$X^p - a$ has only one root in a splitting field where $a$ is not a $p$th power, over a field of char $p$.

Question

Let $F$ be a field of characteristic $p$ and $a \in F$ not a $p$th power. Then the splitting field of $f = X^p - a \in F[X]$ has only one root of $f$. Thus when considering $|\text{Aut}(E/F)| = [E:F]$ it's important that $f$ be separable.

Please help me understand this. What is required to prove the statement in the title?

Answer 1

Let $E$ be a splitting field of $f(X) = X^{p}-a \in F[X]$, and let $\alpha \in E$ be a root of $f$. Then $f(\alpha) = \alpha^{p} - a = 0$, so we can rewrite $f(X)$ as $X^{p} - \alpha^{p} \in E[X]$. Now use the fact that $F$ (and therefore $E$) has characteristic $p$ to rewrite $X^{p}-\alpha^{p}$ in a ''simpler'' way (maybe the word Frobenius will provide insight here).