Show that if $K \subset L$, then the separable closure of $K$ in $L$ is a field 
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.
 A: Let $\alpha,\beta\in K_{s}$, $\beta\neq0$.  Then by theorem, $K(\alpha,\beta)\subset L$ is a separable extension (adjoining a finite number separable elements gives a separable extension).  Then in particular, $\alpha-\beta$, $\alpha\cdot\beta^{-1}$ are separable so that both are in $K_{s}$.  Then you have it, closed under subtraction and multiplication by inverse, so $K_{s}\subset L$ is a subfield.
I believe that does it.  
A: One way to show this could be to first show that if $\alpha_1,\dots,\alpha_n \in L$ are algebraic and separable over $K$, then $K(\alpha_1,\dots,\alpha_n)$ is a separable extension of $K$ (that is, every element of $K(\alpha_1,\dots,\alpha_n)$ is separable over $K$). From this, it would follow that $$K_s = \bigcup K(\alpha_1,\dots,\alpha_n),$$ where the union is taken over all finite subsets $\{\alpha_1,\dots,\alpha_n\}$ of $L$ that consist of algebraic and separable elements over $F$. Then, one can show that $K_s$ is a field by a similar argument as used in showing that algebraic elements are closed under addition and multiplication.
One can use Galois theory to show that if $\alpha_1,\dots,\alpha_n \in L$ are algebraic and separable over $K$, then $K(\alpha_1,\dots,\alpha_n)$ is a separable extension of $K$. You can take a look at this answer of mine for a proof that uses Galois theory.
A: *

*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 $[B(c):B]=\deg minpoly_B(c)=|Hom_B(B(c),\overline{B(c)})|$


*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 $ab,a+b,-a,a^{-1}\in K_s$ ie. $K_s$ is a field.
