202305271605

Type :Note Tags : Number Theory

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Quadratic Reciprocity

Note

Given two odd primes $p,q$, $\left(\frac{q}{p}\right)\left(\frac{p}{q}\right)=(-1{)}^{((p-1)/2)((q-1)/2)}$

Proof 1:

Let $\chi$ be a character of order $2$ on ${\mathbb{F}}_p$. Then by Corollary 1 to Theorem 3 in Jacobi Sums, $$ g(\chi)^{q+1} = \chi(-1)pJ(\chi,\chi, \dots , \chi)

Where there are $q$ $\chi's$ in the right hand side of the equation. This gives $$ J(\chi,\chi, \dots ,\chi) = p ^{-1}(g(\chi)^{2})^{(q+1)/2}\chi(-1)=p ^{-1}(p \chi(-1))^{(q+1)/2} \chi(-1) = p ^{(q-1)/2}(-1)^{((p-1)/2)((q-1)/2)}

Now $J(\chi ,\chi ,\dots ,\chi )=\sum \chi (t_1)\chi (t_2)\dots \chi (t_q)$. If $t_1=t_2=\cdots =t_q=\frac{1}{q}$, we get $\chi {\left(\frac{1}{q}\right)}^q=\chi (q{)}^{-q}=\chi (q)=\left(\frac{q}{p}\right)$. When not all $t_i$ are equal, we can cyclicly rotate them to get $q$ equal terms, and so, mod q they vanish.

Thus $$\begin{align} J(\chi, \dots ,\chi) &= \left( \frac{q}{p} \right) \ (\mathrm{mo d} \ q) \ &= p ^{(q-1)/2}(-1)^{((p-1)/2)((q-1)/2)} (\mathrm{mo d} \ q) \ \implies \left( \frac{q}{p} \right) \left( \frac{p}{q} \right) &= (-1)^{((p-1)/2)((q-1)/2)} (\mathrm{mo d} \ q) \end{align}

and hence, we can drop the mod q since both sides are absolute value 1. --- # References [[Legendre Symbol]] [[Multiplicative Characters]] [[Jacobi Sums]] [[Equations over Finite Fields]]