Show that the extension $\mathbb{R}(X) \supseteq \mathbb{R}\left(X^2+\frac{1}{X^2}\right)$ is Galois and find its Galois group.

Question

I want to show that the extension $\mathbb{R}(X) \supseteq \mathbb{R}\left(X^2+\frac{1}{X^2}\right)$ is Galois, and then find its Galois group $G$ and all subgroups with the corresponding subfields.

Normally when we work with extensions over $\mathbb{Q}$ we find an algebraic element $\alpha \in L \backslash \mathbb{Q}$ and an irrreducible polynomial over $\mathbb{Q}$ such that $p(\alpha)=0$. From there we can prove if the extension is actually a splitting field of a polynomial, find the degree of the extension and its Galois group.

How can I approach the problem in this case? Any tips on where I should start?

Answer 1

Look at $\sigma: \mathbb{R}(X) \mapsto \mathbb{R}(X)$ given by $\sigma(X)=-X$, and $\tau:\mathbb{R}(X) \mapsto \mathbb{R}(X)$ given by $\tau(X)=1/X$.

Show that both $\tau$ and $\sigma$ are self-inverse automorphisms, and that they commute. Then $\langle \sigma, \tau \rangle \cong \mathbb{Z}/{2}\times \mathbb{Z}/{2}.$ Then, $\mathbb{R}(X)^{\langle \sigma \rangle}=\mathbb{R}(X^2)$, while $\mathbb{R}(X)^{\langle \tau \rangle}=\mathbb{R}(X+\frac{1}{X}$).

It also follows that $\mathbb{R}(X)^{\langle \sigma,\tau \rangle} = \mathbb{R}(X^2+\frac{1}{X^2}).$ Therefore, the extension is Galois with Galois Group $\mathbb{Z}/{2}\times \mathbb{Z}/{2}.$