Intermediate fields of a field extension Let $L := \Bbb Q(\sqrt 5,\sqrt 7).$ We have proved that $L/\Bbb Q$ is galois.
I have to find all the intermediate fields of $L/\Bbb Q$.
So far, I have found $\Bbb Q(\sqrt 5)$, $\Bbb Q(\sqrt 7)$, $\Bbb Q(\sqrt{35})$ and $\Bbb Q\left(\sqrt{\frac57}\right)$.
To prove, that these are all intermediate fields:
Let $L \supset E\supset \Bbb Q$ be a intermediate field. Then $[E:\Bbb Q]$ has to divide $[L:\Bbb Q]=4$. If $[E:\Bbb Q]=1$, then $E=\Bbb Q$. If $[E:\Bbb Q]=4$, then $E=L$. So $[E:\Bbb Q]$ has to be $2$.
So my questions are:


*

*Are these all intermediate fields? If no, which have I missed?

*How do I prove that $E$ has to be one of the above fields? Any hints are welcome.

 A: What is the Galois group of the extension? What does the Galois correspondence say about intermediate fields?
A: Firts of all, note that $f(x)=(x^2-5)$$(x^2-7)\in \Bbb{Q}$ then, with roots $\pm \sqrt{5} ,\pm \sqrt{7}$, hence $\Bbb{Q}(\sqrt{5},\sqrt{7})/\Bbb{Q}$   splits $f(x)$ It´s clear that the extension  it's   a Galois extension with degree 4. 
Then $G _\alpha =Gal(\Bbb{Q}(\sqrt{5},\sqrt{7})/\Bbb{Q})=\lbrace \alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4} \rbrace$ consists of 4 elements 
(Autorphisms)
Now let
$\hspace{6cm}$ $\gamma: \sqrt{5} \longmapsto \sqrt{5}$ 
$\hspace{6cm}$ $\gamma:\sqrt{7} \longmapsto -\sqrt{7}$
$\hspace{6cm}$ $\gamma$    fixes $\sqrt{5}$
and
$\hspace{6cm}$ $\sigma: \sqrt{5} \longmapsto  -\sqrt{5}$
$\hspace{6cm}$ $\sigma: \sqrt{7} \longmapsto \sqrt{7}$
$\hspace{6cm}$$\sigma$  fixes $ \sqrt{7}$ 
All these automorphisms satisfy $\lbrace \sigma^2=\gamma^2=(\sigma \gamma)^2=id \rbrace$
For example
$\sigma^2(\sqrt{5})=\sigma \sigma (\sqrt{5})= \sigma(-\sqrt{5})=-\sigma(\sqrt{5})=-(-\sqrt{5})=\sqrt{5}\\$
$\sigma^2(\sqrt{7})=\sigma \sigma (\sqrt{7})= \sigma(\sqrt{7})=\sqrt{7}\\$
Similarly for $\gamma 's$ and for the other conditions,Should not be a problem.
Then we can identify $G _\alpha =Gal(\Bbb{Q}(\sqrt{5},\sqrt{7})/\Bbb{Q})\cong \lbrace id,\gamma,\sigma, \sigma \gamma \rbrace =V_4$ Therefore
$G_{\alpha}\cong V_{4}$; 
The Klein four-group.
Now by The Fundamental theorem of Galois theory There must be a correspondence between the subfields of the extensions and the subgroups of the Galois group of the extension, then we can identify it  by reflection:
$\hspace{6cm}$ $\lbrace id \rbrace  \longleftrightarrow  \Bbb{Q}(\sqrt{5},\sqrt{7})$ 
$\hspace{6cm}$ $\lbrace id, \sigma \rbrace \longleftrightarrow \Bbb{Q}(\sqrt{7})$
$\hspace{6cm}$ $\lbrace id, \gamma \rbrace \longleftrightarrow \Bbb{Q}(\sqrt{5})$
$\hspace{6cm}$ $\lbrace id, \sigma \gamma \rbrace \longleftrightarrow \Bbb{Q}(\sqrt{35})$
$\hspace{6cm}$ $\lbrace id,\sigma, \gamma, \sigma \gamma \rbrace \longleftrightarrow \Bbb{Q}$
Note that there can be no more subfields between the subfields because the degree of each one of these extensions is $p=2$ and we are done.
It's an old question,also  I hope it helps.
