Let $K \subset L$ be a field extension. Consider the separable closure $K_s$ of $K$ in $L$ defined as $$ K_s = \left\{ {x \in L \mid x \text{ is algebraic and separable over } K} \right\} $$ Prove that $K_s$ is a field.
I know how to prove that the algebraic elements are closed under operations. If also the separable elements were closed under addition and multiplication, then I'm done (I think that this happens) but I don't know how to prove it.
If $A/B$ and $B/C$ are separable finite extensions then $A/C$ is a separable finite extension. This follows essentially from that $B(c)/B$ is separable iff $[B(c):B]=\deg\mathrm{minpoly}_B(c)=|\mathrm{Hom}_B(B(c),\overline{B(c)})|$.
If $a,b$ are separable over $K$ then the $K(a)$-minimal polynomial of $b$ (which divides its separable $K$-minimal polynomial) is separable so $K(a,b)/K(a)$ is separable and hence $K(a,b)/K$ is separable.
This proves that if $a,b\in K_s$ then $ab,a+b,-a,a^{-1}\in K_s$, i.e., $K_s$ is a field.