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$\DeclareMathOperator\GF{GF}$Let $R=\GF(q)$ be a finite field with $q=p^r$ elements, where $p$ is a prime number, $S=\GF(q^n)$ be an extension of $R$, where $n\in \mathbb{N}$, $n\geq 2$ and let $K=\GF(q^{mn})$ be an extension of $S$, where $m\in \mathbb{N}$, $m\geq 2$. That is we have a tower of fields $$ R<S<K. $$ For arbitrary nonzero subspace ${}_RW$ of the space ${}_RK$ define $$ W^{-1}:= \{w^{-1} \ \mid \ w\in W\setminus \{0\}\} $$ and $$ W\cdot W^{-1}:= \left\{w\cdot v \ \mid \ w\in W, v\in W^{-1} \right\}, $$ where $\cdot$ is the multiplication operation in $K$. Finally let $\theta$ be a primitive element of $K$, that is $\operatorname{ord}\theta = |K|-1 = q^{mn} - 1$, and let $$ H = \left\{\theta^k \;\middle\vert\; k\in \left\{1,2,\ldots, \frac{q^{mn}-1}{q-1}\right\}\right\}. $$ My question. Does there exist a subspace $_RW<{}_RK$ of dimension $n$ such that $$ H\cap \left(W\cdot W^{-1}\right) = \varnothing? $$

My experiments have shown that the answer is negative. But I can not prove this. I would be grateful for any ideas and guidance.

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  • $\begingroup$ Dimension over what field? $\endgroup$
    – Wlod AA
    Sep 26, 2022 at 14:00
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    $\begingroup$ @WlodAA, dimension over $R$. $\endgroup$ Sep 29, 2022 at 20:57

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The element $\theta^{\frac{q^{mn-1}}{q-1}}$ has order $q-1$, therefore it lies in $R^\times$, the unique subgroup of order $q-1$ of the cyclic group $K^\times$. Now every $r \in R$ can be written as $(rw) \cdot w^{-1} \in W \cdot W^{-1}$.

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  • $\begingroup$ Yes, but $R\cap H = \varnothing$. $\endgroup$ Sep 29, 2022 at 20:56
  • $\begingroup$ @MikhailGoltvanitsa The argument I gave in the answer should have shown that, with the biggest $k$ as given by the construction of $H$, $\theta^k$ must lie in $R$. Could you please elaborate a bit on how you deduced that $R \cap H = \emptyset$, or maybe point out how the above argument fails? $\endgroup$
    – soomakan.
    Oct 1, 2022 at 15:51

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