I need a check on the following question

Let $\alpha$ a primitive element of $\mathbb{F}_{2^n}$. Determine the degree of the minimal polynomial over $\mathbb{F}_2$. What can you say about the splitting field?

From theory I know that the degree $d$ of the min, poly is the minimum integer such that $\alpha^{2^d}=1$. Now, $\alpha$ is primitive, so this means that $\alpha^{2^n} = 1$, hence the the min. poly has degree $2^n$.

Then, since the min. poly is irreducible, by definition, I have that the splitting field of $h$ over $\mathbb{F}_2$ is given by $\mathbb{F}_{2^{2^n}}$

Is everything okay?

  • $\begingroup$ No: you know $\alpha^{2^d-1}=1$ by definition of $d$, and $\alpha^{2^n-1}=1$. So $d | n$. In particular $d$ cannot equal $2^n$. $\endgroup$ – Mindlack Jun 2 at 14:46
  • $\begingroup$ Yes, you're right @Mindlack So, what could be the degree? I can't stil find an answer $\endgroup$ – lukk Jun 2 at 15:05
  • $\begingroup$ Of course, $d=n$... And the splitting field is $\mathbb{F}_{2^n}$. Right @Mindlack ? $\endgroup$ – lukk Jun 2 at 15:06
  • $\begingroup$ Yes, that's it. $\endgroup$ – Mindlack Jun 2 at 16:23

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