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Let's consider the field $\mathbb{R}(x)$ formed by the quotients of $\mathbb{R}[x]$. We know that $A=\begin{pmatrix} \frac{-1}{2} &\frac{-\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{-1}{2} \end{pmatrix}$ -the 120 degree rotation matrix- determines an element $$\varphi_A \in Gal(\mathbb{R}(x)/\mathbb{R})$$ such that $$\varphi_A(x)=\frac{ax+b}{cx+d}.$$ Now I need to prove that the cyclic subgroup generated by the equivalence class of A $$\bar{A} \in M_{2x2}/Ker(\varphi)=Gal(\mathbb{R}(x)/\mathbb{R})$$ has order 3, also calculate the fixed field $$F=\mathbb{R}(x)^{<\bar{A}>},$$ show that $$\mathbb{R}/F$$ is a Galois extention and find $u$ such that $F=\mathbb{R}(u)$. Finally, if $x \in \mathbb{R}(x)$ I need to calculate $\min(x,F) \in F[y]$ and find its roots

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    $\begingroup$ Where are you stuck ? For any $u(x) \in \mathbb{R}(x)$ and non-constant $x \mapsto u(x)$ is a field morphism $\mathbb{R}(x) \to \mathbb{R}(x)$. That morphism is surjective iff $u(x)$ has only one pole and zero. $\endgroup$
    – reuns
    Dec 12, 2018 at 23:31

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Hint: you have your group of three elements. It often happens (but it’s not guaranteed!) that the fixed field is generated by the Trace or the Norm of your special generating element, $x$ in this case. Try both.

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