elements with field trace 0 in Galois field extensions

Question

Assume that $F\subseteq K$ is a Galois field extension. Clearly, for every $\sigma \in \mbox{Gal}(K/F)$ and $a\in K$, $$tr_{K/F}(a-\sigma(a) )=0$$ Is it true that, whenever $x\in K$ is an element such that $tr_{K/F}(x)=0$, there are $a\in K$ and $\sigma\in \mbox{Gal}(K/F)$ such that $x=a-\sigma(a)$?

Answer 1

No. Take a normal basis $K = \sum_{g \in G} g(\beta) F$ then the equation $$\sum_{g \in G} c(g)g(\beta)=\alpha- \sigma(\alpha), \qquad \alpha = \sum_g a(g) g(\beta)$$ yields $$c(g) = a(g)-a(\sigma^{-1} g)$$ having a solution iff for each $g \in \langle\sigma\rangle\setminus G$, $\sum_{n=1}^{ord(\sigma)} c(\sigma^n g)=0$ which is stronger (for $G$ non cyclic) than just $\sum_{g \in G} c(g)=0$.

For example $K = \Bbb{Q}(\beta),\beta=\sqrt{2}+\sqrt{3}+1$, $\scriptstyle\sum_{g \in G} c(g)g(\beta)=1 (\sqrt{2}+\sqrt{3}+1)+2(\sqrt{2}-\sqrt{3}+1)+3(-\sqrt{2}+\sqrt{3}+1)-6(-\sqrt{2}-\sqrt{3}+1)$