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?

Solution 1:

Consider the symmetric group $G$ on $\Omega = \{ 1, 2, 3, 4 \}$, say. The stabilizer $H$ of 4 is isomorphic to $S_3$, and since $G$ acts transitively on $\Omega$, we have $\lvert G : H \rvert$ = 4. Since $G$ acts 2-transitively on $\Omega$, it acts primitively on $\Omega$, so the 1-point stabilizer $H$ is maximal in $G$.

Now consider the splitting field $E$ over $\mathbb{Q}$, say, of a polynomial of degree 4, which has Galois group isomorphic to $G$. The subfield $F$ of $E$ corresponding to $H$ has $\lvert F : \mathbb{Q} \rvert = 4$, and since $H$ is maximal in $G$, there is no subfield between $\mathbb{Q}$ and $F$.

PS One such polynomial is $X^4-X^3+1$.

Solution 2:

Assuming $E/F$ separable, then there is no subfield different than $E,F$ if and only if the Galois group $G$ of the Galois closure $N/F$ is primitive w.r.t. The action on the cosets $G/H$ where $H=Gal(N/E)$. In particular this happens if $G=S_n$ or $A_n$, where $n=[E:F]$.