# Galois group of $\mathcal{Q}(\sqrt{2}, \sqrt{11}) / \mathcal{Q}$

The question:

Determine the Galois group of $K = \mathcal{Q}(\sqrt{2}, \sqrt{11}) / \mathcal{Q}$ and how it works on $\alpha = \sqrt{6 + \sqrt{11}} = > a\sqrt{2} + b\sqrt{11}$.

I'm not sure if I'm approaching this correctly and how to check the way it works on $\alpha$.

My attempt:

$K$ is the splitting field of the polynomial $(X^2 - 2)(X^2 -11)$.Then it is the splitting field of a separable polynomial, thus it is a Galois extension. It is trivial that $[K : \mathcal{Q}] = 4$, so the Galois group has order 4.

Also, since $K$ is a splitting field, the action is transitive so we have a group of order 4 that is a subgroup of $S_4$. Now, I've tried looking at the subgroups of order 4 of $S_4$ but I haven't really been able to see what group the Galois group now is. How do I find this group?

The Galois group is $\cong\Bbb Z/2\times \Bbb Z/2$, its generators are $\sigma, \tau$, commuting field isomorphisms of order two with: \begin{aligned} \sigma\sqrt 2 &= -\sqrt 2\ ,\\ \sigma\sqrt {11} &= +\sqrt {11}\ ,\\[2mm] \tau\sqrt 2 &= +\sqrt 2\ ,\\ \tau\sqrt {11} &= -\sqrt {11}\ . \end{aligned} The elements of the Galois group are thus $1, \sigma,\tau,\sigma\tau$.

The complicated given element can be rewritten $$\sqrt {6+\sqrt{11}} = \frac 1{\sqrt 2}\cdot \sqrt {12+2\sqrt{11}} =\pm\frac 1{\sqrt 2}(1+\sqrt {11})\ ,$$ and it should be clear how $\sigma,\tau$ act on it.

• @Lubin Yes, thanks, (it was too much copy+paste...) – dan_fulea Jun 6 '18 at 18:16

We have the property that if $w$ is a root of the polynomial, then so is $-w$. Therefore, the Galois group is a subgroup of $D_8$. Since it has $4$ elements, it must be $V_4$ instead of $C_4$.

The $4$ elements are just switching $\sqrt2 \leftrightarrow -\sqrt2$ and $\sqrt{11} \leftrightarrow -\sqrt{11}$.

• Alternatively, $\Bbb Q(\sqrt 2, \sqrt 11)/\Bbb Q$ is clearly a degree-4 extension, so the Galois group has four elements. It also clearly has (at least) two distinct degree-2 subextensions, so the Galois group has at least two proper, non-trivial subgroups. Thus $V_4$ instead of $C_4$. – Arthur Jun 6 '18 at 11:15
• Ok I understand why it is $V_4$. But I don't get what all 4 automorphisms are: is there one that sends $\sqrt{2}$ to $\sqrt{11}$? Because there should be right? – xzeo Jun 6 '18 at 11:17
• @xzeo No, there shouldn't. $\sqrt2\mapsto\sqrt 11$ doesn't keep $\Bbb Q$ fixed, since it also sends $2 = (\sqrt2)^2\mapsto (\sqrt{11})^2 = 11$. Not that it's even an automorphism. No, there's the trivial automorphism, then there is the one that sends $\sqrt2\mapsto -\sqrt2$, there is the one that sends $\sqrt{11}\mapsto-\sqrt{11}$, and then finally it's the one that does both at the same time. – Arthur Jun 6 '18 at 11:19
• So the action is not transitive on the roots of the polynomial which I constructed the splitting field for? I thought it would be. Edit: just noticed that the polynomial has to be irreducible. – xzeo Jun 6 '18 at 11:20