In Dummit and Foote 3 ed., Chapter 14, Section 2, Exercise 30, I am asked the following:
Let $ k $ be a field, $ k(t) $ the field of rational functions in the variable $ t $. Define the maps $ \sigma $ and $ \tau \in Aut(k(t)/k) $ by $$ \sigma f(t) = f \left( \frac{1}{1-t} \right) \quad \tau f(t) = f \left( \frac{1}{t} \right) $$ for $ f(t) \in k(t) $. Prove that the fixed field of $ \langle \tau \rangle $ is $ k \left( t + \frac{1}{t} \right) $, the fixed field of $ \langle \tau \sigma^2 \rangle $ is $ k(t(1-t)) $; determine the fixed field of $ \langle \tau \sigma \rangle $ and $ \langle \sigma \rangle $.
The only part of this I am struggling with is the fixed field of $ \langle \sigma \rangle $. Call this fixed field $ E = k(s) $, where $ s = P(t) / Q(t) \in k(t) $ is some rational function. Note, I am making an assumption here that $ E $ is of the form $ k(s) $, and so far cannot justify this a priori. I have shown in a previous exercise from the last chapter that $ [k(t) : k(s)] = \max \left\{ \deg P(t), \deg Q(t) \right\} $, so, since $ k(t)/k(s) $ is a Galois extension ($k(s)$ being the fixed field of a subgroup of automorphisms), I expect $$ \max \left\{ \deg P(t), \deg Q(t) \right\} = [k(t) : k(s)] = |\langle \sigma \rangle| = 3 $$ All I have been able to accomplish at this point was brute-force equation-solving by computer, setting $$ s = \frac{a_3 t^3 + a_2 t^2 + a_1 t + a_0}{b_3 t^3 + b_2 t^2 + b_1 t + b_0} $$ and solving the equations resulting from $ \sigma s = s $. I thereby found the element $ s = \frac{t^3 - 3t + 1}{t(t-1)} $. Hence I am inclined to conclude that $ k \left( \frac{t^3 - 3t + 1}{t(t-1)} \right) $ is the fixed field of $ \langle \sigma \rangle $. This approach feels inelegant, and I would like to know what tools I might have used to avoid an unsatisfying and opaque computer-search.
For $G$ a finite subgroup of $Aut(k(t)/k)$ then the fixed subfield is $k(t)^G=k(a_0(t),\ldots,a_{|G|-1}(t))$ where $\prod_{g\in |G|} (X-g(t))=\sum_{m=0}^{|G|} a_m(t) X^m$.
Then take any non-constant coefficient $a_m(t)$, because each $g(t) = \frac{e_g t+b_g}{c_g t+d_g}$ is a Möbius transformation we get that $a_m(t)$ has at most $|G|$ poles counted with multiplicity (including the pole at $\infty$), thus $[k(t):k(a_m(t))]\le |G|=[k(t):k(t)^G]$ which implies that $$k(t)^G=k(a_m(t))$$
Edit by OP: for this problem, the technique produces the element $ a_2(t) = \frac{t^3 - 3t + 1}{t(t-1)} $, reifying the computer calculations.