TITLE: Splitting subspaces and finite fields
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: 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.