Trace function and linear relations? The setup: Let $n \equiv 0$ mod $3$ and consider the trace function 
$$\text{Tr}\,_{\mathbb{F}_{3^n}/\mathbb{F}_{3^{n/3}}}(\alpha)=\alpha+\alpha^{3^{n/3}}+\alpha^{3^{2n/3}}$$ 
which is a surjective linear map from $\mathbb{F}_{3^n}$ to $\mathbb{F}_{3^{n/3}}$. Clearly, $\text{ker}\,(\text{Tr})$ has dimension $2n/3$ over $\mathbb{F}_3$.

My question: What (if anything) can we say about elements in the kernel of the trace function? For example, must there necessarily exist some element $\beta \in \mathbb{F}_{3^n}$ not contained within any intermediate fields such that $\text{Tr}\,(\beta)=0$? Furthermore, if we have some nontrivial linear relation of the form $\beta+\beta^{3^k}+\beta^{3^m}=0$ with $0 < k,m < n$, does this give us any additional information about what $\text{Tr}\,(\beta)$ might be?

Attempts: Perhaps this is a clever application of additive Hilbert's theorem 90 (below)?

Given a finite cyclic extension $K$ over $F$ with Galois group $\langle \sigma \rangle$ and an element $\beta \in K$, $\text{Tr}\,_{K/F}(\beta)=0$ if and only if $\beta=\alpha-\sigma(\alpha)$ for some $\alpha \in K$.

I also wonder if there is some dimensional argument that shows that the size of all subfields of $\mathbb{F}_{3^n}$ must always be smaller than $\text{ker}\,(\text{Tr})$.
 A: In this case the kernel of the (relative) trace map is a 2-dimensional vector space over the intermediate field $K=\Bbb{F}_{3^{n/3}}$. This is because that trace is $K$-linear. Therefore the kernel, indeed, has cardinality $3^{2n/3}$. Because we are in characteristic three $K$ is itself contained in the kernel.
The first question can be answered in the affirmative by a counting argument.
The subfield $\Bbb{F}_3(\beta)$ fails to be the entire field $L=\Bbb{F}_{3^n}$ if and only if $\beta$ belongs to one of the proper subfields of $L$. Such subfields have $3^d$ elements for some $d\mid n$, and there is exactly one such subfield of each cardinality. As the largest possible factor $d=n/2$ (when $n$ happens to be even), the union of proper subfields has at most
$$
\sum_{d\mid n, d<n}3^d\le\sum_{k=1}^{n/2}3^k=\frac{3^{1+n/2}-1}{3-1}
$$
elements. It is easy to see that this is $<3^{2n/3}$, so there must exist an element $\beta$ such that $\mathrm{Tr}_{L/K}(\beta)=0$ and $\beta$ is not an element of any proper subfield of $L$.
If we know that
$$
\beta+\beta^{3^k}+\beta^{3^m}=0,\qquad(*)
$$
then because $Gal(L/\Bbb{F}_3)$ is cyclic generated by the Frobenius, we can deduce that 
$$
\mathrm{Tr}_{L/K}(\beta+\beta^{3^k}+\beta^{3^m})=0
$$
as well.
We can actually say a bit more about the set of solutions of $(*)$ in $L$.
Not sure you find this interesting, but I do, so I will explain it anyway.
So let
$$
N=\{\beta\in L\mid \beta+\beta^{3^k}+\beta^{3^m}=0\}.
$$
We already saw that $N$ is a vector space over the prime field, and that it is stable under the action of Frobenius automorphism.
Furthermore, the theory of linearized polynomials (see Lidl & Niederreiter or Goss's book Arithmetic in Function Fields or Rosen's book Number Theory in Function Fields or ...) tells us that if we calculate the greatest common divisor
$$
d(\tau)=\gcd(\tau^n-1,\tau^m+\tau^k+1)
$$
in the polynomial ring $\Bbb{F}_3[\tau]$, and denote $r=\deg d(\tau)$, then


*

*$|N|=3^r$, and

*If $d(\tau)=\sum_{i=0}^rd_i\tau^i$ with $d_i\in\Bbb{F}_3$, then
$$
N=\{z\in\overline{L}\mid \sum_{i=0}^r d_iz^{3^i}=0\}.
$$


This also follows from the fact that we can view $L$ (or $\overline{L}$) as an $\Bbb{F}_3[\tau]$-module with $\tau$ acting via Frobenius:
$$\tau\cdot z=z^3.$$
In this language $L$ is the cyclic module $\Bbb{F}_3[\tau]/\langle \tau^n-1\rangle$, and the claims follow from general structure theory of modules over a PID.
Anyway, this description lets you to calculate the size of $N$ very quickly for many interesting choices of $k,m$. I have used this many times (though in characteristic two only).
