# 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.
• Two of the intermediate fields on your list are equal. Commented Jan 22, 2014 at 15:37

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.

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

• The Galois group of $L/\Bbb Q$ is isomorphic to $\Bbb Z/2\Bbb Z \times \Bbb Z/2\Bbb Z$. And with the Galois correspondence the set of intermediate fields is in bijection to the subgroups of the Galois group. Commented Jan 22, 2014 at 15:45
• @AlexR. Great. How many subgroups of $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ are there? Can you name them? That will tell you how many intermediate fields you should have. Then if you can list that many and show they are all distinct, you are done. Commented Jan 22, 2014 at 15:47
• I think the subgroups are $\{(0,0)\}$, $\{(0,0),(0,1)\}$, $\{(0,0),(1,0)\}$ and $\{(0,0),(1,1)\}$, aren't they? Commented Jan 22, 2014 at 15:50
• @AlexR. Yes. Those are the subgroups. The trivial subgroup corresponds to the base field, $\mathbb{Q}$. In order to finish the problem, you need only list three distinct subextensions that all have order 2. You listed four above (but one was repeated). So all you need to do is figure out which is a repeat, then show that the other three are not equal. Then you will be done. Commented Jan 22, 2014 at 15:54
• One remark. The Galois correpondence reverses the order of inclusion. So the subgroup $\{(0,0)\}$ corresponds to the big field $L$, and $\Bbb{Z}/2\Bbb{Z}\times\Bbb{Z}/2\Bbb{Z}$ corresponds to $\Bbb{Q}$. Commented Jan 22, 2014 at 16:49