Timeline for Intersection of two trace equations over finite fields

5 events

when toggle format	what		by	license	comment
Mar 2, 2014 at 21:54	comment	added	Jyrki Lahtonen		Here is a possible way of getting a lower bound for $k$. Study the curve $V$ defined by the pair of equations $y^q-y=x$, $z^q-z=x^k$. It is easy to see that $V$ has $q^2$ $F_{q^n}$-rational points such that $x=0$. If the curve is irreducible, and you can calculate its genus, then Hasse-Weil bound will give a lower bound to the number of rational points. Tally the infinite points also! What remains probably exceeds $q^2$ for small $k$ and/or large $n$ (I think/hope). Then you can conclude that the intersection cannot be trivial for such $k,n$.
Mar 2, 2014 at 13:44	comment	added	Jyrki Lahtonen		A trivial positive case is when $(n,p)=1$ and $k$ is a multiple of $(q^n-1)/(q-1)$. In that case $x^k$ is a non-zero element of $F_q$ whenever $x\neq0$.
Feb 27, 2014 at 22:33	comment	added	Gerry Myerson		Here's a very, very, very special case: let $q$ be prime and one less than a multiple of $4$, and let $n=2$. If a nonzero $x$ has trace zero, then so does $x^k$, if and only if $k$ is odd.
Feb 27, 2014 at 7:47	comment	added	Name		It seems that the trivaility of this intersection, depends on $n$ as well. If $(n,p)>1$ then $1$ is in this intersection.
Feb 27, 2014 at 7:15	history	asked	Joe Zhou	CC BY-SA 3.0