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Difference between revisions of "Quadratic reciprocity"
(→‎Quadratic Reciprocity Theorem)
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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}\right) = \genfrac{(}{)}{}{}{a}{p} \genfrac{(}{)}{}{}{b}{p}</math>.
+
* <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.)

Revision as of 18:42, 5 August 2015

Let $p$ be a prime, and let $a$ be any integer. Then we can define the Legendre symbol \[\genfrac{(}{)}{}{}{a}{p} =\begin{cases} 1 & \text{if } a \text{ is a quadratic residue modulo } p, \\ 0 & \text{if } p \text{ divides } a, \\ -1 & \text{otherwise}.\end{cases}\]

We say that $a$ is a quadratic residue modulo $p$ if there exists an integer $n$ so that $n^2\equiv a\pmod p$.

Equivalently, we can define the function $a \mapsto \genfrac{(}{)}{}{}{a}{p}$ as the unique nontrivial multiplicative homomorphism of $\mathbb{F}_p^\times$ into $\mathbb{R}^\times$, extended by $0 \mapsto 0$.

Quadratic Reciprocity Theorem

There are three parts. Let $p$ and $q$ be distinct odd primes. Then the following hold:

  • $\genfrac{(}{)}{}{}{-1}{p} = (-1)^{(p-1)/2},$
  • $\genfrac{(}{)}{}{}{2}{p} = (-1)^{(p^2-1)/8},$
  • $\genfrac{(}{)}{}{}{p}{q} \genfrac{(}{)}{}{}{q}{p} = (-1)^{(p-1)(q-1)/4} .$

This theorem can help us evaluate Legendre symbols, since the following laws also apply:

  • If $a\equiv b\pmod{p}$, then $\genfrac{(}{)}{}{}{a}{p} = \genfrac{(}{)}{}{}{b}{p}$.
  • $\genfrac{(}{)}{}{}{ab}{p} = \genfrac{(}{)}{}{}{a}{p} \genfrac{(}{)}{}{}{b}{p}$.

There also exist quadratic reciprocity laws in other rings of integers. (I'll put that here later if I remember.)

Proof

Theorem 1. Let $p$ be an odd prime. Then $\genfrac{(}{)}{}{}{-1}{p} = (-1)^{(p-1)/2}$.

Proof. It suffices to show that $(-1)^{(p-1)/2} = 1$ if and only if $-1$ is a quadratic residue mod $p$.

Suppose that $-1$ is a quadratic residue mod $p$. Then $k^2 = -1$, for some residue $k$ mod $p$, so \[(-1)^{(p-1)/2} = (k^2)^{(p-1)/2} = k^{p-1} = 1 = \genfrac{(}{)}{}{}{-1}{p} ,\] by Fermat's Little Theorem.

On the other hand, suppose that $(-1)^{(p-1)/2} = 1$. Then $(p-1)/2$ is even, so $(p-1)/4$ is an integer. Since every nonzero residue mod $p$ is a root of the polynomial \[(x^{p-1} - 1) = (x^{(p-1)/2} + 1)(x^{(p-1)/2} - 1) ,\] and the $p-1$ nonzero residues cannot all be roots of the polynomial $x^{(p-1)/2} - 1$, it follows that for some residue $k$, \[\bigl(k^{(p-1)/2}\bigr)^2 = k^{(p-1)/2} = -1 .\] Therefore $-1$ is a quadratic residue mod $p$, as desired. $\blacksquare$

Now, let $p$ and $q$ be distinct odd primes, and let $K$ be the splitting field of the polynomial $x^q - 1$ over the finite field $\mathbb{F}_p$. Let $\zeta$ be a primitive $q$th root of unity in $K$. We define the Gaussian sum \[\tau_q = \sum_{a=0}^{q-1} \genfrac{(}{)}{}{}{a}{q} \zeta^q .\]

Lemma. $\tau_q^2 = q (-1)^{(q-1)/2}$

Proof. By definition, we have \[\tau_q^2 = \sum_a \sum_b \genfrac{(}{)}{}{}{a}{q} \zeta^a \genfrac{(}{)}{}{}{b}{q} \zeta^b = \sum_{a \neq 0} \sum_b \genfrac{(}{)}{}{}{ab}{q} \zeta^{a+b} .\] Letting $c \equiv a^{-1}b \pmod{q}$, we have \begin{align*} \sum_{a \neq 0} \sum_b \genfrac{(}{)}{}{}{ab}{q} \zeta^{a+b} &= \sum_{a\neq 0} \sum_c \genfrac{(}{)}{}{}{a^2 c}{q} \zeta^{a+ac} \\ &= \sum_c \sum_{a \neq 0} \genfrac{(}{)}{}{}{c}{q} \bigl( \zeta^{1+c} \bigr)^a \\ &= \sum_c \genfrac{(}{)}{}{}{c}{q} \sum_{a \neq 0} \bigl( \zeta^{1+c} \bigr)^a . \end{align*} Now, $\zeta^{c+1}$ is a root of the polynomial \[P(x) = x^q - 1 = (x-1) \sum_{i=0}^{q-1} x^i,\] it follows that for $c\neq -1$, \[\sum_{a \neq 0} \bigl( \zeta^{1+c} \bigr)^a = -1,\] while for $c = -1$, we have \[\sum_{a \neq 0} \bigl( \zeta^{1+c} \bigr)^a = q-1 .\] Therefore \[\sum_c \genfrac{(}{)}{}{}{c}{q} \sum_{a \neq 0} \bigl( \zeta^{1+c} \bigr)^a = q \genfrac{(}{)}{}{}{-1}{q} - \sum_{c=0}^{q-1}\genfrac{(}{)}{}{}{c}{q} .\] But since there are $(q-1)/2$ nonsquares and $(q-1)/2$ nonzero square mod $q$, it follows that \[\sum_{c=0}^{q-1} \genfrac{(}{)}{}{}{c}{q} = 0 .\] Therefore \[\tau_q^2 = q \genfrac{(}{)}{}{}{-1}{q} = q (-1)^{(q-1)/2} ,\] by Theorem 1.

