Let $E\mathop{/}F$ be a finite-degree normal extension of fields. Let $K$ and $L$ be intermediate fields (that is, $F\subseteq K\subseteq E$ and $F\subseteq L\subseteq E$) such that $E$ is separable over both $K$ and $L$. Show that $E$ is separable over $K\cap L$.

So I was thinking that if $f=\min_{K\cap L}(\alpha)$. Then $\min_{K}(\alpha)$ divides $f$. But if $f$ has repeated roots then I don't see what goes wrong.



$E/F$ is normal implies $E/K$ and $E/L$ are both normal. Since they are also separable, we get that $E/K$ and $E/L$ are Galois. Set $G_1=\mathrm{Gal}(E/K)$ and $G_2=\mathrm{Gal}(E/L)$. Note that $E^{G_1}=K$ and $E^{G_2}=L$.

Now consider $G=\mathrm{Aut}(E/K\cap L)$ and $E^G$. Obviously, $K\cap L\leq E^G$. Since obviously $G_1\leq G$ and $G_2\leq G$, we get that $E^G\leq E^{G_1}=K$ and $E^G\leq E^{G_2}=L$, hence $E^G\leq K\cap L$.

So, $E^G=K\cap L$, i.e. $E^{\mathrm{Aut}(E/K\cap L)}=K\cap L$, hence $E/K\cap L$ is Galois, and therefore separable.


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