Let $F$ be a field and let $K$ be the splitting field for the separable polynomial $p(x) \in F[x]$. Prove that $K$ is the splitting field for an irreducible, separable polynomial $q(x)$ in $F[x]$.
Attempt:
My idea is that we can use $p(x)$ to find $q(x)$. If $p(x)$ is irreducible, we are done, because we can take $p(x) = q(x)$. If not, then write $p(x) = p_1(x) p_2(x)$. Now I want to argue that if either $p_1(x)$ or $p_2(x)$ is irreducible, then we are done, since any factor of $p(x)$ is still separable; so suppose $p_1(x)$ is irreducible. However, how can I be sure that the splitting field for $p_1(x)$ is $K$ and not something smaller?
From David's example, the factors of $p(x)$ are not going to be helpful, because while they cover some of the roots of $p(x)$, they may not cover enough to get to the same splitting field.
Use instead the primitive element theorem. The extension $K / F$ is finite and separable (why?) so by the theorem , $K = F[\alpha]$ for some $\alpha \in K$. Think about the minimal polynomial of $\alpha$ over $F$.