Let $K=F\_q$ and $F=F\_{q^3}$, define the set A={$x \in F$ : $Tr\_{F/K} (x)=0$}.
Is it true that for every $x \in A$ there are $y,z \in A$ such that $x=yz$? 



Elements of trace zero in a field extension
In characteristic $3$, yes.


Proof: Let $y=1$.


In characteristic $2$, yes. In all other cases, no.


Proof: The symmetric bilinear form $Tr(yz)$ on the vector space $F$ is nondegenerate (for any finite separable field extension). Its restriction to the codimension one subspace $A$ is also nondegenerate, since the orthogonal complement of $A$ is spanned by $1$, which is not in $A$. Count the nonzero solutions of $x=yz$ with $x,y,z$ all in $A$: the number is $(q^2-1)(q-1)$, because $y$ can be any nonzero vector in $A$ and $z$ can be any nonzero vector in the (one-dimensional) orthogonal complement of $Ky$ in $A$. If we identify the solutions $(x,y,z)$ and $(x,cy,c^{-1}z)$ for $c\in F$ then the count becomes $q^2-1$. This set is mapped to another set $A-0$ of the same size by $(x,y,z)\mapsto x$. To decide whether the map is surjective, look at whether it's injective. 


In characteristic different from $2$, nondegeneracy of the form implies that there is a solution $(x,y,z)$ with $y$ and $z$ linearly independent, so that $(x,z,y)$ is an inequivalent solution.


In characteristic $2$, since $Tr(y^2)$ is universally congruent to $Tr(y)^2$ mod $2$, $A$ is closed under squaring; thus every solution has the form $(x,y,ay)$ with $a\in K$. And injectivity holds because if $(x,y\_1,a\_1y\_1)$ and $(x,y\_2,a\_2y\_2)$ are solutions then $\frac{y\_1}{y\_2}$ is a square root of $\frac{a\_2}{a\_1}$, so an element of $K$.


