Separable algebraic element

An algebraic element $x$ over $K$ is separable when its minimal polynomial over $K$ has no repeated root in a splitting field. For an irreducible polynomial $m_x$, this is equivalent to $m_x^{\prime }\ne 0$, because then $\gcd (m_x,m_x^{\prime })=1$.

If $\alpha$ and $\beta$ are separable algebraic elements, put $K=F(\alpha )$ and $L=K(\beta )$. The minimal polynomial of $\beta$ over $K$ divides its separable polynomial over $F$, and is therefore separable. The extension count for field embeddings gives $[K:F]$ embeddings of $K$ into an algebraic closure, each with $[L:K]$ extensions to $L$. Hence $L$ has $[L:F]$ embeddings. For $E=F(\alpha +\beta )$, each embedding of $E$ has at most $[L:E]$ extensions, so there are at least $[E:F]$ embeddings of $E$ and hence exactly that number. Since $E$ is a simple field extension, its minimal polynomial has $[E:F]$ distinct roots. Thus $\alpha +\beta$ is a separable algebraic element.

For an algebraic element $\alpha$ over a field $F$ of characteristic $p>0$, suppose first that $\alpha$ is a separable algebraic element. Its minimal polynomial over $E=F({\alpha }^p)$ divides its original separable polynomial and also $T^p-{\alpha }^p=(T-\alpha {)}^p$. It is therefore separable with only one distinct root, so has degree one and $\alpha \in E$. Conversely, if the irreducible minimal polynomial $m$ is inseparable, its formal derivative vanishes and $m(T)=g(T^p)$ for some $g$ of degree $\mathrm{deg}(m)/p$. Thus $[F({\alpha }^p):F]\le \mathrm{deg}(m)/p<[F(\alpha ):F]$, proving the converse.