contribs) m |
||
(3 intermediate revisions by 2 users not shown) | |||
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 nontrivial multiplicative [[homomorphism]] of <math>\mathbb{F}_p^\times</math> into <math>\mathbb{R}^ | + | 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 == | ||
Line 15: | Line 15: | ||
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>. | ||
− | * <math>\genfrac{(}{)}{}{}{ab}{p} | + | * <math>\genfrac{(}{)}{}{}{ab}{p} = \genfrac{(}{)}{}{}{a}{p} \genfrac{(}{)}{}{}{b}{p}</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.) | ||
Line 30: | Line 30: | ||
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> | ||
Line 72: | Line 72: | ||
On the other hand, from the lemma, | On the other hand, from the lemma, | ||
− | <cmath> \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 . </cmath> | + | <cmath> \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 . </cmath> |
Since <math>q^{(p-1)/2} = \genfrac{(}{)}{}{}{q}{p}</math>, we then have | Since <math>q^{(p-1)/2} = \genfrac{(}{)}{}{}{q}{p}</math>, we then have | ||
<cmath> \genfrac{(}{)}{}{}{p}{q} \tau_q = \tau_q^p = \genfrac{(}{)}{}{}{q}{p} (-1)^{(p-1)(q-1)/4} \tau_q . </cmath> | <cmath> \genfrac{(}{)}{}{}{p}{q} \tau_q = \tau_q^p = \genfrac{(}{)}{}{}{q}{p} (-1)^{(p-1)(q-1)/4} \tau_q . </cmath> | ||
Line 110: | 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 |
[[Category:Number theory]] | [[Category:Number theory]] |
Latest revision as of 15:46, 28 April 2016
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
.
.
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,
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