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Let $F_q$ be a finite field ($q$ is a power a prime) and irreductible polynomial $f(X)\in F_q[X]$ with degree $n\geq 2$.

I have to see that $F_{q^n}$ is the splitting field of $f$ over $F_q$, and that all the roots of $f$ have the same order under the multiplicative group $F^*_{q^n}$.

What I know so far:

  1. $F_{q^n}$ is the finite field of $q^n$ elements and its elements are the roots of the polynomial $g(X) = X^{q^n}-X \in F_q[X]$. I read here that $g$ is the product of all the monic polynomials of $F_q[X]$ which divide $g$. So $f$ must divide $g$. Because of this, if $f | g$ we have that, as $g$ generates the elements of $F_{q^n}$, thus $f$ splits in $F_{q^n}$.
  2. About the order of the roots, I know that $|F^*_{q^n}| = \phi(q^n)$. But this group is the Galois group of the extension over $F_p$, where $q = p^m$, for some $m\geq 1$. So if you consider the extension $F_{q^n}$ over $F_q$, you have that its degree is $n$, but... here I made a mess in my head and I really don't know how to finish this.

I would appreciate if someone points out something about the second point or anything I could have done wrong on the first one. Many thanks in advance!

2pac
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    The things you listed in item 2 reveal some confusion. For one $|F_{q^n}^*|=q^n-1$ because in a field zero is the only non-invertible element. The Galois group of $F_{q^n}$ over $F_q$, is cyclic of order $n$. It is generated by the Frobenius automorphism $z\mapsto z^q$. The key to success is that if $\alpha$ is a zero of $f(X)$, then $\alpha^q$ must be another. Rinse. Repeat. – Jyrki Lahtonen Jan 29 '19 at 17:41
  • But this question has been answered on our site. – Jyrki Lahtonen Jan 29 '19 at 17:41
  • For example here or here. Not voting to close as a dupe because A) I think I've seen a better version also, B) in such a case I should not pick a target I answered myself. – Jyrki Lahtonen Jan 29 '19 at 21:31

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