Let $F$ be a field of characteristic $p$. Show that if $X^p-X-a$ is reducible in $F[X]$, then it splits into distinct factors in $F[X]$.
I have no problem understanding it's solution except for one part. The solution assumes that it has one root $\alpha$ if it's reducible. Does reducible imply that polynomial has a root? For example:- $$(x+2)^4 - 2 = ((x+2)^2 - \sqrt 2)((x+2)^2 + \sqrt 2)$$ has two factors in $\Bbb Q[\sqrt 2]$ but has no root. Of course my example does not apply here.
How to show that $X^p-X-a$ has root in it's reducible in $F[X]$ for finite field? Does it apply to all reducible polynomials in polynomial rings over finite fields?
First consider the case $a\ne 0$. In an algebraic closure $\bar F$, there will be a root $\alpha$, but we cannot know a priori if it will be $\in F$ (and in fact it cannot be). By the Frobenius homomorphism $\Phi$, $\alpha^p$ is also a root. But $\alpha^p=\alpha+a$. We conclude by induction that the $p$ distinct numbres $\alpha,\alpha+a,\alpha+2a,\ldots,\alpha+(p-1)a$ are the $p$ roots of $X^p-X-a$ in $\bar F$. And $\Phi$ acts transitively on this set of roots. On the other hand, if $X^p-X-a=f(X)g(X)$, then $\Phi$ takes roots of $f$ to roots of $f$ and roots of $g$ to roots of $g$. It follows that one of $f,g$ must have $p$ distinct roots, so the factorization is trivial.
Remains the case $a=0$. But then trivially all elements of the prime field are roots!