The Stacks project

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 $k$. $\square$


Comments (1)

Comment #9530 by Anonymous on

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

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 which is not separable over , we would get an injection from a non-reduced ring into a reduced ring, hence a contradiction.

There are also:

  • 7 comment(s) on Section 9.12: Separable extensions

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