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?

-

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 $X^4-X^3+1$.

-
Thanks @user15464. –  Andreas Caranti Feb 5 '13 at 15:02

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 $n=[E:F]$.

-

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.

-
Take a irreducible quartic with irreducible resolvent and whose discriminat is a square in $\mathbb Q$. The Galois group is then $A_4. – Mariano Suárez-Alvarez Feb 4 '13 at 20:11 In a previous answer, I considered the splitting field$E$of $$f(x) = (x^3 + x + 1)^2 + 1$$ Doing calculation with sage, I determined the Galois group$\text{Gal}(E/F)$has 72 elements. It has subgroups of 8 elements, but no subgroups of 24 elements. Therefore,$E/\mathbb{Q}$contains a degree 9 subextension$K/\mathbb{Q}$, and$K/\mathbb{Q}\$ does not contain any degree 3 subextensions.

-

I don't know how you can find a primitive element abstractly but, for a particular extension with the fields explicitly stated, it must be useful to use the proof of the Theorem of Steinitz on finite extensions. It gives you a method to find a primitive element in the form

$$\alpha + \lambda \beta,\quad \alpha, \beta\in E,\, \lambda \in F.$$

You can find the proof of this theorem in

Bastida, Julio. "Field Extensions and Galois Tehory". Page 155

-