Lemma 9.12.12. Let $E/k$ and $F/E$ be separable algebraic extensions of fields. Then $F/k$ is a separable extension of fields.
Proof. Choose $\alpha \in F$. Then $\alpha $ is separable algebraic over $E$. Let $P = x^ d + \sum _{i < d} a_ i x^ i$ be the minimal polynomial of $\alpha $ over $E$. Each $a_ i$ is separable algebraic over $k$. Consider the tower of fields
\[ k \subset k(a_0) \subset k(a_0, a_1) \subset \ldots \subset k(a_0, \ldots , a_{d - 1}) \subset k(a_0, \ldots , a_{d - 1}, \alpha ) \]
Because $a_ i$ is separable algebraic over $k$ it is separable algebraic over $k(a_0, \ldots , a_{i - 1})$ by Lemma 9.12.3. Finally, $\alpha $ is separable algebraic over $k(a_0, \ldots , a_{d - 1})$ because it is a root of $P$ which is irreducible (as it is irreducible over the possibly bigger field $E$) and separable (as it is separable over $E$). Thus $k(a_0, \ldots , a_{d - 1}, \alpha )$ is separable over $k$ by Lemma 9.12.10 and we conclude that $\alpha $ is separable over $k$ as desired. $\square$
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