$\newcommand{\Z}{\mathbb{Z}}\newcommand{\GF}{{\mathbb{GF}(p)}}\newcommand{\a}{\alpha}\newcommand{\F}{\mathbb{F}}$ Question:
Let $p \in \Z$ be prime, and consider the polynomial $x^p - x + a \in \GF[x]$, where $0 \neq a\in \GF$. Prove that this polynomial is irreducible over $\GF[x]$.
My Proof:
Consider any extension field $\F \supseteq \GF$ in which the polynomial has a root $\a$. We prove that the minimal polynomial of $\a$ over $\GF$ has degree at-least $p$. This will finish the proof. For this, we use the fact that the minimal polynomial of $\a$ over $\GF$ is $(x - \a)(x - \a^p)(x - \a^{p^2}) \cdots (x - \a^{p^{m - 1}})$ where $m$ is the smallest nonnegative number such that $\a^{p^m} = \a$. Using this result, all that remains is to show that $m \geq p$. To prove this, observe the following: $\a^{p^k} = \a - ka$. This is proved by induction. The base cases of $k = 0, 1$ are trivial using $\a^p - \a + a = 0$. If this is true for $k$, $\a^{p^{k + 1}} = \left(\a^{p^k}\right)^p = (\a - ka)^p = \a^p - ka = \a - a - ka = \a - (k + 1)a$, using the fact that $\F$ is of characteristic $p$ so $(a - b)^p = a^p - b^p$ (separately for $p$ odd and $p = 2$), and for $x \in \GF, x^p = x$. To finish the proof, $\a^{p^m} = \a - ma = \a \implies ma = 0$. So $m1_\F = 0$, and thus, $p \mid m \geq p$.
Is this proof correct?
