Exercise 5.3.14

Proof. By hypothesis, $F\subset K\subset L$, and $L$ is separable over $F$.

$\text{∙}$ Every element of $L$ is separable over $F$. A fortiori every element of $K$ is separable over $F$, thus $F\subset K$ is separable.

$\text{∙}$ Let $\alpha$ any element of $L$. As $\alpha$ is separable over $F$, the minimal polynomial $f\in F[x]$ of $\alpha$ over $F$ is separable, thus $f$ has only simple roots in a splitting field $R$ of $f$ over $L$. The minimal polynomial $f_K$ of $\alpha$ over $K$ divides $f$ (since $f(\alpha )=0$ and $f\in F[x]\subset K[x]$). As $f_K\mid f$, the order of multiplicity of a root of $f_K$ is at most the order of multiplicity of this root in $f$, thus all the roots of $f_K$ in the splitting field $R$ are simple, thus $\alpha$ is separable over $K$. Therefore the extension $K\subset L$ is separable. □