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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.

2 Answers 2

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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.