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Hasse-Davenport Relation

Let $F$ be a Finite Field with $q$ elements, and let $F_s$ be a Field containing $F$ such that $[F_s:F]=s$. Let $\chi$ be a nontrivial Multiplicative Character of $F$ and $\chi'=\chi\circ N_{F_s/F}$ a character of $F_s$. Then

\begin{displaymath}
(-g(\chi))^s=-g(\chi'),
\end{displaymath}

where $g(x)$ is a Gaussian Sum.

See also Gaussian Sum, Multiplicative Character


References

Ireland, K. and Rosen, M. ``A Proof of the Hasse-Davenport Relation.'' §11.4 in A Classical Introduction to Modern Number Theory, 2nd ed. New York: Springer-Verlag, pp. 162-165, 1990.




© 1996-9 Eric W. Weisstein
1999-05-25