# Splitting field of an irreductible polynomial $f(X) \in F_{q}[X]$

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!

• 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 at 17:41
• But this question has been answered on our site. – Jyrki Lahtonen Jan 29 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 at 21:31