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?