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Let $R = GF(q), q = p^r$, be a field with identity $e$, where $p$ is a prime number. Let $S=GF(q^n)$ be an extension of $R, n\geq 2$ and $K = GF(q^{mn})$ be an extension of $S$, where $m$ is prime. Let $_RW$ be a subspace of $_RK$ such that $$ \operatorname{dim}_RW =n, e\in W, W\neq S $$ I want to prove that there exist a primitive element of $K$ (that is the element of multiplicative order $q^{mn}-1$) in $W$.

But so far my attempts have not been successful. Experiments on the computer confirm this hypothesis.

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  • $\begingroup$ @Callus. Here we mean by primitive element the element of order $q^{mn}-1$. $\endgroup$ – Mikhail Goltvanitsa May 16 '17 at 17:17
  • $\begingroup$ What if $n=1$? Isn't what you want to prove that every element of $K$ except those in $R=S$ are primitive elements? This is false if $m$ is compound. $\endgroup$ – Ben Webster May 16 '17 at 17:47
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This does not work in general: the problem is you have no control over $m$.

Let $r = 1$. There's a sequence of embeddings of fields $$\mathit{GF}(p)\subset\mathit{GF}(p^2)\subset\mathit{GF}(p^4)\subset\mathit{GF}(p^8)\subset\dotsb.$$

Let $S = \mathit{GF}(p^2)$, so $n = 2$. Let $W$ be any $2$-dimensional subspace of $\mathit{GF}(p^4)$ containing the identity but not equal to $S$.

Let $K = \mathit{GF}(p^8)$. $W$ is still a $2$-dimensional subspace of $K$, but since $\mathit{GF}(p^8)\hookrightarrow K$ as a subfield, any element of $W$ can generate at most $\mathit{GF}(p^4)$, which is not all of $K$.

For $r\ne 1$, you can do the same thing with $$\mathit{GF}(p^r)\subset\mathit{GF}(p^{2r})\subset\mathit{GF}(p^{4r})\subset\mathit{GF}(p^{8r})\subset\dotsb.$$

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  • $\begingroup$ Thank yoy very much. But the most interesting case for me is the case< where $m$ is a prime number. What do you think in this case? $\endgroup$ – Mikhail Goltvanitsa May 16 '17 at 18:28

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