Exercise 11.2.1

Proof. Let $L$ be the splitting field of $f$ over $F$, and $n=[L:F]$. Let $q=|F|$, where $q=p^{\nu }$ is a power of the characteristic $p$. Since $L$ is a vector space with dimension $n$ over $F$, then $|L|=q^n$ for some integer $n$. Therefore the order of every element of the group $L^{\text{∗}}$ divides $q^n-1$, so for every element $\gamma \in L^{\text{∗}}$, ${\gamma }^{q^{n-1}}=1$. Consequently every element of $L$ is a root of $h(x)=x^{q^n}-x$. Since $h^{\prime }(x)=-1$, $\gcd <!--\; FUNCTION\; APPLICATION\; -->(h,h^{\prime })=1$, so $h$ is a separable polynomial. If $\alpha$ is a root of $f$ in the splitting field $L$, $h(\alpha )=0$ and $f$ is irreducible over $F$, so $f$ divides $h$, and $h$ is separable, therefore $f$ is a separable polynomial. □