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Let $L=\mathbb{C}(t)$ (where $t$ is an indeterminate). Let $\phi\colon L\to L$ be the $\mathbb{C}$-automorphism of $L$ given by $\phi(t) = \frac{3t-2}{4t-3},$ and let $G$ be the group generated by $\phi$. Find the fixed field of $G$.

Could really use some help with this. Thanks!

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    hint to get started: if you're lucky, the fixed field will be a finite extension of $\mathbb C$. In that case, you'd be looking for a polynomial over $\mathbb C$ satisfied by the element $\phi(t)$. I claim that you are lucky and you can in fact find a polynomial over $\mathbb C$ satisfied by the automorphism $\phi$ itself (hint #2: try computing $\phi^n$ for various $n$ and see if the computation reminds you of any other situation where you find a polynomial satisfied by an operator); (if you were unlucky, then I have no idea since transcendental Galois theory is currently beyond me).2011-03-07

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Note that $\phi^2$ = identity on $L$. Let $K$ be the fixed field of $G = \langle\phi\rangle$. Then $[L : K] = 2$. Also $t$ satisfies the polynomial $(X - t)(X - \phi(t)) \in C(t+\phi(t), t\phi(t))[X] $. So the fixed field is $K = C(t+\phi(t), t\phi(t))$

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    It's better to use `\langle` and `\rangle` for $\langle$ and $\rangle$, rather than `<` and `>`; the spacing is all wrong for the latter.2011-03-07