Zero trace elements in finite fields

Question

There is so much literature on the relation between the multiplicative structure of a finite field and elements having zero trace, that I am hoping that the following is known.

Let $q$ be a prime power, let $n$ be an odd prime number, let $\mathbb{F}_{q^{2n}}$ be the field of cardinality $q^{2n}$ and let $\mathrm{Tr}:\mathbb{F}_{q^{2n}}\to \mathbb{F}_{q^2}$ be the trace map. Let $\mathcal{A}$ be the subgroup of $\mathbb{F}_{q^{2n}}^\times$ having order $(q^n+1)/(q+1)$.

Is there an element $a\in \mathcal{A}$ and an element $y$ in the multiplicative group $\mathbb{F}_{q^n}^\ast$ of the field $\mathbb{F}_{q^n}$ with $\mathrm{Tr}(y)\ne 0$ and $\mathrm{Tr}(ay)=0$?

It seems to me that $(n,q)=(3,2)$ is the only exception.

Answer 1

I can show that exceptions occur at most for $n=3$. (Primality of $n$ is never used.)

Since $n$ is odd, $\mathbb F_{q^{2n}} = \mathbb F_{q^2} \otimes_{\mathbb F_q} \mathbb F_{q^n}$. The trace map $\operatorname{Tr}:\mathbb F_{q^{2n}} \to \mathbb F_{q^2}$ is obtained by tensoring the identity map $\mathbb F_{q^2} \to \mathbb F_{q^2}$ with the trace map $\operatorname{tr} : \mathbb F_{q^n} \to \mathbb F_q$.

Thus, choosing an arbitrary basis of $\mathbb F_{q^2}$, we can write any $a$ as a pair of elements $a_1,a_2 \in \mathbb F_{q^n}$, and your condition that $y \in \mathbb F_{q^n}$ satisfies $\operatorname{Tr}(y)\neq 0$ but $\operatorname{Tr}(ay)=0$ is equivalent to the condition that $\operatorname{tr} (y) \neq 0$ but $\operatorname{tr} (a_1 y ) =\operatorname{tr}(a_2 y)=0$. (We can ignore the condition that $y\neq 0$ as it is implied by the condition that $y$ has trace zero.)

Since the trace map of a product is a perfect $\mathbb F_q$-linear pairing on $\mathbb F_q^n$, such a $y$ exists unless $1$ is an $\mathbb F_q$-linear combination of $a_1$ and $a_2$.

I will show there must exist a member of $\mathcal A$ that has this unusual property by bounding the number of members of $\mathcal A$ that do have this unusual property.

Note that every member of $\mathcal A$ is in the subgroup of order $q^n+1$, thus has norm to $\mathbb F_{q^n}$ equal to $1$. This is a nonsingular quadratic equation in $a_1,a_2$. For each $\lambda_1,\lambda_2$ in $\mathbb F_q$, not both zero, $\lambda_1 a_1 + \lambda_2 a_2 =1$ is a linear equation. There can be at most two solutions to a linear equation together with an nonsingular quadratic equation in two variables, since it gives a nontrivial quadratic equation in one variable.

Summing over possible choices of $\lambda_1,\lambda_2$, the number of members of $\mathcal A $ with this unusual property is at most $2 (q^2-1)$. So we can only have all members of $\mathcal A$ with this property if

$$ \frac{q^n+1}{q+1} \geq 2 (q^2-1)$$ i.e.

$$q^n+1 \geq 2 (q^2-1) (q+1).$$

For $n\geq 5$, the left side dominates the right side for any $q$.

Answer 2

Note that $\mathcal{A}$ is contained in $\def\F{\mathbb{F}}\F_{q^2}$ if and only if $(q^n+1)/(q+1)$ divides $q^2-1$, and it is easy to check with a bit of case analysis that this happens only if $(n, q) = (3,2)$. In all other cases we have some $a \in \mathcal{A} \setminus \F_{q^2}$.

Let $\langle u, v\rangle = \mathrm{Tr}_{\F_{q^{2n}} / \F_{q^2}}(uv)$ denote the trace form on $\F_{q^{2n}}$, so $\langle, \rangle$ is a nondegenerate $\F_{q^2}$-bilinear form. Note that the trace of $\F_{q^{2n}} / \F_{q^2}$ restricts to the trace of $\F_{q^n} / \F_q$: indeed both are just $x + x^{q^2} + \cdots + x^{q^{2(n-1)}}$. Hence $\langle, \rangle$ restricted to $\F_{q^n}$ takes values in $\F_q$.

Let $\def\eps{\varepsilon}\eps \in \F_{q^2} \setminus \F_q$. Then $\F_{q^{2n}} = \F_{q^n}(\eps) = \F_{q^n} + \F_{q^n}\eps$.

Since $a \notin \F_{q^2}$ we can find a hyperplane containing $1$ but not $a$, so there is some $v$ such that $\langle 1, v\rangle = 0$ but $\langle a, v \rangle \neq 0$. Since $\F_{q^{2n}} = \F_{q^n} + \F_{q^n}\eps$ we have $v = x + y \eps$ for some $x, y \in \F_{q^n}$. Hence $\langle 1, x\rangle + \langle 1, y\rangle \eps = 0$ and $\langle a, x \rangle + \langle a, y \rangle \eps \neq 0$. This implies $\langle 1, x \rangle = \langle 1, y \rangle = 0$ (since both are elements of $\F_q$) but at least one of $\langle a, x\rangle$, $\langle a, y\rangle$ is nonzero.

This achieves what you want but with the roles of $=0, \neq0$ the wrong way around, i.e., $\langle 1, y\rangle = 0$ but $\langle a, y \rangle \neq 0$. Maybe there's a some way of fixing this.