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Quadratic reciprocity
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===Polynomials over a finite field=== Let ''F'' be a [[finite field]] with ''q'' = ''p<sup>n</sup>'' elements, where ''p'' is an odd prime number and ''n'' is positive, and let ''F''[''x''] be the [[Polynomial ring|ring of polynomials]] in one variable with coefficients in ''F''. If <math>f,g \in F[x]</math> and ''f'' is [[Irreducible polynomial|irreducible]], [[Monic polynomial|monic]], and has positive degree, define the quadratic character for ''F''[''x''] in the usual manner: :<math>\left(\frac{g}{f}\right) = \begin{cases} 1 & \gcd(f,g)=1 \text{ and } \exists h,k \in F[x] \text{ such that }g-h^2 = kf \\ -1 & \gcd(f,g)=1 \text{ and } g \text{ is not a square}\bmod{f}\\ 0 & \gcd(f,g)\ne 1 \end{cases} </math> If <math>f=f_1 \cdots f_n</math> is a product of monic irreducibles let :<math>\left(\frac{g}{f}\right) = \left(\frac{g}{f_1}\right) \cdots \left(\frac{g}{f_n}\right). </math> Dedekind proved that if <math>f,g \in F[x]</math> are monic and have positive degrees,<ref>Bach & Shallit, Thm. 6.7.1</ref> :<math>\left(\frac{g}{f}\right) \left(\frac{f}{g}\right) = (-1)^{\frac{q-1}{2}(\deg f)(\deg g)}.</math>
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