# 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.

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?

• Another equivalent definition of being Galois: $K/F$ is Galois iff $F$ is the fixed field of a subgroup of $\operatorname{Aut}(K/F)$. Can you find the subgroup fixing $X^2 + 1/X^2$? I can see at least two automorphisms that do. Sep 22 '21 at 0:20
• Let $\alpha=X^2+1/X^2$ then you can see that $X$ is a root of some polynomial with coefficients in $\mathbb {R} (\alpha)$. This is also the minimal polynomial for $X$. Sep 22 '21 at 1:58

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 $$\mathbb{Z}/{2}\times \mathbb{Z}/{2}.$$