## xᵖ-x+a is irreducible and separable over ZZ/(p)

Prove that for any prime $p$ and any nonzero $a \in \mathbb{F}_p$, $q(x) = x^p-x+a$ is irreducible and separable over $\mathbb{F}_p$.

Note that $D_x(q) = px^{p-1}-1 = -1$, so that $q$ and $D(q)$ are relatively prime. So $q$ is separable.

Now let $\alpha$ be a root of $q$. Using the Frobenius endomorphism, $(\alpha+1)^p-(\alpha+1)+a = \alpha^p - \alpha + a = 0$, so that $\alpha+1$ is also a root. By induction, $\alpha_k$ is a root for all $k \in \mathbb{F}_p$, and since $q$ has degree $p$, these are all of the roots.

Now $\mathbb{F}_p(\alpha+k) = \mathbb{F}_p(\alpha)$, and in particular the minimal polynomials of $\alpha$ and $\alpha+k$ have the same degree over $\mathbb{F}_p$ – say $d$. Since $q$ is the product of the minimal polynomials of its roots, we have $p = dt$ for some $t$. Since $p$ is prime, we have either $d=1$ (so that $\alpha \in \mathbb{F}_p$, a contradiction) or $d=p$, so that $q$ itself is the minimal polynommial of $\alpha$, hence is irreducible.