My question is:
If $f \in \mathbb{F}_q$ is irreducible, is $\mathbb{F}_q[X]/(f)$ automatically the splitting field of f?
Since i don't study in English my terminology might be inaccurate. What i mean by a splitting field is a field extension $\mathbb{L}$ of $\mathbb{F}_q$ such that $f = \prod(X-a_i)$ for $a_i \in \mathbb{L}$.
What I know so far: We have been discussing finite fields lately and while I was working through a problem set I realized that seems like a very probable result, since I can already prove the following:
The finite field $\mathbb{F}_q[X]/(f)$ already has one root $[x]$ of $f$ per construction.
$[x]^{(p^m)} \in \mathbb{F}_q[X]/(f)$ are also roots of $f$ for all $m\in \mathbb{N}$.
The smallest $m$ for which $[x]^{(p^m)} = [x]$ is $n := Grad(f)$
My proof for the 3rd statement felt a little shaky and circular, but essentially I argued that if such an $m$ smaller than $n$ existed, the Frobenius automorphism of the field $\mathbb{F}_q[X]/(f)$ would have order $m$ instead of $n$ which contradicts a rule we already proved in a previous lecture. It is obvious, that the statement is true, if the powers $a^{(p^m)}$ are all distinct for $m \leq n$, but I can't seem to prove this. It seems a little natural for this to follow from the 3rd statement above, but since I can't prove it I am not 100% sure if it is true.
To Sum up: The statement in my question is obviously true if for the elemen $[x] \in \mathbb{F}_q/(f)$ where f is irreducible, the powers $[x]^{(p^m)}$ are all distinct for $m \leq Grad(f)$. Is this always true?
