What are the subfields of $\mathbb{Q}(\sqrt{3},\sqrt[7]{5})$? I'm trying to find the subfields of $\mathbb{Q}(\sqrt{3},\sqrt[7]{5})$. This is easily seen to be a degree $14$ extension of $\mathbb{Q}$. 
I found that there is a unique subfield of degree $2$ over $\mathbb{Q}$. If $E$ and $F$ are distinct subfields of degree $2$ over $\mathbb{Q}$, then 
$$
[FE:\mathbb{Q}]=[FE:E][E:\mathbb{Q}]\leq[F:\mathbb{Q}][E:\mathbb{Q}]=4.
$$
In particular, $[FE:E]\leq 2$. Since $FE\neq E$, I must get $[FE:E]=2$. But then this implies $[FE:\mathbb{Q}]=4$, a contradiction since $4\nmid 14$. So there is at most one subfield of degree $2$over $\mathbb{Q}$, and it is $\mathbb{Q}(\sqrt{3})$. 
The only part left is to find the subfields of degree $7$ over $\mathbb{Q}$. Certainly $\mathbb{Q}(\sqrt[7]{5})$ is one, but I'm having trouble finding if there are others. I know the Galois closure of is $\mathbb{Q}(\sqrt{3},\sqrt[7]{5},\zeta_7)$, as this is a splitting field of the separable polynomial $(X^2-3)(X^7-5)$. Then the subfields will be the fixed fields of the subgroups in the Galois group containing the subgroup fixing $\mathbb{Q}(\sqrt{3},\sqrt[7]{5})$, but this seems computationally difficult.
Is there a way to do it without having to look at the Galois group?
 A: Let $E=\mathbb{Q}(\sqrt[7]{5})$ and suppose that you have another field $F$ of degree 7. Then $[F(\sqrt[7]{5}):F][F:\mathbb{Q}]\leq 14$ so we must have $[F(\sqrt[7]{5}):F]=2$. It follows that $\alpha=\sqrt[7]{5}$ satisfies an irreducible polynomial of degree 2 over $F$.
There is now a general fact that states that if a polynomial $x^p-a$ is reducible over some field F for some $p$ prime, then it has a root in F.
More precisely, this polynomial is of the form $(x-\zeta^i \alpha)(x-\zeta^j \alpha)=x^2 -(\zeta^i+\zeta^j)\alpha +\zeta^{i+j}\alpha^2$ where $\zeta$ is a primitive 7 root of unity and $0\leq i<j<7$ - this is because it divides $x^7-5=\prod_{i=1}^7 (x-\zeta^i \alpha)$.
You now have that $\zeta^{i+j}\alpha^2 \in F$ so by taking the 4th power you also have that $\zeta^{4i+4j}\alpha^8=5\zeta^{4i+4j}\alpha \in F$ and therefore $\zeta^{4i+4j}\alpha \in F$. We now have two options - either $\zeta^{4i+4j}=1$ but then $E=F$, or it is some primitive 7 root of unity, which then must be in $F(\alpha)$. But this means that $6\mid [F(\alpha):\mathbb{Q}]$ - contradiction.
