Fixed Field of Automorphisms of $k(x)$

Question

Fixed field of automorphisms of $k(x)$, with $k$ a field, induced by $I(x)=x$, $\varphi_1(x) = \frac{1}{1-x}$, $\varphi_2 (x)=\frac{x-1}{x}$?

Since $I(x)=x$, $\varphi_1(x)=\frac{1}{1-x}$, $\varphi_2 (x)=\frac{x-1}{x}$ form a group of order 3 the group is cyclic, so it is generated by $\varphi_1$ then I have to find the fixed field of $\varphi_1$.

If $a(x)=\frac{f(x)}{g(x)} \in k(x)$ with $(f,g)=1$ and $\varphi_1$ fix to $a(x)$ then $a(x)=a(\frac{1}{1-x}) \Rightarrow \frac{f(x)}{g(x)} = \frac{f(\frac{1}{1-x})}{g(\frac{1}{1-x})} \Rightarrow f(x) \mid f(\frac{1}{1-x)})$ and by th same reason $ f(\frac{1}{1-x}) \mid f(x)$ so $f(\frac{1}{1-x})=f(x)$ so $\varphi_1$ fix to $a(x)$, $f(x)$ then $\varphi_1$ fix to $g(x)$. So $\varphi_1$ fix to $\frac{f(x)}{g(x)}$.

Someone can tell me if it is correct.

Answer 1

Hint:

• The sum $$t=x+\frac1{1-x}+\frac{x-1}x=\frac{x^3-3x+1}{x^2-x}$$ is stable under the action of your group of transformations.
• Can you show that $[k(x):k(t)]=3$?