# Zero trace elements in finite fields

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.

• Do you mean that $\mathcal{A}$ is a subgroup of $\mathbb{F}_{q^{2n}}^\times$? Commented Mar 24, 2022 at 14:17
• ops, sorry, $\mathcal{A}$ is a subgroup of the multiplicative group $\mathbb{F}_{q^{2n}}^\ast$ of $\mathbb{F}_{q^{2n}}$. Commented Mar 24, 2022 at 14:20

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 $$q$$.

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.

• sorry, but there is something I do not understand, the trace map in my case has domain $\mathbb{F}_{q^{2n}}$ and not $\mathbb{F}_{q^n}$. The element $a$ is required to lie in a subgroup having order $(q^n+1)/(q+1)$, which is relatively prime to the order of the multiplicative group of $\mathbb{F}_{q^n}$. The element $y$ instead is restricted to be in $\mathbb{F}_{q^n}$, Commented Mar 24, 2022 at 15:05
• @PabloSpiga I said "more generally", because the situation you describe has needless restrictions. Read the first paragraph of my answer with $q$ replaced by $q^2$ and assume $n$ is prime if you want. Commented Mar 24, 2022 at 15:13
• I missed the restriction that $y$ should be in a subfield. Commented Mar 24, 2022 at 15:14