Intermediate fields of a field extension

Question

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.

Answer 1

What is the Galois group of the extension? What does the Galois correspondence say about intermediate fields?

Answer 2

First of all, note that $f(x)=(x^2-5)$$(x^2-7)\in \Bbb{Q}[x]$ 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 is a Galois extension with degree 4.

Then $G _\alpha =\mathrm{Gal}(\Bbb{Q}(\sqrt{5},\sqrt{7})/\Bbb{Q})=\lbrace \alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4} \rbrace$ consists of 4 elements (automorphisms).

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 $\gamma$.

Then we can identify $G _\alpha =\mathrm{Gal}(\Bbb{Q}(\sqrt{5},\sqrt{7})/\Bbb{Q})\cong \lbrace id,\gamma,\sigma, \sigma \gamma \rbrace =V_4$. 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 extension 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.