I have a question concerning certain elements having zero trace in a finite field extension and I do have the feeling that additive characters should play a role, but I am not sure how. I am stating the problem slightly more generally than what I am really interested in.


Let $q$ be prime power, let $\mathbb{F}\_{q}$ the finite field of cardinality $q$ and let $n$ be a prime with $n$ dividing $q+1$. I denote with $\mathrm{Tr}\_{\mathbb{F}\_{q^{2n}}/\mathbb{F}\_{q^2}}:\mathbb{F}\_{q^{2n}}\to \mathbb{F}\_{q^2}$ the trace map.


I am interested in the cardinality of the set $$\{x\in \mathbb{F}\_{q^{2n}}|\mathrm{Tr}\_{\mathbb{F}\_{q^{2n}}/\mathbb{F}\_{q^2}}(x)=0 \hbox{ and } x=y^n \textrm{ for some }y\in \mathbb{F}\_{q^{2n}}\}.$$
For instance when $n=3$, I have a bit of evidence that that this set has cardinality $(q+1)^2(q^2-1)/3$.



Number of certain elements in a finite field having zero trace
We have
$$ \# \{x\in \mathbb{F}\_{q^{2n}}|\mathrm{Tr}\_{\mathbb{F}\_{q^{2n}}/\mathbb{F}\_{q^2}}(x)=0 \hbox{ and } x=y^n \textrm{ for some }y\in \mathbb{F}\_{q^{2n}}\} = \frac{\#\{y\in \mathbb{F}\_{q^{2n}}|\mathrm{Tr}\_{\mathbb{F}\_{q^{2n}}/\mathbb{F}\_{q^2}}(y^n)=0 \} +n-1}{ n}$$
since $\mathbb F\_{q^n}$ contains all the $n$th roots of unity, so it suffices to calculate.


\begin{align\*} 
\# \{y\in \mathbb{F}\_{q^{2n}}|\mathrm{Tr}\_{\mathbb{F}\_{q^{2n}}/\mathbb{F}\_{q^2}}(y^n)=0 \} &= \frac{1}{q^2} \sum\_{\lambda \in \mathbb F\_{q^2} } \sum\_{ y \in \mathbb F\_{q^{2n}} }\psi\_2 (\lambda \mathrm{Tr}\_{\mathbb{F}\_{q^{2n}}/\mathbb{F}\_{q^2}}(y^n) ) \\ &= \frac{1}{q^2} \sum\_{\lambda \in \mathbb F\_{q^2} } \sum\_{ y \in \mathbb F\_{q^{2n}} }\psi\_{2n} (\lambda y^n) ) \\& = q^{2n-2} + \frac{1}{q^2} \sum\_{\lambda \in \mathbb F\_{q^2}^\times } \sum\_{ y \in \mathbb F\_{q^{2n}} }\psi\_{2n} (\lambda y^n) ) \\ &=q^{2n-2} + \frac{1}{q^2} \sum\_{\lambda \in \mathbb F\_{q^2}^\times } \sum\_{ x \in \mathbb F\_{q^{2n}} } \sum\_{\substack{ \chi \colon \mathbb F\_{q^{2n}}^\times \to \mathbb C^\times \\ \chi^n=1 } } \psi\_{2n} (\lambda x) \chi(x) \\ &=q^{2n-2} + \frac{1}{q^2} \sum\_{\lambda \in \mathbb F\_{q^2}^\times } \sum\_{ x \in \mathbb F\_{q^{2n}} } \sum\_{\substack{ \chi \colon \mathbb F\_{q^{2n}}^\times \to \mathbb C^\times \\ \chi^n=1 } } \psi\_{2n} (x) \chi(\lambda^{-1} x) \end{align\*}


where $\psi\_2 \colon \mathbb F\_{q^2} \to \mathbb C^\times$ is a nondegenerate additive character and $\psi\_{2n} = \psi\_2 \circ \mathrm{Tr}\_{\mathbb{F}\_{q^{2n}}/\mathbb{F}\_{q^2}}$


Now since $\chi$ has order $n$ and the quotient group $\mathbb F\_{q^{2n}}^\times / \mathbb F\_{q^2}^\times $ has order $q^{2n-2} + q^{2n-4} + \dots + 1$ which is divisible by $n$, $\chi$ must factor through this quotient group so $\chi(\lambda^{-1})=1$, giving


$$\#\{y\in \mathbb{F}\_{q^{2n}}|\mathrm{Tr}\_{\mathbb{F}\_{q^{2n}}/\mathbb{F}\_{q^2}}(y^n)=0 \} = q^{2n-2} + \frac{q^2-1}{q^2} \sum\_{\substack{ \chi \colon \mathbb F\_{q^{2n}}^\times \to \mathbb C^\times \\ \chi^n=1 } } \sum\_{ x \in \mathbb F\_{q^{2n}} } \psi\_{2n} (x) \chi( x) .$$


Here the inner sum $\sum\_{ x \in \mathbb F\_{q^{2n}} } \psi\_{2n} (x) \chi( x)$ vanishes for $\chi$ trivial and is a Gauss sum of absolute value exactly $q^{n}$ for $\chi$ trivial. This gives an estimate with main term and error term


$$ \left| \#\{y\in \mathbb{F}\_{q^{2n}}|\mathrm{Tr}\_{\mathbb{F}\_{q^{2n}}/\mathbb{F}\_{q^2}}(y^n)=0 \} -q^{2n-2} \right| \leq (n-1) (q^2-1) q^{n-2} $$


