Let $k$ be a field and consider $k(t)$. Define $\sigma, \tau : k(t) \rightarrow k(t)$
by $\sigma f(t)= f(1/1-t)$ and $\tau f(t)= f(1/t)$ where $f(t) \in k(t)$. Find $\mathcal F( <\tau>)$ and $\mathcal F( <\sigma>)$.
For the first one , I supposed $g(t) \in \mathcal F( <\tau>) $. Then $\tau (g(t) = g(t) \implies g(1/t)= g(t)$ .From here I guessed $ g(t )= f(t+1/t) $ and it worked. So I can conclude $\mathcal F( <\tau>) = k(t+1/t)$. But this process is not correct.
For the second one the second one , I supposed $g(t) \in \mathcal F( <\sigma>) $. Then $\sigma( g(t))= g(t) \implies g(1/1-t )= g(t).$ What can I conclude from here?
Thanks in advance.
