Ok,this problem might appear a bit trivial but I have some doubts..If it's not a burden take a look and comment!
Let $F$ be a finite field of characteristic equal to $p$ and $ƒ(x)=x^p-α$ $∊F[x]$.Show that $ƒ(x)$ either has one root of multiplicity equal to $p$ or that $ƒ(x)$ is irreducible over $F$.
My answer:
Let $x₁$ and $x₂$ be two different roots of $ƒ(x)$. Then it follows: $x₁^p-α=x₂^p-α=0$ ⇒$x₁^p-x₂^p=0$ But since the characteristic of the field is $p$,it derives from Euler's theorem that in general: $(b-c)^p=b^p-c^p$ for any $b,c∊F$
Thus:$(x₁-x₂)^p=0⇒x₁=x₂$. So $ƒ(x)$ can have no distinct roots. If $α$ is a root,then indeed $(x-α)^p=0$ and $α$ is a root of multiplicity equal to $p$. Since no other factorization of $ƒ(x)$ exists-that is one that does not include the $(x-α)$ factor,it follows that if $α$ is not a root,then $ƒ(x)$ is irreducible.