Let $F_q$ be a finite field with $q$ elements. Let $n$ be an integer and $Tr:F_{q^n} \rightarrow F_q$ the trace function. My question is: For which integer $k$, $$\{x: Tr(x)=0\}\cap\{x: Tr(x^k)=0\}=\{0\}.$$ Furthermore, is there any result concerning the size of $\{x: Tr(x)=0\}\cap\{x: Tr(x^k)=0\}$ for arbitrary $k$.
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$\begingroup$ It seems that the trivaility of this intersection, depends on $n$ as well. If $(n,p)>1$ then $1$ is in this intersection. $\endgroup$– NameCommented Feb 27, 2014 at 7:47
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$\begingroup$ 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. $\endgroup$– Gerry MyersonCommented Feb 27, 2014 at 22:33
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$\begingroup$ 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 $x\neq0$. $\endgroup$– Jyrki LahtonenCommented Mar 2, 2014 at 13:44
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1$\begingroup$ 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 $k,n$. $\endgroup$– Jyrki LahtonenCommented Mar 2, 2014 at 21:54
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