Question. When does a irreducible polynomial contains linearly independent roots over a finite field?
Motivation. For a finite cyclic Galois extension $E/F$, if $\alpha\in E$ generates a normal basis, then for any intermediate extension $E/M/F$, $M=F\left[\operatorname{Tr}_{E/M}(\alpha)\right],$ where $\operatorname{Tr}_{E/M}(\alpha)$ is the trace of $\alpha$.
In particular, for any extension $E/F$ of finite fields, let $|F|=p^n$. It's a standard fact that $E/F$ is Galois and $\operatorname{Gal}(E/F)=\left<\sigma_{p^n}\right>,$ where $\sigma_{p^n}(x)=x^{p^n}.$ If there is a canonical way to choose a normal basis in some nice cases, one could then compute the corresponding intermediate extensions from the above theorem.
Let $|F|=p^n$ and $\left[E:F\right]=d$. Let $\omega$ be a generator of the (cyclic) multiplicative group $E^{\times}$ so that $E=\{0,\omega,...,\omega^{p^{nd}-1}=1\}$. The minimal polynomial $p_{\omega_k}$ of $\omega^{k}$ over $F$ is given by $(x-\omega^k)(x-\omega^{kp^n})(x-\omega^{kp^{2n}})...$ by conjugating $\omega^k$ by the automorphism $\sigma_{p^n}$. Note that $\deg p_{\omega^k}=d$ iff $(k,p^{nd}-1)=1$ and such $\omega^k$ is exactly the roots of the cyclotomic polynomial $\Phi_{p^{nd}-1}$ and $\Phi_{p^{nd}-1}$ factors into irreducible factors of degree $d$ over $F$.
To determine whether an element $\omega^k$ that generates a normal basis, it is necessary and sufficient to determine a irreducible factor of $\Phi_{p^{nd}-1}$ with linearly independent roots. It's not hard too see that every irreducible polynomial over a finite field $F$ is a factor of some cyclotomic polynomial over $F$. Thus goes the above question.
Example. For $\Bbb F_{8}/\Bbb F_{2}$ the polynomial $\Phi_7=x^6+...+x+1$ factors into $(x^3+x^2+1)(x^3+x+1)$ over $\Bbb F_2$. The roots of $x^3+x+1$ cannot be linearly independent as the trace of a root is $0$, which is just the sum of all the three roots. So $x^3+x^2+1$ gives a normal basis in this case. Naturally one asks when does the linearly independency of the roots is equivalent to a non-zero trace. This happens when $|E| = |F|^{|F|^m}$ for some $m$. See this answer.
Example. Let $\operatorname{char}(F)=p$ and let $m=p^{nd}-1$. Then $\Phi_m$ is irreducible over $F$ if and only if $m$ is a prime and $p$ is a generator for the multiplicative group $(\Bbb Z/m\Bbb Z)^{\times}$. In this case roots of $\Phi_m$ clearly give a normal basis for $E/F$.
Update. I quoted the motivation part below again:
Motivation. For a finite cyclic Galois extension $E/F$, if $\alpha\in E$ generates a normal basis, then for any intermediate extension $E/M/F$, $M=F\left[\operatorname{Tr}_{E/M}(\alpha)\right],$ where $\operatorname{Tr}_{E/M}(\alpha)$ is the trace of $\alpha$.
This is not really a good motivation for us to find a normal basis element for cyclic cyclotomic extensions over $\Bbb Q$, as the primitive root of unity is automatically a normal basis element.
It is neither a good motivation for us to compute a primitive element for intermediate extension between two finite fields. Since the multiplicative group of every finite field is cyclic, the primitive element for the intermediate extension is then clear after one finds a cyclic generator $\omega$ and makes a suitable exponent.
The motivation part, however, still could be useful for other scenarios. Also note that the conclusion could be sharpened as "$\operatorname{Tr}_{E/M}(\alpha)$ is a normal basis element of $M/F$."
