# Listing all the subfield and corresponding subgroup

Here is an exercise in the book of Dummit Foote :

Find the Galois group of the splitting field of $(x^2-2)(x^2-3)(x^2-5)$ over $\mathbb{Q}$. Then list all the subgroups and the corresponding subfield

Here is my argument :

It is not hard to see that $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})=K$ is the splitting field for the above polynomial. Since the degree of the extension $K/\mathbb{Q}$ is 8, so the order of the Galois group is 8. The automorphism in $Gal(K/\mathbb{Q})$ are :

\begin{cases} \sqrt{2}\mapsto \sqrt{2}\\ \sqrt{3}\mapsto \sqrt{3}\\ \sqrt{5}\mapsto -\sqrt{5}\\ \end{cases} \begin{cases} \sqrt{2}\mapsto \sqrt{2}\\ \sqrt{3}\mapsto -\sqrt{3}\\ \sqrt{5}\mapsto -\sqrt{5}\\ \end{cases} \begin{cases} \sqrt{3}\mapsto \sqrt{3}\\ \sqrt{5}\mapsto \sqrt{5}\\ \sqrt{2}\mapsto -\sqrt{2}\\ \end{cases} \begin{cases} \sqrt{3}\mapsto \sqrt{3}\\ \sqrt{2}\mapsto -\sqrt{2}\\ \sqrt{5}\mapsto -\sqrt{5}\\ \end{cases} \begin{cases} \sqrt{5}\mapsto \sqrt{5}\\ \sqrt{2}\mapsto -\sqrt{2}\\ \sqrt{3}\mapsto -\sqrt{3}\\ \end{cases} \begin{cases} \sqrt{5}\mapsto \sqrt{5}\\ \sqrt{3}\mapsto -\sqrt{3}\\ \sqrt{2}\mapsto -\sqrt{2}\\ \end{cases} \begin{cases} \sqrt{5}\mapsto -\sqrt{5}\\ \sqrt{2}\mapsto -\sqrt{2}\\ \sqrt{3}\mapsto -\sqrt{3}\\ \end{cases} \begin{cases} \sqrt{5}\mapsto \sqrt{5}\\ \sqrt{2}\mapsto \sqrt{2}\\ \sqrt{3}\mapsto \sqrt{3}\\ \end{cases}

From there, we see that except the identity map(the last map) are of order 2. But I do not know any group of order 8 having that property.

My first question is : what is the Galois group in this case ?

Here is the other argument :

Any element of $K$ can be represented in the form : $a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}+e\sqrt{5}+m\sqrt{10}+n\sqrt{15}+p\sqrt{30}$ Under the map : \begin{cases} \sqrt{2}\mapsto -\sqrt{2}\\ \sqrt{3}\mapsto -\sqrt{3}\\ \sqrt{5}\mapsto -\sqrt{5}\\ \end{cases} this element turns to : $a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6}-e\sqrt{5}+m\sqrt{10}+n\sqrt{15}-p\sqrt{30}$.

This element is fixed under that map iff $b=c=e=p=0$ or it must has the form : $a+b\sqrt{6}+c\sqrt{10}+d\sqrt{15}$.

But from that I can not deduce the corresponding subfield with the subgroup generated by the above map. My second question is : What is that corresponding subfield ?

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wait I'm not seeing how $\sqrt{5}$ is in the splitting field... did you make a typo? –  dinoboy Jan 5 '13 at 8:20
+1 for the formatting effort. –  user1551 Jan 5 '13 at 8:25
Thanks @dinoboy for [pointing it out. I have edited it. –  knot Jan 5 '13 at 8:29

For the first question, consider $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. Note this relationship for "independent" automorphisms.

Alright, since I don't know how to do the larger tables in latex here (usually I use xymatrix!) lets (diagrammatically) look at the example at the beginning of 14.2 in Dummit and Foote. The Galois group of $\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}$ gives the table

$$\begin{array}{cccccccc} & \{1\} & \\ \nearrow & \uparrow & \nwarrow \\ \{1,\tau\} & \{1,\sigma\tau\} & \{1,\sigma\} \\ \nwarrow & \uparrow & \nearrow \\ & \{1,\sigma,\tau,\sigma\tau\} \end{array}$$ In particular, when both $\sqrt{2}$ and $\sqrt{3}$ are permuted, the fixed field is $\mathbb{Q}(\sqrt{6})$.

Now, in our larger example, if $\sqrt{6}$ is fixed, the automorphism $\sigma_2\sigma_3$ is being applied (where $\sigma_n:\sqrt{n}\mapsto -\sqrt{n}$). Similarly, when $\sqrt{10}$ and $\sqrt{15}$ are fixed, the respective automrophisms are $\sigma_2\sigma_5$ and $\sigma_3\sigma_5$. Here's where a complete description of the Galois group is useful: $$\operatorname{Gal}(\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})/\mathbb{Q}) \simeq \{1,\sigma_2,\sigma_3,\sigma_5,\sigma_2\sigma_3,\sigma_2\sigma_5,\sigma_3\sigma_5,\sigma_2\sigma_3\sigma_5\}$$

The subgroup you're interested in is then $\{1,\sigma_2\sigma_3,\sigma_2\sigma_5,\sigma_3\sigma_5\}$.

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Ok, I get your idea, but could you please show me what was wrong in my second argument? (the argument of the subfield) –  knot Jan 5 '13 at 8:39
So, what is the fixed field for the group generated by that map ($\sqrt{2}\mapsto-\sqrt{2},$\sqrt{3}\mapsto-\sqrt{3}, $\sqrt{5}\mapsto-\sqrt{5})$ ? –  knot Jan 5 '13 at 9:01
@knot, alright....... hopefully I'm not making any sleep deprived mistakes this time! I completely misinterpreted something from my initial reading of the question (my fault, not yours!) but I hope it's OK now. –  Alex Jan 5 '13 at 9:30