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

Question

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.

Answer 1

Three methods that do not directly refer to counting field embeddings are in http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/separable2.pdf. The first two ways involve tensor products of fields and the third way uses derivations.

Answer 2

Hint: Consider the splitting field of $f$ over $F$, where $f$ is the minimal polynomial of $\alpha$ (and hence separable).