quadratic reciprocity for polynomialsQuadraticReciprocityForPolynomials2002-01-17 03:59:392003-09-15 15:53:37Theorem% this is the default PlanetMath preamble. as your knowledge
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% define commands hereLet $F$ be a finite field of characteristic $p$, and let $f$ and $g$ be distinct monic irreducible (non-constant) polynomials in the polynomial ring $F[X]$. Define the {\em Legendre symbol} $\left(\frac{f}{g}\right)$ by
$$
\left(\frac{f}{g}\right) :=
\begin{cases}
1 & \text{ if $f$ is a square in the quotient ring $F[X]/(g)$,} \\
-1 & \text{ otherwise.}
\end{cases}
$$
The quadratic reciprocity theorem for polynomials over a finite field states that
$$
\left(\frac{f}{g}\right) \left(\frac{g}{f}\right) = (-1)^{\frac{p-1}{2} \deg(f) \deg(g)}.
$$
\begin{thebibliography}{9}
\bibitem{feng-ying}
Feng, Ke Qin and Ying, Linsheng,
{\em An elementary proof of the law of quadratic reciprocity in $F\sb q(T)$.}
Sichuan Daxue Xuebao {\bf 26} (1989), Special Issue, 36--40.
\bibitem{merrill-walling}
Merrill, Kathy D. and Walling, Lynne H.,
{\em On quadratic reciprocity over function fields.}
Pacific J. Math. {\bf 173} (1996), no. 1, 147--150.
\end{thebibliography}