I am in the process of solving the following problem: $x^{p^n}-x+1$ is irreducible only when $n=1$ or $n=p=2$ over $F_p$.
The hint is to note that if $\alpha$ is a root, then $\alpha +a$, $a \in F_{p^n}$ is a root. I am supposed to use this to show that $F_p(\alpha)$ contains $F_{p^n}$ and that $[F_p(\alpha):F_{p^n}]=p$.
I was thinking about this problem and it got me very confused, mainly the first part. I am torn because assuming the polynomial is irreducible, $F_p(\alpha)$ is a finite extension so it must be Galois, so normal and so $F_p(\alpha)$ is a splitting field. This means that $F_p(\alpha)$ has all the zeroes of $f$.
My question is, are they of the "form" $\alpha +a$? This would imply $F_{p^n}$ is contained in the splitting field. My idea is just to say take two roots, subtract them, and you get an element of $F_{p^n}$ but i am not sure how to formalize this. Hopefully my confusion is clear: roots are unique, but their representations are not.
The map $x \to x^{p^n}$ is a Frobenius automorphism. It is an automorphism of any field of characteristic $p$. It's fixed field is the field $\mathbb{F}_{p^n}$.
All equalities are equalities $\mod p$.
Let $\alpha$ be a root, and $a \in \mathbb{F}_{p^n}$. Write $q := p^n$. Then we have
$$(\alpha + a)^{q} = \sum_{i=0}^q \binom{q}{i} \alpha^i \cdot a^{q-i} = \alpha^q + a^q$$ since for every i with $0<i<q$, $p$ divides $\binom{q}{i}$ hence $\binom{q}{i} = 0 \mod p$.
Now, since $a\in \mathbb{F}^q$, we have $a^q=a$. Hence, $$(\alpha + a)^q -(\alpha + a) +1 = \alpha^q + a^q - \alpha - a +1 = \alpha^q - \alpha + 1 =0$$ QED