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 $k$. $\square$

Comment #9530 by Anonymous on

Maybe this is pedantic, but do you need some argument on why $k(\alpha)$ are separable over $k$ (i.e. that every element of $k(\alpha)$ is separable over $k$), and similarly for $k(\beta)$?

For example, one could use Tags 9.12.11 and 9.12.4.

Alternative way to say it (but maybe worse exposition): if there were an element $x \in k(\alpha)$ which is not separable over $k$, we would get an injection $k(x) \otimes_k \overline{k} \rightarrow k(\alpha) \otimes_k \overline{k}$ from a non-reduced ring into a reduced ring, hence a contradiction.

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