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 $\text{ker}\,(\text{Tr})$.
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
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).