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Let $k$ be a field, and let $K = k(x)$ be the rational function field in one variable over $k$. Let $\sigma$ and $\tau$ be the automorphisms of $K$ defined by $\sigma(f(x)/g(x)) = f (1/x)/g(1/x)$ and $\tau (f(x)/g(x)) = f(1—x)/g(1—x)$, respectively. Determine the fixed field $F$ of {$\sigma, \tau $ }, and determine $Gal(K/F)$. Find an $h\in F$ so that $F = k(h)$.

The fixed field of {$\sigma, \tau $ } is $F(\left \{ \sigma ,\tau \right \})=\left \{ a\in K:\varphi (a)=a, \varphi \in \left \{ \sigma ,\tau \right \} \right \}$ and I come to that $f(x)/g(x)=\tau(f(x)/g(x))=f(1-x)/g(1-x)$ and that $f(x)/g(x)=\sigma(f(x)/g(x))=f(1/x)/g(1/x)$, so that $f(1-x)/g(1-x)= f(1/x)/g(1/x)$ then $F(\left \{ \sigma ,\tau \right \})=\left \{f(x)/g(x)\in K: f(1-x)/g(1-x)= f(1/x)/g(1/x)\right \}$ Is this correct? I know $\sigma, \tau,id, \in Gal(K/F)$ but I do not know how to find the other automorphisms, could someone please help me? And how can I find that $h\in F$?, Thank you.

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This is standard beginning complex-variable theory, at least if you learn it from Ahlfors. The trick is to start with the nonsingular matrix $$ \begin{pmatrix}a&b\\c&d\end{pmatrix} $$ with entries in $k$, and associate to it the substitution (automorphism of $k(x)\,$) $$ x\mapsto \frac{ax+b}{cx+d}\,. $$ For instance, your automorphism $\sigma$ is just what comes from the matrix $\begin{pmatrix}0&1\\1&0\end{pmatrix}$. Then you show that this association (matrix $\mapsto$ automorphism of $k$) is a homomorphism of groups, with an obvious nontrivial kernel. Now composing two rational functions becomes very easy, since you just multiply their matrices. In particular, it’s easy to get all the elements of the group generated by your $\sigma$ and $\tau$. Should be six of them in all, giving you an $S_3$.

I’m not sure of the appropriate generator of the fixed field in this case, but you can very often get away with the Norm of a generator of the upper field. You should try that first.

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