# Fixed field of a subgroup of a Galois group

For the Galois group $Gal(\mathbb{Q}(\sqrt2, \sqrt3, \sqrt5)/\mathbb{Q})$, I'm trying to understand how to find the permutations of the roots and how the subgroups of the Galois group are related to their fixed fields.

Take, for example, the permutation that takes $\sqrt3$ to $-\sqrt3$, $\sqrt5$ to $-\sqrt5$, and fixes the other roots (lets call the permutation $\alpha$). Is $\{\varepsilon, \alpha\}$ a subgroup of the Galois group? If so, is its fixfield $\mathbb{Q}(\sqrt2, \sqrt{15})$? Or is it $\mathbb{Q}(\sqrt2)$? Or something else entirely?

If I used any terminology incorrectly, or used incorrect logic, please correct me. Thanks in advance!

## 2 Answers

By definition, Galois group $\operatorname{Gal}(L/K)$ consists of all $K$-linear automorphisms of $L$, where $K$-linearity is in the sense of linear algebra. For your example, you have $\mathbb Q$-linear maps acting on $L = \mathbb Q(\sqrt 2, \sqrt 3, \sqrt 5)$ and $L$ is $\mathbb Q$-algebra generated by set $\{1,\sqrt 2, \sqrt 3, \sqrt 5\}$, and that means that we only need to know where $\{1,\sqrt 2, \sqrt 3, \sqrt 5\}$ are sent to fully determine automorphism of $L$. $\mathbb Q$-linearity precisely means that $\varphi(1) = 1$ for any such automorphism $\varphi$. But where do other roots go? Let's say we want to check where $\alpha$ goes, and let $f$ be it's minimal polynomial over $\mathbb Q$. Then $0 = \varphi(f(\alpha)) = f(\varphi(\alpha))$ thus $\varphi(\alpha)$ is the root of the same polynomial as $\alpha$ ($f$ and $\varphi$ commute because $f$ is polynomial and $\varphi$ is ring homomorphism). So, in your case, $\sqrt 2$ goes to $\pm \sqrt 2$ because minimal polynomial of $\sqrt 2$ is $x^2 - 2$ and similarly, $\sqrt 3$ maps to $\pm \sqrt 3$ and $\sqrt 5$ maps to $\pm\sqrt 5$. Thus, there are only $8$ possible automorphisms, so Galois group is group with $8$ elements.

Now, for the subgroups, Fundamental theorem of Galois theory tells us that any subgroup of Galois group corresponds to a subextension of fields. How do you get this field? Well, for a subgroup $H$, intermediate field $L'$ is given by set $\{ x\in L \ | \ \varphi(x) = x,\ \forall \varphi\in H\}$. And yes, to your question, subgroup spanned by $\varphi = [\sqrt 2\mapsto \sqrt 2, \sqrt 3\mapsto -\sqrt 3, \sqrt 5\mapsto -\sqrt 5]$ is $\{\mathrm{id}, \varphi\}$ because $\varphi^2 = \mathrm{id}$.

Timbuc already wrote explicit basis for $L$ and $\varphi$ fixes $\{1,\sqrt 2, \sqrt {15}, \sqrt {30}\}$ which corresponds to field $\mathbb Q(1,\sqrt 2, \sqrt {15}, \sqrt {30}) = \mathbb Q(\sqrt 2, \sqrt 15)$ since other generators can be obtained by multiplication.

It's not hard to see that $\;G:=Gal(\Bbb Q(\sqrt2,\sqrt3,\sqrt5)/\Bbb Q)\cong C_2\times C_2\times C_2\;$

Now, with your notation, clearly $\;|\langle\,\alpha\,\rangle|=2\;$ , which means its fixed field has dimension $\;\frac82=4\;$ over $\;\Bbb Q\;$ , and since

$$Q(\sqrt2,\sqrt3,\sqrt5)=\text{Span}_{\Bbb Q}\left\{1,\sqrt2,\sqrt3,\sqrt5,\sqrt6,\,\sqrt10,\sqrt{15},\sqrt{30}\right\}$$

you can find what exactly this fixed field is by checking the action of $\;\alpha\;$ on this (or any other) basis:

$$x\in\left\{1,\sqrt2,\sqrt3,\sqrt5,\sqrt6,\,\sqrt10,\sqrt{15},\sqrt{30}\right\}\implies$$

$$\implies\alpha(x)=x\iff x\in\text{Span}\,\{\sqrt2\,,\,\sqrt{15}\}$$

Just check now the degrees/dimensions fit.