Lemma 9.12.13. Let $E/k$ be a field extension. Then the elements of $E$ separable over $k$ form a subextension of $E/k$.

**Proof.**
Let $\alpha , \beta \in E$ be separable over $k$. Then $\beta $ is separable over $k(\alpha )$ by Lemma 9.12.3. Thus we can apply Lemma 9.12.12 to $k(\alpha , \beta )$ to see that $k(\alpha , \beta )$ is separable over $k$.
$\square$

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Comment #9530 by Anonymous on

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