Field extension of composite degree has a non-trivial sub-extension Let $E/F$ be an extension of fields with $[E:F]$ composite (not prime). Must there be a field $L$ contained between $E$ and $F$ which is not equal to either $E$ or $F$? To prove this is true, it suffices to produce an element $x\in E$ such that $F(x) \not= E$, but I cannot find a way to produce such an element. Any ideas?

Let $L/\mathbb Q$ be a Galois extension with Galois group $A_4$, this is certainly possible although off the top of my head I can't remember a polynomial that gives you this extension. Now $A_4$ has a subgroup $H$ of index $4$ namely, the subgroup generated by a cycle, but this subgroup is not properly contained in another proper subgroup of $A_4$. In particular let $K$ be the fixed field of $H$, then $H$ has no non-trivial proper subfields because this would imply that $H$ was properly contained in a proper subgroup. Of course we also have $[K:\mathbb Q]=4$.
In general to find such an example one can find a group $G$ which has a maximal subgroup with the desired index. Then one can realize $G$ as a Galois group of a field of rational functions and look at the fixed field of the subgroup to get a general counterexample.
