Splitting subspaces and finite fields Hellow. I'm sure that the following is truth, but I can't prove it.
Let $R<S<K, R=\mathrm{GF}(q),\  S= \mathrm{GF}(q^n), \ K= \mathrm{GF}(q^{mn})$ be a chain of finite fields and
$A = \{\theta\in K: \mathrm{ord}\theta = q^{mn}-1\}$ be the set of primitive elements. I want to prove the following statement. If $_RW$ be a subspace of $_RK$, $\mathrm{dim}_RW = n$  such that for every $\theta \in A$
$$
W\oplus W\theta\oplus\ldots\oplus W\theta^{m-1} = K,
$$
then $W = aS$ for some $a\in K$.
I will be grateful for the help.
 A: Clearly, the statement is invariant under multiplication by $a\in K$, so we may assume that $W\ni 1$. This implies that $W\supseteq R$, and we want to show that $W=S$. 
Suppose that $t\in W$. I claim that $t$ can not be written as a value of a  non constant rational function $f(x)/g(x)$ with coefficients in $R$ and degrees of numerator and denominator $<m$ at a primitive root $x=\theta$.
Indeed, if we have $tg(\theta)-f(\theta)=0$, then this translates into
linear dependence on $W,W\theta,\ldots,W\theta^{m-1}$, since $1,t\in W$.
It remains to argue that the complement in $K$ to the set of non constant "degree $\leq m-1$" rational functions in primitive roots is contained in (and is equal to) $S$. This sounds plausible, but I don't have a proof.
A: The statement obviously holds for $W=S$. Take now any nonzero element $a\in K$ and take $W=S\cdot a$. Then since everything is commutative, and multiplication by $a$ is invertible, the statement is still true for $W$ even though $W\neq S$ if $a\notin S$. 
If the question is: are all subspaces $W$ are of this form, I have the following partial answer:
If for example $m$ is a prime and $n$ is a power of $m$ we can argue in the following way: there cannot be two elements $w_1$ and $w_2$ such that $w_1 = \theta w_2$ where $\theta$ is primitive, because that will contradict the assumption. This means that for a given $0\neq w\in W$, all the elements in $Ww^{-1}$ will not be primitive. But the fact that they are not primitive already means that they belong to $S$, by the assumption on $m$ and $n$. I am not sure how further it is possible to use this argument. 