Theorem 2. $\genfrac{(}{)}{}{}{p}{q} \genfrac{(}{)}{}{}{q}{p} = (-1)^{(p-1)(q-1)/4}$.

Proof. We compute the quantity $\tau_q^p$ in two different ways.

We first note that since $p=0$ in $K$, \[\tau_q^p = \biggl( \sum_{a=0}^{q-1} \genfrac{(}{)}{}{}{a}{q} \zeta^a \biggr)^p = \sum_{a=0}^{q-1} \genfrac{(}{)}{}{}{a}{q}^p \zeta^{ap} = \sum_{a=0}^{q-1} \genfrac{(}{)}{}{}{a}{q} \zeta^{ap} .\] Since $\genfrac{(}{)}{}{}{p}{q}^2 = 1$, \[\sum_{a=0}^{q-1} \genfrac{(}{)}{}{}{a}{q} \zeta^{ap} = \genfrac{(}{)}{}{}{p}{q} \sum_{a=0}^{q-1} \genfrac{(}{)}{}{}{pa}{q} \zeta^{ap} = \genfrac{(}{)}{}{}{a}{q} \tau_q .\] Thus \[\tau_q^p = \genfrac{(}{)}{}{}{p}{q} \tau_q .\]

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 $q^{(p-1)/2} = \genfrac{(}{)}{}{}{q}{p}$, we then have \[\genfrac{(}{)}{}{}{p}{q} \tau_q = \tau_q^p = \genfrac{(}{)}{}{}{q}{p} (-1)^{(p-1)(q-1)/4} \tau_q .\] Since $\tau_q$ is evidently nonzero and \[\genfrac{(}{)}{}{}{q}{p}^2 = 1,\] we therefore have \[\genfrac{(}{)}{}{}{p}{q} \genfrac{(}{)}{}{}{q}{p} = (-1)^{(p-1)(q-1)/4},\] as desired. $\blacksquare$

Theorem 3. $\genfrac{(}{)}{}{}{2}{p} = (-1)^{(p^2 - 1)/8}$.

Proof. Let $K$ be the splitting field of the polynomial $x^8-1$ over $\mathbb{F}_p$; let $\zeta$ be a root of the polynomial $x^4+1$ in $K$.

We note that \[(\zeta + \zeta^{-1})^2 = \zeta^2 + 2 + \zeta^{-2} = 2 + \zeta^{-2} (\zeta^{4} + 1) = 2 .\] So \[(\zeta + \zeta^{-1})^p = (\zeta + \zeta^{-1}) 2^{(p-1)/2} = (\zeta + \zeta^{-1}) \genfrac{(}{)}{}{}{2}{p}.\]

On the other hand, since $K$ is a field of characteristic $p$, \[(\zeta + \zeta^{-1})^p = \zeta^p + \zeta^{-p} .\] Thus \[\zeta^p + \zeta^{-p} = (\zeta + \zeta^{-1})^p = (\zeta + \zeta^{-1} \genfrac{(}{)}{}{}{2}{p} .\] Now, if $p \equiv 4 \pm 1 \pmod{8}$, then \[\zeta^{p} + \zeta^{-p} = - ( \zeta + \zeta^{-1} )\] and $p^2 - 1 \equiv 8 \pmod{16}$, so $(-1)^{(p^2-1)/8} = -1$, and \[\genfrac{(}{)}{}{}{2}{p} = -1 = (-1)^{(p^2 - 1)/8} .\] On the other hand, if $p \equiv \pm 1 \pmod{8}$, then \[\zeta^p + \zeta^{-p} = \zeta + \zeta^{-1},\] and $p^2 -1 \equiv 0 \pmod{16}$, so \[\genfrac{(}{)}{}{}{2}{p} = 1 = (-1)^{p^2-1} .\] Thus the theorem holds in all cases. $\blacksquare$


References

  • Helmut Koch, Number Theory: Algebraic Numbers and Functions, American Mathematical Society 2000. ISBN 0-8218-2054-0
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