Lemma 9.15.4. Let $E/F$ be a normal algebraic field extension. Then the subextension $E/E_{sep}/F$ of Lemma 9.14.6 is normal.

**Proof.**
If the characteristic is zero, then $E_{sep} = E$, and the result is clear. If the characteristic is $p > 0$, then $E_{sep}$ is the set of elements of $E$ which are separable over $F$. Then if $\alpha \in E_{sep}$ has minimal polynomial $P$ write $P = c(x - \alpha )(x - \alpha _2) \ldots (x - \alpha _ d)$ with $\alpha _2, \ldots , \alpha _ d \in E$. Since $P$ is a separable polynomial and since $\alpha _ i$ is a root of $P$, we conclude $\alpha _ i \in E_{sep}$ as desired.
$\square$

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