Set-up: Let $\mathbb{Q}(\zeta)$ be an extension of $\mathbb{Q}$ where $\zeta$ is the 5-th root of $1$ and define $\psi:\mathbb{Q}(\zeta)\to\mathbb{C}$, $\zeta^{x}\mapsto1/\zeta^{x}$.
Aim: I need to describe the subfield of real elements in $\psi(\mathbb{Q}(\zeta))$ (which I'll label $R$), show that it is a simple extension, and then find the generator.
My Attempt: So far I have that $\mathbb{Q}(\zeta)/\mathbb{Q}$ is Galois and $\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})=\mathbb{Z}/4\mathbb{Z}$. I also have that $\psi$ is an injective mapping.
As far as I can work out, $R$ is everything in $\mathbb{Q}(\zeta)$ which is mapped to the real line under $\psi$. But isn't it that only $\zeta^{5y}=1$ are mapped to the real line? Clearly, this makes no real sense for the question so I'm guessing not.
To show it is a simple extension I just need to show that it has only one generator - how do I do that?
You know from Galois theory that there if you call $K$ your cyclotomic extension it is true that $[K: K \cap \mathbb{R}]=[K\mathbb{R}:\mathbb{R}]=2$ so that your real subextension has degree $2$ over the rational.(This is a general trick to be sure of the degree of your real subextension). Note also that $\mathbb{Q}(\zeta+\zeta^{-1})$ is a real subextension of $K$ and that it has degree $2$ over $\mathbb{Q}$. To prove the its degree it's actually $2$ there are lots of ways. Maybe try to prove this.