Exercise 5.3.6

Proof. Let $F\subset L$ be the splitting field of a separable polynomial $f\in F[x]$ : $L=F({\alpha }_1,\dots ,{\alpha }_n)$, where ${\alpha }_1,\dots ,{\alpha }_n$ are the distinct roots of $f$ in $L$, and

Let $f_i$ be the minimal polynomial of ${\alpha }_i$ over $F$. Then $f_i$ divides $f$, thus $f_i\in F[x]$ is a separable polynomial, since the unicity of the decomposition in irreducible factors in $L[x]$ shows that the only irreducible factors of $f_i$ in $L[x]$ are associate to $x-{\alpha }_j$. Consequently the ${\alpha }_i$ are separable for all $i,1\le i\le n$. Part (a) of Theorem 5.3.15 shows then that $F\subset L$ is a separable extension. □