If $\alpha$ separable over $F$ then $F(\alpha )/F$ is a separable extension.

Let $K/F$ be a field extension and $\alpha \in K$ is algebraic over the field $F.$ Now suppose $\alpha$ is separable over $F.$ Then how can I show that $F(\alpha)/F$ is a separable extension, i.e., an arbitrary element of $F(\alpha)$ is a root of a separable polynomial over $F$ ?

I don't want to use embeddings and it is also clear when $F$ is a perfect field. I also know that if $\operatorname{char}(F)=p>0$ then $F(\alpha)=F(\alpha ^p).$ Using this only I want to get the result. Any help will be appreciated, many thanks.

• I don’t know of any slick way of doing this without bringing embeddings in. Maybe somebody with a better viewpoint will do it for you, in which case I will slip away in embarrassment and delete this comment. – Lubin Jan 21 '18 at 5:46
• Sir is it not possible to prove for any $\beta \in F(\alpha) ,$ $F(\beta)=F(\beta ^p)$ ? – user371231 Jan 21 '18 at 5:55
• In characteristic $p$, that is surely true in a separable extension. In $F(\alpha)$, when $\alpha$ is a separable element, that will require a proof. And of course, one will have to prove that your criterion implies that the whole extension is separable. – Lubin Jan 22 '18 at 0:22