 Solution

Put $\gamma =\alpha +\beta$, $E=F(\gamma )$ and $E_0=F({\gamma }^p)$. By the preceding separability criterion using a pth power, it suffices to prove $E=E_0$.

Since $\alpha$ is separable, $\alpha \in F({\alpha }^p)$. But ${\alpha }^p={\gamma }^p-{\beta }^p\in E_0(\beta )$, so $\alpha \in E_0(\beta )$. ConsequentlyLet $h$ and $m$ be the monic minimal polynomials of $\beta$ over $E_0$ and $E$, respectively. Both divide its separable minimal polynomial over $F$, so both are separable. Also $m\mid h$ in $E[T]$.

Every $c\in E=F(\gamma )$ has $c^p\in F({\gamma }^p)=E_0$. Hence $m(T{)}^p$ has coefficients in $E_0$ and vanishes at $\beta$, forcing $h\mid m^p$ in $E_0[T]$. Since $h$ remains a square-free polynomial over $E$, this divisibility implies $h\mid m$ in $E[T]$. Together with $m\mid h$, monicity gives $h=m$. ThereforeThe tower law makes $[E:E_0]=1$. Part (c) now proves that $\gamma =\alpha +\beta$ is separable over $F$. This establishes closure of separable algebraic elements under addition without presupposing that closure as a field theorem.