Let $E$ and $K$ be fields such that $K\subseteq E.$
It is well-known that if the extension field $E/F$ is finite and separable, then $E/F$ is simple, that is, there exists $\alpha \in E$ such that $$E = F(\alpha).$$ I am interested in its converse, that is,
Question: If $E/F$ is a finite extension and $E = F(\alpha)$ for some $\alpha\in E,$ is it true that $E/F$ separable?
Certainly not. Take the following standard example: let $F= \mathbb{F}_p(t)$ be the rational function field over $\mathbb{F}_p$ and let $E=F[x]/(x^p−t)=F[\sqrt[p]{t}]$. The extension $E/F$ is simple but is purely inseparable.
In general, a simple finite extension $E = F(\alpha)$ is separable iff the minimal polynomial $m$ of $\alpha$ is squarefree. This holds iff $\gcd(m, m') = 1$, where $m'$ its formal derivative. You can see how this fails in the example above, since $m' = p x^{p-1} = 0$.