# separable closure , the separable elements are closed under operations

Let $K \subset L$ be a field extension. Consider the separable closure in $L$ $$K_s = \left\{ {x \in L|x\,\,algebraic\,and\,separable\,over\,K} \right\}$$ 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 this , then I'm done (I think that this happens) but I don't know how to prove it .

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As far as I know, there is no simple proof of this fact. –  user18119 Oct 2 '12 at 23:43

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.