# Galois group and field correspondence

I struggle to understand the correspondence between groups and fields that is given by Galois theory

I know that one formulation is given as:

For any subgroup $H$ of $Gal(E/F)$, the corresponding fixed field, $E^H$, is the set of elements of $E$ which are fixed by every automorphism in $H$

I realise this is abstract algebra, yet I have trouble 'visualising', or even understanding this concept

Would really appreciate if someone could help me gain some intuition on this, perhaps by example

The Galois group $G = \textrm{Gal}(E/F)$ is the set of field automorphisms of $E$ which fix every element of $F$. It becomes a group under the operation of function composition. Suppose you have a subgroup $H$ of $G$. Since $H$ is not necessarily the whole group, it's possible that there are some elements in $E \setminus F$ fixed by all elements of $H$. The set of elements in $E$ which are fixed by every automorphism in $H$ form a field, called the fixed field of $H$ and denoted $E^H$.

Let's look at the example $E = \mathbb{Q}(\sqrt{2},i), F = \mathbb{Q}$. You can check that the automorphisms of $E$ which fix $\mathbb{Q}$ are \begin{equation*} \begin{aligned} \textrm{id}&: \sqrt{2} \mapsto \sqrt{2}, \quad i \mapsto i\\ \sigma&: \sqrt{2} \mapsto -\sqrt{2}, \quad i \mapsto i \\ \tau&: \sqrt{2} \mapsto \sqrt{2}, \quad i \mapsto -i \\ \sigma\tau&: \sqrt{2} \mapsto -\sqrt{2}, \quad i \mapsto -i. \end{aligned} \end{equation*} It is not hard to check that $\textrm{Gal}(E/F)$ is the Klein group $V_4$ with generators $\sigma,\tau$. The subgroups are thus the ones generated by each element. Let's look at each one.

$H = <\textrm{id}>$. Clearly all of $E$ is fixed by the identity, so the corresponding fixed field is just $E$.

$H = <\sigma>$. We see that only $i$ is fixed by $\sigma$, so the corresponding fixed field is $\mathbb{Q}(i)$.

$H = <\tau>$. We see that only $\sqrt{2}$ is fixed by $\sigma$, so the corresponding fixed field is $\mathbb{Q}(\sqrt{2})$.

$H = <\sigma\tau>$. Now here neither $\sqrt{2}$ nor $i$ is fixed by $\sigma\tau$. But they are both negated by $\sigma\tau$, hence since $\sigma\tau$ fixes their product $i\sqrt{2}$, i.e. $$\sigma\tau(i\sqrt{2}) = \sigma\tau(i)\sigma\tau(\sqrt{2}) = -i(-\sqrt{2}) = i\sqrt{2}.$$ The corresponding fixed field is $\mathbb{Q}(i\sqrt{2})$.

Now how exactly can we be sure we've found the fixed fields? Well, suppose $H$ is a subgroup of $\textrm{Gal}(E/F)$. Since $E/F$ is Galois, so is $E/E^H$, and so $H = \textrm{Gal}(E/E^H)$ and the order of $H$ is equal to $[E : E^H]$. Therefore $[E^H : F] = [E : F]/[E : E^H] = \vert \textrm{Gal}(E/F)\vert/[E : E^H]$. So if you can find what appears to be the fixed field of a subgroup, you can use degrees to check if actually is the whole fixed field.

• Thanks for this, this is a great answer. I will try and practice some problems of this type and see how I get on – amiz9 Apr 17 '16 at 18:00