Prove implications of irreducible polynomial

Question

Suppose $\alpha, \beta \in \mathbb{F}_{q}$ and $x^q - \alpha x - \beta $ is irreducible in $\mathbb{F}_{q}[x]$ then I have to show that $\beta \ne 0, \alpha = 1$ and $q$ is prime. I clearly see the $\beta \ne 0$ implication, but I don't see the other 2. Maybe if $\alpha \ne 1$, then $x^q - \alpha x - \beta $ contains a root. For the third statement, if $q$ not prime, then $q$ is a prime power, and maybe one can derive a contradiction.

Answer 1

As already explained in the comments it's clear that $\beta\ne 0$ and $\alpha=1$ otherwise the polynomial has a zero.

I just came up with a proof of one direction:

Let $p$ be prime. We want to show that $f(X):=X^p-X-\beta$ is irreducible. Let $\Phi:\mathbb F_p[X]\to\mathbb F_p[X]$ be the automorphism sending $X$ to $X+1$. It preserves the degree of polynomials. We have $\Phi^p=\operatorname{id}$. Since $\Phi(f)=f$ the automorphism $\Phi$ permutes the factors of $f$. The orbit of each factor has either length $p$ or length $1$. Length $p$ implies that there are at least $p$ factors hence $f$ decomposes into a product of linear factors which can't be the case since $f$ has no root in $\mathbb F_p$. So every factor of $f$ is a fixpoint of $\Phi$. Being a fixpoint $g$ of $\Phi$ means that as function $g:\mathbb F_p\to \mathbb F_p$ is constant. Hence $\deg(g)=0$ or $\deg(g) \ge p$. So $f$ has only one factor, itself, and hence is irreducible.