Exercise 5.1.10

Proof. If $F\subset K\subset L$, and if $L$ is the splitting field of $f$ over $K$, then $L$ is the splitting field of the same polynomial $f$ over $K$.

Indeed, $L$ contains the roots ${\alpha }_1,\dots ,{\alpha }_n$ of $f$, and $L=F({\alpha }_1,\dots {\alpha }_n)$. Moreover $f=c(x-{\alpha }_1)\cdots (x-{\alpha }_n),c\in F$.

$F\subset K\subset L$ and ${\alpha }_1,\dots ,{\alpha }_n\in L$, thus $K({\alpha }_1,\dots ,{\alpha }_n)\subset L=F({\alpha }_1,\dots ,{\alpha }_n)\subset K({\alpha }_1,\dots ,{\alpha }_n)$. Consequently, $L=K({\alpha }_1,\dots ,{\alpha }_n)$, and $f$ splits completely over the extension $K\subset L$ since $c\in F\subset K$. The conditions (a), (b) of definition 5.1.1 are filled: $L$ is the splitting field of $f$ over $F$.

Note: therefore, if $F\subset K\subset L$, and if $F\subset L$ is a normal extension, so is $K\subset L$. □