Let $F$ be a field and $\operatorname{char} F = p$. If $x^p – x – a$ is reducible in $F[X]$, I am to prove that the irreducible factors of the polynomial have at most degree 1. The case for $p = 2$ is easy. I have not been able to progress any further. All I know is that $F[X]$ is an Euclidean Domain, hence a PID and hence a UFD, implying that the polynomial can be written as a product of irreducible factors.
Factorisation of a polynomial.
abstract-algebra
Best Answer
If $\alpha$ is a root we have that $(\alpha+1)^p -(\alpha+1) +a=\alpha^p+1-\alpha -1 +a=0$ as well, so that the roots of the polynomial in a splitting field are $\alpha+r$ for $r=0,1,\dots,p-1$.
Suppose then that $\phi(x)$ is an irreducible divisor of degree $s$ of our polynomial in $F[X]$. The coefficient of $-X^{s-1}$ in $\phi(X)$ is $s\alpha+t$ where $t$ is an integer; hence $s\alpha\in F$. Either $s=p$, contrary to the assumption that $X^p-X+a$ is reducible; or $\alpha\in F$. In the latter case each $X-\alpha-r\in F[X]$, and $X^p-X+a=\prod_{r=0}^{p-1}(X-\alpha-r)$.